Lecture Notes: Chet Nath Tiwari Department of Mathematics ...

Lecture Notes:

Chet Nath Tiwari

Department of Mathematics

Tri-Chandra Multiple Campus

Tribhuwan University, Nepal

List of Figures

1.1 Direction field and equlibrium solution for (1.17) . . . . . . . . . . . 12



1.4 Graphs of (1.25), for different values of y0. . . . . . . . . . . . . . . . 16

1.5 Graph of (1.27), for different values of y0. . . . . . . . . . . . . . . . 17

1.6 Graphs of solutions (1.29), for different values of c. . . . . . . . . . . 19

1.7 Graph of (1.30) for different value of a . . . . . . . . . . . . . . . . . 20

1.8 Graph of (1.31) for different value of b . . . . . . . . . . . . . . . . . 21

1.9 Graph of (1.30) for different value of b . . . . . . . . . . . . . . . . . 22

2.1 f(y) versus y for dydx = f(y) = r

(1− y

K

)y . . . . . . . . . . . . . . . 72

2.2 Logistic growth dydx = r

(1− y

k

)y (a) The phase line (b) plot of y

versus t. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

2.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

2.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

9.1 Triangular wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510

9.2 Graph of Example 330 . . . . . . . . . . . . . . . . . . . . . . . . . . 513

9.3 Graph of the function . . . . . . . . . . . . . . . . . . . . . . . . . . 519

9.4 The graph of given function . . . . . . . . . . . . . . . . . . . . . . . 526

9.5 Even function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535

9.6 Odd function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536

9.7 Odd Periodic Extension . . . . . . . . . . . . . . . . . . . . . . . . . 544

9.8 Even Periodic Extension . . . . . . . . . . . . . . . . . . . . . . . . . 544

10.1 Dirichlet Problem for a rectangle . . . . . . . . . . . . . . . . . . . . 603

10.2 Dirichlet Problem for a rectangle . . . . . . . . . . . . . . . . . . . . 606

iii

iv LIST OF FIGURES

List of Tables

v

vi LIST OF TABLES

Contents

List of Figures iii

List of Tables v

1 Introduction of Differential Equations 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Classification of Differential equations . . . . . . . . . . . . . . . . . 2

1.3 Order and Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 Linear and Nonlinear Equation . . . . . . . . . . . . . . . . . . . . . 3

1.5 Solution of Differential Equations . . . . . . . . . . . . . . . . . . . . 5

1.6 Direction Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.7 Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.8 Construction of Mathematical Models . . . . . . . . . . . . . . . . . 22

2 First Order Linear and Nonlinear Differential Equations 27

2.1 Linear Equation; Method of Integrating Factors . . . . . . . . . . . . 27

2.1.1 Linear equation . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.2 Exact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.1.3 Integrating Factor of Linear Equation . . . . . . . . . . . . . 28

2.2 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.3 Modeling with First Order Equations . . . . . . . . . . . . . . . . . . 46

2.3.1 Mixture Problem . . . . . . . . . . . . . . . . . . . . . . . . . 46

2.3.2 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . . 54

2.3.3 Newton’s Law of Cooling . . . . . . . . . . . . . . . . . . . . 58

2.4 Difference Between Linear and Nonlinear Equations . . . . . . . . . 60

2.4.1 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . 66

2.5 Autonomous Equation and Population Dynamics . . . . . . . . . . . 69

2.5.1 Exponential Growth . . . . . . . . . . . . . . . . . . . . . . 69

2.5.2 Logistic Growth . . . . . . . . . . . . . . . . . . . . . . . . . 70

2.5.3 A Critical Threshold . . . . . . . . . . . . . . . . . . . . . . . 75

2.5.4 Logistic Growth with a Threshold . . . . . . . . . . . . . . . 78

2.6 Exact Equation and Integrating Factor . . . . . . . . . . . . . . . . . 91

vii

viii CONTENTS

2.6.1 Exact Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 91

2.6.2 Integrating Factor . . . . . . . . . . . . . . . . . . . . . . . . 102

2.7 Numerical Aproximations: Euler’s Method . . . . . . . . . . . . . . . 108

2.8 First Order Difference Equation . . . . . . . . . . . . . . . . . . . . . 118

2.8.1 Linear difference equation . . . . . . . . . . . . . . . . . . . . 118

3 Second Order Linear Equations 123

3.1 Second Order Linear Equation . . . . . . . . . . . . . . . . . . . . . 123

3.1.1 Non Linear Equation . . . . . . . . . . . . . . . . . . . . . . . 124

3.1.2 Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . 124

3.1.3 Homogeneous and Non-homogeneous . . . . . . . . . . . . . . 124

3.1.4 Homogeneous Equation with Constant Coefficients and its So-

lution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

3.2 Solution of Linear Homogeneous Equations; the Wronskian . . . . . 131

3.2.1 Derivative Operator and Differential Operators . . . . . . . . 132

3.2.2 The Wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . 136

3.3 Complex Roots of the Characterstics Equation . . . . . . . . . . . . 149

3.3.1 Euler’s Form . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

3.3.2 Solution of Differential Equation with Complex Roots . . . . 150

3.3.3 Euler’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . 159

3.4 Repeated Roots; Reduction of Order . . . . . . . . . . . . . . . . . . 163

3.4.1 Repeated Roots . . . . . . . . . . . . . . . . . . . . . . . . . . 163

3.4.2 Reduction of order . . . . . . . . . . . . . . . . . . . . . . . . 170

3.5 Nonhomogeous Equatio; Method of Undetermined Coefficients . . . 176

3.5.1 Method of Undetermined Coefficients . . . . . . . . . . . . . 178

3.6 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . 192

3.7 Mechanical and Electrical Vibration . . . . . . . . . . . . . . . . . . 202

3.7.1 Spring Problem: Undamped Free Vibrations . . . . . . . . . 205

3.7.2 Spring Problem: Free Vibration with Damping . . . . . . . . 213

3.7.3 Electric Vibration . . . . . . . . . . . . . . . . . . . . . . . . 219

3.8 Force Vibration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

3.8.1 Forced Vibration with Damping . . . . . . . . . . . . . . . . 223

3.8.2 Forced Vibration with out Damping . . . . . . . . . . . . . . 233

4 Higher Order Linear Equations 239

4.1 Higher Order Linear Equation . . . . . . . . . . . . . . . . . . . . . 239

4.2 Solution of Linear Homogeneous equations; the Wronskian . . . . . . 240

4.2.1 Linear Dependence and Independence . . . . . . . . . . . . . 245

4.3 Homogeneous Equations with Constant Coefficients . . . . . . . . . . 247

4.4 Nonhomogeneous Equations; Method of Undetermined Coefficients 256

4.5 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . 267

CONTENTS ix

5 System Of First Order Linear Equations 281

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

5.2 Linear Independence and Linear Dependence . . . . . . . . . . . . . 291

5.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . 298

5.4 Basic Theory of System of First Order Linear Equations . . . . . . . 309

6 Differential Equation of First Order but not of the First degree 317

6.1 Solvable for p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317

6.2 Equation Solvable for y . . . . . . . . . . . . . . . . . . . . . . . . . 329

6.3 Equation Solvable for x . . . . . . . . . . . . . . . . . . . . . . . . . 332

6.4 Equation Homogeneous in x and y . . . . . . . . . . . . . . . . . . . 335

6.5 Clairaut’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

6.5.1 Reducible to Clairaut’s Form . . . . . . . . . . . . . . . . . . 341

7 Partial Differential Equations of the first order 347

7.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . 347

7.1.1 Order of a Partial Differential Equation . . . . . . . . . . . . 348

7.1.2 Degree of a Partial Differential Equation . . . . . . . . . . . . 348

7.1.3 Linear and Non-linear Partial Differential Equations . . . . . 348

7.2 Origin Of First Order Partial Differential Equation . . . . . . . . . 348

7.2.1 By Elimination of Arbitrary Constants . . . . . . . . . . . . . 348

7.2.2 By the Elimination of Arbitrary Functions . . . . . . . . . . . 352

7.3 Cauchy’s Problem for First-Order Equations . . . . . . . . . . . . . . 357

7.4 Linear Equation of the First Order . . . . . . . . . . . . . . . . . . . 358

7.5 Lagrange’s Method for More than Two Variables . . . . . . . . . . . 369

7.6 Integral Surface Passing through a Given Curve . . . . . . . . . . . . 370

7.7 Geometrical Interpretation of Pp+Qq = R . . . . . . . . . . . . . . 378

7.8 Charpit’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

7.8.1 Non-Linear Partial Differential Equation . . . . . . . . . . . . 385

7.8.2 Types of Solutions . . . . . . . . . . . . . . . . . . . . . . . . 385

7.8.3 General Method of Solution of a Non-linear Partial Differen-

tial Equation of Order One with Two Independent Variable

(Charpit’s method) . . . . . . . . . . . . . . . . . . . . . . . . 386

7.9 Special Types of Equations . . . . . . . . . . . . . . . . . . . . . . . 395

7.9.1 Type I: Equations That Tnly Involve p and q . . . . . . . . . 396

7.9.2 Type II: Equations not Involving the Independent Variables . 398

7.9.3 Type III: Separable Equation . . . . . . . . . . . . . . . . . . 402

7.9.4 Type IV: Clairaut’s Equation . . . . . . . . . . . . . . . . . 405

8 Partial Differential Equations of Second Order 409

8.1 The Origin of Second Order Partial Differential Equation . . . . . . 409

8.2 Linear Partial Differential Equation with Constant Coefficients . . . 415

x CONTENTS

8.2.1 Linear Homogeneous and Non-homogeneous Equation with

Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . 416

8.2.2 Linear Differential Operator . . . . . . . . . . . . . . . . . . . 417

8.2.3 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417

8.2.4 Method of Finding the Complementary Function of PDE with

Constant Coefficients. . . . . . . . . . . . . . . . . . . . . . . 419

8.2.5 Determination of Particular Integral (P.I.) . . . . . . . . . . . 422

8.3 Non-homogeneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432

8.3.1 Reducible Partial Differential Equations . . . . . . . . . . . . 433

8.3.2 Irreducible Partial Differential Equation . . . . . . . . . . . . 436

8.4 Second Order Partial Differential Equation with Variable Coefficients 442

8.4.1 Canonical Forms (Method of Transformations) . . . . . . . . 442

8.5 General Method of Solving Rr + Ss+ Tt = V . . . . . . . . . . . . . 461

8.5.1 Monge’s Method for Rr + Ss+ Tt+ U(rt− s2) = V . . . . . 473

9 Partial Differential Equations and Fourier Series 479

9.1 Boundary Value Problem . . . . . . . . . . . . . . . . . . . . . . . . 479

9.1.1 Two Points Boundary Balue Problem . . . . . . . . . . . . . 479

9.1.2 Homogeneous and Non-homogeneous Boundary Value Probelms479

9.1.3 Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . 489

9.2 Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500

9.2.1 Periodic Functions . . . . . . . . . . . . . . . . . . . . . . . . 500

9.2.2 Orthogonality of sine and cosine Functions . . . . . . . . . . 502

9.2.3 Trigonometric Series . . . . . . . . . . . . . . . . . . . . . . . 506

9.2.4 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 507

9.2.5 Determination of Fourier Coefficients . . . . . . . . . . . . . . 507

9.3 The Fourier Convergence Theorem . . . . . . . . . . . . . . . . . . . 523

9.4 Odd and Even Functions . . . . . . . . . . . . . . . . . . . . . . . . 535

9.4.1 Fourier sine series or Fourier series of odd functions . . . . . 540

9.4.2 Fourier cosine series or Fourier series of even functions . . . . 540

9.4.3 Even and Odd Extension . . . . . . . . . . . . . . . . . . . . 542

10 Separation of Variables 549

10.1 One Dimensional Heat Equation . . . . . . . . . . . . . . . . . . . . 549

10.1.1 Fourier ’s Law of Heat Conduction . . . . . . . . . . . . . . . 549

10.1.2 Derivation of One Diamensional heat Equation . . . . . . . . 549

10.1.3 Initial and boundary conditions . . . . . . . . . . . . . . . . . 551

10.1.4 Separation of Variables . . . . . . . . . . . . . . . . . . . . . 551

10.1.5 Solution of One Diamensional Heat Equation by Separation

of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . 554

10.2 Other Heat Conduction Problem . . . . . . . . . . . . . . . . . . . . 566

10.2.1 Non-homogeneous Boundary conditions, and its Solution . . 567

CONTENTS xi

10.2.2 Bar with Insulated Ends . . . . . . . . . . . . . . . . . . . . . 573

10.3 The Wave Equation: Vibrations of an Elastic String . . . . . . . . . 580

10.3.1 General Solution of One Dimensional Wave Equation . . . . . 582

10.3.2 Vibration of Elastic String with Non-zero Initial Displacement 583

10.3.3 Vibration of Elastic String with Non-zero Initial Velocity . . 594

10.4 Laplace Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602

10.4.1 Dirichlet Problem for a Rectangule . . . . . . . . . . . . . . . 602

10.4.2 Laplace’s Equation in Polar form . . . . . . . . . . . . . . . . 608

10.4.3 Dirichlet Problem for a Circle . . . . . . . . . . . . . . . . . . 610

10.4.4 Laplace Equation with Numann Boundary Conditions . . . . 616

Bibliography 621

xii CONTENTS

Contents

xiii

Chapter 1

Introduction of Differential

Equations

In this chapter we will study definition and classification of differential equation ,

solution and some mathematical models.

1.1 Introduction

Many of the principles, or laws, underlying the behavior of the natural world are

statements or principle or relations involving rates at which the things happen.

When expressed in mathematical terms, the relations are equations and the rates

are derivatives. Equations containing derivatives are differential equations.

A differential equation that describes some physical process is called a mathematical

model of the process and many such models are discussed.

Example 1.

Suppose that an object is falling in the atmosphere near sea level. Formulate a

differential equation that describes the motion.

Solution: Let v represent the velocity of a falling object of mass m at any time t.

Then v is function of time and we will assume positive in downward direction. From

the Newton’s second law of motion, the net force F on this object is

F = ma (1.1)

where a is acceleration of the object and it is given by the relation

a =dv

dt

Thus from (1.1),

F = mdv

dt(1.2)

1

2 CHAPTER 1. INTRODUCTION OF DIFFERENTIAL EQUATIONS

Now we consider the forces that acts on the objects as it falls.

1. Gravity exerts a force equal to the weight of the object mg, where g is acceleration

due to gravity.

2.There is also a force due to air resistance or drag. This drag force is directly

proportinal to velocity. Thus, drag force is equal to γv, where γ is drage coefficient

and act upwward direction.

Hence net force

F = mg − γv

From (1.2),

mg − γv = mdv

dt

or, mdv

dt= mg − γv

which is a mathematical model of an object falling in the atmosphere near to the

sea level.

1.2 Classification of Differential equations

The classification of differntial equation is based on wheather the unknown functions

depends on a single independent variable or on the several variables.

(a) Ordinary Differential Equation: If the unkown function depend on one

independent variable, only ordinary derivatives appear in the differential equation,

and it is called an ordinary differential equation. For example

1. mdy

dt+ γv = mg

2. x2 d2y

dx2+ 2x

dy

dx+ y = tanx

3. x2 d2y

dx2+ 2x

(dy

dx

)3

+ y2 = tanx

(1.3)

(b) Partial Differential Equation If the unkown function depend on more than

one independent variables, then partial derivatives appear in the differential equa-

tion, and it is called a partial differential equation.

1.∂2z

∂x2+∂z

∂y= x+ y

2. x2 ∂z

∂x+∂z

∂y= z

1.3. ORDER AND DEGREE 3

1.3 Order and Degree

The order of a differential equation is the order of the higest derivative that appear

in the equation. Degree of a differential equation is the power of the highest ordered

derivative involved in the equation, which is free from the radical sign. For example

1.d3y

dt3+ et

d2

dt2+ y

dy

dt= t3

is third order differential equation with degree one.

2.d3y

dt3=√t+ 2y

or,

(d3y

dt2

)2

= t+ 2y

is third order differential equation with degree two.

1.4 Linear and Nonlinear Equation

The ordinary ordinary differential equation

F (t, y′, y′′, · · · , y(n)) = 0

is said to be linear if F is linear function of y, y′, y′′, · · · , y(n). The general linear

ordinary equation of order n is

a0(t)dny

dtn+ a1(t)

dn−1y

dtn−1+ · · ·+ an−1(t)

dy

dt+ an(t)y = g(t) (1.4)

where a0(t) 6= 0. Thus, a differential equation is called linear if

1. dependent variables and derivatives occur only in the first degree.

2. product of dependent variables do not occur.

3. product of derivatives do not occur.

4. the product of dependent variable and derivative do not occur.

Examples:

1.d2y

dt2+ sin t

dy

dt+ 2ty = t2 + 1

2. td3y

dt3+ t

dy

dt+ 2ty = t2 + 1

3,∂2z

∂2x+

∂2z

∂y∂x+∂z

∂y= z

A nonlinear equation is an equation which is not linear. Foer examplr

1. (x+ y)dy

dt+ y2 = t


Example 2.

Determine the order of the given differential equations; also state whether the

equations is linear or not.

1. (1 + y2)d2y

dt2+ t

dy

dt+ y = et

2.dy

dt+ ty2 = 0

3.d3y

dt3+ t

dy

dt+ cos2 t y = t3

4. t2d2y

dt2+ t

dy

dt+ 3y = sin t

Solution: 1. The given differential equation

(1 + y2)d2y

dt2+ t

dy

dt+ y = et

This is second order and present of y2 the equation is nonlinear.

2. Here

dy

dt+ ty2 = 0

This is first order and present of y2 the equation is nonlinear.

3. Here

d3y

dt3+ t

dy

dt+ cos2 t y = t3 (1.5)

This is third order and linear equation.

4. Here

t2d2y

dt2+ t

dy

dt+ 3y = sin t

is linear equation of order three.

Example 3.

Determine the order of the given partial differential equations; also state whether

the equations is linear or not.

1. uxx + uyy + uzz = 0

2. uxx + uyy + uux + uuy + u = 0

3. ut + uux = 1 + uxx


uxx + uyy + uzz = 0

1.5. SOLUTION OF DIFFERENTIAL EQUATIONS 5

This is second order and linear equation.

2. Here

uxx + uyy + uux + uuy + u = 0

This is second order and present of uux, the equation is nonlinear.

3. Here

ut + uux = 1 + uxx

This is second order and present of uux, the equation is nonlinear.

1.5 Solution of Differential Equations

A solution of a differential equation is a function that satisfies the differential equa-

tion on some open interval. For example, the function y1 = sin t is a solution of the

differential equation

y′′ + y = 0 (1.6)

for all t. Since

y1 = sin t

y′1 = cos t

y′′1 = − sin t

∴ y′′1 + y1 = − sin t+ sin t = 0

Example 4.

Verify that each given function is a solution of the differential equation

1. ty′ − y = t2; y = 3t+ t2

2. y′′ + 2y′ − 3y = 0; y1(t) = e−3t, y2(t) = et

3. t2y′′ + 5ty′ + 4y = 0; t > 0, y1(t) = t−2, y2(t) = t−2 ln t

4. y′′ + y = sec t; 0 < t < π/2, y = cos t ln cos t+ t sin t

5. y′ − 2ty = 1; y = et2

∫ t

0e−s

2ds+ et

2

Solution:

1. Here the given equation is

ty′ − y = t2 (1.7)


The given function is

y = 3t+ t2

Differentiating,

y′ = 3 + 2t

Substituting, the values of y and y′ in (1.7)

t(3 + 2t)− (3t+ t2) = t2

or, t2 = t2

Hence, y = 3t+ t2 is solution of (1.7).


y′′ + 2y′ − 3y = 0 (1.8)


y = e−3t

Differentiating,

y′ = −3e−3t, y′′ = 9e−3t

Substituting, the values of y, y′ and y′′ in (1.8)

9e−3t − 6e−3t − 3e−3t = 0

or, 0 = 0

Hence, y = e−3t is solution of (1.8).

Again, another fuction is

y = et

Differentiating,

y′ = et, y′′ = et


et + et − 3et = 0

or, 0 = 0

Hence, y = et is solution of (1.8).


t2y′′ + 5ty′ + 4y = 0; t > 0, (1.9)



y = t−2

Differentiating,

y′ = −2t−3 = − 2

t3, y′′ = 6t−4 =

6

t4


6t2 · 1

t4− 10t · 1

t3+ 4

1

t2= 0

or,10

t2− 10

t2= 0

or, 0 = 0

Hence, y = t−2 is solution of (1.9).

Again, another fuction is

y = t−2 ln t

Differentiating,

y′ = −2t−3 ln t+ t−2 1

t= −2 ln t

t3+

1

t3

y′′ = 6t−4 ln t− 2t−3 1

t− 3

t4

y” =6 ln t

t4− 5

t4


t2(

6 ln t

t4− 5

t4

)+ 5t

(−2 ln t

t3+

1

t3

)+

4 ln t

t3= 0

or, 0 = 0

Hence, y = t−2 ln t is solution of (1.9).


y′′ + y = sec t; 0 < t < π/2, (1.10)


y = cos t ln cos t+ t sin t

Differentiating,

y′ = − sin t ln cos t− cos tsin t

cos t+ t cos t+ sin t

or, y′ = − sin t ln cos t+ t cos t

y′′ = − cos t ln cos t+sin2 t

cos t− t sin t+ cos t



− cos t ln cos t+sin2 t

cos t− t sin t+ cos t+ cos t ln cos t+ t sin t = sec t

or,sin2 t

cos t+ cos t = sec t

or,sin2 t+ cos2 t

cos t= sec t

or,1

cos t= sec t

or, sec t = sec t

Hence, y = cos t ln cos t+ t cos t is solution of (1.10).


y′ − 2ty = 1 (1.11)


y = et2

∫ t

0e−s

2ds+ et

2

Differentiating,

y′ = 2tet2

∫ t

0e−s

2ds+ et

2e−t

2+ 2tet

2

or, y′ = 2tet2

∫ t

0e−s

2ds+ 1 + 2tet

2


2tet2

∫ t

0e−s

2ds+ 1 + 2tet

2 − 2t

(et

2

∫ t

0e−s

2ds+ et

2

)= 1

or, 1 = 1

Hence, y = et2 ∫ t

0 e−s2ds+ et

2is solution of (1.11).

Example 5.

Determine the value of r for which the given differential equation has the solution

of the form y = ert

1. y′ + 2y = 0

2. y′′ − y = 0

Solution: 1. Here the given equation is

y′ + 3y = 0 (1.12)


If y = ert is a solution of (1.12), then it must be satisfy the equation (1.12). There-

fore,

rert + 3ert = 0

or, ert(r + 3) = 0

or, r + 3 = 0

or, r = −3

2.. Here the given equation is

y′ − y = 0 (1.13)

If y = ert is a solution of (1.12), then it must be satisfy the equation

r2ert − ert = 0

or, ert(r2 − 1) = 0

or, (r − 1)(r + 1) = 0

or, r = −1, 1

Example 6.

Determine the value of r for which the differential equation

t2y′′ − 4ty′ + 4y = 0

has the solution of the form y = tr for t > 0.

Solution: Here the given differential equation is

t2y′′ − 4ty′ + 4y = 0 (1.14)

Let y = tr be solution of the equation (1.14), Then putting

y = tr, y′ = rtr−1, y′′ = r(r − 1)tr−2

in (1.14).

t2r(r − 1)tr−2 − 4trtr−1 + 4tr = 0

or, tr(r2 − 5r + 4) = 0

or, r2 − 5r + 4 = 0

or, r2 − 4r − r + 4 = 0

or, r(r − 4)− 1(r − 4) = 0

or, r = 4, r = 1

Example 7.


Verify that y1 = cosx cosh y and u2 = ln(x2 + y2) are solutions of uxx + uyy = 0.

Solution: The given partial differential equation is

uxx + uyy = 0 (1.15)

Here u = cosx cosh y. Differentiating u partially with respect with respect to x and

y,

ux =∂u

∂x= − sinx coshy

uxx =∂2u

∂x2= − cosx coshy

uy =∂u

∂y= cosx sinh y

uxx =∂2u

∂x2= cosx coshy

Putting the value of uxx and uyy in (1.15), we get

− cosx coshy + cosx coshy = 0

or, 0 = 0

Thus u = cosx coshy is solution of (1.15).

Again, u = ln(x2 + y2). Differentiating u partially with respect with respect to x

and y

ux =∂u

∂x=

2x

x2 + y2

uxx =∂2u

∂x2=

2(x2 + y2)− 4x2

(x2 + y2)2=

2y2 − 2x2

(x2 + y2)2

uy =∂u

∂y=

2y

x2 + y2

uxx =∂2u

∂x2=

2(x2 + y2)− 4y2

(x2 + y2)2=

2x2 − 2y2

(x2 + y2)2

Putting the values of uxx and uyy in (1.15), we get

2y2 − 2x2

(x2 + y2)2+

2x2 − 2y2

(x2 + y2)2= 0

or, 0 = 0

Thus, u = ln(x2 + y2) is solution of (1.15).

1.6 Direction Fields

Direction field are valuable tools in the studying of differential equation of the form

dy

dt= f(t, y) (1.16)

1.6. DIRECTION FIELDS 11

where f(t, y) is rate function, which is function of two variables t and y.

To find the direction field.

1. The direction field of equation (1.16) can be constructed by evaluting f at each

point of grid.

2. At each point of the grid, a short line segment is drawn whose slope is the value

of f at that point. Thus each line segment is tangent to the graph of the solution

passing through that point.

A direction field drawn on a fairly fine grid gives a good picture of the overal be-

haviour of the solution of the differential equation. Usually a grid consisting of a

few hundred point is sufficient. The construction of a direction field is often a useful

first step in the investigation of differential equation.

Two observation are worth particular mention. First, in constructing a direction

field, we do not have to solve equation (1.16), but evaluate the given function f(t, v)

many times. Thus the direction field can be readily constructed even for equations

that may be quit difficult to solve. Second repeated evaluation of the given function

is a task for which a computer programmin (MatLab, Mathematica) is well suited,

and you should usually use a computer to draw a direction field.

Example 8.

Draw direction field for the differential equation y′ = 3 − 2y. Based on the

direction field, determined the behavior of y as t→∞. If the behavior depends on

the initial value of y at t = 0, describe the dependency.

Solution: The given differential equation is

dy

dt= 3− 2y

or,dy

dt= −2

(y − 3

2

)(1.17)

From (1.17), we getdy

dt= 0 at y =

3

2

Hence the equilibrium solution is y = 32 .

For

y >3

2, we get

dy

dt< 0

the slope of tangent is negative, inclination tangent is greater than π2 and y is

decreasing and indecreasing reason we give downward arrow.

For

y <3

2, we get

dy

dt> 0

, the slope of tangent is positive, inclination tangent is less than π2 and y is increasing

and we give arrow in upward direction.

The figure of direction field As we see that as t → ∞, direction field converges to


Figure 1.1: Direction field and equlibrium solution for (1.17)

y = 32 . We obesrve that the converges is faster if y0 is near to the equilibrium solu-

tion y = 1.5.

Example 9.

Write down the differential equation of the form dydt = ay+b whose solution have

the following required behavior as t→∞.

1. all solutions approach y = 3.

2. All other solutions diverge from y = 13 .

Solution: 1. Here we wish to construct a differential equation

dy

dt= ay + b (1.18)

where all solutions approaches to y = 3.

But For the equlibrium solution

dy

dt= 0

or, ay + b = 0

or, y = − ba

Hence the equilibrium solution is

y = − ba.

1.6. DIRECTION FIELDS 13

For the equilibrium solution y = 3, one of the possibility is b = 3 and a = −1. Then

Equation (1.18) becomes

dy

dt= −y + 3 = −(y − 3) (1.19)

Clearly , the solution decreases for y > 3 and increases for y > 3. Hence direction

field of (1.19) is shown as figure.


2. Again we wish to construct a differential equation

dy

dt= ay + b (1.20)

where all solutions diverges to from y = 13 . But the equlibrium solution is

y = − ba

For the equilibrium solution y = 13 , one of the possibility is b = −1 and a = 3. Then

Equation (1.18) becomes

dy

dt= 3y − 1 = 3

(y − 1

3

)(1.21)

Clearly , the solution increase for y > 13 and decrease for y < 1

3 . Hence the direction

field of (1.21) is shown as figure.



1.7 Initial Value Problem

Let us consider a first order differential equation

dy

dt= f(t, y) (1.22)

where y is a function of independent variables t.

Let the value of y(t) at initial point t0 be known.

y(t0) = y0 (1.23)

The differential equation (1.22) with the initial condition (1.23), is called initial value

problem. The solution of a initial value problem gives the integral curve through

(t0, y0) point in ty−plane.

Example 10.

Solve the initial value problem

dy

dy= −y + 10, y(0) = y0

and plot the solutions of several value of y0.

1.7. INITIAL VALUE PROBLEM 15

Solution: Here the given initial value problem.

dy

dt= −y + 10, y(0) = y0

or,dy

y − 10= −dt

Integrating

or, ln |y − 10| = −t+ C

or, |y − 10| = eCe−t

or, y − 10 = ±ece−t

or, y − 10 = ce−t

or, y = 10 + ce−t (1.24)

Using the initial conditions y(0) = y0, we get

y(0) = 10 + c

or, y0 = 10 + c

or, c = y0 − 10

From (1.24)

y = 10 + (y0 − 10)e−t (1.25)

This solution is plotted for the different value of y0 : -20,-10,0,10,20, 30. The solu-

tions are converging to the equilibrium solution y = 10.


Figure 1.4: Graphs of (1.25), for different values of y0.

Example 11.


dy

dy= y − 3, y(0) = y0

and plot the solutions of several value of y0.

Solution: Here the given initial value problem.

dy

dt= y − 3, y(0) = y0

or,dy

y − 3= dt

Integrating

or, ln |y − 3| = t+ C

or, |y − 3| = eCet

or, y − 3 = ±ecet

or, y − 3 = cet

or, y = 3 + cet (1.26)


Using the initial conditions y(0) = y0, we get

y(0) = 3 + c

or, y0 = 3 + c

or, c = y0 − 3

From (1.26)

y = 3 + (y0 − 3)et (1.27)

This solution is plotted for the different value of y0. The solutions are diverging

away from the equilibrium solution y = 3.

Figure 1.5: Graph of (1.27), for different values of y0.

Example 12.

Consider the differential equation

dy

dt= −ay + b

where a and b are positive numbers.

1. Solve the differential equation.

2. Sketch the solution for several different initial condtions.


3. Describe how the solutions change under each of the following conditions.

(a) a increases.

(b) b increases.

(c) Both a and b increase, but the ratio ba remains the same.

Solution: 1. Here the given initial value problem.

dy

dt= −ay + b, y(0) = y0

or,dy

ay − b= −dt

Integrating

or,1

aln |ay − b| = −t+

C

aor, ln |ay − b| = −at+ C

or, |ay − b| = eCe−at

or, ay − b = ±eCe−at

or, ay − b = ce−at

or, y =1

a(b+ ce−at) (1.28)

which is required solution.

2. Let

dy

dt= 0

or, − ay + b = 0

or, y =b

a

Thus, y = ba is equilibrium solution.

In particular we take a = 1 and b = 2. Then, the equilibrium solution is

y =b

a= 2

and the solution (1.28) becomes

y = 2 + ce−t (1.29)

The graph the solution (1.29) for different values of c : 3, 2, 1, 0,−1,−2 and −3.

These solution converge to the solution y = 2.


Figure 1.6: Graphs of solutions (1.29), for different values of c.

3. (i) Here we fix b = 2 and c = 3. The equation becomes

y =1

a

(2 + 3e−at

)(1.30)

We plot the solutions for different value of a. We plot the solutions for different

value of a

When a = 1, the equilibrium solution is

y =b

a= 2

and equation (1.30) become

y = 2 + 3e−t

When a = 2, the equilibrium solution is

y =b

a= 1


y =1

2

(2 + 3e−2t

)When a = 3, the equilibrium solution is

y =b

a=

2

3



y =1

3

(2 + 3e−3t

)

When the value of a increased, the equilibrium is lowered and is approached must

faster (for lower value of t).

Figure 1.7: Graph of (1.30) for different value of a

3(ii) Let us fix a = 1 and c = 3. Then the solution becomes


y(t) = b− 3e−t (1.31)

and the equilibrium solutions becomes

y =b

a= b

We plot the solutions for different values for b = 2, b = 3 and b = 4. As b increases,

the equilibrium solution gets higher i. e. y = 2, y = 3, and y = 4 respectively for

b = 2, b = 3 and b = 4.

Figure 1.8: Graph of (1.31) for different value of b

3(iii) Let us fix c = 3, The values of (a, b) are choosen (1, 2), (2, 4) and (3, 4) so

that the values increase together but the ratio remains same. So, all three solutions

approache the equilibrium solution y = 2 and is approached more rapidly.


Figure 1.9: Graph of (1.30) for different value of b

1.8 Construction of Mathematical Models

In applying differential equation to any of the numerous field in which they are

useful, it is neccesary first to formulate the appropriate differential equation that

describes , or models, the problem being investigated. The following steps are useful

for the formulation of problem:

1. Identify the independent and dependent variables and assign letters to repre-

sent them. In many cases the independent variable is time.

2. Choose the units of measurement for each variable. In a sense the choice of

units is arbitrary, but some choices may be more convenient than others.

3. Articulate the basic principle that underlines or governs the problem you are

investigating. That may be physical law or it may be a more a speculative

assumption that may be based on your own experience or observations.

4. Express the principle or law in step 3 in terms of variable you choose in step

1. It may be require the introduction of physical constants or parameters and

determination of appropriate value of them.

5. Make sure that each term in your equation has the same units. If the unit

agree, then your equation at least is dimensionally consistent.

6. In the problem considered here, the result of step 4 is a single differential

equation which consitute the desired mathematical model. Keep in mind,

through, that in more complex problems the resulting mathematical model

1.8. CONSTRUCTION OF MATHEMATICAL MODELS 23

may be much complicated, perhaps involving a system of several differential

equations.

Example 13.

Newton’s law of cooling state that the temperature of an object changes at a rate

proportinal to the difference between the temperature of the object itself and the

temperature of its surrounding. Suppose that the ambient teperature is 75oF and

that the rate constant is 0.05min−1. Write a differential equation for the temperature

of the object at any time.

Solution: Let T be the temperature of the object at any time t and the surrounding

temerature be T0. By Newton’s law of cooling,

dT

dt= r(T − T0)

where r is proportionality constants. Again r = 0.05min−1 and T0 = 75oF , thus the

equation becomes

dT

dt= 0.05(T − 75)

Example 14.

A certain drug is being administered intravenously to a hospital patient. Fluid

containing 5mg/cm3 of the drug enters the patient’s bloodstream at a rate of

100cm3/h. The drug is absorbed by body tissue or otherwise leaves the bloodstream

at a rate proportional to the amount present, with a rate constant of 0.4(h)−1.

(a) Assuming that the drug is always uniformly distributed throughout the blood-

strem, write a differential equation for the amount of the drug that is present in the

bloodstrem at any time.

(b) How much of the drug is present in the bloodstream after a long time?

Solution: (a) Let q mg be amount of drug in the patient’s body at any time t in

hours. Then

rate of change in drug = rate of drug enter body− rate of drug leave or absorbed(1.32)

Since a fluid containing 5mg/cm3 of the drug enter the patient bloodstrem at a rate

of 100cm3/h , the rate at drug is enter into body 5× 100mg/h = 500mg/h.

Again the quantity of drug in body is q(t) leave or absorbed at constant rate 0.4(h)−1,

so the rate at which the drug leave or absorbed = q(t)× 0.4 = 0.4q(t)mg/h.

Hence from (1.32), we get

dq

dt= 500− 0.4q


(b) After the long time i.e. steady state , we have

dq

dt= 0

or, 500− 0.4q = 0

or, q =500

0.4= 1250mg

Hence 1250 mg of the drug is present in the bloodstream after a long time.

Example 15.

Consider a population p of field mice that grows at a rate proportinal to the

current population, so that dpdt = rp.

(a) Find the rate constant r if the population doubles in 40 days.

(b) Find r if the population doubles in N days.


dp

dt= rp

or,dp

p= rdt

Integrating,

ln p = rt+ ln c

or, p = cert (1.33)

Initially let p(0) = p0, so from (1.33), we get

p0 = c

Putting the value of c in (1.33)

p(t) = p0ert (1.34)

(a) Let the population double in 40 days. Then take t = 40 and p(t) = 2p0, from

(1.34)

2p0 = p0e40r

or, ln 2 = 40r

or, r =1

40ln 2 per day (1.35)

(b) Let the population double at t = N days. Then from (1.33), we get

2p0 = p0erN

or, 2 = erN

or, ln 2 = rN

or, r =1

Nln 2

1.8. CONSTRUCTION OF MATHEMATICAL MODELS 25

Example 16.

A radioactive material, such as the isotope thorium-234, disintegrate at a rate

proportional to the amount currently present. If Q(t) is the amount present at time

t, then dQdt = −rQ, where r > 0 is the decay rate.

(a) If 100 mg of thorium-234 decays to 82.04 mg in 1 week, determine the decay rate

r.

(b) Find an expression for the amount of thorium-234 present at any time t.

(c) Find the required for the thorium-234 to decay to one -half its original amount.


dQ

dt= −rQ

or,dQ

Q= −rdt

Integrating,

lnQ = −rt+ ln c

or, Q = ce−rt (1.36)

Initially let Q(0) = Q0, so from (1.36), we get

Q0 = c


Q(t) = Q0e−rt (1.37)

(a) If Q0 = 100, t = 7 day and p(t) = 82.04mg, from (1.37)

82.04 = 100e−7r

or, e7r =100

82.04

or, 7r = ln

(100

82.04

)or, r =

1

7ln

(100

82.04

)= 0.028 per day. (1.38)

(b) The required expression for the amount at t days is

Q(t) = 100e−0.028t (1.39)

(b) Let the t = T days be half-life . Then from (1.39), we get

50 = 100e−0.028T

or,1

2= e−0.028T

or, ln1

2= −0.028T

or, T = 24.5 days

Chapter 2

First Order Linear and

Nonlinear Differential Equations

IntroductionThis chapter deals with differential equations of first order

dy

dt= f(t, y) (2.1)

where f is a given function of two variables t and y. Any differential function

y = φ(t) that satisfies this equation for all t in some interval is called a solution. Our

object is to determine wheather such function exists and if so, to develop methods

for finding them. We also discuss some of important applications of first order

differential equation, introduce the idea ,of aprroximating a solution by numerical

computation.

2.1 Linear Equation; Method of Integrating Factors

2.1.1 Linear equation

If the differential equation

dy

dt= f(t, y) (2.2)

depends linearly on the dependent variable y, then the equation (2.2), is called a

first order linear equation. i.e.

dy

dx= −p(t)y + g(t)

Thus, the first order linear equation can be written as

dy

dt+ p(t)y = g(t) (2.3)

27

28CHAPTER 2. FIRST ORDER LINEAR ANDNONLINEARDIFFERENTIAL EQUATIONS

For exampledy

dt+ tan t y = sec t

If p and q are constants then the equation (2.3) is called the linear equation with

constant coefficients. i.e.

dy

dt+ ay = b

2.1.2 Exact

The general first order equation is

dy

dt= f(t, y) (2.4)

which can be written as

Mdy +Ndt = 0 (2.5)

where M and N are functions of t, y or constants. The differential equation (2.5) is

said to be exact if it takes the form

dG(t, y) = 0

Integrating, we get solution G(t, y) = c.

2.1.3 Integrating Factor of Linear Equation

Let us consider a linear equation

dy

dt+ p(t)y = g(t) (2.6)

where p and g are functions of t or constants. To determine appropriate integrating

factor, we multiply (2.6) by an yet undetermined function µ(t), so that

µ(t)dy

dt+ µ(t)p(t)y = µ(t)g(t) (2.7)

be exact. Then LHS of (2.7) is the derivative of the product µ(t)y if µ(t) satisfies

d

dt[µ(t)] = p(t)µ(t)

or,dµ(t)

µ(t)= p(t)dt

Integrating, by assuming µ(t) positive for all t,

ln[µ(t)] =

∫p(t)dt+ c

2.1. LINEAR EQUATION; METHOD OF INTEGRATING FACTORS 29

By choosing the arbitrary constant c to be zero, we obtained the simplest possible

function for µ

ln[µ(t)] =

∫p(t)dt

or, µ(t) = e∫p(t)dt

Note that µ(t) is positive for all t, as we assumed. Hence equation (2.7) can be

written as

d

dt[µ(t)y] = µ(t)g(t)

Integrating,

µ(t)y =

∫µ(t)g(t)dt+ k

where k is an arbitrary constant. However, in general this is not possible, so the

general solution of the equation (2.6)

y =1

µ(t)

[∫ t

t0µ(s)g(s)ds+ k

](2.8)

where t0 is some convenient lower limit of the integration.

Method of solving linear differential equation

Let us consider a linear equation

dy

dt+ p(t)y = g(t) (2.9)

where p and g are functions of t or constants.

Integrating factor

I.F. = e∫pdt

Multiplying the equation (2.9) by I.F., we get

e∫pdtdy

dt+ e

∫pdtp(t)y = g(t)e

∫pdt

or,d

dt

(e∫pdty

)= g(t)e

∫pdt

or, d(e∫pdty

)= g(t)e

∫pdtdt

Integrating,

e∫pdty =

∫ (g(t)e

∫pdtdt

)dt+ c

which gives the solution.

Example 17.


Solve the differential equation dydt = −ay + b.


dy

dt= −ay + b

or,dy

dt+ ay = b (2.10)

Integrating factor

I.F. = e∫pdt = e

∫adt = eat

Multiplying the equation (2.10), we get

eatdy

dt+ ayeat = beat

or,d

dt

(eaty

)= beat

or, d(eaty

)= beatdt

Integrating,

eaty = b

∫eatdt+ c

or, eaty =b

aeat + c

or, y =b

a+ ce−at

Example 18.

Solve the differential equation dydt − 2y = 4− t. Discuss the behavior of solution

as t→∞.


dy

dt− 2y = 4− t (2.11)

Integrating factor

I.F. = e∫pdt = e

∫−2dt = e−2t


e−2tdy

dt− 2ye−2t = 4e−2t − te−2t

or,d

dt

(e−2ty

)= 4e−2t − te−2t

or, d(e−2ty

)= 4e−2tdt− te−2tdt


Integrating,

e−2ty = 4

∫e−2tdt−

∫te−2tdt+ c

or, e−2ty =4

−2e−2t −

t

∫e−2t −

∫ (dt

dt

∫e−2tdt

)dt

+ c

or, e−2ty = −2e−2t −− t

2e−2t −

∫−1

2e−2tdt

dt+ c

or, e−2ty = −2e−2t +t

2e−2t +

1

4e−2t + c

or, e−2ty = −7

4e−2t +

1

2te−2t + c

or, y = −7

4+

1

2t+ ce2t

As t→∞, e2t →∞ for a positive value of c and y(t)→∞.

Example 19.

Find the general solution of the differential equation dydt + 3y = t+ e−2t. Discuss

the behavior of solutions as t→∞.


dy

dt+ 3y = t+ e−2t (2.12)

Integrating factor

I.F. = e∫pdt = e

∫3dt = e3t


e3tdy

dt+ 3ye3t = te3t + e3te−2t

or,d

dt

(e3ty

)= te3t + et

or, d(e3ty

)= te3t + et

Integrating,

e3ty =

∫te3tdt+

∫etdt+ c

or, e3ty = t

∫e3t −

∫ (dt

dt

∫e3tdt

)dt+ et + c

or, e3ty =1

3te3t −

∫1

3e3tdt+ et + c

or, e3ty =1

3te3t − 1

9e3t + et + c

or, y =t

3− 1

9+ e−2t + ce−3t


Since the exponential functions e−2t and e−3t decrease rapidly as compare to the

linear function t3 −

19 , so e−2t → 0 and e−3t → 0 as t → ∞. Thus, solution y is

asymptotic to t3 −

19 and y →∞ as t→∞.

Example 20.

Find the general solution of the differential equation dydt − 2y = t2e2t. Discuss

the behavior of solutions as t→∞.


dy

dt− 2y = t2e2t (2.13)

Integrating factor

I.F. = e∫pdt = e

∫−2dt = e−2t


e−2tdy

dt− 2ye−2t = t2e−2te2t

or,d

dt

(e−2ty

)= t2

or, d(e−2ty

)= t2dt

Integrating,

e−2ty =

∫t2dt+ c

or, e−2ty =t3

3+ c

or, y =1

3t3e2t + ce2t

AS y →∞ as t→∞ for all positive c.

Example 21.

Find the general solution of the differential equation tdydt + 2y = sin t, t > 0,

and use it to determine how solutions behave as t→∞.


tdy

dt+ 2y = sin t t > 0

or,dy

dt+

2

ty =

sin t

t(2.14)

Integrating factor

I.F. = e∫pdt = e

∫2tdt = e2 ln t = eln t2 = t2



t2dy

dt+ 2ty = t sin t

or,d

dt

(t2y)

= t sin t

or, d(t2y)

= t sin tdt

Integrating,

t2y =

∫t sin tdt+ c

or, t2y = t

∫sin tdt−

∫ (dt

dt

∫sin t dt

)+ c

or, t2y = −t cos t+

∫cos t dt+ c

or, t2y = −t cos t+ sin t+ c

or, y =c+ sin t− t cos t

t2

For behaviour of solution as t→∞

limt→∞

y = limt→∞

c+ sin t− t cos t

t2

= limt→∞

(c

t2+

sin t

t2− cos t

t

)= 0

Example 22.

Find the general solution of the differential equation (1 + t2)dydt + 4ty = 1(1+t2)2

,

and use it to determine how solutions behave as t→∞.


(1 + t2)dy

dt+ 4ty =

1

(1 + x2)2

or,dy

dt+

4t

1 + t2y =

1

(1 + t2)3(2.15)

Integrating factor

I.F. = e∫pdt = e

∫4t

1+t2dt

= e2∫

2t1+t2

dt= e2 ln(1+t2) = eln(1+t2)2 = (1 + t2)2


(1 + t2)2dy

dt+ 4t(1 + t2)y =

1

1 + t2

or,d

dt

((1 + t2)2y

)=

1

1 + t2

or, d((1 + t2)2y

)=

1

1 + t2dt


Integrating,

(1 + t2)2y =

∫1

1 + t2dt+ c

or, (1 + t2)2y = tan−1 t+ c

or, y =c+ tan−1 t

(1 + t2)2

For behaviour of solution as t→∞

limt→∞

y = limt→∞

c+ tan−1 t

(1 + t2)2= lim

t→∞

1

4t(1 + t2)2= 0

using the L-Hopital rule.

Example 23.

Solve the initial value problem tdydt + 2y = 4t2, y(1) = 2.


tdy

dt+ 2y = 4t2, y(1) = 2

or,dy

dt+

2

ty = 4t (2.16)

Integrating factor

I.F. = e∫pdt = e

∫2tdt = e2



t2dy

dt+ 2ty = 4t3

or,d

dt

(t2y)

= 4t3

or, d(t2y)

= 4t3 dt

Integrating,

t2y = 4

∫t3dt+ c

or, t2y = t4 + c

or, y = t2 +c

t2(2.17)

which is general solution.

Using the initial condition y(1) = 2 in (2.17), we get

1 + c = 2

or, c = 1

From (2.17), the required solution is

y = t2 +1

t2


Example 24.

Solve the initial value problem 2dydt + ty = 2, y(0) = 1.


2dy

dt+ ty = 2, y(0) = 1

or,dy

dt+t

2y = 1 (2.18)

Integrating factor

I.F. = e∫pdt = e

∫t2dt = e

t2

4


et2

4dy

dt+t

2et2

4 y = et2

4

or,d

dt

(ye

t2

4

)= e

t2

4

or, d

(ye

t2

4

)= e

t2

4 dt

Integrating,

yet2

4 =

∫et2

4 dt+ c

or, yet2

4 =

∫et2

4 dt+ c

which gives the general solution.

The integral on the right sides can not be evaluated in term of the usual elementary

functions, so we leave the integral as unevaluate. However, by choosing the lower

limit of the integral as initial point t0 = 0. Thus the solution is

yet2

4 =

∫ t

0es2

4 ds+ c (2.19)

Using the initial conditions t = 0, y = 0, from (2.19), we get

e0 1 =

∫ 0

0es2

4 ds+ c

or, c = 1

From (2.19), we get

et2

4 y =

∫ t

0es2

4 ds+ 1

or, y = e−t2

4

∫ t

0es2

4 ds+ e−t2

4


Example 25.

Solve the initial value problem tdydt + 2y = t2 − t+ 1, y(1) = 12 .


tdy

dt+ 2y = t2 − t+ 1, y(1) =

1

2

or,dy

dt+

2

ty = t− 1 +

1

t(2.20)

Integrating factor

I.F. = e∫pdt = e



t2dy

dt+ 2ty = t3 − t2 + t

or,d

dt

(t2y)

= t3 − t2 + t

or, d(t2y)

= (t3 − t2 + t) dt

Integrating,

t2y =

∫(t3 − t2 + t)dt+ c

or, t2y =t4

4− t3

3+t2

2+ c

or, y =t2

4− t

3+

1

2+c

t2(2.21)


or,1

4− 1

3+

1

2+ c =

1

2

or, c =1

12


y =t2

4− t

3+

1

2+

1

12

is solution of the initial value problem.

Example 26.

Solve the initial value problem dydt − 4y = e4t, y(0) = 4.


dy

dt− 4y = e4t, y(0) = 4

or,dy

dt− 4y = e4t (2.22)


Integrating factor

I.F. = e∫pdt = e

∫−4dt = e−4t


e−4tdy

dt− 4e−4ty = e−4te4t

or,d

dt

(e−4ty

)= 1

or, d(e−4ty

)= dt

Integrating,

e−4ty = t+ c

or, y = (t+ c)e4t (2.23)



c = 2


y = (t+ 2)e4t

Example 27.

Solve the initial value problem t3 dydt + 4t2y = e−t, y(−1) = 0, t < 0.


t3dy

dt+ 4t2y = e−t, y(−1) = 0, t < 0

or,dy

dx+

4

t=e−t

t3(2.24)

Integrating factor

I.F. = e∫pdt = e



t4dy

dt+ 4t3y = te−t

or,d

dt

(t4y)

= te−t

or, d(t4y)

= te−tdt


Integrating,

t4y =

∫te−tdt+ c

t4y = t

∫e−tdt−

∫ (dt

dt

∫e−tdt

)dt+ c

or, t4y = −te−t +

∫e−tdt+ c

or, t4y = −te−t − e−t + c (2.25)


Using the initial condition y(−1) = 0 in (2.25), we get

(−1)4 · 0 = −(−1)e1 − e1 + c

or, c = 0


t4y = −te−t − e−t

or, y = −(t+ 1)e−t

t4

Example 28.

Solve the initial value problem tdydt + (t+ 1)y = t, y(ln 2) = 1, t > 0.


tdy

dt+ (t+ 1)y = t, y(ln 2) = 1, t > 0

or,dy

dx+

(1 +

1

t

)y = 1 (2.26)

Integrating factor

I.F. = e∫pdt = e

∫(1+ 1

t )dt = et+ln t = eteln t = tet


tetdy

dt+ tet

(1 +

1

t

)y = tet

or,d

dt

(tety

)= tet

or, d(tety

)= tetdt

Integrating,

tety =

∫tetdt+ c

tety = t

∫etdt−

∫ (dt

dt

∫etdt

)dt+ c

or, tety = tet −∫etdt+ c

or, tety = tet − et + c (2.27)



Using the initial condition t = ln 2, y(ln 2) = 1 in (2.27), we get

ln 2 eln 2 · 1 = ln 2 eln 2 − eln 2 + c

or, 0 = −2 + c

or, c = 2


tety = tet − et + 2

or, y =tet

tet− et

tet+

2

tet

or, y = 1− 1

t+

2e−t

t

or, y =t− 1 + 2e−t

t

Example 29.

Consider the initial value problem

dy

dx− 3

2y = 3t+ 2et, y(0) = y0

Find the value of y0 that separates solutions that grow positively as t → ∞ from

those that grow negatively. How does the solution that corresponds to the critical

value of y0 behaves as t→∞.


dy

dt− 3

2y = 3t+ 2et, (2.28)

Integrating factor

I.F. = e∫pdt = e

∫− 3

2dt = e−

3t2


e−3t2dy

dt− 3

2e−

3t2 y = 3te−

3t2 + 2ete−

3t2

or, d(e−

3t2 y)

=(

3te−3t2 + 2e−

t2

)dt

Integrating,

e−3t2 y = 3

∫te−

3t2 dt+ 2

∫e−

t2dt

e−3t2 y = 3t

∫e−

3t2 dt− 3

∫ (dt

dt

∫e−

3t2 dt

)dt− 4e−

t2 + c

or, e−3t2 y = −2te−

3t2 + 2

∫e−

3t2 dt− 4e−

t2 + c

or, e−3t2 y = −2te−

3t2 − 4

3e−

3t2 − 4e−

t2 + c

or, y = −2t− 4

3− 4et + ce3t/2 (2.29)



Using the initial condition t = 0, y(0) = y0 in (2.29), we get

y0 = −4

3− 4 + c

or, y0 = −16

3+ c

or, c = y0 +16

3

The critical value of y0 that separates solutions that grow positively as t→∞ from

those that grow negatively is when c = 0. i.e.

y0 = −16

3

For this critical value of y0 the solution is

y = −2t− 4

3− 4et

As −2t and −4et both diverge to−∞ as t→∞. Also, y → −∞ as t→∞.

2.2 Separable Equations

The general first order differential equation is

dy

dx= f(x, y) (2.30)

If f(x, y) has in the form f(x, y) = M(x)N(y) then the differntial equation is called

variable separeted form. The equation(2.30) becomes

dy

dx=M(x)

N(y)

or, N(y)dy = M(x)dx

Integrating, we get the solution∫N(y)dy =

∫M(x)dx+ c

Example 30.

Solve the given differential equations

(a) y′ =x2

1− y2, (b) y′ =

3x2 − 1

3 + 2y

(c) y′ =x2

y, (d) y′ =

x2

1 + y2

2.2. SEPARABLE EQUATIONS 41

(e) y′ = cos2 x cos2 y (f) y′ =x− e−x

y + ey

Solution: (a) The given differential equation is

dy

dx=

x2

1− y2

or, (1− y2)dy − x2dx = 0

Integrating,

y − y3

3− x3

3= c

(b) The given differential equation is

dy

dx=

3x2 − 1

3 + 2y

or, 3dy + 2ydy − 3x2dx+ dx = 0

Integrating,

3y + y2 − x3 + x = c

(c) The given differential equation is

dy

dx=x2

y

or, ydy − x2dx = 0

Integrating,

y2

2− x3

3= c

or, 3y2 − 2x3 = c

(d) The given differential equation is

dy

dx=

x2

1 + y2

or, dy + y2dy − x2dx = 0

Integrating,

y +y3

3− x3

3=c

3or, 3y + y3 − x3 = c


(e)The given differential equation is

dy

dx= cos2 x cos2 y

or, sec2 ydy − cos2 xdx = 0

or, sec2 ydy −(

1

2+

1

2cos 2x

)dx = 0

Integrating,

tan y − 1

2x− 1

4sin 2x = c

(d) The given differential equation is

dy

dx=x− e−x

y + ey

or, ydy + eydy − xdx+ e−xdx = 0

Integrating,

y2

2+ ey − x2

2− e−x = c

Example 31.


dy

dx=

4x− x3

4 + y3

Also determine the solution passing through (0, 1).

Solution: The given differential equation

dy

dx=

4x− x3

4 + y3

or, (4 + y3)dy = (4x− x3)dx

Integrating,

4y +y4

4= 2x2 − x4

4+c

4or, y4 + 16y + x4 − 8x2 = c (2.31)

where c is arbitrary constant. Using the initial conditions x = 0, y = 1 in (2.31), we

get

1 + 16 = c, c = 16

Putting the value of (2.31)

y4 + 16y + x4 − 8x2 = 16


Example 32.


dy

dx=

3x2 + 4x+ 2

2(y − 1), y(0) = −1

and determine the interval in which the solution exists.


dy

dx=

3x2 + 4x+ 2

2(y − 1)

or, 2(y − 1)dy = (3x2 + 4x+ 2)dx

Integrating,

y2 − 2y = x3 + 2x2 + 2x+ c (2.32)

where c is arbitrary constant. Using the initial conditions x = 0, y = −1 in (2.32)

1 + 2 = c, c = 3


y2 − 2y = x3 + 2x2 + 2x+ 3

or, y2 − 2y − (x3 + 2x2 + 2x+ 3) = 0

Solving for y

y =2±

√4 + 4(x3 + 2x2x+ 3)

2

or, y = 1±√x3 + 2x2 + 2x+ 4

Thus we get two solutions

y = 1 +√x3 + 2x2 + 2x+ 4 (2.33)

y = 1−√x3 + 2x2 + 2x+ 4 (2.34)

When x = 0, from (2.33), we get y = 1 +√

4 = 3, which is not solution. When

x = 0,from (2.34), we get y = 1−√

4 = −1 i.e. (2.34) satisfied the initial condition.

Thus the solution of the initial value problem is

y = 1−√x3 + 2x2 + 2x+ 4

or, y = 1−√x2(x+ 2) + 2(x+ 2)

or, y = 1−√

(x2 + 2)(x+ 2)

(2.35)

When x > −2 the fuction (x2+2)x−(−2) is positive and y = 1−√

(x2 + 2)(x+ 2)

is defined. Thus the initial value problem has solution in [−2,∞).


Example 33.


y′ = 2y2 + xy2, y(0) = 1

and determine where the solution attains its minimum value.


dy

dx= 2y2 + xy2

or,dy

dx= y2(x+ 2)

or,1

y2

dy

dx= x+ 2

or,1

y2dy = xdx+ 2dx

Integrating, ∫1

y2

dy

dx=

∫xdx+ 2

∫dx+ c

or, − 1

y=x2

2+ 2x+ c (2.36)

Using initial condition y(0) = 1, in (2.36) ,

−1 = c

Now, from the equation (2.36)

−1

y=x2

2+ 2x− 1

or, y =−2

x2 + 4x− 2(2.37)

is the required solution.

For a minimum value of y,

y′ = 0

or, y2(x+ 2) = 0

or, x = −2

Also,

y′ = y2(x+ 2)

Differentiating

y′′ = 2yy′(x+ 2) + y2 (2.38)

or, y′′ = 4y3(x+ 2)3 + y2 (2.39)


When x = −2, y = −24−8−2 = 1

3 6= 0 and from (2.39), we get

y′′ = 4y2 · 0 + y2 = y2 > 0

Hence, the solution attains minimum value at x = −2.

Example 34.


y′ = 2(1 + x)(1 + y2), y(0) = 0

and determine where the solution attains its minimum value.


dy

dx= 2(1 + x)(1 + y2)

or,dy

dx= 2(1 + y2)(x+ 1)

or,1

1 + y2

dy

dx= 2(1 + x)

or,1

1 + y2dy = 2(1 + x)dx

Integrating, ∫1

1 + y2dy =

∫2xdx+ 2

∫dx+ c

or, tan−1 y = x2 + 2x+ c (2.40)

Using initial condition y(0) = 0, in (2.40) ,

0 = c

Now, from the equation (2.40)

tan−1 y = x2 + 2x

or, y = tan(x2 + 2x) (2.41)

is the required solution.

For a minimum value of y,

y′ = 0

or, (1 + y2)(x+ 1) = 0

or, x = −1


Also,

y′ = 2(1 + y2)(x+ 1)

Differentiating

y′′ = 4yy′(x+ 1) + 2(1 + y2)

or, y′′ = 8y(1 + y2)(x+ 1)2 + 2(1 + y2) (2.42)

When x = −1, y = tan(−1) 6= 0 and from (2.42), we get

y′′ = 0 + 2(1 + y2) > 0

Hence, the solution attains minimum value at x = −1.

2.3 Modeling with First Order Equations

Some examples that are applications of first order differential equations.

2.3.1 Mixture Problem

Example 35.

At time t = 0 a tank contains Q0 lb of salt dissolved in 100 gal of water. Assume

that water containing 14 lb of salt/gal is entering the tank at a rate of r gal/min and

that well-strirred mixture is draining from the tank at the same rate.

1. Set up the initial value problem that describes this flow process.

2. Find the amount of salt Q(t) in the tank at any time.

3. Find the limiting amount QL that is present after a very long time.

4. If r = 3 and Q0 = 2QL, find the time T after the salt level is within 2% of QL.

5. Find the flow rate that is required if the T is not to exceeds 45 min.

Solution: Here Q(t) is amount of salt at any time t.

1. The rate of change of salt in the tank dQdt , is equal to the rate at which salit

flowing in minus the rate at which it is flowing out.

dQ

dt= rate in− rate out (2.43)

Again, rate in = concentration × rate in= 14 × r = r

4 lb/min.

Since the rates of flow in and out are equal, the volume of water in the tank remains

constant at 100 gal and well-stirred, the concentration throughout the tank is same,

so concentration outQ(t)

100lb/gal

Therefore, the rate at which salt leaves the tank is

= concentration× rate out =rQ(t)

100lb/min

2.3. MODELING WITH FIRST ORDER EQUATIONS 47

Thus, the differential equation governing this process, from (2.43) is

dQ

dt=r

4− rQ(t)

100(2.44)

The initial condition is

Q(0) = Q0 (2.45)

2. Rewritting the equation (2.44),

dQ

dt+rQ(t)

100=r

4(2.46)

Thus, the integrating factor is

I.F. = e∫pdt = er/100

∫dt = ert/100


ert/100dQ

dt+

r

100ert/100y =

r

4tert/100

or, d(ert100Q

)=r

4ert/100dt

Intrgrating,

Qert/100 =r

4

∫ert/100dt+ c

or, Qert/100 =r

4

100

rert/100 + c

or, Qert/100 = 25ert/100 + c

or, Q(t) = 25 + ce−rt/100 (2.47)

Using the initial condition Q(0) = Q0, in (2.47), we get

Q0 = 25 + c =⇒ c = Q0 − 25

From (2.47)

Q(t) = 25 + (Q0 − 25)e−rt/100 (2.48)

which gives the concentration of salt at any time t.

3. As t→∞, Q(t)→ 25. Therefore, QL = 25.

4. Now suppose that r = 3 and Q0 = 2QL = 50, then equation (2.48) becomes

Q(t) = 25 + (50− 25)e−3t/100

or, Q(t) = 25 + 25e−0.03t (2.49)


Again Q(t) = 25 + 2%of 25 = 25.5. Thus from (2.49)

25.5 = 25 + 25e−0.003t

or, 0.5 = 25e−0.003t

or, e0.03t =25

0.5= 50

or, 0.03t = ln 50 taking log on both sides

or, t =ln 50

0.03= 130.4min.

5. To determine r so that t = 45 min. we return to the equation (2.48), take

Q(t) = 25.5, Q0 = 50, we get

25.5 = 25 + 25 e−45r100

or, 0.5 = 25 e−45r100

or, e45r100 = 50

or,45r

100= ln 50

or, r =100

45ln 50 = 8.89gal/min.

Example 36.

Consider a tank used in certain hydrodynamic experiments. After one experi-

ment, the tank contains 150L of a dye solution with a concentration of 1 g/L. To

prepare for the next experiment, the tank is to be rinsed with fresh water flowing

in at a rate of 2L/min, the well-stirred solution flowing out at the same rate. Find

the time that will elapse before the concentration of the dye in the tank reaches 1%

of its original value.

Solution: Initially Q(0) = Q0 = 1g/L × 150L = 150g. Let Q(t)g be the amout of

dye in the tank at any time t in min.

The fresh water has the concerntration 0%.

Rate in = Concentration× rate = 0× 2 = 0

At any time amout of dye is Q(t), in 150 L of water, so concentration of dye at any

timeQ(t)

150

Since the solution is flowing out at the same rate as it enter i.e. the solution is

flowing out at rate 2L/min.

Rate out = Concentration× rate =Q(t)

150× 2 =

Q(t)

75


∴dQ


or,dQ

dt= 0− Q(t)

75

or,dQ

Q= − 1

75dt

or,

∫dQ

Q= − 1

75

∫dt+ c

or, lnQ = − t

75+ ln c

or, Q = ce−t75 (2.51)

Using the initial condition Q(0) = 150, we get

150 = c

Putting, value of c, in (2.51)we get

Q = 150 e−t75 (2.52)

Again, Q = 1% of 150=1.5, so from (2.52),

1.5 = 150 e−t75

or, et75 = 100

or,t

75= ln 100

or, t = 75 ln 100 = 345.4 min. (2.53)

Example 37.

A tank is initally contains 120L of pure water. A mixture containing a con-

centration γg/L, of salts enters the tank at a rate of 2L/min and the well-stirred

mixture leave out at the same rate. Find an expression in term of γ for the amount

of salt in the tank at any time t. Also find the limiting amount of salt in the tank

as t→∞.Solution: Initially Q(0) = Q0 = 0g/L× 120L = 0g. Let Q(t) be the amout of salt

in the tank at any time t in min.

.

Rate in = Concentration× rate = γ × 2 = 2γ

At any time amout of dye is Q(t), in 120 L of water, so concentration of salt at any

timeQ(t)

120Since the solution is flowing out at the same rate as it enter i.e. the solution is



120× 2 =

Q(t)

60


dQ

dt= rate in− rate out

or,dQ

dt= 2γ − Q(t)

60

or,dQ

Q+Q(t)

60= 2γ (2.54)

Which is linear,

I.F. = e∫pdt = e

160

∫dt = e

t60

Multiplying (2.54) by I.F.

et60dQ

Q+Q(t)

60et60 = 2γe

t60 (2.55)

or,d

dt

(Qe

t60

)= 2γe

t60 (2.56)

or, d(Qe

t60

)= 2γe

t60dt (2.57)

Integrating,

Q et60 = 2γ

∫et60dt+ c

Q et60 = 120γe

t60 + c (2.58)


c = −120γ


Q et60 = 120γe

t60 − 120γ

Q(t) = 120γ − 120γe−t60

Q(t) = 120γ(

1− e−t60

)(2.59)

As t→∞, Q(t) = 120γ.

Example 38.

A tank originally contains 100 gal of water. Then the water containing 12 lb of

salt per gallon is poured into a rate of 2gal/min, and the mixture is allowed to leave

at the same rate. After 10 min the process is stopped, and fresh water is poured

into the tank at a rate of 2gal/min, with the mixture again leaving at the same rate.

Find the amount of salt in the tank at the end of additional 10 min.

Solution: For first 10 minutes

Initially, for fresh water concerntration is 0, Q(0) = Q0 = 0 lb/gal × 100gal = 0lb.

Let Q(t) lb be the amout of dye in the tank at any time t in min.

Rate in = Concentration× rate =1

2lb/gal× 2gal/min = 1 lb/min


At any time amout of salt is Q(t) lb, in 100 gal of water, so concentration of dye at

any timeQ(t)

100




100× 2 =

Q(t)

50

dQ


or,dQ

dt= 1− Q(t)

50

or,dQ

dt=

50−Q50

or,dQ

50−Q=

1

50dt

or,

∫dQ

50−Q=

1

50

∫dt+ c

or, − ln(50−Q) =t

50+ c (2.61)


− ln 50 = c


− ln(50−Q) =t

50− ln 50

or, ln 50− ln(50−Q) =t

50

or, ln50

50−Q=

t

50

or,50

50−Q= e

t50

or,50−Q

50= e−

t50

or, 50−Q = 50e−t50

or, Q = 50(

1− e−t50

)(2.62)

When t = 10 min.

Q = 50(

1− e−1050

)= 9.0635lb


For next additional 10 min.

Initially, concerntration is 0, Q(0) = Q0 = 9.0635 lb. Let Q(t) lb be the amout of

salt in the tank at any time t in min.

Rate in = Concentration× rate = 0 lb/gal× 2gal/min = 0 lb/min

At any time amout of salt is Q(t) lb, in 100 gal of water, so concentration of dye at

any timeQ(t)

100




100× 2 =

Q(t)

50

dQ


or,dQ

dt= 0− Q(t)

50

or,dQ

Q= − dt

50

or, lnQ = − t

50+ ln c

or, Q = ce−t50 (2.64)

Using the initial condition Q(0) = 9.0635 lb, we get

9.0635 = c


Q = 9.0635e−t50 (2.65)

After the 10 min additional time

Q = 9.0635e−1050 = 7.4205lb

Example 39.

A tank with capacity of 500 originally contains 200 gal of water with 100 lb of salt

in solution. Water containing 1lb of salt per gallon is entering at a rate of 3gal/min,

and the mixture is allowed to flow out of the tank at the rate of 2 gal/min. Find

the amount of salt in the tank at any time prior to the instant when the solution

begin to overflow. Find the concentration of salt in the tank when it is on the point

of overflow. Compare this concentration with the theoritical limiting concentration


if the tank had infinite capacity.

Solution: Initially Q(0) = Q0 = 200lb. Let Q(t)g be the amout of dye in the tank

at any time t in min.

Rate in = Concentration× rate = 1lb/gal × 3gal/min = 3 lb/min.

Let us assume that the solution overflow after time t. Since the solution is collected

at the rate of 1gal/min. Hence, in t minutes the collected solution is 200 + t gallons.

Therefore the concentration isQ(t)

200 + t


200 + t× 2 =

2Q(t)

200 + t

dQ

dt= rate in− rate out

or,dQ

dt= 3− Q(t)

200 + t

or,dQ

dt+

2Q(t)

200 + t= 3 (2.66)

Thus, the integrating factor is

I.F. = e∫pdt = e

∫2

200+tdt = e2 ln(200+2) = eln(200+t)2 = (200 + t)2


(200 + t)2dQ

dt+ 2(200 + t)y = 3(200 + t)2

or, d((200 + t)2Q

)= 3(200 + t)2dt

Intrgrating,

(200 + t)2Q = (200 + t)3 + c

or, Q = 200 + t+c

(200 + t)2(2.67)

Using the initial condition Q(0) = 100, in (2.67), we get

100 = 200 +c

40000

c = −4000000 = −4× 106

From (2.67)

Q = 200 + t− 4× 106

(200 + t)2


But the total collection 200 + t, if t is time of follow the

200 + t = 500

or, t = 300

‘So the required solution is

Q = 200 + t− 4× 106

(200 + t)2if t < 300 (2.68)

At the point of overflow i.e. t = 300, the concerntation of salt

Q(300) = 200 + 300− 4× 106

(200 + 300)2= 484lb

The concentration of the salt at that time

c(300) =Q(300)

volume=

484

500= 0.968 lb/gal

Also, the concerntration of salt any time t is

c(t) =Q(t)

volume at time t=

200 + t− 4×106

(200+t)2

200 + t

If the tank had infinite capacity it takes infinite minutes to fill up the tank.Thus,

taking t→∞

limt→∞

c(t) = limt→∞

(1− 4× 106

(200 + t)3

)= 1

2.3.2 Compound Interest

The value S(t) of the investment at any time t depends on the frequency with

which interest is compounded as well as on the interest rate. If we assume that

compounding takes place continuously, then we can set up a simple initial value

problem.

The rate of change of the value of the investment is dSdt , and this quantity is equal

to the rate at which interest accrues which is the interest rate r times the current

value of the investment S(t). Thus

dS

dt= rS (2.69)

Suppose that we also know the value of investment at some time particular time,

say

S(0) = S0 (2.70)


From the equation (2.69),

dS

S= rdt

or, lnS = rt+ ln c

or, lnS − ln c = rt

or, lnS

c= rt

or,S

c= ert

or, S = cert (2.71)

Using the initial condition S(0) = S0 in 2.71, we get

S0 = c

From (2.71), we get

S = S0ert

Again, let us suppose that there may be deposites or with drawals in addition to

the accrual interest, dividents, or capital gain. If we assume that the deposite or

withdrawals take place at a constant rate k, then the equation (2.69) is replaced by

dS

dt= rS + k

or,dS

dt− rS = k (2.72)

where k is positive for deposits and negative for withdrawls. The equation (2.72) is

linear . Integrating factor

I.F. = e∫

(−r)dt = e−rt

Multiplying the equation (2.72) by I.F.

e−rtdS

dt− rS = ke−rt

or,d

dt

(Se−kt

)= ke−rt

Integrating,

or, Se−rt = −krke−rt + c

S = cert − k

r(2.73)

Let the initial investment be S0. Then we get

c = S0 +k

r


Thus, the solution of the IVP (3.4) is

S(t) = S0ert +

k

r

(ert − 1

)(2.74)

If there is no initial investment i.e. S0 = 0, the equation (2.74) becomes

S(t) =k

r

(ert − 1

)(2.75)

Example 40.

A young person with no initial capital invest k dollars per year at of return r.

Assume that investments are made continuously and that return is compounded

continuosly.

(a) Determine the sum S(t) accumulated at any time t.

(b) If r = 7.5%, determine k so that $1 milion will be available for retirement in 40

years.

(c) If k = $2000/year, determine the return rate r that must be obtained to have $1

million avialable in 40 years.

Solution: 1.Let S(t) be amount accumulate at any time t. Also $k/year be annual

deposite and r be the annual rate of return. Then

dS

dt= rS + k

or,dS

dt− rS = k (2.76)

where k is positive for deposits and negative for withdrawls. The equation (2.76) is

linear . Integrating factor

I.F. = e∫

(−r)dt = e−rt


e−rtdS

dt− rS = ke−rt

or,d

dt

(Se−kt

)= ke−rt

Integrating,

or, Se−rt = −krke−rt + c

S = cert − k

r(2.77)

Since initial investment be S(0) = 0. Then from (2.77)

0 = c− k

r

or, c =k

r(2.78)


S(t) =k

r

(ert − 1

)(2.79)

2. Here r = 7.5% = 0.075 S(t) = $1000000, t = 40 years and k =?. Then from

(2.79, we get

1000000 =k

0.075

(e0.075×40 − 1

)or, 100000× 0.075 = k(20.0855− 1)

or, k =1000000

19.0855= $49787

3.Here k−20000/year S(t) = $1000000, t = 40 years and r =?. Then from (2.79,

we get

1000000 =20000

r

(e40r − 1

)or, 100 = k(20.0855− 1)

Solving the graphic calculator or Matlab we get r = 0.0977 = 9.77%.

Example 41.

Suppose that a sum S0 is investeted at an annual rate of return r compounded

continuously.

1. Find the time T required for the original sum to double in value as a function of

r.

2. Determine T if r = 8%.

3. Find the return rate that must be achived if the initial investment is to double

in 8 years.

Solution: Let S(t) be the investment at any time t. Here a sum S0 is investeted

at an annual rate of return r% compounded continuously.

∴ S(0) = S0

The rate of change of invsetment dSdt is equal to the rate of interest times the current

value .

dS

dt= rS

dS

S= rdt

or, lnS = rt+ ln c

or, lnS − ln c = rt

or, lnS

c= rt

or,S

c= ert

or, S = cert (2.80)


Using the initial condition S(0) = S0 in (2.80), we get

S0 = c

From (2.71), we get

S = S0ert (2.81)

1. When t = T, S = 2S0, from (2.80), we get

2S0 = S0erT

or, 2 = erT

or, ln 2 = rT ln e

or, T =ln 2

r(2.82)

2. if r = 8% per annum = 0.08 then (2.82),

T =ln 2

r=

ln 2

0.08= 8.66 years

3. If T = 8 years and S = 2S0 then from (2.82)

2S0 = S0e8r

or, 2 = erT

or, ln 2 = 8r ln e

or, r =ln 2

8= 0.0866 = 8.66%

2.3.3 Newton’s Law of Cooling

Newton’s law of cooling state that the temperature of an object changes at a rate

proportional to the difference between its temperature and that of its surrounding.

Example 42.

Suppose that the temperature of a cup of coffee obeys Newton’s Law of cooling.

If the coffee has a temperature of 200oF when freshly poured and 1 min later has

cooled to 190oF in a room at 70oF , determine when the coffee reaches a temperature

of 150oF .

Solution: Let T oF be the temperature of coffee at any time t min. Then according

to Newton’s law of cooling, we have

dT

dt= k(T − Ts)

where Ts is the temperature of the surrounding and k is the constant of proportion-

ality. But Ts = 70oF and T (0) = 200oF . Hence initial value problem becomes

dT

dt= k(T − 70), T (0) = 200oF (2.83)


From (2.83),

dT

T − 70= kdt

or, ln(T − 70) = kt+ ln c

or, ln(T − 70)− ln c = kt

or, lnT − 70

c= kt

or,T − 70

c= ekt

or, T (t) = 70 + cekt (2.84)

Using the initial condition T (0) = 200oF , we get

200 = 70 + c

or, c = 130

Putting the value of c = 130 in (2.84), we get

T (t) = 70 + 130 ekt (2.85)

When t = 1 min, T = 190oF , so from (2.85), we get

190 = 70 + 130ek

or, 120 = 130 ek

or, ek =120

130

or, k = ln12

13= −0.08

Substituting the value of (2.85), we get

T (t) = 70 + 130 e−0.08t (2.86)

which gives the temperature of coffee at time t. When T = 150, from (2.86)

150 = 70 + 130 e−0.08t

or, 80 = 130 e−0.08t

or, e0.08t =130

80

or, 0.08t = ln130

80

or, t =1

0.08ln

130

80= 6.07 min.


2.4 Difference Between Linear and Nonlinear Equations

We have discussed a number of initial value problems, each had a solution and

apparently only one solution. This raises the question of wheather this is true of all

initial value problems for first order equations. The following fundamental theorem

gives the sufficient condition for the existence and uniqueness of solution for an

initial value problem involving a linear differential equation of first order.

Theorem 1.

If the functions p(t) and g(t) are continuous on an open interval I = (α, β)

containing the point t = t0, then there exists a unique function y = φ(t) that

satisfies the differential equation

dy

dt+ p(t)y = g(t) (2.87)

for each t ∈ I and that also satisfies the initial condition

y(t0) = y0 (2.88)

where y0 is an arbitrary prescribed initial value.

Example 43.

Find an interval in which the solution of the given initial value problem is certain

to exist.

1. tdy

dt+ 2y = 4t2, y(1) = 2

2. (t− 3)dy

dt+ (ln t)y = 2t, y(1) = 2

3. t(t− 5)dy

dt+ y = 0, y(2) = 1

4.dy

dt+ (tan t)y = sin t, y(π) = 0

5. (4− t2)dy

dt+ 2ty = 3t2, y(−3) = 1

6. (ln t)dy

dt+ y = cot t, y(2) = 3

Solution: 1. The given initial value problem is

tdy

dt+ 2y = 4t2, y(1) = 2 (2.89)

Rewritting the equation in the standard form, we have

dy

dt+

2

ty = 4t

2.4. DIFFERENCE BETWEEN LINEAR AND NONLINEAR EQUATIONS 61

Comparing this equation with

dy

dt+ p(t) y = g(t)

p(t) =2

t, g(t) = 4t

Thus, g(t) = 4t is continuous for all t, while p(t) = 2t is not continuous at t = 0

and continuous only for t < 0 or t > 0. The interval t > 0 contains the initial

point t = 1; consequently, the initial value problem (2.89) has unique solution on

the interval 0 < t <∞.

2. The given initial value problem is

(t− 3)dy

dt+ ln t y = 2t, y(1) = 2 (2.90)


dy

dt+

ln t

t− 3y =

2t

t− 3


dy

dt+ p(t) y = g(t)

p(t) =ln t

t− 3, g(t) =

2t

t− 3

Thus, g(t) = 2tt−3 is continuous for all t except t = 3, while p(t) = ln t

t−3 is not

continuous at t = 3 and t ≤ 0; and is continuous only for 0 < t < 3 or t > 3. The

interval 0 < t < 3 contains the initial point t = 1; consequently, the initial value

problem (2.90) has unique solution on the interval 0 < t < 3.


t(t− 5)dy

dt+ y = 0, y(1) = 2 (2.91)


dy

dt+

1

t(t− 5)y = 0


dy

dt+ p(t) y = g(t)

p(t) =1

t(t− 3), g(t) = 0

Thus, g(t) = 0 is continuous for all t, while p(t) = 1t(t−5) is not continuous at t = 0, 5

; and continuous for t < 0 or 0 < t < 5 or t > 5. The interval 0 < t < 5 contains the


initial point t = 1; consequently, the initial value problem (2.91) has unique solution

on the interval 0 < t < 5.


dy

dt+ tan t y = sin t (2.92)


dy

dt+ p(t) y = g(t)

p(t) = tan t, g(t) = sin t

Thus, g(t) = sin t is continuous for all t, while p(t) = tan t is not continuous at

t = (2n+1)π2 ; and continuous for

R−

(2n+ 1)π

2, n ∈ Z

The interval π

2 < t < 3π2 contains the initial point t = π; consequently, the initial

value problem (2.92) has unique solution on the interval π2 < t < 3π

2 .


(4− t2)dy

dt+ 2ty = 3t2, y(−3) = 1 (2.93)


dy

dt+

2t

4− t2y =

3t2

4− t2


dy

dt+ p(t) y = g(t)

p(t) =2t

4− t2, g(t) =

3t2

4− t2

Thus, g(t) = 2t4−t2 and p(t) = 3t2

4−t2 are continuous for all t except t = ±2. The

interval t < −2 contains the initial point t = −3; consequently, the initial value

problem (2.93) has unique solution on the interval −∞ < t < −2.


(ln t)dy

dt+ y = cot t, y(2) = 1 (2.94)


dy

dt+

1

ln ty =

cot t

ln t



dy

dt+ p(t) y = g(t)

p(t) =1

ln t, g(t) =

cot t

ln t

Thus, g(t) = 1ln t is continuous on 0 < t < ∞ except t = 1 and p(t) = cot t

ln t is

continuous for all t > 0, except t = nπ : n ∈ N and t = 1. The interval 1 < t < π

contains the initial point t = 2; consequently, the initial value problem (2.94) has

unique solution on the interval 1 < t < π.

Turning to the nonlinear differential equation, the existance and uniqueness theorem

is

Theorem 2.

Let the functions f and ∂f∂y be continuous in some rectangle α < t < β, γ < y < δ

containing the point (t0, y0). Then, in some interval t0 − h < t < t0 + h containing

in α < t < β, there is a unique solution y = φ(t) of the initial value problem

dy

dt= f(t, y), y(t0) = y0

Example 44.

State where in the ty−plane, the following equations have solutions

1.dy

dx= (1− t2 − y2)1/2

1.dy

dx=

t− y2t+ 5y

3.dy

dx=

ln |ty|1− t2 + y3

Solution: 1. The given differential equation is

dy

dt= (1− t2 − y2)1/2 (2.95)

Comparing (2.95) with dydt = f(t, y) we get

f(t, y) = (1− t2 − y2)1/2

Thus, f(t, y) is continuous if 1− t2 − y2 ≥ 0 i.e.

t2 + y2 ≤ 1 (2.96)

Also

∂f

∂y=

1

2(1− t2 − y2)−1/2(−2y) =

2y√1− t2 − y2


which is continuous if 1− t2 − y2 > 0, i.e.

t2 + y2 < 1 (2.97)

Hence from (2.96) and (2.97), f and ∂f∂t are continuous in t2 + y2 < 1. Hence, the

given equation has unique solution in interior of unit circle t2 + y2 < 1.

2. The given differential equation is

dy

dt=

t− y2t+ 5y

(2.98)

Comparing (2.98) with dydx = f(t, y) we get

f(t, y) =t− y

2t+ 5y

Thus, f(t, y) is continuous if 2t+ 3y 6= 0 i.e.

2t+ 3y > 0 or, 2t+ 3y < 0 (2.99)

Also

∂f

∂y=

(2t+ 5y)(−1)− 5(t− y)

(2t+ 5y)2=−7t+ 10y

(2t+ 5y)2

. Thus, ∂f∂y is continuous if 2t+ 3y 6= 0 i.e.

2t+ 3y > 0 or, 2t+ 3y < 0 (2.100)

Hence f and ∂f∂t are continuous in 2t + 3y > 0 or 2t + 3y < 0. Hence, the given

equation has unique solution in 2t+ 3y > 0 or 2t+ 3y < 0.

3. The given differential equation is

dy

dt=

ln |ty|1− t2 + y2

(2.101)

Comparing (2.98) with dydx = f(t, y) we get

f(t, y) =ln |ty|

1− t2 + y2

Thus, f(t, y) is continuous if 1− t2 + y2 6= 0 i.e.

1− t2 + y2 > 0 or, 1− t2 + y2 < 0 (2.102)

and t 6= 0, y 6= 0. Also

∂f

∂y=

(1− t2 + y2)(

1ty

)− 2y ln |ty|

(1− t2 + y2)2(2.103)

which is continuous if 1− t2 − y2 6= 0, and t 6= 0, y 6= 0 i.e. Hence from (2.102) and

(2.103), f and ∂f∂t are continuous in 1 − t2 − y2 6= 0, and t 6= 0, y 6= 0. Hence, the

given equation has unique solution in 1− t2 − y2 6= 0, and t 6= 0, y 6= 0.


Example 45.

Solve the initial value problem and determine how the interval in which the

solution exists depends on the initial value y0,dydt = −4t

y .

Solution: Here,

dy

dt= −4t

y, y(0) = y0

or, ydy = −4tdt

or,y2

2= −4t2

2+c

2or, y2 = c− 4t2 (2.104)

When t = 0, y(0) = y0, from (2.104), we get

y20 = c


y2 = y20 − 4t2

or, y = ±√y2

0 − 4t2, y0 6= 0 (2.105)

Thus, the value of y given by (2.105) is defined and continuous if and only if

y20 − 4t2 > 0

or, t2 <y2

0

4

or, |t| < |y0|2

So, the required interval is

0 < |t| < |y0|2

Example 46.

Solve the initial value problem and determine how the interval in which the

solution exists depends on the initial value y0,dydt = t2

y(1+t3).

Solution: Here,

dy

dt=

t2

y(1 + t3), y(0) = y0

or, ydy =t2

1 + t3dt

or, 3ydy =3t2

1 + t3dt

or,3y2

2= ln(t3 + 1) + c (2.106)


When t = 0, y(0) = y0, from (2.106), we get

3y20

2= c


3y2

2= ln(t3 + 1) +

3y20

2

or, y2 =2

3ln(t3 + 1) + y2

0,

or, y =

√2

3ln(t3 + 1) + y2

0, y0 6= 0 (2.107)

Thus, the value of y given by (2.105) is defined and continuous if and only if

2

3ln(t3 + 1) + y2

0 > 0

or,2

3ln(t3 + 1) > −y2

0

or, ln(t3 + 1) > −3

2y2

0

or, t3 + 1 > e−32y20

or, t > −1− e−32y201/3

or, −1− e−32y201/3 < t <∞

2.4.1 Bernoulli Equation

Sometimes it is possible to solve a nonlinear equation by making a change of the

dependent variable that convert it into a linear equation. The most important such

equation has the form

y′ + p(t)y = q(t)yn

and is called a Bernoulli equation.

Example 47.

1. Solve Bernoulli equation y′ + p(t)y = q(t)yn when n = 0; and n = 1.

2. Show that if n 6= 0, 1, then substitution v = y1−n reduces Bernoulli equation to

a linear equation.

Solution: We have Bernoulli’s equation

dy

dt+ p(t)y = q(t)yn (2.108)

1(a) When n = 0, then the equation (2.108) becomes

dy

dt+ p(t)y = q(t) (2.109)


which is linear.

Integrating factor

I.F. = e∫pdt


e∫pdtdy

dt+ pe

∫p(t)y = q(t)e

∫p(t)dt

or,d

dt

(e∫p(t)dty

)= q(t)e

∫p(t)dt

or, d(e∫p(t)dty

)= e

∫p(t)dt dt

Integrating,

e∫p(t)dty =

∫q(t)e

∫p(t)dtdt+ c

or, e∫p(t)dty =

∫q(t)e

∫p(t)dtdt+ c


(b) When n = 1, then the equation (2.108) becomes

dy

dt+ p(t)y = q(t)y

or,1

y

dy

dt= q(t)− p(t)

or,1

ydy = (p(t)− q(t))dt

∴ ln y =

∫(p(t)− q(t))dt+ ln c

or, y(t) = ce∫

(p(t)−q(t))dt

2. Again, the Bernoulli equation is

dy

dt+ p(t)y = q(t)yn

or,1

yndy

dt+

1

yn−1p(t) = q(t) (2.110)

Let

z =1

yn−1= y−(n−1)

. Thendz

dt= −(n− 1)

yndy

dt

− 1

n− 1

dz

dt=

1

yndy

dt


From (2.110)

− 1

n− 1

dz

dt+ p(t)z = q(t)

or,dz

dt+ (1− n)z = (1− n)q(t)

Example 48.

Solve the Bernoulli’s equation y′ = ε y − σy3 and σ > 0. (This equation occurs

in the study of stability of fluid flow)

Solution: The Bernoulli equation is

dy

dt− ε y = −σy3

or,1

y3

dy

dt− ε 1

y2= −σ (2.111)

Let

z =1

y2

. Thendz

dt= − 2

y3

dy

dt

−1

2

dz

dt=

1

y3

dy

dt

From (2.110)

−1

2

dz

dt− εz = −σ

or,dz

dt+ 2εz = 2σ

which is linear.

Integrating factor

I.F. = e∫

2ε dt = e2εt


e2εtdz

dt+ 2ε e2εtz = 2σe2εt

or,d

dt

(e2εtz

)= 2σe2εt

or, d(e2εtz

)= 2σe2εt dt

2.5. AUTONOMOUS EQUATION AND POPULATION DYNAMICS 69

Integrating,

e2εtz = 2σ

∫e2εtdt+ c

or, e2εtz =2σ

2εe2εt + c

or, z =σ

ε+ ce−2εt

or,1

y2=

1

ε(σ + cεe−2εt)

or, y = ±√

ε

σ + cεe−2εt(2.112)


2.5 Autonomous Equation and Population Dynamics

An important class of first order equation consists of those in which the independent

variable does not appear. Such equations are called autonomous and have the form

dy

dt= f(y) (2.113)

Such type of equations occurs in the contest of the growth or decay of population

of a species, economic and medicine.

2.5.1 Exponential Growth

The simplest hypothesis concerning the variation of population is that the rate of

change of y proportional to the current value of y, that is

dy

dt= ry (2.114)

where the constant of proportionality r is called the rate of growth or decline, de-

pending on whether it is positive or negative. Let r > 0. i.e. the population is

increasing. Let us consider the initial condition

y(0) = y0 (2.115)

From (2.114)

dy

y= rdt

or, ln y = rt+ ln c (integrating)

or, y = cert (2.116)


Using the initial condition y(0) = y0, we get

c = y0

From (2.116), we get

y = y0 ert (2.117)

Thus the mathematical model consisting of the initial value problem (2.114), (2.115),

with r > 0 predicts that the population will grow exponentially for all time. In

practice, this can not continue indefinitely, limitations on space, food supply or other

resourses will reduce the growth rate and bring an end to uninhibited exponetial

growth.

2.5.2 Logistic Growth

The growth rate depends on the current population y

dy

dt= ry (2.118)

Let us replace the constant r by a function h(y) and therefore the modified equation

dy

dt= h(y) y (2.119)

We want to choose h(y) such that

h(y) =

r > 0 when y is small

decrease as y grow large and

h(y) < 0 when y is sufficiently large

(2.120)

The simplest function that has these properties is

h(y) = r − ay

where a is also positive constant. Using this in the Eq. (2.119), we get

dy

dt= (r − ay)y (2.121)

Equation (2.121) is known as the Verhult equation or the Logistic equation. The

equation (2.121) can be written as

dy

dt= r

(1− a

ry)y

or,dy

dt= r

(1− y

K

)y where K =

r

a(2.122)


The constant r is called the intrinsic growth rate that is the growth rate in the

absence of limiting factors.

First we seek simple solution of (2.122). Let y be constant. Then

dy

dt= 0

or, r(

1− y

K

)y = 0

=⇒ y = 0, y = K

Thus the constant solutions are y = φ1(t) = 0 and y = φ2(t) = K. These solutions

are called the equalibrium solutions of Eq. (2.122), beacues the correspond to no

change or variation in the value of y as t increases.

The equilibrium solutions can be obtained by locating the roots of f(y) = 0 of the

differential equation dydt = f(y). The zeros of f(y) are also called critical points.

Thus the equalibrium points of (2.122) are (0, 0) and (K, 0).

Let

f(y) = r(

1− y

K

)y

= ry − ry2

K

= − r

K(y2 −Ky)

= − r

K

(y2 − 2 · y · K

2+K2

4

)+

r

K

K2

4

= − r

K

(y − K

2

)2

+rK

4(2.123)

Here f(y) is a parabola and the coefficient of y2 is negative, so it is concave downward

and vertex(K2 ,

rK4

). Again

dy

dt= (K − y)

ry

K=

> 0 for 0 < y < K

< 0 for y > K(2.124)

Hence y is an increasing function of t when 0 < y < K and y is a decreasing

function of t, when y > K.

In this context the y−axis is often called the phase line and it is reproduced in its

more customary vertical orientation. The dots at y = 0 and y = K are the critical

points or equilibrium solutions. The arrows in figure indicates that y is increasing

for 0 < y < K and y is decreasing for y > K.

Further if near to y = 0 or y = K, then the slope f(y) is near to zero, so the

solution curve are relatively flat. They becomes steeper as the value of y leaves the

neighborhood of zero or K.

To sketch the graph of solution of (2.122) in ty−plane, we start with the equilibrium


Figure 2.1: f(y) versus y for dydx = f(y) = r

(1− y

K

)y

solutions y = 0 and y = K, then we draw other curves that are increasing when

0 < y < K, decreasing when y > K, and flatten out as y approaches either of the

values 0 or K. Thus the graph of solutions of equation (2.122) must have the general

shape is shown in Figure.

Figure 2.2: Logistic growth dydx = r

(1− y

k

)y (a) The phase line (b) plot of y versus

t.

Again, f(y) = ry−ay2 is a quadractic polynomial in y and ∂f∂y = r−2ay is linear.

Both f and ∂f∂y are continuous for all t ∈ R. Thus the equation (2.122) has only

one solution passes through a given point in ty−plane. So the curve does intersect

each other. Althogh other solutions may be asymptotic to the equilibrium solution


as t→∞, they can not intersect at a finite time.

We can determine the concavity of the solution curves and locate the point of in-

flection.

We have

dy

dt= f(y)

or,d2y

dt2=

d

dt(f(y)) =

d

dy(f(y))

dy

dt= f ′(y)f(y)

=

> 0 if both f(y) and f ′(y) have the same sign

< 0 if both f(y) and f ′(y) have the opposite signs

=

Concave upward if both f(y) and f ′(y) have the same sign

Concave downward if both f(y) and f ′(y) have the opposite signs

Hence

f(y) =dy

dt= (K − y)

ry

K=

> 0 for 0 < y < K

< 0 for y > K(2.125)

f ′(y) = r − 2ry

K=rK − 2ry

K

=2r

K

(K

2− y)

=

> 0 for 0 < y < K

2

< 0 for y > K2

(2.126)

Interval f(y) from Eq.(2.125) f ′(y) from Eq.(2.126) y′′ = f(y)f ′(y) y is concave

0 < t < K2 +ve +ve +ve upwards

K2 < t < K +ve −ve −ve downwards

t > K −ve −ve +ve upwards

The line y = K2 ,separates the graph of the solution from concave upward to from

downward. So, the graphs are sigmoidal (S-shaped). Since K is the upper bound

that is approached, but not exceeded, by growing populations starting below this

value. Thus, it is natural to refer to K as saturated level. for the given species. The

last row of the table is not applicable.

Analytic soltion The logistic model is from (2.122)

dy

dt= r

(1− y

K

)y where K =

r

a

or,dy(

1− yK

)y

= rdt

or,

1

y+

1/k

1− yk

dy = rdt

(2.127)

Integrating,

ln |y| − ln∣∣∣1− y

k

∣∣∣ = rt+ ln c (2.128)


where the constant is determined from the initial conditions y(0) = y0, the initial

ppopulation of the species.

Case I: For 0 < y0 < K, 0 < y < K, both y and 1− Ky are positive. From (2.128),

we get

ln y − ln(

1− y

k

)= rt+ ln c

or, ln

(y

1− yk

)= rt+ ln c

or,y

1− yk

= cert (2.129)


y0

1− y0k

= c

From (2.129)

y

1− yk

=y0

1− y0k

ert

or,1− y

k

y=

1− y0k

y0e−rt

or,1

y− 1

K=

(1

y0− 1

K

)e−rt

or,1

y=

1

K+

(1

y0− 1

K

)e−rt

or,1

y=y0 + (K − y0)e−rt

ky0

or, y =ky0

y0 + (K − y0)e−rt(2.130)

which gives the solution of initial value problem for 0 < y0 < K.

For the long time

limt→∞

y = limt→∞

Ky0

y0 + (K − y0)e−rt=Ky0

y0= K

Thus, for each y0 > 0, the solution approaches the equilibrium solution y = K as

t → ∞. So, the constant solution y = K is an asymptotically stable solution of

the logistic model. As r increases, the solutions approach the equilibrium solution

y = K, more rapidly.

Case II: For y0 > K, y > K, y is positive and 1− Ky is negative. From (2.128), we

get

ln y − ln(yk− 1)

= rt+ ln c

or, ln

(yyk

− 1

)= rt+ ln c

or,y

yk − 1

= cert (2.131)



y0y0k − 1

= c

From (2.129)

yyk − 1

=y0

y0k − 1

ert

or,yk − 1

y=

y0k − 1

y0e−rt

or,1− y

k

y=

1− y0k

y0e−rt

or,1

y− 1

K=

(1

y0− 1

K

)e−rt

or,1

y=

1

K+

(1

y0− 1

K

)e−rt

or,1

y=y0 + (K − y0)e−rt

ky0

or, y =ky0

y0 + (K − y0)e−rt

which gives the solution of initial value problem for 0 < y0 < K.

For the long time

limt→∞

y = limt→∞

Ky0

y0 + (K − y0)e−rt=Ky0

y0= K

Thus, for each y0 > 0, the solution approaches the equilibrium solution y = K as

t → ∞. So, the constant solution y = K is an asymptotically stable solution of

the logistic model. As r increases, the solutions approach the equilibrium solution

y = K, more rapidly.

2.5.3 A Critical Threshold

We consider the another logistic equation

dy

dt= −r

(1− y

T

)y (2.132)

where r and T are given positive constants, T is the critical amplitude. This equation

is differs from the logistic equation (2.122) only in the presence of the minus sign.

The graph of f(y) = −r(1− y

T

)y versus y is the downward parabola .

f(y) = 0

or, y = 0 = φ1, y = T = φ2(t)


as the critical points or the equilibrium soutions . Thus, (0, 0) and (0, T ) are critical

points of Eq. (2.132).

f(y) = −r(

1− y

T

)y

= −ry +ry2

T

=r

K(y2 − Ty)

=r

T

(y2 − 2 · y · T

2+T 2

4

)− r

T

T 2

4

=r

T

(y − T

2

)2

− rT

4(2.133)

Hefe f(y) is a parabola and the coefficient of y2 is positive, so it is concave upward

and vertex(T2 ,−

rT4

). Again

dy

dt= − (T − y)

ry

K=

< 0 for 0 < y < T

> 0 for y > T(2.134)

Figure 2.3:

Hence y is an decreasing function of t when 0 < y < T and y is a increasing

function of t, when t > T .

In this context the y−axis is often called the phase line and it is reproduced in its


more customary vertical orientation. The dots at y = 0 and y = K are the critical

points or equilibrium solutions. The arrows in fig indicate that y is decreasing for

0 < y < T and y is increasing for t > T . Hence y = 0 is asymptotically stable and

y = K


flection.

We have

dy

dt= f(y)

or,d2y

dt2=

d

dt(f(y)) =

d

dy(f(y))

dy

dt= f ′(y)f(y)

=



=



Hence

f(y) =dy

dt= − (T − y)

ry

T=

< 0 for 0 < y < T

> 0 for y > T(2.135)

f ′(y) = −r +2ry

T=

2ry − rTT

=2r

T

(y − T

2

)=

< 0 for 0 < y < T

2

> 0 for y > T2

(2.136)

Interval f(y) from Eq.(2.125) f ′(y) from Eq.(2.126) y′′ = f(y)f ′(y) y is concave

0 < t < T2 −ve −ve +ve upwards

T2 < t < T −ve +ve −ve downwards

t > T +ve +ve +ve upwards

The line y = T2 ,separates the graph of the solution from concave upward to from

downward.

Also y > T , the curve is increasing and concave upward. We conclude that as time

inceases, y either approaches to 0 or grows without bound, depending on wheather

the initail value y0 is less than or greater than T . Hence T is a thresthold level,

below which the population dies out.

However we get solution of (2.132) as replacing −r by r, and K by T in (2.130)

or, y =Ty0

y0 + (T − y0)ert(2.137)

If 0 < y0 < T ,as t → ∞, y → 0. This agrees with our qualitative geomeric

analysis.


Figure 2.4:

If y0 > T then let

y0 + (T − y0)ert′

= 0

or ert′

=y0

y0 − T> 1

or, rt′ = ln

(y0

y0 − T

)or, t′ =

1

rln

(y0

y0 − T

)> 0

If y0 > T then the denominator of the equation (2.137), is zero and the population

becomes unbounded in a finite value of t.

2.5.4 Logistic Growth with a Threshold

The threshold model

dy

dt= −r

(1− y

T

)y (2.138)

may need to modified so that unbounded growth does not occur when y is above

the threshold T . The simplest way to do this to introduce another factor that will

have the effect of making dydt negative when y is large. For this we consider

dy

dt= −r

(1− y

T

)(1− y

K

)y = f(y) (2.139)

where r > 0 and 0 < T < K.


Example 49.

Sketch the graph of f(y) versus y, determine the critical points and classify each

one as asymptotically stable or unstable. Draw the phase line and draw graph of

several graphs of solution in the ty−plane for

dy

dt= ay + by2, a > 0, b > 0, −∞ < y0 <∞


dy

dt= ay + by2 = f(y), a > 0, b > 0, −∞ < y0 <∞ (2.140)

Let

dy

dt= 0

or ay + by2 = 0

or, y = 0, y = −ab

Hence critical points are y = 0 and y = − ba .

Interval dydt = f(y) = a(y − 0)y −

(−ab

) y is

−∞ < y < −ab

dydt = f(y) = a(y − 0)−vey −

(ab

)−ve = +ve increasing

−ab < y < 0 dy

dt = f(y) = a(y − 0)−vey −(ab

)+ve = −ve decreasing

y > 0 dydt = f(y) = a(y − 0)+vey −

(ab

)+ve = +ve increasing


flection.

We have

dy

dt= f(y)

or,d2y

dt2=

d

dt(f(y)) =

d

dy(f(y))

dy

dt= f ′(y)f(y)

=



=



Hence

f(y) =dy

dt= ay(a+ by) =

> 0 for −∞ < y < −a

b

< 0 for − ab < y < 0

> 0 for y > 0

(2.141)

f ′(y) = a+ 2by = 2b[y −

(− a

2b

)]=

< 0 for y < − a

2b

> 0 for y > − a2b

(2.142)


Sign change points are −ab , −

a2b , 0

Interval f(y) from (2.141) f ′(y) from(2.142) y′′ = f(y)f ′(y) y is concave

−∞ < y < −ab +ve −ve −ve downward

−ab < y < − a

2b −ve −ve +ve upward

− a2b < y < 0 −ve +ve −ve downward

y > 0 +ve +ve +ve upward

Using all these, the graph of y(t) in ty−plane, the of f(y) versus y and phase diagram

is shown as figure.

Figure 2.5:

Example 50.




dy

dt= y(y − 1)(y − 2), −∞ < y0 <∞


dy

dt= y(y − 1)(y − 2)quad−∞ < y0 <∞ (2.143)

Let

dy

dt= 0

or y(y − 1)(y − 2) = 0

or, y = 0, y = 1, y = 2


Hence critical points are y = 0, y = 1 and y = 2.

Interval dydt = f(y) = (y − 0)(y − 1)(y − 2) y is

−∞ < y < 0 dydt = f(y) = (y − 0)−ve(y − 1)−ve(y − 2)−ve = −ve decreasing

0 < y < 1 dydt = f(y) = (y − 0)+ve(y − 1)−ve(y − 2)−ve = +ve increasing

1 < y < 2 dydt = f(y) = (y − 0)+ve(y − 1)+ve(y − 2)−ve = −ve decreasing

y > 2 dydt = f(y) = (y − 0)+ve(y − 1)+ve(y − 2)+ve = +ve increasing


flection.

We have

dy

dt= f(y)

or,d2y

dt2=

d

dt(f(y)) =

d

dy(f(y))

dy

dt= f ′(y)f(y)

=



=



Hence

f(y) =dy

dt= (y − 0)(y − 1)(y − 2) =

< 0 for −∞ < y < 0

> 0 for 0 < y < 1

< 0 for 1 < y < 2

> 0 for y > 2

(2.144)

f(y) = y3 − 3y2 + 2y

f ′(y) = 3y2 − 6y + 2

= 3

(y2 − 2y +

2

3

)= 3

(y2 − 2y · 1 + 12 − 1 +

2

3

)= 3

(y − 1)2 − 1

3

= 3

y −

(1− 1√

3

)y −

(1 +

1√3

)

=

> 0 for −∞ < y < 1− 1√

3

< 0 for 1− 1√3< t < 1 + 1√

3

> 0 for t > 1 + 1√3

(2.145)


Sign change points are 0, 1− 1√3, 1, 1 + 1√

3, 2


−∞ < y < 0 −ve +ve −ve downward

and decreasing

0 < y < 1− 1√3

+ve +ve +veupward

and increasing

1− 1√3< y < 1 +ve −ve −ve downward

and increasing

1 < y < 1 + 1√3

−ve −ve +veupward

and decreasing

1 + 1√3< y < 2 −ve +ve −ve downward

and decreasing

y > 2 +ve +ve +veupnward

and increasing

Using all these, the graph of y(t) in ty−plane, the of f(y) versus y and phase diagram

is shown as figure.figure y = 1 is stable but y = 0 and y = 2 are unstable.

Figure 2.6:

Example 51.





dy

dt= y2(y2 − 1), −∞ < y0 <∞


dy

dt= y2(y2 − 1), −∞ < y0 <∞ (2.146)

Let

dy

dt= 0

or y2(y2 − 1) = 0

or, y = 0, y = 1, y = −1

Hence critical points are y = −1, y = 0 and y = 1.

Interval dydt = f(y) = y2 [y − (−1)] (y − 1) y is

−∞ < y < −1 dydt = f(y) = y2

+ve [y − (−1)]−ve(y − 1)−ve = +ve increasing

−1 < y < 0 dydt = f(y) = y2

+ve [y − (−1)]+ve(y − 1)−ve = −ve decreasing

0 < y < 1 dydt = f(y) = y2

+ve [y − (−1)]+ve(y − 1)−ve = −ve decreasing

y > 1 dydt = f(y) = y2

+ve [y − (−1)]+ve(y − 1)+ve = +ve increasing


flection.

We have

dy

dt= f(y)

or,d2y

dt2=

d

dt(f(y)) =

d

dy(f(y))

dy

dt= f ′(y)f(y)

=



=




Hence

f(y) =dy

dt= y2 [y − (−1)] (y − 1) =

> 0 for −∞ < y < −1

< 0 for − 1 < y < 0

< 0 for 0 < y < 1

> 0 for y > 1

(2.147)

f(y) = y4 − y2

f ′(y) = 4y3 − 2y

= 4y

(y2 − 1

2

)= 3

y −

(− 1√

2

)(y − 1√

2

)

=

> 0 for −∞ < y < − 1√

2

< 0 for − 1√2< t < 1√

2

> 0 for t > 1√2

(2.148)

Sign change points are −1,− 1√2, 0, 1√

2, 1


−∞ < y < −1 +ve +ve +vedownward

& increasing

−1 < y < − 1√2

−ve +ve −ve upward

& decreasing

− 1√2< y < 0 −ve −ve +ve

downward

& decreasing

0 < y < 1√2

−ve −ve +veupward

and decreasing

1√2< y < 1 −ve +ve −ve downward

and decreasing

y > 1 +ve +ve +veupnward

and increasing


See figure and check a latter

Figure 2.7:

Example 52.

Suppose that a certain population obeys that the logistic equation dydt = ry

(1− y

K

)(a) If y0 = K

4 , find the time τ at which the initial population has doubled. Find the

value of τ corresponding to r − 0.025.

(b) If y0K = α, find the time T at which y(T )

K = β where 0 < α, β < 1. Observe that

T →∞ as α→ 0 or as β → 1. Find the value of T for r = 0.025 per year, α = 0.1,

and β = 0.9.

Solution: The given logistic model is

dy

dt= r

(1− y

K

)y y(0) = y0

or,dy(

1− yK

)y

= rdt

or,

1

y+

1/k

1− yk

dy = rdt

(2.149)


Integrating,

ln |y| − ln∣∣∣1− y

k

∣∣∣ = rt+ ln c (2.150)


population of the species.

(a)For 0 < y0 = K4 < K, 0 < y < K, both y and 1 − K

y are positive. From (2.154),

we get

ln y − ln(

1− y

k

)= rt+ ln c

or, ln

(y

1− yk

)= rt+ ln c

or,y

1− yk

= cert (2.151)


y0

1− y0k

= c

From (2.151)

y

1− yk

=y0

1− y0k

ert

or,1− y

k

y=

1− y0k

y0e−rt

or,1

y− 1

K=

(1

y0− 1

K

)e−rt

or,1

y=

1

K+

(1

y0− 1

K

)e−rt

or,1

y=y0 + (K − y0)e−rt

Ky0

or, y =Ky0

y0 + (K − y0)e−rt

or,y

K=

y0

y0 + (K − y0)e−rt(2.152)


If y0 = K4 , then we have to find the time τ when y = 2y0 = K

2 . From (2.152),

1

2=

K4

K4 + (K − K

4 )e−rτ

or,1

2=

14

14 + 3

4e−rτ

or,1

4+

3

4e−rτ =

1

2

or,3

4e−rτ =

1

4

or, e−rτ =1

3or, erτ = 3

or, rτ = ln 3

or, τ =1

rln 3 =

1

0.025ln 3 = 43.95 years

(b) Again, from (2.152)

y

K=

y0

y0 + (K − y0)e−rt

or,y

K=

y0K

y0K +

(1− y0

K

)e−rt

But y0K = α, t = T and β = y

K .

β =α

α+ (1− α)e−rT

or,α

β= α+ (1− α)e−rT

or,α

β− α = (1− α)e−rT

or,α(1− β)

β= (1− α)e−rT

or, erT =β(1− α)

α(1− β)

or, rT = ln

(β(1− α)

α(1− β)

)Hence, as α→ 0 or β → 1, then T →∞.

Also, taking α = 0.1, β = 0.9 and r = 0.025 per year.

T =1

0.025ln

(0.9× 0.9

0.1× 0.1

)= 175.78 years

Example 53.


The logistic model has been applied to the natural growth of the haibut popula-

tion in certain areas of the pacific Ocean. Let y, measure in kilograms, be the total

mass or biomass, of the halibut population at time t. The parameters in the logistic

equation are estimated to have the values r = 0.71/ year and K = 80.5 × 106 kg.

If the initial biomass is y0 = 0.25K, find the biomass 2 years latter. Also find the

time τ for which y(τ) = 0.75K.

Solution: The given logistic model is

dy

dt= r

(1− y

K

)y y(0) = y0

or,dy(

1− yK

)y

= rdt

or,

1

y+

1/k

1− yk

dy = rdt

(2.153)

Integrating,

ln |y| − ln∣∣∣1− y

k

∣∣∣ = rt+ ln c (2.154)


population of the species.

(a)For 0 < y0 = K4 < K, 0 < y < K, both y and 1 − K

y are positive. From (2.154),

we get

ln y − ln(

1− y

k

)= rt+ ln c

or, ln

(y

1− yk

)= rt+ ln c

or,y

1− yk

= cert (2.155)


y0

1− y0k

= c


From (2.151)

y

1− yk

=y0

1− y0k

ert

or,1− y

k

y=

1− y0k

y0e−rt

or,1

y− 1

K=

(1

y0− 1

K

)e−rt

or,1

y=

1

K+

(1

y0− 1

K

)e−rt

or,1

y=y0 + (K − y0)e−rt

Ky0

or, y =Ky0

y0 + (K − y0)e−rt

or,y(t)

K=

y0K

y0K + (1− y0

K )e−rt(2.156)

Using the given data y0K = 0.25, r = 0.71/year, t = 2 years and K = 80.5× 106 kg.

y(2)

80.5× 106=

0.25

0.25 + 0.75e−1.42= 0.5797

or, y(2) = 0.5797× 80.5× 106 = 46.7× 106kg

To find τ when y(τ) = 0.75K, solving (2.156)

y(y0 + (K − y0)e−rτ

)= Ky0

or, 0.75K(0.25K + (K − 0.25K)e−rτ

)= 0.25K2

or, 0.75K2(0.25 + 0.75e−0.71τ ) = 0.25K2

or, 0.75(0.25 + 0.75e−0.71τ ) = 0.25

or, 9 = e0.71τ

or, ln 9 = 0.71τ

or, τ =ln 9

0.71= 3.095 years (2.157)

Example 54.

A pound form as water collects in a conical depression of radius a and depth h.

Suppose that water flows in at a constant rate K and is lost through evaporation at

the rate proportional to the surface area.

(a) Show that the volume V (t) of water in the pond at time t satisfies the differential

equation

dV

dt= k − α

(3a

πh

)2/3

V 2/3

where α is the coefficient of evaporation.

(b) Find the equilibrium depth of water in the pound. Is the equilibrium asymtoti-

cally stable?


(c) Find a condition that must be satisfied if the pond is not to overflow?

Solution: Here we take a pound form as water collects in a conical depression of

radius a and depth h. Let V be volume, r be the radius of top and l be the depth

of water at any time t in the pound.

Figure 2.8:

Since the water is colecting at a constant rate k and the rate of decrease of

evaporation is porportinal to the upper surface area of the water . Thus the rate of

change of the volume of the water is given by

dV

dt= k − απr2 (2.158)

But

r

l=a

h=⇒ l =

rh

a

Again,

V =1

3πr2l =

1

3πr2 rh

a=

1

3aπr3

or, r3 =3V a

πh

or, r =

(3V a

πh

)1/3

2.6. EXACT EQUATION AND INTEGRATING FACTOR 91

Putting the value of r in (2.158)

dV

dt= k − απ

(3V a

πh

)2/3

(b) For equilibrium

dV

dt= 0

or, k − απr2 = 0

or, r2 =k

απ

or, r =

√k

απ

∴ l =rh

a=h

a

√k

απ

When the water depth increase, more area is exposed for evaporation which again

tends to decrease the water depth. When the water depth decrease, due to less

evaporation, the depth increase. Hence the equilibrium depth is asymptotically

stable.

(c) For the pond is not to overflow, we must have

l ≤ h

=⇒ h

a

√k

απ≤ h

=⇒ k2

απ≤ a2

=⇒ k

α≤ πa2

which is a required condition.

2.6 Exact Equation and Integrating Factor

For the first order equations there are a number of integration methods are applicable

to various problem. Here we consider a class of equations, which are exact.

2.6.1 Exact Equation

Let a differential equation

M(x, y) +N(x, y)dy

dx= 0 (2.159)

be given. Suppose that we can identify a function ψ(x, y) such that

∂ψ

∂x= M(x, y) and

∂ψ

∂y= N(x, y) (2.160)


and such that ψ(x, y) = c defines y = φ(x) implicitly as a differential function of x.

Then

M(x, y) +N(x, y)dy

dx=∂ψ

∂x+∂ψ

∂y

dy

dx=

d

dxψ(x, y)

From the differential equation (2.160), becomes

d

dxψ(x, y) = 0 (2.161)

Integrating,

ψ(x, y) = c (2.162)

whcin is solution of (2.159).

Thus, a first order differential equation

M(x, y) +N(x, y)dy

dx= 0

is said to be exact if there exist a function ψ(x, )y such that

d

dxψ(x, y) = M(x, y) +N(x, y)

dy

dxor, d(ψ(x, y)) = M(x, y)dx+N(x, y)dy

Let a differential equation is not exact. If it is possible to make exact by multi-

plying it an appropriate factor u(x, y), which is called an integrating factor for the

differential equation.

Theorem 3.

Let the functions

M,N,∂M

∂y, and

∂N

∂x

be continuous in the rectangular region R : α < x < β; γ < y < δ. Then

M(x, y) +N(x, y)dy

dx= 0

is an exact differential equation in R if and only if

∂M

∂y=∂N

∂x

That is , there exists a function ψ(x, y) satisfying such that

∂ψ

∂x= M(x, y),

∂ψ

∂y= N(x, y)

if and only if M and N satisfy the equation

M(x, y) +N(x, y)dy

dx= 0

Steps for solving the exact differentail equation of first order


1. Make the given equation in the form

M(x, y) +N(x, y)dy

dx= 0

2. if ∂M∂y = ∂N

∂x then it is exact. Otherwise, do not follow it.

3. Consider∂ψ

∂x= M(x, y),

∂ψ

∂y= N(x, y)

4. Integrate ∂ψ∂x = M(x, y) with respect to x, to get ψ.

ψ = φ+ h(y) (2.163)

Here h is an unknown function of y and which is constant of integration.

5. Differentiating, (2.163) partially with respect to y, we get

∂ψ

∂y= φy + h′(y)

6. Equate

∂ψ

∂y= N(x, y)

or, φy + h′(y) = N(x, y)

or, h′(y) = N(x, y)− φy

Integrating, we get h(y).

7. Putting the value of h(y) in (2.163), we get the required solution.

Example 55.

Solve the differential equation

y cosx+ 2xey + (sinx+ x2ey − 1)dy

dx= 0

Solution: Comparing the equation

y cosx+ 2xey + (sinx+ x2ey − 1)dy

dx= 0 (2.164)

with

M(x, y) +N(x, y)dy

dx= 0 (2.165)

we get

M = y cosx+ 2xey, N = sinx+ x2ey − 1


∴∂M

∂y= cosx+ 2xey,

∂N

∂x= cosx+ 2xey

∂M

∂y=∂N

∂x(2.166)

Hence, the given differential equation is exact. Thus, thier exists a function ψ(x, y)

such that

∂ψ

∂x= M(x, y),

∂ψ

∂y= N(x, y)

or,∂ψ

∂x= y cosx+ 2xey, (2.167)

∂ψ

∂y= sinx+ x2ey − 1 (2.168)

Integrating with respect to x, the equation (2.167), we get

ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

= y

∫cosxdx+ 2ey

∫xdx+ h(y)

= y sinx+ x2ey + h(y) (2.169)

But from (2.168), we get

N(x, y) =∂ψ

∂y

or, sinx+ x2ey − 1 = sinx+ x2ey + h′(y)

=⇒ h′(y) = −1

=⇒ h(y) = −y + c

Hence, from (2.169), we get

ψ(x, y) = y sinx+ x2ey − y + c = 0


Example 56.

Determine whether the equation

(2x+ 4y) + (2x− 2y)dy

dx= 0

is exact. If it is exact, find the solution.


(2x+ 4y) + (2x− 2y)dy

dx= 0 (2.170)


with

M(x, y) +N(x, y)dy

dx= 0

we get

M = 2x+ 4y, N = 2x− 2y

∴∂M

∂y= 4,

∂N

∂x= 2

∂M

∂y6= ∂N

∂x

Hence, the given differential equation is not exact.

Example 57.


dy

dx= −ax− by

bx− ay



dy

dx= −ax− by

bx− ay

(ax− by) + (bx− ay)dy

dx= 0 (2.171)

with

M(x, y) +N(x, y)dy

dx= 0

we get

M = ax− by, N = bx− ay

∴∂M

∂y= −b, ∂N

∂x= b

∂M

∂y6= ∂N

∂x


Example 58.



(2x+ 3) + (2y − 2)dy

dx= 0



(2x+ 3) + (2x− 2)dy

dx= 0 (2.172)

with

M(x, y) +N(x, y)dy

dx= 0

we get

M = 2x+ 3, N = 2x− 2

∴∂M

∂y= 0,

∂N

∂x= 0

∂M

∂y=∂N

∂x

Hence, the given differential equation is exact. Thus, there exists a function ψ(x, y)

such that

∂ψ

∂x= M(x, y),

∂ψ

∂y= N(x, y)

or,∂ψ

∂x= 2x+ 3, (2.173)

∂ψ

∂y= 2y − 2 (2.174)


ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

=

∫(2x+ 3)dx+ h(y)

= x2 + 3x+ h(y) (2.175)


N(x, y) =∂ψ

∂y

or, 2y − 2 = h′(y)

=⇒ h′(y) = 2y − 2

=⇒ h(y) = y2 − 2y + c


ψ(x, y) = x2 + 3x+ y2 − 2y + c = 0



Example 59.


(2xy2 + 2y) + (2x2y + 2x)dy

dx= 0



(2xy2 + 2y) + (2x2y + 2x)dy

dx= 0 (2.176)

with

M(x, y) +N(x, y)dy

dx= 0

we get

M = 2xy2 + 2y, N = 2x2y + 2x

∴∂M

∂y= 4xy + 2,

∂N

∂x= 4xy + 2

∂M

∂y=∂N

∂x


such that

∂ψ

∂x= M(x, y),

∂ψ

∂y= N(x, y)

or,∂ψ

∂x= 2xy2 + 2y, (2.177)

∂ψ

∂y= 2x2y + 2x (2.178)


ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

=

∫(2xy2 + 2y)dx+ h(y)

= x2y2 + 2xy + h(y) (2.179)


N(x, y) =∂ψ

∂y

or, 2x2y + 2x = 2x2y + 2x+ h′(y)

=⇒ h′(y) = 0

=⇒ h(y) = c



ψ(x, y) = x2y2 + 2xy + c = 0


Example 60.


(x ln y + xy) + (y lnx+ xy)dy

dx= 0



(x ln y + xy) + (y lnx+ xy)dy

dx= 0 (2.180)

with

M(x, y) +N(x, y)dy

dx= 0

we get

M = x ln y + xy, N = y lnx+ xy

∴∂M

∂y=x

y+ x,

∂N

∂x=y

x+ y

∂M

∂y6= ∂N

∂x


Example 61.

Solve the initial value problem and determine at least aprroximately where the

solution is valid

(2x− y)dx+ (2y − x)dy = 0, y(1) = 3


(2x− y) + (2y − x)dy

dx= 0

with

M(x, y) +N(x, y)dy

dx= 0

we get

M = 2x− y, N = 2y − x


∴∂M

∂y= −1,

∂N

∂x= −1

∂M

∂y=∂N

∂x


such that

∂ψ

∂x= M(x, y),

∂ψ

∂y= N(x, y)

or,∂ψ

∂x= 2x− y, (2.181)

∂ψ

∂y= 2y − x (2.182)


ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

=

∫(2x− y)dx+ h(y)

= x2 − xy + h(y) (2.183)


N(x, y) =∂ψ

∂y

or, 2y − x = −x+ h′(y)

=⇒ h′(y) = 2y

=⇒ h(y) = y2 + c‘


ψ(x, y) = c

or, x2 − xy + y2 + c = 0 (2.184)

Using the initial condition y(1) = 3 i.e. x = 1, y = 3 in (2.184),

1− 3 + 9 + c = 0

or, c = −7



x2 − xy + y2 − 7 = 0

or, y2 − xy + x2 − 7 = 0

or, y =x±

√x2 − 4 · 1 · (x2 − 7)

2

or, y =x±√

28− 3x2

2

Either, y =x−√

28− 3x2

2(2.185)

or y =x+√

28− 3x2

2(2.186)

When x = 1, from (2.185), we get y = 1−52 = −2. Thus, (2.185) does not satisfied

the initial condition. Which is not solution of the the initial value problem. Again

when x = 1, from (2.186), we get y = 1+52 = 3. Thus, (2.186) satisfied the initial

condition. Which is solution of the the initial value problem. Thus solution is

y =x+√

28− 3x2

2(2.187)

The solution is valid when

28− 3x2 ≥ 0

or, x2 ≤ 28

3

or, |x|2 ≤ 28

3

or, |x| ≤√

28

3

or, −√

28

3≤ x ≤

√28

3

This interval contains the initial point x = 1. Hence the solution of the given

equation is valid in −√

283 ≤ x ≤

√283 .

Example 62.

Find the value of b for which the differential equation

(ye2xy + 4x3)dx+ bxe2xydy = 0

is exact and solve it using that value of b.


(ye2xy + 4x3) + bxe2xy dy

dx= 0


with

M(x, y) +N(x, y)dy

dx= 0

we get

M = ye2xy + 4x3, N = bxe2xy

∴∂M

∂y= e2xy + 2xye2xy,

∂N

∂x= bye2xy + 2bxye2xy

Since the equation is exact

∂M

∂y=∂N

∂x

or, e2xy + 2xye2xy = b(e2xy + 2xye2xy

)or, b = 1

Hence, the exact equation is

(ye2xy + 4x3) + xe2xy dy

dx= 0 (2.188)

. Thus, thier exists a function ψ(x, y) such that

∂ψ

∂x= M(x, y),

∂ψ

∂y= N(x, y)

or,∂ψ

∂x= ye2xy + 4x3, (2.189)

∂ψ

∂y= xe2xy (2.190)


ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

= y

∫e2xydx+ 4

∫x3dx+ h(y)

= y1

2ye2xy + x4 + h(y)

=1

2e2xy + x4 + h(y) (2.191)


N(x, y) =∂ψ

∂y

or, xe2xy =1

22xe2xy + h′(y)

=⇒ h′(y) = 0

=⇒ h(y) = c′



ψ(x, y) = c′

or,1

2e2xy + x4 = −c′

or, e2xy + 2x4 = c

2.6.2 Integrating Factor

Let a differential equation is not exact. If it is possible to make exact by multi-

plying it an appropriate factor µ(x, y), which is called an integrating factor for the

differential equation. Let the differential equation

M(x, y) +N(x, y)dy

dx= 0

or, M(x, y)dx+N(x, y)dy = 0 (2.192)

is not exact. Let µ(x, y) such that

µ(x, y)M(x, y)dx+ µ(x, y)N(x, y)dy = 0 (2.193)

is exact. But the equation (2.193) is exact if and only if

(µM)y = (µN)x

or, µyM + µMy = µxN + µNx

or, Mµy −Nµx + (My −Nx)µ = 0 (2.194)

If a function µ satisfying equation (2.194) can be found then the equation (2.193)

will be exact. In general, finding out µ from the equations like (2.194) is not easy.

We consider some special cases as µ is function of x only or y only.

Let us consider µ = µ(x). Then

(µM)y = µMy and (µN)x = µNx +Ndµ

dx

From condition of exactness (µM)y = (µN)x implies

µMy = µNx +Ndµ

dx

or, Ndµ

dx= µMy − µNx

or,dµ

µ=My −Nx

Ndx (2.195)

which is a linear ordinary differential equation of first order with variables separated

ifMy−Nx

N is a function of x alone. Futher µ(x) can be found by solving (2.195),


which is both linear and separable.

Let us consider µ = µ(y). Then

(µM)y = µMy +Mdµ

dyand (µN)x = µNx

From condition of exactness (µM)y = (µN)x implies

µMy = µNx +Ndµ

dy

or, Mdµ

dy= µ(Nx −My)

or,dµ

µ=Nx −My

Mdy (2.196)

which is a linear ordinary differential equation of first oder with variables separated

ifNx−My

M is a function of y alone. Futher µ(y) can be found by solving (2.196),

which is both linear and separable.

Example 63.

Find an integrating factor for the equation

(3xy + y2) + (x2 + xy)dy

dx= 0

and then solve the equation.


(3xy + y2) + (x2 + xy)dy

dx= 0 (2.197)

with

M(x, y) +N(x, y)dy

dx= 0

we get

M = 3xy + y2, N = (x2 + xy)

or, My =∂M

∂y= 3x+ 2y, Nx =

∂N

∂x= 2x+ y

∴My −Nx

N=

3x+ 2y − (2x+ y)

x2 + xy=

x+ y

x(x+ y)=

1

x

which is function of x only. Let µ is integrating factor. Then

dµ

µ=My −Nx

Ndx

or,dµ

µ=

1

xdx

∴ lnµ = lnx

=⇒ µ = x


Multiplying the equation (2.197) by integrating factor, we get

(3x2y + xy2) + (x3 + x2y)dy

dx= 0 (2.198)

which is exact. Comparing the equation with

M1(x, y) +N1(x, y)dy

dx= 0

we get

M1 = 3x2y + xy2, N1 = x3 + x2y

Thererfore, we get a function ψ(x, y) such that

∂ψ

∂x= M1(x, y),

∂ψ

∂y= N1(x, y)

or,∂ψ

∂x= 3x2y + xy2, (2.199)

∂ψ

∂y= x3 + x2y (2.200)


ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

= 3y

∫x2dx+ y2

∫xdx+ h(y)

= x3y +1

2x2y2 + h(y) (2.201)


N1(x, y) =∂ψ

∂y

or, x3 + x2y = x3 + x2y + h′(y)

=⇒ h′(y) = 0

=⇒ h(y) = c‘


ψ(x, y) = 0

or, x3y +1

2x2y2 + c = 0

Example 64.

Find an integrating factor for the equation

dx+

(x

y− sin y

)dy = 0




1 +

(x

y− sin y

)dy

dx= 0 (2.202)

with

M(x, y) +N(x, y)dy

dx= 0

we get

M = 1, N =

(x

y− sin y

)or, My =

∂M

∂y= 0, Nx =

∂N

∂x=

1

y

∴My −Nx

N=

0− 1y(

xy − sin y

)which is not function of x only. Again

Nx −My

M=

1y − 0

1=

1

y

Let µ is integrating factor. Then

dµ

µ=Nx −My

Mdy

or,dµ

µ=

1

ydy

∴ lnµ = ln y

=⇒ µ = y


y + (x− y sin y)dy

dx= 0 (2.203)



dx= 0

we get

M1 = y, N1 = x− y sin y


∂ψ

∂x= M1(x, y),

∂ψ

∂y= N1(x, y)

or,∂ψ

∂x= y, (2.204)

∂ψ

∂y= x− y sin y (2.205)



ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

= y

∫dx+ h(y)

= xy + h(y) (2.206)


N1(x, y) =∂ψ

∂y

or, x− y sin y = x+ h′(y)

=⇒ h′(y) = −y sin y

=⇒ h(y) = −∫y sin ydy‘

= −y

∫sin ydy −

∫ (dy

dy

∫sin ydy

)dy

= −

−y cos y +

∫cos ydy

= y cos y − sin y


ψ(x, y) = 0

or, xy + y cos y − sin y + c = 0

Example 65.

Find an integrating factor for the equation(4x3

y2+

3

y

)dx+

(3x

y2+ 4y

)dy = 0


Solution: Comparing the equation(4x3

y2+

3

y

)+

(3x

y2+ 4y

)dy

dx= 0 (2.207)

with

M(x, y) +N(x, y)dy

dx= 0

we get

M =4x3

y2+

3

y, N =

3x

y2+ 4y

or, My =∂M

∂y= −8x3

y3− 3

y2, Nx =

∂N

∂x=

3

y2

(2.208)


Now,

Nx −My

M=

3y2

+ 8x3

y3+ 3

y2

4x3

y2+ 3

y

=2

y

Let µ is integrating factor. Then

dµ

µ=Nx −My

Mdy

or,dµ

µ=

2

ydy

∴ lnµ = 2 ln y = ln y2

=⇒ µ = y2


(4x3 + 3y) + (3x+ 4y3)dy

dx= 0 (2.209)



dx= 0

we get

M1 = 4x3 + 3y, N1 = 3x+ 4y3


∂ψ

∂x= M1(x, y),

∂ψ

∂y= N1(x, y)

or,∂ψ

∂x= 4x3 + 3y, (2.210)

∂ψ

∂y= 3x+ 4y3 (2.211)


ψ(x, y) =

∫∂ψ

∂xdx+ h(y)

=

∫(4x3 + 3y)dx+ h(y)

= x4 + 3xy + h(y) (2.212)


N1(x, y) =∂ψ

∂y

or, 3x+ 4y3 = 3x+ h′(y)

=⇒ h′(y) = 4y3

=⇒ h(y) = 4

∫y3dy = y4 + c



ψ(x, y) = 0

or, x4 + 3xy + y4 + c = 0

2.7 Numerical Aproximations: Euler’s Method

Let us consider a first order initial velue problem

dy

dt= f(t, y), y(t0) = y0 (2.213)

If f and ∂f∂y are contionuous, the the initial value problem (2.213) has unique solution

y = φ(t) in some in interval containing the initial points t = t0. In some case it

can not be solved analytically. Therefore, it is important to be able to aprroach the

problem in other ways. i.e. numerical approach. Now we will discuss about Euler’s

method or Tangent line method.

Let y = φ(t) be exact solution. We know that the solution of the equation (2.213)

Figure 2.9:

passes through the initial point (t0, y0) and from the differential equation, the slope

of the tangent is

m =

(dy

dt

)t0,y0

= f(t0, y0)

The equation of tangent is

y − y0 = f(t0, y0)(x− x0)

or, y = y0 + f(t0, y0)(x− x0) (2.214)

2.7. NUMERICAL APROXIMATIONS: EULER’S METHOD 109

The tangent line is good approximation to the actual solution curve on a short

interval so that the slope of the solution does not change. If t1 is close to t0, we

can approximate φ(t1) by the value y1 determined by substituting t = t1 into the

tangent line approximattion (2.214) at t = t0, thus

y1 = y0 + f(t0, y0)(t1 − t0)

Again, we construct the line through (t1, y1) with slope f(t1, y1)

y = y1 + f(t1, y1)(t− t1) (2.215)

To approximate the value of φ(t) at a nearby the point t2, we use the equation

(2.215)

y2 = y1 + f(t1, y1)(t2 − t1)

Continuing the same process, the general expression for the tangent line at (tn, yn)

is

y = yn + f(tn, yn)(t− tn) (2.216)

hence the approximate value yn+1 at tn+1 in terms of tn, tn+1 and yn is

yn+1 = yn + f(tn, yn)(tn+1 − tn), n = 0, 1, 2, 3 · · · (2.217)

Finally, we assume that there is uniform step size h between the points t0, t1, · · · , tnthen tn+1 − tn = h for all n = 1, 2, 3, · · · , so the Euler’s formula becomes

yn+1 = yn + hf(tn, yn), n = 0, 1, 2, 3 · · · (2.218)

Example 66.

Find approximate values of the solution of the given initial value probelm at

t = 0.2, 0.4, 0.6, 0.8 and 1 and compare the values obtained by exact solution.

y′ = 3− 2t− 1

2y, y(0) = 1


dy

dx+

1

2y = 3− 2t (2.219)

Integrating factor

I.F. = e∫pdt = e

∫1/2dt = e

t2


et/2dy

dx+ yet/2 = 3et/2 − 2tet/2

or,d

dt

(et/2y

)= 3et/2 − 2tet/2

or, d(et/2y

)= 3et/2dt− 2tet/2


Integrating,

et/2y = 3

∫et/2dt− 2

∫tet/2dt+ c

or, et/2y = 6et/2 − 2

t

∫et/2dt−

∫ (dt

dt

∫et/2dt

)dt

+ c

or, et/2y = 6et/2 − 4tet/2 + 4

∫et/2dt+ c

or, et/2y = 6et/2 − 4tet/2 + 8et/2 + c

or, et/2y = 14et/2 − 4tet/2 + c

or, y = 14− 4t+ ce−t/2

Using the initial condition t = 0, y = 1, we get

1 = 14 + c

or, c = −13

Thus exact solution is

y = 14− 4t− 13e−t/2 (2.220)

Exact valuet value of y = 14− 4t− 13e−t/2

0 y = 14− 4× 0− 13e−0 = 1

0.2 y = 14− 4× 0.2− e−0.1 = 1.43711

0.4 y = 14− 4× 0.4− e−0.2 = 1.75650

0.6 y = 14− 4× 0.6− e−0.3 = 1.96936

0.8 y = 14− 4× 0.8− e−0.4 = 2.08584

1 y = 14− 4× 1− e−0.5 = 2.11510

To find approximate solution by Euler’s method:

We have

f(t, y) = 3− 2t− y/2

For approximate solution at t1 = 0.2 :

When t0 = 0, y0 = 1,

f0 = f(t0, y0) = f(0, 1) = 3− 0− 1/2 = 2.5

The equation of tangent line near t = 0 is

y = y0 + f(t0, y0)(t− t0)

or, y = 1 + 2.5(t− 0)

or, y = 1 + 2.5t


Taking t = t1 = 0.4, the approximate value of y1 of the solution is

y1 = 1 + 2.5× 0.2 = 1.5

For approximate solution at t = 0.2 :

f1 = f(t1, y1) = f(0.2, 1.5) = 3− 2× 0.2− 1.5/2 = 1.85

The equation of tangent line near t = 0.4 is

y = y1 + f(t1, y1)(t− t1)

or, y = 1.5 + 1.85(t− 0.4)

or, y = 1.13 + 1.85t

Taking t2 = 0.4, the approximate value of y2 of the solution is

y2 = 1.13 + 1.85× 0.4 = 1.87


f2 = f(t2, y2) = f(0.4, 1.87) = 3− 2× 0.4− 1.87/2 = 1.265

The equation of tangent line near t2 = 0.6 is

y = y2 + f(t2, y2)× (t− t2)

or, y = 1.87 + 1.265(t− 0.6)

or, y = 1.364 + 1.265t


y3 = 1.364 + 1.265× 0.6 = 2.123


f3 = f(t3, y3) = f(0.6, 2.123) = 3− 2× 0.6− 2.123/2 = 0.7385


y = y3 + f(t3, y3)× (t− t3)

or, y = 2.123 + 0.7385(t− 0.6)

or, y = 1.6799 + 0.7385t


y4 = 1.6799 + 0.7385× 0.8 = 2.77070


Again

f4 = f(t4, y4) = f(0.8, 2.77070) = 3− 2× 0.8− 2.77070/2 = 0.26467

The tangent line at 0.8 is

y = y4 + f(t4, y4)× (t− t4)

or, y = 2.77070 + 0.26467(t− 0.8)

or, y = 2.05898 + 0.26467t

Solution at t5 = 1

2.05898 + 0.26467× 1 = 2.32363

Comarision of exact and Euler’s method for the step size h = 0.1

t Exact Euler Approximation Tangent line

0 1 1 y = 1 + 2.5t

0.2 1.43711 1.5 y = 1.13 + 1.85t

0.4 1.7565 1.87 y = 1.364 + 1.265t

0.6 1.96936 2.123 y = 1.6799 + 0.7385t

0.8 2.08584 2.27070 y = 2.05898 + 0.26465t

1 2.11510 2.32363

Example 67.


t = 0.1, 0.2, 0.3 and 0.4 and compare the values obtained by exact solution.

y′ = 3 + t− y, y(0) = 1


dy

dx+ y = 3 + t (2.221)

Integrating factor

I.F. = e∫pdt = e

∫dt = et


etdy

dx+ yet = 3et + tet

or,d

dt

(ety)

= 3et + tet

or, d(ety)

= 3etdt+ tet


Integrating,

ety = 3

∫etdt+

∫tetdt+ c

or, ety = 3et + t

∫etdt−

∫ (dt

dt

∫etdt

)dt+ c

or, ety = 3et + tet −∫etdt+ c

or, ety = 3et + tet − et + c

or, ety = tet + 2et + c

or, y = t+ 2 + ce−t


1 = 0 + 2 + c

or, c = −1


y = t+ 2− e−t (2.222)

Exact valuet value of y = t+ 2− e−t

0 y = 0 + 2− e−0 = 1

0.1 y = 0.1 + 2− e−0.1 = 1.1952

0.2 y = 0.2 + 2− e−0.2 = 1.3813

0.3 y = 0.3 + 2− e−0.3 = 1.5592

0.4 y = 0.4 + 2− e−0.4 = 1.7297


We have

f(t, y) = 3 + t− y


When t0 = 0, y0 = 1,

f0 = f(t0, y0) = f(0, 1) = 3 + 0− 1 = 2


y = y0 + f(t0, y0)(t− t0)

or, y = 1 + 2(t− 0)

or, y = 1 + 2t


y1 = 1 + 2× 0.1 = 1.2



f1 = f(t1, y1) = f(0.1, 1.2) = 3 + 0.1− 1.2 = 1.9


y = y1 + f(t1, y1)(t− t1)

or, y = 1.2 + 1.9(t− 0.1)

or, y = 1.01 + 1.9t


y2 = 1.01 + 1.9× 0.2 = 1.39


f2 = f(t2, y2) = f(0.2, 1.39) = 3 + 0.2− 1.39 = 1.81


y = y2 + f(t2, y2)× (t− t2)

or, y = 1.39 + 1.81(t− 0.2)

or, y = 1.028 + 1.81t


y3 = 1.028 + 1.81× 0.3 = 1.571


f3 = f(t3, y3) = f(0.3, 1.571) = 3 + 0.3− 1.571 = 1.729


y = y3 + f(t3, y3)× (t− t3)

or, y = 1.571 + 1.729(t− 0.3)

or, y = 1.0523 + 1.729t


y4 = 1.0523 + 1.729× 0.4 = 1.7439

Again

f4 = f(t4, y4) = f(0.4, 1.7439) = 3 + 0.4− 1.7439 = 1.6561



y = y4 + f(t4, y4)× (t− t4)

or, y = 1.7439 + 1.6561(t− 0.4)

or, y = 1.08146 + 1.6561t



0 1 1 y = 1 + 2t

0.1 1.1952 1.2 y = 1.01 + 1.9t

0.2 1.3813 1.39 y = 1.028 + 1.81t

0.3 1.5592 1.571 y = 1.0523 + 1.729t

0.4 1.7297 1.7439 y = 1.08146 + 1.6561t

Example 68.


t = 0.1, 0.2, 0.3 and 0.4 and compare the values obtained by exact solution.

y′ = 2y − 1, y(0) = 1


dy

dx− 2y = −1 (2.223)

Integrating factor

I.F. = e∫−2dt = e

∫dt = e−2t


e−2t dy

dx− 2ye−2t = −e−2t

or,d

dt

(e−2ty

)= −e−2t

or, d(e−2ty

)= e−2tdt

Integrating,

e−2ty = −∫e−2tdt+ c

or, e−2ty =1

2e−2t + c

or, y =1

2+ ce2t


1 =1

2+ c


or, c =1

2


y =1

2+

1

2e2t =

1

2(1 + e2t)

Exact valuet value of y = 1

2(1 + e2t)

0 y = 12(1 + e2×0) = 1

0.1 y = 12(1 + e2×0.1) = 1.1107

0.2 y = 12(1 + e2×0.2) = 1.2459

0.3 y = 12(1 + e2×0.3) = 1.4111

0.4 y = 12(1 + e2×0.4) = 1.6128


We have

f(t, y) = 2y − 1


When t0 = 0, y0 = 1,

f0 = f(t0, y0) = f(0, 1) = 2× 1− 1 = 1


y = y0 + f(t0, y0)(t− t0)

or, y = 1 + 1(t− 0)

or, y = 1 + t


y1 = 1 + 0.1 = 1.1


f1 = f(t1, y1) = f(0.1, 1.1) = 2× 1.1− 1 = 1.2.


y = y1 + f(t1, y1)(t− t1)

or, y = 1.1 + 1.2(t− 0.1)

or, y = 0.98 + 1.2t


y2 = 0.98 + 1.2× 0.2 = 1.22



f2 = f(t2, y2) = f(0.2, 1.22) = 2× 1.22− 1 = 1.44


y = y2 + f(t2, y2)× (t− t2)

or, y = 1.22 + 1.44(t− 0.2)

or, y = 0.932 + 1.441t


y3 = 0.932 + 1.44× 0.3 = 1.364


f3 = f(t3, y3) = f(0.3, 1.364) = 2× 1.364− 1 = 1.728


y = y3 + f(t3, y3)× (t− t3)

or, y = 1.364 + 1.728(t− 0.3)

or, y = 0.8456 + 1.728t


y4 = 0.8456 + 1.728× 0.4 = 1.5368

Again

f4 = f(t4, y4) = f(0.4, 1.5368) = 2× 1.5368− 1 = 2.0736


y = y4 + f(t4, y4)× (t− t4)

or, y = 1.5368 + 2.0736(t− 0.4)

or, y = 0.7074 + 2.0736t



0 1 1 y = 1 + t

0.1 1.1107 1.1 y = 0.98 + 1.2t

0.2 1.2459 1.22 y = 0.932 + 1.44t

0.3 1.4111 1.364 y = 0.8456 + 1.728t

0.4 1.6128 1.5368 y = 0.7074 + 2.0736t


2.8 First Order Difference Equation

A continuous model leading to a differential equation is resonable and atrractive for

many problems, there are some cases in which a discreate model may be more nat-

ural. Some times popultion growth may be described more accuretly by a discreate

than by a continuous model. This is true of the species whose generations do not

overlap and that propagate at regular intervals such as a particular times of the

year. Then the population yn+1 of the species in the year n+ 1 is some function of

n and the population yn in the preceding year.

yn+1 = f(n, yn), n = 1, 2, 3, · · · . (2.224)

The equation (2.224) is called a first order difference equation. It is called first

order because the value of yn+1 depends on the value of yn but not the values

yn−1, yn−2, · · ·The difference equation (2.224) is called linear if f is a linear function of yn; other-

wise, it is nonlinear.

A solution of the difference equation (2.224) is a sequence of numbers y1, y2, · · · that

satisfy the equation for each value of n. In addition to the difference equation itself

, there may also be an initial condition

y0 = α (2.225)

that prescribes the value of the first term of the sequence.

Let us assume that the function f in the equation (2.224) depends only on yn. Then

yn+1 = f(yn), n = 1, 2, · · · (2.226)

with a given initial condiition y0. Thus

y1 = f(y0)

y2 = f(y1) = f(f(y0)) = f2(y0)

y3 = f(y2) = f(f2(y0)) = f3(y0)

...

yn = fn(y0)

Solutions for which yn has the same value for all n is called equilibrium solutions.

If the equilibrium solution exist then we can get it by solving the equation

yn = f(yn) (2.227)

2.8.1 Linear difference equation

Let the population of a certain species in a given region in year n + 1, denoted by

yn+1 be given by

yn+1 = ρnyn, n = 0, 1, 2, 3 · · · (2.228)

2.8. FIRST ORDER DIFFERENCE EQUATION 119

where ρn is reproduction rate, which may differ from year to year. Then

y1 = ρ0y0

y2 = ρ1y1 = ρ1ρ0y0

y3 = ρ2y2 = ρ2ρ1ρ0y0

...

yn = ρn−1yn−1 = ρn−1ρn−2 · · · ρ0y0 n = 1, 2, · · · (2.229)

If the reproduction rate ρn is constant and has the same value ρ for each n then the

difference equation (2.228) becomes

yn+1 = ρyn, n = 0, 1, 2, 3 · · · (2.230)

and cosequetly,

yn = ρny0 (2.231)

Thus, yn is a geometric sequence with commom ratio ρ and

limn→∞

yn =

0 if |ρ| < 1

y0 if ρ = 1

does not exists otherwise

(2.232)

In otherwords, the equilibrium solution yn = 0 ia asymptotically stable for |ρ| < 1

and unstable if ρ > 1. Now we will modify the population model represented by

(2.228) to include the effect of of immigration or emigration. If bn is the net increase

in population in year n due to immigration, then the population in year n+ 1 is the

sum of those due to natural reproduction and those due to immigration. Thus

yn+1 = ρyn + bn n = 0, 1, 2, · · · (2.233)

where we assuming that the reproduction rate ρ is constant. We can solve the

equation (2.228)

y1 = ρy0 + b0

y2 = ρy1 + b1 = ρ(ρy0 + b0) + b1 = ρ2y0 + ρb0 + b1

y3 = ρy2 + b2 = ρ(ρ2y0 + ρb0 + b1) + b2 = ρ3y0 + ρ2b0 + ρb1 + b2...

yn = ρny0 + ρn−1b0 + ρn−2b1 + · · ·+ ρbn−2 + bn−1

yn = ρny0 +n−1∑j=0

ρn−1−jbj (2.234)

In the special case where bn = b 6= 0 for all n, the difference equation (2.233) becomes

yn+1 = ρyn + b (2.235)


and from (2.234)

yn = ρny0 + (1 + ρ+ ρ2 + · · ·+ ρn−1)b (2.236)

If ρ 6= 1, then (2.236) becomes

yn = ρny0 +1− ρn

1− ρb

or, yn = ρny0 +b

1− ρ− bρn

1− ρ

or, yn = ρn(y0 −

b

1− ρ

)+

b

1− ρ(2.237)

where the two terms on the right are effect ot the original population and immigra-

tion. For long time behavior, of yn, From the equation (2.237)

limn→∞

yn =

b

1−ρ if |ρ| < 1

does not exists if |ρ| > 1, ρ = −1(2.238)

So, for ρ 6= 1 the equilibrium solution is b1−ρ .

If ρ = 1, the from (2.236)

yn = y0 + nb

and yn becomes unbounded for n→∞.

Example 69.

Solve the difference equation yn+1 = −0.8yn interm of the initial value y0. De-

scribe the behavior of the solution as n→∞.

Solution: Here

yn+1 = −0.8yn

Taking, n = 0, 1, 2, · · ·

y1 = −0.8y0

or, y2 = −0.8y1 = (−1)2(0.8)2y0

or, y3 = −0.8y2 = (−1)3(0.8)3y0

...

or, yn = (−1)n(0.8)ny0

As n→∞, yn → 0.

Example 70.

2.8. FIRST ORDER DIFFERENCE EQUATION 121

Sove the difference equation yn+1 =√

n+3n+1 yn interm of the initial value y0.

Describe the behavior of the solution as n→∞.

Solution: Here

yn+1 =

√n+ 3

n+ 1yn

Taking, n = 0, 1, 2, · · ·

y1 =√

3 y0 =√

3y0 =√

1 + 2y0

or, y2 =√

2 y1 =√

2√

3y0 =√

6 y0 =√

1 + 2 + 3 y0

or, y3 =

√5

3y2 =

√5

3

√6 y0 =

√1 + 2 + 3 + 4 y0

...

or, yn =√

1 + 2 + 3 + · · ·+ n+ (n+ 1) y0

=

√n(n+ 1)

2+ (n+ 1) y0

=

√(n+ 1)(n+ 2)

2y0

(2.239)

As n→∞, yn →∞.

Example 71.

Sove the difference equation yn+1 = −0.5 yn + 6 interm of the initial value y0.

Describe the behavior of the solution as n→∞.

Solution: Here

yn+1 = −0.5 yn + 6

Taking, n = 0, 1, 2, · · ·

y1 = −0.5 y0 + 6

or, y2 = −0.5 y1 + 6 = −0.5(−0.5 y0 + 6) + 6 = (−0.5)2 y0 + 6× (−0.5) + 6

or, y3 = −0.5 y2 + 6 = −0.5((−0.5)2 y0 + 6× (−0.5) + 6) + 6

= (−0.5)3 y0 + 6[1 + (−0.5) + (−0.5)2]

...

or, yn = (−0.5)ny0 + 6[1 + (−0.5) + (−0.5)2 + · · ·+ (−0.5)n−1

]= (−0.5)ny0 +

6(1− (0.5)n)

1 + 0.5= (−0.5)ny0 + 4[1− (−0.5)n]

As n→∞, yn → 4 since (−0.5)n →∞


Example 72.

An investor deposites $1000 in an account paying interest at a rate of 7% com-

pounded monthly and also make additional deposits of $25 per month. Find the

balance in the account after 4 years.

Solution: Let yn be the balance at the end of nth months , then

yn+1 =(

1 +r

12

)yn + 25

yn+1 = ρyn + 25

where ρ = 1 + r12 = 1 + 7/100

12 = 1.006. Therefore,

y1 = ρ y0 + 25

y2 = ρ y1 + 25

or, y2 = ρ(ρy0 + 25) + 25 = ρ2y0 + 25ρ+ 25

y3 = ρy2 + 25

= ρ(ρ2y0 + 25ρ+ 25) + 25

= ρ3y0 + 25ρ2 + 25ρ+ 25

= ρ3y0 + 25(ρ2 + ρ+ 1)

...

yn = ρny0 + 25(1 + ρ+ ρ2 + · · ·+ ρn−1)

= ρny0 + 251− ρn

1− ρ

= ρny0 −25ρn

1− ρ+

25

1− ρ

= ρn(y0 −

25

1− ρ

)+

25

1− ρ(2.240)

In 4 years n = 4× 12 = 48 months, y0 = $1000 and ρ = 1.006, from (2.240)

y48 = (1.006)48

(1000− 25

1− 1.006

)+

25

1− 1.006

= 6871.67− 4166.67

= $2705

Chapter 3

Second Order Linear Equations

3.1 Second Order Linear Equation

A second order differential equation has form

d2y

dt2= f

(t, y,

dy

dt

)(3.1)

where f is some given function.

The equation (3.1) is said to be linear if the function f has form

f

(t, y,

dy

dt

)= g(t) = p(t)

dy

dt− q(t)y (3.2)

i.e. if f is linear function in y and dydt . In equation (3.2) g, p and q are specified

functions of the independent variable y but not dependent variable y. Now the

equation (3.1) becomes

d2y

dt2= g(t)− p(t)dy

dt

or,d2y

dt2+ p(t)

dy

dt+ q(t)y = g(t)

or, y′′ + p(t)y′ + q(t)y = g(t) (3.3)

where the primes denots differentiation with respect to t. Instead of equation3.3 we

often see the equation

P (t)y′′ +Q(t)y′ +R(t)y = G(t) (3.4)

where P (t) 6= 0.Dividing both sides by P (t) we get

y′′ +Q(t)

P (t)y′ +

R(t)

P (t)y =

R(t)

P (t)

123

124 CHAPTER 3. SECOND ORDER LINEAR EQUATIONS

y′′ + p(t)y′ + q(t)y = g(t)

where,

p(t) =Q(t)

P (t), q(t) =

R(t)

P (t), g(t) =

G(t)

P (t)

which is in the form 3.3.

3.1.1 Non Linear Equation

If the equation 3.1 is not the form of 3.3 and 3.4 is called non linear.

3.1.2 Initial Value Problem

An initail value problem of second order differential equation consists of a differential

equationd2y

dt2= f

(t, y,

dy

dt

)with a pair of initial conditions, like

y(t0) = y0, y′(to) = y0

3.1.3 Homogeneous and Non-homogeneous

A second order linear equation

y′′ + p(t)y′ + q(t)y = g(t)

is said to be homogeneous if g(t)=0 for all t and non-homogeneous if g(t) 6= 0 for

some value of t.

3.1.4 Homogeneous Equation with Constant Coefficients and its

Solution

A linear second order differential equations of the form

ay′′ + by′ + cy = 0 (3.5)

where a, b, c are constant is called homogeneous second order linear equation with

constant coefficients. Let y = ert be a solution of (3.5). Then

y′ = rert, y′′ = r2ert

Substituting the values of y, y′ and y′′ in (3.5) we get

(ar2 + br + c)ert = 0

3.1. SECOND ORDER LINEAR EQUATION 125

Since, ert 6= 0 for all t ∈ R

ar2 + br + c = 0 (3.6)

The equation (3.6) is called the auxiliary or characteristics equation for the differ-

ential equation (3.5). The roots of the (3.6) are

r1 =−b+

√b2 − 4ac

2a, r2 =

−b−√b2 − 4ac

2a

Now we will have the following cases

1. r1 and r2 real and distinct if b2 − 4ac > 0.

2. r1 and r2 and equal if b2 − 4ac = 0

3. r1 and r2 conjugate complex if b2 − 4ac < 0.

Case I Auxiliary equatio having real and different roots. Let r=r1, r2 be the two

real and unequal roots of the auxiliary equation. Then er1t and er2t are two solution

of (3.5). The general solution of (3.5) is

y = c1er1t + c2e

r2t

where c1 and c2 are two arbitary constants.

Case II Auxiliary equation having real and equal roots. Let r = r1, r2 be the two

real and equal roots of the auxiliary equation. Then the general equation of (3.5) is

y = (c1 + c2t)er1t

where c1 and c2 are constants.

Case III Auxiliary equation having imaginary and conjugate roots. Let r = α+ iβ

and r = α − iβ be roots of the auxiliary equation. Then the generall solution of

(3.5) is

y = eαt(c1 cosβt+ c2 sinβt)


Example 73.

Find the general solution of

y′′ + 5y′ + 6y = 0

Solution: The given second degree equation is

y′′ + 5y′ + 6y = 0


Its auxiliary equation is

r2 + 5r + 6 = 0

or, (r + 2)(r + 3) = 0

or, r = −2,−3

Thus the general solution is

y = c1er1t + c2e

r2t = c1e−2t + c2e

−3t

Example 74.


6y′′ − y′ − y = 0


6y′′ − y′ − y = 0


6r2 − r − 1 = 0

or, (3r + 1)(2r − 1) = 0

or, r = −1

3,

1

2


y = c1er1t + c2e

r2t = c1e− 1

3t + c2e

12t

Example 75.


y′′ − 16y = 0

Solution: The given second degree homogeneous equation is

y′′ − 16y = 0


r2 − 16 = 0

or, (r + 4)(r − 4) = 0

or, r = −4, 4


y = c1er1t + c2e

r2t = c1e−4t + c2e

4t


Example 76.


y′′ − 9y′ + 9y = 0


y′′ − 9y′ + 9y = 0


r2 − 9r + 9 = 0

or, r =−b±

√b2 − 4ac

2a

or, r =−(−9)±

√(−9)2 − 4 · 1 · 92 · 1

=9± 3

√3

2


y = c1er1t + c2e

r2t = c1e9+3√

32

t + c2e9−3√3

2t

Example 77.


y′′ + 5y′ + 6y = 0, y(0) = 2, y′(0) = 3


y′′ + 5y′ + 6y = 0


r2 + 5r + 6 = 0

or, (r + 2)(r + 3) = 0

or, r = −2,−3


y(t) = c1er1t + c2e

r2t = c1e−2t + c2e

−3t (3.7)

Differentiating with respect to t

y′(t) = −2c1e−2t − 3c2e

−3t (3.8)

Also, using the initial conditions y(0) = 2, y′(0) = 3. Then from (3.7) and (3.8)

we get

y(0) = 2 =⇒ c1 + c2 = 2 (3.9)

y′(0) = 3 =⇒ −2c1 − 3c2 = 3 (3.10)


Solving the equations (3.9) and (3.10), we get

c1 = 9, c2 = −7

Putting these values in the equation (3.7), we obtained the solution

y = 9e−2t − 7e−3t

Example 78.


y′′ + 4y′ + 3y = 0, y(0) = 2, y′(0) = −1


y′′ + 4y′ + 3y = 0


r2 + 4r + 3 = 0

or, (r + 1)(r + 3) = 0

or, r = −1,−3


y(t) = c1er1t + c2e

r2t = c1e−t + c2e

−3t (3.11)


y(t) == −c1e−t − 3c2e

−3t (3.12)

Also, using the initial conditions y(0) = 2, y′(0) = −1. Then from (3.11) and

(3.12) we get

y(0) = 2 =⇒ c1 + c2 = 2 (3.13)

y′(0) = 3 =⇒ −2c1 − 3c2 = −1 (3.14)


c1 =5

2, c2 = −1

2


y =5

2e−t − 1

2e−3t

Example 79.



4y′′ − 8y′ + 3y = 0, y(0) = 2, y′(0) = −1

2


4y′′ − 8y′ + 3y = 0


4r2 − 8r + 3 = 0

or, (3r − 2)(2r − 1) = 0

or, r =2

3,

1

2


y(t) = c1er1t + c2e

r2t = c1e23t + c2e

12t (3.15)


y(t) =2

3c1e

23t +

1

2c2e

12t (3.16)

Also, using the initial conditions y(0) = 2, y′(0) = −12 . Then from (3.15) and

(3.16) we get

y(0) = 2 =⇒ c1 + c2 = 2 (3.17)

y′(0) = −1

2=⇒ 3

2c1 +

1

2c2 = −1

2(3.18)


c1 = −1

2, c2 =

5

2


y = −1

2e

32t +

5

2et2

Example 80.


2y′′ − 3y′ + y = 0, y(0) = 2, y′(0) = −1

2

Then determine the maximum value of the solution and also find the point where

the solution is zero.


2y′′ − 3y′ + y = 0



2r2 − 3r + 1 = 0

or, (2r − 1)(r − 1) = 0

or, r =1

2, 1


y(t) = c1er1t + c2e

r2t = c1et + c2e

12t (3.19)


y(t) = c1et +

1

2c2e

12t (3.20)

Also, using the initial conditions y(0) = 2, y′(0) = 12 . Then from (3.19) and (3.20)

we get

y(0) = 2 =⇒ c1 + c2 = 2 (3.21)

y′(0) = 3 =⇒ c1 +1

2c2 =

1

2(3.22)


c1 = −1, c2 = 3


y = −et + 3et2

For stationary point:

Differentiating y with respect to t,

y′ = −et +3

2et2

y′′ = −et +3

4et2

For maxima and minima:

y′ = 0

or, − et +3

2et2 = 0

or, et =3

2et2

or, et2 =

3

2

3.2. SOLUTIONOF LINEAR HOMOGENEOUS EQUATIONS; THEWRONSKIAN131

Taking log on both sides

ln et2 = ln

3

2

or,t

2= ln

3

2

or, t = 2 ln3

2= ln

9

4

Putting the value of t = ln 94 in y′′, we get

y′′ = −eln 94 +

3

4eln 9

8t = −9

4+

9

8= −9

8< 0

Hence y has maximum value at x = ln 94 and the maximum value is

ymax = −eln 94 + 3e

12

ln 94 = −9

4+ 3eln 3

2 = −9

4+

9

2=

9

4

For zero solution:

Let

y = 0

or, − et + 3et2 = 0

or, et = 3et2

or, et2 = 3

or, ln et2 = ln 3

or,t

2= ln 3

or, t = 2 ln 3 = ln 9

Hence, the solution y is zero when t = ln 9.

3.2 Solution of Linear Homogeneous Equations; the Wron-

skian

In the previous section, we discuss about the process of solving the homogeneous

linear second order differential equation of the form

ay′′ + by′ + cy = 0

where a, b, c are constants.

In this section, we discuss the homogeneous equation of second order of the form

y′′ + p(t)y′ + q(t)y = 0 (3.23)

where p and q are continuous real valued functions on an open interval I = (α, β),

where α and β are finite or α =∞ or both are infinite.


3.2.1 Derivative Operator and Differential Operators

In differential calculus the derivative operator D defined as

D =d

dt

∴dy

dt= Dy

and it transform a differential function into another function. For example

D(tan t) =d

dttan t = sec2 t

The higher order derivatives can be expressed in terms of D as

d2y

dt2=

(d

dt

)2

y = D2y,d3y

dt3=

(d

dt

)3

y = D3y

dny

dtn=

(d

dt

)ny = Dny

Let φ be a twice differential function operator on open interval I, we define the

differential operator L by the equation

L[φ] = φ′′ + pφ′ + qφ (3.24)

Here L[φ] is a function on I. The value of L[φ] at a point t is

L[φ](t) = φ′′(t) + p(t)φ′(t) + q(t)φ(t) (3.25)

For example, if p(t)− t3, q(t) = (1 + t2) and φ(t) = sin 2t. Then

L[φ](t) = (sin 2t)′′ + t3(sin 2t)′ + (1 + t2) sin 2t

= −4 sin 2t+ 2t3 cos 2t+ (1 + t2) sin 2t

Again from (3.24)

L[φ] = φ′′ + pφ′ + qφ

= D2φ+ pDφ+ qφ

= (D2 + pD + q)φ

Hence we write

L = D2 + pD + q

Now in this section the seond order homogeneous equation can be written as

φ′′ + pφ′ + qφ = 0

or, L[φ] = 0

or, L[y] = y′′ + py′ + q = 0 where y = φ (3.26)


where equation (3.26), we consider initial conditions

y(t0) = y0, y′(t0) = y0 (3.27)

where t0 ia any point in the interval I, and y(t0) and y′(t0) are given real numbers.

Theorem 4. Existence and Uniquness Consider the initial value proble

y′′ + p(t)y′ + q(t)y = g(t) y(t0) = y0, y′(t0) = y′0 (3.28)

where p, q and φ(t) are continuous on an open interval I that contaiins the point t0.

Then there is exactly one solution y = φ(t) of this problem, and the solution exists

throughout the interval I

The theorems says

1. The initial value problem has a solution; that is a solution exists.

2. The initial value problem has only one solution; that is the solution is unique.

3. The solution φ is defined throughout the interval I where the coefficient are

continuous and is at least twice differentiable there.

Example 81. Find the longest interval in which the solution of the initial value

problem

(t2 − 2t)y′′ + ty′ − (t+ 3)y = 0, y(1) = 2, y′(1) = 1

Solution: The given initial value problem is

(t2 − 3t)y′′ + ty′ − (t+ 3)y = 0, y(1) = 2, y′(1) = 1

or, y′′ +t

t2 − 3ty′ − t+ 3

t2 − 3ty = 0

or, y′′ +1

t− 3y′ − t+ 3

t(−3)y = 0


y′′ + p(t)y′ + q(t)y = g(t)

we get

p(t) = − 1

t− 3, q(t) = − t+ 3

t(t− 3), g(t) = 0

When t = 0, p(t) =∞ and when t = 0, 3 q(t) =∞. These the points of discontin-

uty are t = 0 and t = 3. Therefore, the longest open interval containing the initial

point t = 1, in which all the coefficients are continuous is 0 < t < 3. Thus (0, 3) is

the largest interval in which the solution of the initial value problem exists.

Example 82.


Find the longest interval in which the solution of the initial value problem

ty′′ + 3y = 0, y(2) = 1, y′(2) = 3


ty′′ + 3y = 0, y(2) = 1, y′(2) = 3

or, y′′ +3

ty = 0


y′′ + p(t)y′ + q(t)y = g(t)

we get

p(t) =3

t, q(t) = 0, g(t) = 0

When t = 0, p(t) = ∞. These the pointof discontinuty is t = 0. Therefore, the

longest open interval containing the initial point t = 2, in which all the coefficients

are continuous is 0 < t <∞. Thus (0,∞) is the longest interval in which the solution

of the initial value problem exists.

Example 83.


(t− 1)y′′ − 3ty′ + 4y = sin t, y(−2) = 2, y′(−2) = 1


(t− 1)y′′ − 3ty′ + 4y = sin t, y(1) = 2, y′(1) = 1

or, y′′ − 3t

t− 1y′ +

4

t− 1y = sin t


y′′ + p(t)y′ + q(t)y = g(t)

we get

p(t) = − 3t

t− 1, q(t) =

4

(t− 1), g(t) =

sin t

t− 1

When t = 1, p(t), q(t), g(t) = ∞. These the point of discontinuty is t = 1. There-

fore, the longest open interval containing the initial point t = −2, in which all the

coefficients are continuous is −∞ < t < 1. Thus (−∞, 1) is the largest interval in

which the solution of the initial value problem exists.

Example 84.



t(t− 4)y′′ + 3ty′ + 4y = 2, y(3) = 0, y′(3) = −1


t(t− 1)y′′ + 3ty′ + 4y = 2, y(3) = 0, y′(3) = −1

or, y′′ +3

(t− 4)y′ +

4

t(t− 4)y =

2

t(t− 4)


y′′ + p(t)y′ + q(t)y = g(t)

we get

p(t) =3

t− 4, q(t) =

4

t(t− 4), g(t) =

2

t(t− 4)

When t = 4, p(t), g(t) = ∞ and at t = 0, t = 4, q(t) = ∞.Thus the points of

discontinuty are t = 0, 3. Therefore, the longest open interval containing the initial


the longest interval in which the solution of the initial value problem exists.

Example 85.


(t− 3)y′′ + ty′ + ln |t|y = 2, y(1) = 0, y′(1) = 1


(t− 3)y′′ + ty′ + ln |t|y = 2, y(1) = 0, y′(1) = 1

or, y′′ +t

(t− 3)y′ +

ln |t|(t− 3)

y =2

(t− 3)


y′′ + p(t)y′ + q(t)y = g(t)

we get

p(t) =t

t− 3, q(t) =

ln |t|(t− 3)

, g(t) =2

(t− 3)

When t = 3, p(t), g(t) = ∞ and at t = 0, t = 3, q(t) = ∞. Thus the points of

discontinuty are t = 0, 3. Therefore, the longest open interval containing the initial


the longest interval in which the solution of the initial value problem exists.

Theorem 5. Principle of Superposition


If y1 and y2 are two solutions of the differential equation

L[y] = y′′ + p(t)y′ + q(t)y = 0

then the linear combination c1y1 + c2y2 is also a solution for any values of c1 and c2.

Proof: Since y1 and y2 are solutions of L[y] = y′′ + p(t)y′ + q(t)y = 0, so

L[y1] = y′′1 + p(t)y′1 + q(t)y1 = 0

L[y2] = y′′2 + p(t)y′2 + q(t)y2 = 0

Let c1 and c2 be any constants. Then

L[c1y1 + c2y2]

= (c1y1 + c2y2)′′ + p(t)(c1y1 + c2y2)′ + q(t)(c1y1 + c2y2)

= c1y′′1 + c2y

′′2 + p(t)(c1y

′1 + c2y

′2) + q(t)(c1y1 + c2y2)

= c1(y′′1 + p(t)y′1 + q(t)y1) + c2(y′′2 + p(t)y′2 + q(t)y2)

= c1l[y1] + c2L[y2]

= c1 · 0 + c2 · 0 = 0

This shows that c1y1 + c2y2 is a solution of L[y] = 0. Proved

Note Let y1, y2, · · · , yn be solutions of L[y] = 0. Then c1y1 + c2y2 + · · · + cnyn is

solution of L[y] = 0.

3.2.2 The Wronskian

Let us consider the initial value problem

L[y] = y′′ + p(t)y′ + q(t)y = 0 with initial condition y(t0) = y0, y′(t0) = y0(3.29)

Let y1(t) and y2(t) be two solutions of (3.29). Then we can costruct an infinitely

family of solutions by

y(t) = c1y1(t) + c2y2(t)3tw (3.30)

By the initial condition y(t0) = y0 gives from (3.30)

c1y1(t0) + c2y2(t0) = y0 (3.31)

and from the initial condition y′(t0) = y′0 gives from (3.30)

c1y′1(t0) + c2y

′(t0) = y′0 (3.32)

The determinant of the coefficient of c1 and c2 of the equations (3.31) and (3.32) is

W =

∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣ = y1(t0)y′2(t0)− y′1(t0)y2(t0)


When W 6= 0, the equations (3.31) and (3.32) have a unique solution (c1, c2). Solving

the equations by Cramer’s rule

c1 =

∣∣∣∣∣ y0 y2(t0)

y′0 y′2(t0)

∣∣∣∣∣∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣c2 =

∣∣∣∣∣ y1(t0) y0

y′1(t0) y′0

∣∣∣∣∣∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣With these values of c1 and c2 the expression (3.30) satisfies the initial conditions

y(t0) = y0 and y′(t0) = y0. The determinant

W =

∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣is called the Wronskian Determinat or Wronskian of the solutions y1 and y2.

Some time it is denoted by

W = W (y1, y2)(t0)

Theorem 6. Let y1 and y2 be two solutions of

L[y] = y′′ + p(t)y′ + q(t)y = 0 (3.33)

and that the initial condition

y(t0) = y0, y′(t0) = y′0 (3.34)

Then it is always possible to choose the constants c1, c2 so that

y = c1y1(t) + c2y2(t)

satisfies the differential equation (3.33) and the initial condition (3.34) if and only

if the Wronskian

W =

∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣ 6= 0

Example 86. Find the Wronskian of the solutions y1(t) = e−2t and y2 = e−3t of

the differential equation

y′′ + 5y′ + 6y = 0

Solution: The Wronskian of two solutions y1(t) = e−2t and y2 = e−3t is

W =

∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣=

∣∣∣∣∣ e−2t e−3t

−2e−2t −3e−3t

∣∣∣∣∣ = −3e−2te−3t + 2e−2te−3t = −e−3t 6= 0

for all value of t, thus the functions y1 and y2 can be used to construct solutions

of the given differential equation, together with initial conditions for any prescribed

value of t.


Example 87. Find the Wronskian of the given pair of functions

1. cos t, sin t 2. e−2t, te−2t, 3. x, xex, 4. et sin t, et cos t

Solution: 1. Let y1 = cos t and y2 = sin t. Then y′1 = − sin t and y′2 = cos t.

Therefore, the Wronskian of y1 and y2 is

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ cos t sin t

− sin t cos t

∣∣∣∣∣ = cos2 t+ sin2 t = 1

2. Let y1 = e−2t and y2 = te−2t. Then y′1 = −2e−2t and y′2 = e−2t − 2te−2t.


W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ e−2t te−2t

−2e−2t e−2t − 2te−2t

∣∣∣∣∣= e−2t

(e−2t − 2te−2t

)+ 2e−2tte−2t = e−4t

3. Let y1 = x and y2 = xex. Then y′1 = −1 and y′2 = xex + ex. Therefore, the

Wronskian of y1 and y2 is

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ x xex

1 ex + xex

∣∣∣∣∣= xex + x2ex − xex = x2ex

4. Let y1 = et sin t and y2 = et cos t. Then y′1 = et sin t+ et cos t and y′2 = et cos t−et sin t. Therefore, the Wronskian of y1 and y2 is

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ et sin t et cos t

et sin t+ et cos t et cos t− et sin t

∣∣∣∣∣= e2t

∣∣∣∣∣ sin t cos t

sin t+ cos t cos t− sin t

∣∣∣∣∣= e2t(sin t(cos t− sin t)− cos t(sin t+ cos t))

= −e2t(sin2 t+ cos2 t) = −e2t


Theorem 7.

Let y1 and y2 be two solutions of the differntial equation

L[y] = y′′ + p(t)y′ + q(t)y = 0

Then the family of solutions

y = c1y1 + c2y1

with arbitrary coefficients c1 and c2 includes every solution if and only if there is a

point t0 where the Wronskian of y1 and y2 is not zero.

Example 88. Suppose that y1 = eµt and y2 = eλt are two fundamental solutions of

the form

L[y] = y′′ + p(t)y′ + q(t)y = 0

Show that they form a fundamental solutions if λ 6= µ.

Solutions: We have y1 = eµt and y2 = eλt. Then y′1 = µeµt and y′2 = λeλt.


W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ eµt eλt

µeµt λeλt

∣∣∣∣∣= eµteλt

∣∣∣∣∣ 1 1

µ λ

∣∣∣∣∣ = e(λ+µ)t(λ− µ) 6= 0

(3.35)

for all t when λ 6= µ. Hence y1 and y2 forms a fundamental solutions.

Example 89.

Show that y1(t) = t1/2 and y2 = t−1 form a fundamental set of solution of

2t2y′′ + 3ty′ − y = 0, t > 0


2t2y′′ + 3ty′ − y = 0, t > 0 (3.36)

Here y1(t) = t1/2. Then

y′1(t) =1

2t−1/2, y′′1(t) = −1

4t−3/2


From (3.36)

2t2(−1

4t−3/2

)+ 3t

(1

2t−1/2

)− t1/2 = 0

or, 0 = 0

Hence, y1 = t1/2 is a solution of (3.36).

Here y2(t) = t−1. Then

y′2(t) = −t−2, y′′2(t) = 2t−3

From (3.36)

2t2(2t−3

)+ 3t

(−t−2()

)− t−1 = 0

or, 0 = 0

Hence, y2 = t−1 is a solution of (3.36). The Wronskian of y1 and y2 is

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ t1/2 t−1

12 t−1/2 −t−2

∣∣∣∣∣= t1/2(−t2)− 1

2t−1/2t−1 = −3

2t−3/2 6= 0, for t > 0

Thus, W (y1, y2) 6= 0 for t > 0. Hence y1, y2 form a fundamental set of solutions.

Example 90.

Show that y1(t) = t2 and y2 = t−1 form a fundamental set of solution of

t2y′′ − 2y = 0, t > 0

Then show that y = c1t+ c2t−1 is also solution of this equation for any c1 and c2.


t2y′′ − 2y = 0, t > 0 (3.37)

Here y1(t) = t2. Then

y′1(t) = 2t, y′′1(t) = 2

From (3.36)

t2 (2)− 2t2 = 0

or, 0 = 0

Hence, y1 = t2 is a solution of (3.37).

Here y2(t) = t−1. Then

y′2(t) = −t−2, y′′2(t) = 2t−3


From (3.36)

t2(2t−3

)− 2t−1 = 0

or, 0 = 0

Hence, y2 = t−1 is a solution of (3.37). The Wronskian of y1 and y2 is

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ t2 t−1

2t −t−2

∣∣∣∣∣= t2(−t−2)− 2t(t−1) = −3 6= 0


Also y = c1y1 + c2y2 is solution for all t > 0.

Example 91. Verify that y1(t) = 1 and y2(t) = t12 are solutions of the differential

equation yy′′ + (y′)2 = 0 for t > 0. Then show that y = c1 + c2t12 is not general

solution of this equation. Explain.


yy′′ + (y′)2 = 0 for t > 0 (3.38)

Here y1(t) = 1. Then

y′1 = 0, y′′1 = 0

From (3.38)

0− 2 · 0 = 0

or, 0 = 0

Again, we have y2(t) = 12 t

12 . Then

y′2 =1

2t−1/2, y′′2 = −1

4t−3/2

From (3.38)

t1/2(−1

4t−3/2

)+

(1

2t1/2)2

= 0

or, 0 = 0

Thus, y2 = t1/2 is a solution of (3.38) If y = c1y1 + c2y2 = c1 + c2t1/2 and

y′ =1

2c2t−1/2, y′′ = −1

4c2t−3/2


substituted in the given equation, we get

yy′′ + (y′)2 =(c1 + c2t

1/2)(−1

4c2t−3/2

)+

(1

2c2t−1/2

)2

= −c1c2

4t−3/2 − c2

2

4t−1 +

c22

4t−1 = −c1c2

4t−3/2

which is zero only when c1 = 0 or c2 = 0. Thus the linear combination of two

solutions is not, in general, a solution. The Wronskian of y1 and y2 is

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣ =

∣∣∣∣∣ 1 t1/2

0 12 t−1/2

∣∣∣∣∣ =1

2t−1/2for t > 0.

Thus, W (y1, y2) 6= 0 for t > 0. This contradiction due to the fact the differential

equation is not linear.

Theorem 8.

Consider the differential equation

L[y] = y′′ + p(t)y′ + q(t)y = 0 (3.39)

whose coefficients p(t) and q(t) are continuous on some interval I. Choose some

point t0 ∈ I. Let y1 be solution of the equation (3.39) that also satisfies the initial

conditions

y1(t0) = 1, y′1(t0) = 0

and y2 be solution of the equation (3.39) that also satisfies the initial conditions

y2(t0) = 0, y′2(t0) = 1

Then y1 and y2 form a fundamental set solution of equation (3.39).

Proof: To show that they form a fundamental set of solution, we need to calcuate

their Wronskian at t0

W (y1, y2)(t0) =

∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣ =

∣∣∣∣∣ 1 0

0 1

∣∣∣∣∣ = 1 6= 0

That is W (y1, y2)(t0) 6= 0. Thus, their Wronskian is not zero at the point t0, the

functions y1 and y2 do form a fundamental solutions.

Example 92.

Le p(t) and q(t) be continuous and that the functions y1 and y2 are solutions of

the differential equation y′′ + p(t)y′ + q(t)y = 0 on an open interval I.

1. Prove that if y1 and y2 are zero at same point in I, they can not be a funda-

mental set of solution on that interval.


2. Prove that if y1 and y2 have maximum or minimum at the same point in I,

then they can not be a fundamental set solutions on that interval.

Solution: 1. Let y1 and y2 are solutions of the differential equation

y′′ + p(t)y′ + q(t)y = 0

on an open interval I, such that y1(t0) = 0 and y2(t0) = 0 where t0 ∈ I. Now the

Wronskian at t0

W (y1, y2)(t0) =

∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣ =

∣∣∣∣∣ 0 0

y′1(t0) y′2(t0)

∣∣∣∣∣ = 0

That is W (y1, y2)(t0) = 0. Thus, their Wronskian is zero at the point t0, the

functions y1 and y2 do not form a fundamental soution.

2. Let y1 and y2 are solutions of the differential equation

y′′ + p(t)y′ + q(t)y = 0

on an open interval I, such that y1 and y2 have maxima or minima at t0. Then

y′1(t0) = 0 and y′2(t0) = 0 where t0 ∈ I. Now the Wronskian at t0

W (y1, y2)(t0) =

∣∣∣∣∣ y1(t0) y2(t0)

y′1(t0) y′2(t0)

∣∣∣∣∣ =

∣∣∣∣∣ y1(t0) y2(t0)

0 0

∣∣∣∣∣ = 0

That is W (y1, y2)(t0) = 0. Thus, their Wronskian is zero at the point t0, the

functions y1 and y2 do not form a fundamental soution.

Example 93.

Find the fundamental set of solution specified by theorem 8 for the differential

equation

y′′ − y = 0

using the initial point t0 = 0.


y′′ − y = 0 initial point t0 = 0 (3.40)

The auxiliary equation of (3.40) is

r2 − 1 = 0

and roots are r = ±1. So the solutions of (3.40) are

y1(t) = et, y2(t) = e−t

The general solution is

y = c1et + c2e

−t


The Wronskian of the solutions is

W (y1, y2) =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣ =

∣∣∣∣∣ et e−t

et −e−t

∣∣∣∣∣= −ete−t − ete−t = −1− 1 = −2 6= 0

and hence the exponential form a fundamental set of solutions . Howevere,

y1(t0) = y1(0) = e0 = 1

y′1(t0) = y′1(0) = e0 = 1

y2(t0) = y2(0) = e0 = 1

y′2(t0) = y′2(0) = −e0 = −1

and hence they are not the fundamental solutions indicated by Theorem (8). We

need to find the solutions satisfying the initial coonditions

y3(t0) = 1, y′3(t0) = 0, y4(t0) = 0, y′4(t0) = 1

Since general solution of equation (3.40) is c1et + c2e

−t, so let

y3(t) = c1et + c2e

−t, and y4 = d1et + d2e

−t

∴ y′3(t) = c1et − c2e

−t, and y4 = d1et − d2e

−t

Since, y3(0) = 1, so

c1 + c2 = 1 (3.41)

and y′3(0) = 0, so

c1 − c2 = 0 (3.42)


c1 =1

2, c2 =

1

2

Thus,

y3(t) =1

2et +

1

2e−t = cosh t


d1 + d2 = 0 (3.43)

and y′4(0) = 1, so

d1 − d2 = 1 (3.44)



d1 =1

2, d2 = −1

2

Thus,

y4(t) =1

2et − 1

2e−t = sinh t

Also, y′3(t) = sinh t and y′4(t) = cosh t, therefore,

W (y3, y4)(t) =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣ =

∣∣∣∣∣ cosh t sinh t

sinh t cosh t

∣∣∣∣∣= cosh2 t− sinh2 t = 1 6= 0

Hence y3(t) and y4(t) forms a fundamental set of solution of the the differential

equation (3.40) and the general solution is

y = k1y3 + k2y4 = k1 cosh t+ k2 sinh t

where k1 and k2 are arbitrary constants.

Example 94.

Find the fundamental set of solution specified by theorem 8 for the differential

equation

y′′ + 4y′ + 3y = 0

using the initial point t0 = 1.


y′′ + 4y′ + 3y = 0 initial point t0 = 1 (3.45)


r2 + 4r + 3 = 0

and roots are r = −1, r = −3. So the solutions of (3.45) are

y1(t) = e−3t, y2(t) = e−t


y = c1e−3t + c2e

−t

The Wronskian of the solutions is

W (y1, y2) =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣ =

∣∣∣∣∣ e−3t e−t

−3e−3t −e−t

∣∣∣∣∣= −e−3te−t + 3e−3te−t = 2e−4t 6= 0 at t0 = 1 where t0 = 1


and hence the exponential functions form a fundamental set of solutions . Howevere,

y1(t0) = y1(1) = e−3 =1

e3

y′1(t0) = y′1(1) = −3e−3 = − 3

e3

y2(t0) = y2(1) = e−1 =1

e

y′2(t0) = y′2(0) = −e−1 = −1

e

and hence they are not the fundamental solutions indicated by Theorem (8). We

need to find the solutions satisfying the initial coonditions

y3(t0) = 1, y′3(t0) = 0, y4(t0) = 0, y′4(t0) = 1

Since general solution of equation (3.40) is c1e−3t + c2e

−t, so let

y3(t) = c1e−3t + c2e

−t, and y4 = d1e−3t + d2e

−t

∴ y′3(t) = −3c1e−3t − c2e

−t, and y4 = −3d1e−3t − d2e

−t


c1e−3 + c2e

−1 = 1

or, c1 + c2e2 = e3 (3.46)

and y′3(0) = 0, so

−3c1e−3 − c2e

−1 = 0

or, 3c1 + c2e2 = 0 (3.47)


c1 = −e3

2, c2 =

3e

2

Thus,

y3(t) = −e3

2e−3t +

3e

2e−t = −1

2e−3(t−1) +

3e

2e−t = −1

2e−3(t−1) +

3

2e−(t−1)


d1e−3 + d2e

−1 = 0

or, d1 + d2e2 = 0 (3.48)

and y′4(0) = 1, so

−3d1e−3 − d2e

−1 = 1

or, − 3d1 − d2e2 = e3 (3.49)



d1 = −e3

2, d2 =

e

2

Thus,

y4(t) = −e3

2e−3t +

e

2e−t = −1

2e−3(t−1) +

1

2e−(t−1)

Also, y′3(t) = 32e−3(t−1) − 3

2e−(t−1) and y′4(t) = 3

2e−3(t−1) − 1

2e−(t−1), therefore,

W (y3, y4)(t) =

∣∣∣∣∣ y3 y4

y′3 y′4

∣∣∣∣∣=

∣∣∣∣∣ −12e−3(t−1) + 3

2e−(t−1) −1

2e−3(t−1) + 1

2e−(t−1)

32e−3(t−1) − 3

2e−(t−1) 3

2e−3(t−1) − 1

2e−(t−1)

∣∣∣∣∣= −e−4(t−1) 6= 0

Hence y3(t) and y4(t) forms a fundamental set of solution of the the differential

equation (3.40) and the general solution is

y = k1y3 + k2y4 = k1

(−1

2e−3(t−1) +

3

2e−(t−1)

)+ k2

(−1

2e−3(t−1) +

1

2e−(t−1)

)where k1 and k2 are arbitrary constants.

The following theorem, gives a simple explicit formula for the Wronskial of any two

solutions of any homogeneous equation, even if the solutions themselves are not

known.

Theorem 9. Abel’s Theorem

If y1 and y2 are two solutions of the differential equation

L[y] = y′′ + p(t)y′ + q(t)y = 0 (3.50)

where p(t) and q(t) are continuous on an open interval I, then the Wronskian

W (y1, y2)(t) is given by

W (y1, y2)(t) = ce∫p(t)dt (3.51)

where c is a certain constant that depends on y1 and y2 but not on t. Further,

W (y1, y2)(t) either is zero foe all t in I or else is never zero in I (if c 6= 0).


L[y] = y′′ + p(t)y′ + q(t)y = 0 (3.52)

Let y1 and y2 be solutions of (3.52). Then

y′′1 + p(t)y′1 + q(t)y1 = 0 (3.53)

y′′1 + p(t)y′2 + q(t)y2 = 0 (3.54)


Multiplying the equation (3.53) by y2 and (3.54) by y1 and subtracting, we get

(y1y′′2 − y′′1y2) + p(t)(y1y

′2 − y′1y2) = 0 (3.55)

We have

W = W (y3, y4)(t) =

∣∣∣∣∣ y3 y4

y′3 y′4

∣∣∣∣∣ = y1y′2 − y′1y2

Differentiating both sides with respect to t,

dW

dt= y1y

′′2 + y′1y

′2 − y′′1y2 − y′1y′2 = y1y

′′2 − y′′1y2

Using these values in (3.55)

dW

dt+ p(t)W = 0

or,dW

W= −p(t)dt

Integrating, ∫dW

W= −

∫p(t)dt+ ln c

or, lnW = −∫p(t)dt+ ln c

or, W = e−∫p(t)dt+ln c

or, W = ce−∫p(t)dt

where c is a constant. The value of c depends on which pair of solution (3.52).

However the exponential function is never zero, W (t) is not zero unless c = 0, in

this case W (t) = 0 for all t.

Example 95.

Let y1(t) = t1/2 and y2(t) = t−1 be solutions of the equation 2t2y′′ + 3ty′ − y =

0, t > 0. Find the Wronskian W (y1, y2)(t) and also verify it by Abel’s formula

W (y1, y2) = ce−∫p(t)dt


2t2y′′ + 3ty′ − y = 0, t > 0 (3.56)

Given that y1(t) = t1/2 and y2(t) = t−1 are solution of (3.56). The Wronskian of y1

and y2 is

W =

∣∣∣∣∣ y3 y4

y′3 y′4

∣∣∣∣∣=

∣∣∣∣∣ t1/2 t−1

12 t−1/2 −t−2

∣∣∣∣∣= t1/2(−t2)− 1

2t−1/2t−1 = −3

2t−3/2 6= 0 for t > 0

3.3. COMPLEX ROOTS OF THE CHARACTERSTICS EQUATION 149


The equation (3.56) can be written as

y′′ +3

2ty′ − 1

2t2y = 0

Comparing it with y′′ + p(t)y′ + q(t)y = 0, we get

p(t) =3

2t

From Abel’s theorem

W (y1, y2)(t) = ce−∫p(t)dt

= ce−∫

32tdt

= ce−32

ln t

= celn t−3/2

= ct−3/2

Choose c = −32 , we get

W (y − 1, y2) = −3

2t−3/2

Hence, verified.

3.3 Complex Roots of the Characterstics Equation

3.3.1 Euler’s Form

We have Euler’s form of complex number

eiθ = cos θ + i sin θ

and

e−iθ = cos(−θ) + i sin(−θ) = cos θ − i sin θ

Example 96.

Use Euler’s Formula to write the given expression in the form of a+ ib.

1.e1+2i, 2. 21−i, 3.π−1+2i

Solution: 1.

e1+2i = e · e2i = e(cos 2 + i sin 2)

2. We have

x = elnx


Putting x = 2−i

2−i = eln(2−i) = e−i ln 2 = cos(ln 2)− i sin(ln 2)

Now,

21−i = 2 · 2−i = 2(cos ln 2− i sin ln 2)

3. We have

x = elnx

Putting x = 2−i

π2i = eln(π2i) = eπ ln 2 = cos(π ln 2) + i sin(π ln 2)

Now,

π−1+2i = π−1π2i =1

π(cos(π ln 2) + i sin(π ln 2))

3.3.2 Solution of Differential Equation with Complex Roots

Let us consider the second order homogeneous equation of constant coefficients

ay′′ + by′ + cy = 0 (3.57)

Let y = ert be a solution of (3.57). Then the auxiliary equation is

ar2 + br + c = 0 (3.58)

Solving (3.58), we get

r =−b±

√b2 − 4ac

2a

Let b2 − 4ac < 0. Then roots of (3.58)

r =−b± i

√4ac− b2

2a= α± iβ, where α = − b

2a, β =

√4ac− b2

2a(3.59)

where α and β are real numbers. Thus solutions of the equation (3.57) are

y1 = e(α+iβ)t, y2 = e(α−iβ)t

We have by Euler’s formula

eiθ = cos θ + i sin θ

e−iθ = cos θ − i sin θ

Hence

y1 = e(α+iβ)t = eαteiβt = eαt(cosβt+ i sinβt)

y2 = e(α−iβ)t = eαte−iβt = eαt(cosβt− i sinβt)


and

y′1 = (α+ iβ)e(α+iβ)t, y′2 = (α− iβ)e(α−iβ)t

Now Wronskian is

W (y1, y2) =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ e(α+iβ)t e(α−iβ)t

(α+ iβ)e(α+iβ)t (α− iβ)e(α−iβ)t

∣∣∣∣∣= e(α+iβ)te(α−iβ)t

∣∣∣∣∣ 1 1

(α+ iβ) (α− iβ)

∣∣∣∣∣= e2iβ[α− iβ − (α+ iβ)]

= −2iβe2αt 6= 0

hence y1 and y2 forms a fundamental set of solution of (3.57) and hence the general

solution is the linear combination

c1y1 + c2y2

where c1 and c2 are arbitrary constants. In particular

1

2(y1 + y2),

1

2(y1 − y2)

are solutions. Now1

2(y1 + y2) =

1

2

eαt(cosβt+ i sinβt) + eαt(cosβt− sinβt)

= eαt cosβt

1

2(y1 − y2) =

1

2

eαt(cosβt+ i sinβt)− eαt(cosβt− sinβt)

= eαt sinβt

which are real solutions. Let

u(t) = eαt cosβt, v(t) = eαt sinβt

Then

u′(t) = αeαt cosβt− βeαt sinβt = eαt(α cosβt− β sinβt)

v′(t) = αeαt sinβt+ βeαt cosβt = eαt(α cosβt+ β sinβt)

Now, the Wronskian of u and v is

W (u, v) =

∣∣∣∣∣ u v

u′ v′

∣∣∣∣∣=

∣∣∣∣∣ eαt cosβt eαt sinβt

eαt(α cosβt− β sinβt) eαt(α cosβt+ β sinβt)

∣∣∣∣∣= eαteαt

∣∣∣∣∣ cosβt sinβt

(α cosβt− β sinβt) (α sinβt+ β cosβt)

∣∣∣∣∣= e2αt(α cosβt sinβt+ β cos2 βt− α sinβt cosβt+ β sin2 βt)

= e2αtβ 6= 0 as β 6= 0


Thus, u(t) and v(t) form a fundamental set of solution and hence the general solution

is

y = c1u+ c2v = eαt(c1 cosβt+ c2 sinβt)

Example 97.

Find the general solution of the differential equation

y′′ + y′ + 9.25y = 0

by finding out the fundamental set of solution. Also, find the solution that satisfies

the initial conditions

y(0) = 2, y′(0) = 8

Solution: The given second order homogeneous equation of constant coefficients

y′′ + y′ + 9.25y = 0 (3.60)

The auxiliary equation is

r2 + r + 9.25 = 0

Solving (??), we get

r =−b±

√b2 − 4ac

2a

=−1±

√12 − 4 · 1 · 9.25

2 · 1

=−1±

√1− 37

2=−1± 6i

2= −1

2± 3i

Thus solutions of the equation (3.60) are

y1 = e(− 12

+3i)t = e−12te3it = e−

12t(cos 3t+ i sin 3t)

y2 = e(− 12−3i)t = e−

12te−3it = e−

12t(cos 3t− i sin 3t)

and

y′1 =

(−1

2+ 3i

)e(− 1

2+3i)t, y′2 =

(−1

2− 3i

)e(− 1

2−3i)t

Now Wronskian is

W (y1, y2) =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ e(− 12

+3i)t e(− 12−3i)t(

−12 + 3i

)e(− 1

2−3i)t

(−1

2 + 3i)e(− 1

2−3i)t

∣∣∣∣∣= e(− 1

2+3i)te(− 1

2+3i)t

∣∣∣∣∣ 1 1

(−12 + 3i) (−1

2 − 3i)

∣∣∣∣∣= e−t[−1

2− 3i+

1

2− 3i]

= −6ie−t 6= 0


hence y1 and y2 forms a fundamental set of solution of (3.60) and hence the general

solution is the linear combination

c1y1 + c2y2

where c1 and c2 are arbitrary constants. In particular

1

2(y1 + y2),

1

2(y1 − y2)

are solutions. Now

1

2(y1 + y2) =

1

2

e−

12t(cos 3t+ i sin 3t) + e−

12t(cos 3t− sin 3t)

= e−

12t cos 3t

1

2(y1 − y2) =

1

2

e−

12t(cos 3t+ i sin 3t)− e−

12t(cos 3t− sin 3t)

= e−

12t sin 3t

which are real solutions. Let

u(t) = e−12t cos 3t, v(t) = e−

12t sin 3t

Then

u′(t) = −1

2e−

12t cos 3t− 3e

12t sin 3t = −e−

12t

(1

2cos 3t+ 3 sin 3t

)

v′(t) = −1

2e−

12t sin 3t+ 3e−

12t cos 3t = e−

12t

(−1

2cos 3t+ 3 sin 3t

)Now, the Wronskian of u and v is

W (u, v) =

∣∣∣∣∣ u v

u′ v′

∣∣∣∣∣=

∣∣∣∣∣ e−12t cos 3t e−

12t sin 3t

−e−12t(

12 cos 3t+ 3 sin 3t

)e−

12t(−1

2 cos 3t+ 3 sin 3t) ∣∣∣∣∣

= e−12te−

12t

∣∣∣∣∣ cos 3t sin 3t

−12 cos 3t− 3 sin 3t −1

2 cos 3t+ 3 sin 3t

∣∣∣∣∣= e−t(−1

2cos 3t sin 3t+ 3 cos2 3t+

1

2sin 3t cos 3t+ 3 sin2 3t)

= 3e−t 6= 0

Thus, u(t) and v(t) form a fundamental set of solution and hence the general solution

is

y = c1u+ c2v = e−12t(c1 cos 3t+ c2 sin 3t)

where c1 and c2 are arbitrary constants. Using the initial condition

y(0) = 2

=⇒ 1[c1 cos 0 + c2 sin 0] = 2

=⇒ c1 = 2


Again,

y′(t) = −1

2e−

12t(c1 cos 3t+ c2 sin 3t) + 3e−

12t(c1 sin 3t+ c2 cos 3t)

Again y′(0) = 8 gives

8 = −1

2c− 1 + 3c2

or, 8 = −1 + 3c2

or, c2 = 3

Hence, the required solution of the given initial value problem is

y = e−t/2(2 cos 3t+ 3 sin 3t)

Example 98.

Find the general solution of the following equations

1. y′′ + 9y = 0

2. y′′ + 2y′ + 2y = 0

3. y′′ + 6y′ + 13y = 0

4. 9y′′ + 9y′ − 4y = 0

Solution: 1. The given differetial equation is

y′′ + 9y = 0 (3.61)

The charaterstics equation of (3.61) is

r2 + 9 = 0

Solving it we get r = 0± 3i. Thus, the general solution is

y = e0t(cos 3t+ c2 sin 3t) = c1 cos 3t+ c2 sin 3t

2. The given differetial equation is

y′′ + 2y′ + 2y = 0 (3.62)


r2 + 2r + 2 = 0

Solving it we get

r =−2±

√22 − 4 · 1 · 22 · 1

=−2±

√−4

2= −1± 2i


Thus, the general solution is

y = e−t(cos 2t+ c2 sin 2t) = e−t(c1 cos 2t+ c2 sin 2t)


y′′ + 6y′ + 13y = 0 (3.63)


r2 + 6r + 13 = 0

Solving it we get

r =−6±

√62 − 4 · 1 · 13

2 · 1=−6±

√−16

2= −3± 2i


y = e−3t(cos 2t+ c2 sin 2t) = e−3t(c1 cos 2t+ c2 sin 2t)


9y′′ + 9y′ − 4y = 0 (3.64)


9r2 + 9r − 4 = 0

Solving it we get

r =−9±

√92 − 4 · 9 · (−4)

2 · 9=−9±

√225

18=−9± 15

18

∴ r =1

3, r = −4

3


y = c1et/3 + c2e

−4t/3

Example 99.

Find the solution of the initial value problem of the following differential equation

1. y′′ + 4y = 0, y(0) = 0, y′(0) = 1.

2. 16y′′ − 8y′ + 145y = 0 = 0, y(0) = −2, y′(0) = 1.

3. y′′ − 6y′ + 13y = 0, y(π/2) = 0, y′(pi/2) = 2.

4. y′′ + y′ + 1.25y = 0, y(0) = 4, y′(0) = 1.



y′′ + 4y = 0 (3.65)

The characterstics equation of (3.65) is

r2 + 4 = 0

Solving it we get r = 0± 2i. Thus, the general solution is

y = e0t(c1 cos 2t+ c2 sin 2t) = c1 cos 2t+ c2 sin 2t (3.66)

where c1 and c2 are arbitrary constants, is general solution.

Using the initial condition y(0) = 0, we get

c1 = 0

Also,

y′ = −2c1 sin 2t+ 2c2 cos 2t

Using y′(0) = 0, we get

2c2 = 1

∴ c2 =1

2

Putting the values of c1 and c2 in (3.66), we get

y =1

2sin 2t


16y′′ − 8y′ + 145y = 0, y(0) = −2, y′(0) = 1 (3.67)


16r2 − 8r + 145 = 0

Solving,

r =8±

√(−8)2 − 4 · 16 · 145

2 · 16

=8± 96i

32=

1

4± 3i


y = e1/4t(c1 cos 3t+ c2 sin 3t) (3.68)



Using the initial condition y(0) = −2, we get

c1 = −2

Also,

y′ = et/4 (−3c1 sin 3t+ 3c2 cos 3t) +1

4et/4(c1 cos 3t+ c2 sin 2t)


3c2 +1

4c1 = 1

or, 3c2 −1

2= 1

or, c2 =1

2


y = et/4(−2 cos 3t+

1

2sin 3t

)Solution: 3. The given differential equation is

y′′ − 6y′ + 13y = 0, y(pi/2) = 0, y′(pi/2) = 2 (3.69)


r2 − 6r + 13 = 0

Solving,

r =6±

√(−6)2 − 4 · 1 · 13

2 · 1=

6± 4i

2= 3± 2i


y = e3t(c1 cos 2t+ c2 sin 2t) (3.70)


Using the initial condition y(π/2) = 0, we get

c1 = 0


y = e3tc2 sin 2t (3.71)


Also,

y′ = 3c2e3t sin 2t+ 2c2e

3t cos 2t

Using y′(π/2) = 2, we get

3c2e3π/2 sinπ + 2c2e

3π/2 cosπ = 2

or, − 2c2e3π/2 = 2

or, c2 = −e−3π/2


y = −e3te−3π/2 sin 2t = −e3(t−π2 ) sin 2t


y′′ + y′ + 1.25y = 0, y(0) = 4, y′(0) = 1 (3.72)


r2 + r + 1.25 = 0

4r2 + 4r + 5 = 0

Solving,

r =−4±

√(4)2 − 4 · 4 · 52 · 4

=−4± 8i

2= −1

2± i


y = e−t2 (c1 cos t+ c2 sin t) (3.73)


Using the initial condition y(0) = 4, we get

c1 = 4

Also,

y′ = −1

2e−

t2 (c1 cos t+ c2 sin t) + e−

t2 (−c1 sin t+ c2 cos t)


c2 −1

2c1 = 1

or, c2 − 2 = 1

or, c2 = 3


y = e−t/2(4 cos t+ 3 sin t)


3.3.3 Euler’s Equation

An equation of the form

t2d2y

dt2+ αt

dy

dt+ βy = 0 (3.74)

where α and β are real constants, is called an Euler’s Equation. To solve the

equation (3.74). let us put

t = ez that is z = ln t

Thendz

dt=

1

t

Now,

dy

dt=dy

dz

dz

dt=

1

t

dy

dz

or, tdy

dt=dy

dz

Again,

d2y

dt2=

d

dt

(dy

dt

)=

d

dt

(1

t

dy

dz

)= − 1

t2dy

dt+

1

t

d

dt

(dy

dz

)= − 1

t2dy

dt+

1

t

d

dz

(dy

dz

)dz

dt

= − 1

t2dy

dt+

1

t2d2y

dz2

∴ t2d2y

dt2= −dy

dt+d2y

dt2

The equation (3.74) becomes

−dydt

+d2y

dz2+ α

dy

dz+ βy = 0

or,d2y

dz2+ (α− 1)

dy

dz+ βy = 0

The equation (3.75) has constant coefficents. If y1(z) and y2(z) form fundamental

set of solution of the equation (3.75), then y1(ln t) and y2(ln t) form fundamental set

of solution of the equation (3.74).

Example 100.


Find the general solution to the following Euler’s equations

1. t2y′′ + ty′ + y = 0.

2. t2y′′ + 4ty′ + 2y = 0.

3. t2y′′ − 4y′ − 6y = 0.

Solution: 1.The given equation is

t2d2y

dt2+ t

dy

dt+ y = 0 (3.75)

To solve the equation (3.75). let us put


Thendz

dt=

1

t

Now,

dy

dt=dy

dz

dz

dt=

1

t

dy

dz

or, tdy

dt=dy

dz

Again,

d2y

dt2=

d

dt

(dy

dt

)=

d

dt

(1

t

dy

dz

)= − 1

t2dy

dt+

1

t

d

dt

(dy

dz

)= − 1

t2dy

dt+

1

t

d

dz

(dy

dz

)dz

dt

= − 1

t2dy

dt+

1

t2d2y

dz2

∴ t2d2y

dt2= −dy

dt+d2y

dt2


−dydt

+d2y

dz2+dy

dz+ y = 0

or,d2y

dz2+ y = 0



r2 + 1 = 0

or, r = 0± i


y = e0z(c1 cos z + c2 sin z)

or, y = c1 cos(ln t) + c2 sin(ln t)


t2d2y

dt2+ 4t

dy

dt+ 2y = 0 (3.76)



Thendz

dt=

1

t

Now,

dy

dt=dy

dz

dz

dt=

1

t

dy

dz

or, tdy

dt=dy

dz

Again,

d2y

dt2=

d

dt

(dy

dt

)=

d

dt

(1

t

dy

dz

)= − 1

t2dy

dt+

1

t

d

dt

(dy

dz

)= − 1

t2dy

dt+

1

t

d

dz

(dy

dz

)dz

dt

= − 1

t2dy

dt+

1

t2d2y

dz2

∴ t2d2y

dt2= −dy

dt+d2y

dt2


−dydt

+d2y

dz2+ 4

dy

dz+ 2y = 0

or,d2y

dz2+ 3

dy

dz+ 2y = 0



r2 + 3r + 2 = 0

or, r2 + 2r + r + 2 =

or, (r + 2)(r + 1) = 0

or, r = −1,−2


y = c1e−2z + c2e

−z

or, y = c1e−2 ln t + c2e

− ln t

or, y =c1

t2+c2

t


t2d2y

dt2− 4t

dy

dt− 6y = 0 (3.77)



Thendz

dt=

1

t

Now,

dy

dt=dy

dz

dz

dt=

1

t

dy

dz

or, tdy

dt=dy

dz

Again,

d2y

dt2=

d

dt

(dy

dt

)=

d

dt

(1

t

dy

dz

)= − 1

t2dy

dt+

1

t

d

dt

(dy

dz

)= − 1

t2dy

dt+

1

t

d

dz

(dy

dz

)dz

dt

= − 1

t2dy

dt+

1

t2d2y

dz2

∴ t2d2y

dt2= −dy

dt+d2y

dt2

3.4. REPEATED ROOTS; REDUCTION OF ORDER 163


−dydt

+d2y

dz2− 4

dy

dz− 6y = 0

or,d2y

dz2− 5

dy

dz− 6y = 0


r2 − 5r − 6 = 0

or, r2 − 6r + r − 2 =

or, (r − 6)(r + 1) = 0

or, r = 6,−1


y = c1e6z + c2e

−z

or, y = c1e6 ln t + c2e

− ln t

or, y = c1t6 +

c2

t

3.4 Repeated Roots; Reduction of Order

3.4.1 Repeated Roots

Let us consider the second degree homogeneous differential equation of constant

coefficients

ay′′ + by′ + cy = 0 (3.78)

where a, b, c are constants. The characterstics equation of the given equation is

ar2 + brc = 0 (3.79)

Let b2 − 4ac = 0. Then the roots of (3.78) are given by

r =−b±

√b2 − 4ac

2a= − b

2a

hence r1 = r2 = − b2a . Therefore, y1 = e−

bt2a is one of the solution (3.78). To find

the general solution of (3.78), we need to find another independent solution.

Let

y = v(t)e−bt2a (3.80)


Differentiating,

y′ = v′(t)e−bt2a − b

2av(t)e−

bt2a (3.81)

y′′ = v′′(t)e−bt2a − b

2av′(t)e−

bt2a − b

2av′(t)e−

bt2a +

b2

4a2v(t)e−

bt2a

= v′′(t)e−bt2a − b

av′(t)e−

bt2a +

b2

4a2v(t)e−

bt2a (3.82)

Putting the values of y, y′ and y′′ in (3.78), we get

a

(v′′(t)e−

bt2a − b

av′(t)e−

bt2a +

b2

4a2v(t)e−

bt2a

)+ b

(v′(t)e−

bt2a − b

2av(t)e−

bt2a

)+cv(t)e−

bt2a = 0

or,

(av′′(t)− bv′(t) +

b2

4av(t) + bv′(t)− b2

2av(t) + cv(t)

)e−

bt2a = 0

or, av′′(t)− bv′(t) +b2

4av(t) + bv′(t)− b2

2av(t) + cv(t) = 0

or, av”(t)− b2

4a+ cv(t) = 0

or, av′′(t)− b2 − 4ac

2av(t) = 0

or,av”(t) = 0 as b2 − 4ac = 0

or, v′′(t) = 0

Integrating,

v′(t) = c1

or, v(t) = c1t+ c2

From (3.80),

y = (c1t+ c2)e−bt2a

= c1te− bt

2a + c2e− bt

2a

hence, y is linear combinations of two functios y1 = te−bt2a and y2 = e−

bt2a .

For Wronskian of y1 and y2

y′1 = e−bt2a − b

2ate−

bt2a , y′2 = − b

2ae−

bt2a


Now, the Wronskian of y1 and y2 is

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ te−bt2a e−

bt2a

e−bt2a − b

2a te− bt

2a − b2ae− bt

2a

∣∣∣∣∣= e−

bt2a e−

bt2a

∣∣∣∣∣ t 1

1− bt2a − b

2a

∣∣∣∣∣= e−

bta

(− bt

2a− 1 +

bt

2a

)= −e−

bta 6= 0 for all t.

Hence y1 and y2 are fundamental set of solutions of (3.78) and the general solutions

is

y = (c1t+ c2)e−bt2a

Example 101.

Find the general solution of the following equations

1. 4y′′ + 12y′ + 9y = 0

2. 25y′′ − 20y′ + 4y = 0

3. 4y′′ − 4y′ − 3y = 0

4. y′′ − 2y′ + 10y = 0


4y′′ + 12y′ + 9y = 0 (3.83)

The characterstic equation is

4r2 + 4r + 9 = 0

or, (2r)2 + 2 · 2r · 3 + 32 = 0

or, (2r + 3)2 = 0

or, 2r + 3 = 0

∴ r1 = r2 = −3

2


y = (c1 + c2t)er1t = (c1 + c2t)e

− 3t2


25y′′ − 20y′ + 4y = 0 (3.84)



25r2 − 20r + 4 = 0

or, (5r)2 − 2 · 5r · 2 + 22 = 0

or, (5r − 2)2 = 0

or, 5r − 2 = 0

∴ r1 = r2 = −2

5


y = (c1 + c2t)er1t = (c1 + c2t)e

2t5


4y′′ − 4y′ + 3y = 0 (3.85)


4r2 − 4r − 3 = 0

or, 4r2 − 6r + 2r − 3

or, 2r(2r − 3) + 1(2r − 3) = 0

or, (2r − 3)(2r + 1) = 0

∴ r1 =3

2r2 = −1

2


y = c1er1t + c2e

r2t = c1e3t2 + c1e

− t2


y′′ − 2y′ + 10y = 0 (3.86)


r2 − 2r + 10 = 0

(3.87)

or, r =−b±

√b2 − 4ac

2a

=2±√

4− 40

2= 1± 3i


y = eαt(c1 cosβt+ c2 sinβt) = et(c1 cos 3t+ c2 sin 3t)


Example 102.

Solve the following initial value problem

1. y′′ − y′ + 0.25y = 0, y(0) = 2, y′(0) = 13

2. y′′ − 6y′ + 9y = 0, y(0) = 0, y′(0) = 2

3. 9y′′ − 12y′ + 4y = 0, y(0) = 2, y′(0) = −1


y′′ − y′ + 0.25y = 0, y(0) = 2, y′(0) =1

3


r2 − r + 0.25 = 0

or, r2 − 2 · (0.5)r · 2 + (0.5)2 = 0

or, (r − 0.5)2 = 0

or, r − 0.5 = 0

∴ r1 = r2 =1

2


y = (c1 + c2t)er1t = (c1 + c2t)e

t2 (3.88)

From first initial condition

y(0) = 2 =⇒ c1 = 2

Again digfferentiating both sides with respect to t,

y′(t) = c2et2 +

1

2(c1 + c2t)e

t2

Using the second initial condition

y′(0) =1

3

=⇒ c2 +1

2c1 =

1

3

=⇒ c2 + 1 =1

3

=⇒ c2 = −2

3

Putting the value of c1 and c2 in (3.88), we get the solution of initial value problem

y =

(2− 2t

3

)et2



y′′ − 6y′ + 9y = 0, y(0) = 0, y′(0) = 2


r2 − 6r + 9 = 0

or, r2 − 2 · 3 · r + (3)2 = 0

or, (r − 3)2 = 0

or, r − 3 = 0

∴ r1 = r2 = 3


y = (c1 + c2t)er1t = (c1 + c2t)e

3t (3.89)


y(0) = 0 =⇒ c1 = 0


y′(t) = c2e3t + 3(c1 + c2t)e

3t


y′(0) = 2

=⇒ c2 + 3c1 = 2

=⇒ c2 = 2


y = 2te3t


9y′′ − 12y′ + 4y = 0, y(0) = 2, y′(0) = −1


9r2 − 12r + 4 = 0

or, (3r)2 − 2 · ·(3r) · 2 + (2)2 = 0

or, (3r − 2)2 = 0

or, 3r − 2 = 0

∴ r1 = r2 =2

3



y = (c1 + c2t)er1t = (c1 + c2t)e

2t3 (3.90)


y(0) = 2 =⇒ c1 = 2


y′(t) = c2e2t3 +

2

3(c1 + c2t)e

3t2


y′(0) = −1

=⇒ c2 +2

3c1 = −1

=⇒ c2 +4

3= −1

=⇒ c2 = −7

3


y =

(2− 7t

3

)e

2t3

Example 103.


y′′ − y′ + 0.25y = 0, y(0) = 2, y′(0) = b

Find the solution as a function of b and determine the critical value of b that separate

the grow positively from those that eventually grow negatively.


y′′ − y′ + 0.25y = 0, y(0) = 2, y′(0) = b


r2 − r + 0.25 = 0

or, r2 − 2 · (0.5)r · 2 + (0.5)2 = 0

or, (r − 0.5)2 = 0

or, r − 0.5 = 0

∴ r1 = r2 =1

2


y = (c1 + c2t)er1t = (c1 + c2t)e

t2 (3.91)



y(0) = 2 =⇒ c1 = 2


y′(t) = c2et2 +

1

2(c1 + c2t)e

t2


y′(0) = b

=⇒ c2 +1

2c1 = b

=⇒ c2 + 1 = b

=⇒ c2 = b− 1


y = (2 + (b− 1)t) et2

Since the second term dominates, the long term solution depends on the coefficient

b− 1. The critical value is b = 1.

3.4.2 Reduction of order

Let y1(t) be a solution not everywhere zero of

y′′ + p(t)y′ + q(t)y = 0 (3.92)

Then

y′′1 + p(t)y′1 + q(t)y1 = 0 (3.93)

To find the second solution, let

y(t) = v(t)y1(t) (3.94)

Then

y′ = v′(t)y1(t) + v(t)y′1(t)

y′′ = v′′(t)y1(t) + v′(t)y′1(t) + v′(t)y′1(t) + v(t)y′′1(t)

y′′ = v′′(t)y1(t) + 2v′(t)y′1(t) + v(t)y′′1(t)

Putting the values of y, y′, y′′ in (3.92),

v′′(t)y1(t) + 2v′(t)y′1(t) + v(t)y′′1(t) + p(t)(v′(t)y1(t) + v(t)y′1(t)) + q(t)v(t)y(y) = 0

or, v′′y1 + (2y1 + py1)v′ + (y′′1 + p(t)y′1 + q(t)y1) = 0

or, v′′y1 + (2y1 + py1)v′ = 0

which is a first order differential equation for the function the derivative of v i.e. in

v′ and can be solved either as first order linear equation or as a separable equation.

After finding v′, v can be obtained by integration. y is obtained from (3.94).


Example 104.

Use the method of reduction of order to find second solution of the given differ-

etial equation

1. t2y′′ + 2ty′ − 2y = 0, t > 0, y1(t) = t

2. t2y′′ + 3ty′ + y = 0, t > 0, y1(t) = t−1

3. t2y′′ − 4ty′ + 6y = 0, t > 0, y1(t) = t2


t2y′′ + 2ty′ − 2y = 0, t > 0, (3.95)

Also one of solution is

y1(t) = t (3.96)

To obtain a second solution, we consider

y(t) = v(t)y1(t) = tv(t) (3.97)

Then

y′ = tv′(t) + v(t)

y′′ = v′′(t)t+ v′(t) + v′′(t)

y′′ = v′′(t)t+ 2v′(t)


t2(tv′′ + 2v′) + 2t(tv′ + v)− 2tv = 0

or, t3v′′ + 4t2v′ = 0

or,v′′

v′= −4

t

or,1

v′dv′

dt= −4

t

or,dv′

v′==

4

tdt

Integrating, we get ∫dv′

v′=

∫4

tdt

or, ln v′ = −4 ln t+ ln c1

or, ln v′ = ln t−4 + ln c1

or, ln v′ = ln(c1t−4)

or, v′ = c1t−4

or,dv

dt= c1t

−4

or, dv = c1

∫t−4dt


Integrating, ∫dv = c1

∫t−4dt+ c2

or, v = c1t−3

(−3)+ c2

or, v = c2 −c1

3t3

Substituting the value of v(t) in (3.97), we get

y = t(c2 −

c1

3t2

)= c2t−

c1

3t2

Hence, y is linear combination of two solutions

y1 = t, y2 =1

t2


t2y′′ + 3ty′ + y = 0, t > 0, (3.98)


y1(t) =1

t(3.99)


y(t) = v(t)y1(t) =v(t)

t= t−1v(t) (3.100)

Then

y′ = t−1v′(t)− t−2v(t)

y′′ = v′′(t)t−1 − t−2v′(t)− t−2v′(t) + 2vt−3

or, y′′ = t−1v′′ − 2v′t−2 + 2vt−3


t2(t−1v′′ − 2v′t−2 + 2vt−3) + 3t(t−1v′(t)− t−2v(t)) + t−1v = 0

or, tv′′ + v′ = 0

or,v′′

v′= −1

t

or,1

v′dv′

dt= −1

t

or,dv′

v′= −1

tdt


Integrating, we get ∫dv′

v′= −

∫1

tdt

or, ln v′ = − ln t+ ln c1

or, ln v′ = ln t−1 + ln c1

or, ln v′ = ln(c1t−1)

or, v′ = c1t−1

or,dv

dt= c1t

−1

or, dv = c1

∫t−1dt

Integrating, ∫dv = c1

∫1

tdt+ ln c1

or, v = c1 ln t+ c2


y =1

t(c1 ln t+ c2) = c1

ln t

t+

1

tc2

Hence y is linear combination of two solutions

y1(t) =1

t, y2 =

ln t

t


t2y′′ − 4ty′ + 6y = 0, t > 0, (3.101)


y1(t) = t2 (3.102)


y(t) = v(t)y1(t) = t2v(t) (3.103)

Then

y′ = t2v′ + 2tv

y′′ = t2v′′ + 2tv′ + 2tv + 2v′

or, y′′ = t2v′′ + 4tv + 2v′



t2(t2v′′ + 4tv + 2v′)− 4t(t2v′ + 2tv) + 6t2v = 0

or, t2v′′ = 0

or, v′′ = 0

Integrating, we get

v′ = c1

or,dv

dt= c1

or, dv = c1dt

or, v = c1t+ c2 (3.104)


y = t2 (c1t+ c2) = c1t3t+ c2t

2


y1(t) = t3, y2 = t2

Example 105.

Given that y1(t) = t−1 is a solution of

2t2y” + 3ty′ − y = 0, t > 0

Find a fundamental solution.


t2y′′ + 3ty′ − y = 0, t > 0, (3.105)


y1(t) = t−1 (3.106)


y(t) = v(t)y1(t) = t−1v(t) (3.107)

Then

y′ = t−1v′ − t−2v

y′′ = t−1v′′ − t−2v′ − t−2v′ + 2t−3v

or, y′′ = t−1v′′ − 2t−2v′ + 2t−3v



2t2(t−1v′′ − 2t−2v′ + 2t−3v) + 3t(t−1v′ − t−2v)− t−1v = 0

or, 2tv′′ − v′ = 0

or,v′′

v′=

1

2t

Integrating, we get

ln v′ =1

2ln t+ ln c1

or, ln v′ = ln t1/2 + ln c1

or, ln v′ = ln(t1/2c1)

or, v′ = c1t1/2

or,dv

dt= c1t

1/2

or, dv = c1t1/2dt

or, v =2c1

3t3/2 + c2


y = t−1

(2c1

3t3/2 + c2

)=

2c1

3t1/2 +

c2

t


, y1(t) =1

t, y2(t) = t1/2

Differentiating,

y′1(t) = − 1

t2, y′2(t) =

1

2t−1/2

W =

∣∣∣∣∣ y1 y2

y′1 y′2

∣∣∣∣∣=

∣∣∣∣∣ 1t t1/2

− 1t2

12 t−1/2

∣∣∣∣∣=

1

2t−3/2 + t−3/2 =

3

2t3/26= 0 for t > 0

Hence

y1 =1

t, y2 = t1/2

forms a fundamental set of a solution.


3.5 Nonhomogeous Equatio; Method of Undetermined

Coefficients

Let us consider the second order nonhomogeneous equation

L[y] = y′′ + p(t)y′ + q(t)y = g(t) (3.108)

where p, q and g are functions continuous on the open interval I. Then the equation

L[y] = y′′ + p(t)y′ + q(t)y = 0 (3.109)

in which g(t) = 0 ans p and q are same as in (3.108) is called homogeneous equation

corresponding to the equation (3.108).

The following two theorems describes the structure of the general solutions of non-

homogeous equation (3.108).

Theorem 10.

If Y1 and Y2 are two solutions of the nonhomogeous equation

L[y] = y′′ + p(t)y′ + q(t)y = g(t)

Then their different Y1 − Y2 is a solution of the corresponding homogeous equation

L[y] = y′′ + p(t)y′ + q(t)y = 0

If, in addition, y1 and y2 are a fundamental set of solution of the equation

L[y] = y′′ + p(t)y′ + q(t)y = 0

then

Y1(t)− Y2(t) = c1y1 + c2y2

where c1 and c2 are certain constants.

Proof: We have, Y1 and Y2 are two solutions of the nonhomogeous equation

L[y] = y′′ + p(t)y′ + q(t)y = g(t)

so

L[Y1] = Y ′′1 + p(t)Y ′1 + q(t)Y1 = g(t) (3.110)

L[Y2] = Y ′′2 + p(t)Y ′2 + q(t)Y2 = g(t) (3.111)

Subtracting the equation (3.111) from (3.111), we get

L[Y1]− L[Y2] = Y ′′1 − Y ′′2 + p(t)(Y ′1 − Y ′2) + q(t)(Y1 − Y2) = (g(t)− g(t))

or, L[Y1]− L[Y2] = (Y1 − Y2)′′ + p(t)(Y1 − Y2)′ + q(t)(Y1 − Y2) = 0 (3.112)

3.5. NONHOMOGEOUS EQUATIO; METHODOF UNDETERMINED COEFFICIENTS177

using the linearity of the derivative,

(Y1 − Y2)′′ = Y ′′1 − Y ′′2 , (Y1 − Y2)′ = Y ′1 − Y ′2

Thus, from (3.112) shows that Y1 − Y2 is a solution of

L[y] = y′′ + p(t)y′ + q(t)y = 0

Again y1 and y2 are fundamental solutions of L[y] = 0, so all solutions of L[y] = 0

can be expressed as linear combination of y1 and y2. Hence

Y1 − Y2 = c1y1 + c2y2


Theorem 11.

The general solution of the nonhomogeous equation

L[y] = y′′ + p(t)y′ + q(t)y = g(t) (3.113)

can be written in the form

y = φ(t) = c1y1(t) + c2y2(t) + Y (t)

where y1 and y2 are a fundamental solutions of the correspondiing homogeneous

equation

L[y] = y′′ + p(t)y′ + q(t)y = 0

where c1 and c2 are arbitrary constants, and Y is some specific solution of the

nonhomogeneous equation (3.113).

Proof: Here y1, y2 are fundamental solutions of

L[y] = y′′ + p(t)y′ + q(t)y = 0

Therefore,

y′′1 + p(t)y′1 + q(t)y1 = 0 (3.114)

y′′2 + p(t)y′2 + q(t)y2 = 0 (3.115)

Multiplying (3.114) by c1 and (3.115) by c2 and adding,

c1y′′1 + c2y

′′2 + p(t)(c1y

′1 + c2y

′2) + q(t)(c1y1 + c2y2) = 0 (3.116)

Again Y (t) is a solution of

L[y] = y′′ + p(t)y′ + q(t)y = g(t)


so

Y ′′ + p(t)Y ′ + q(t)Y = g(t) (3.117)

Now, we show that

y = y = φ(t) = c1y1(t) + c2y2(t) + Y (t)

is solution of

L[y] = y′′ + p(t)y′ + q(t)y = g(t)

Now,

y′′ + p(t)y′ + q(t)y = (c1y1(t) + c2y2(t) + Y (t))′′ + p(t)(c1y1(t) + c2y2(t) + Y (t))′

+q(t)(c1y1(t) + c2y2(t) + Y (t))

= (c1y′′1(t) + c2y

′′2(t) + Y (t)′′) + p(t)(c1y

′1(t) + c2y

′2(t) + Y ′(t))

+q(t)(c1y1(t) + c2y2(t) + Y (t))

= c1y′′1 + c2y

′′2 + p(t)(c1y

′1 + c2y

′2) + q(t)(c1y1 + c2y2) +

+Y ′′ + p(t)Y ′ + q(t)Y

= 0 + Y ′′ + p(t)Y ′ + q(t)Y form (3.116).

= g(t) from (3.117) (3.118)

Thus, y = y = φ(t) = c1y1(t) + c2y2(t) + Y (t) is a solution of L[y] = g(t).

Note: The theorem states that to solve the nonhomogeous equation, we must have

1. Find the general solution c1y1+c2y2 of the corresponding homogeous equation.

This solution is called the coplemetrary solution and may be represented by

yc(t).

2. Find some singular solution Y (t) of the nonhomogeous equation. Obtain this

solution is called particular solution.

3. Add together the functions found in the two proceding steps.

3.5.1 Method of Undetermined Coefficients

We know that yc(t) is solution of homogeneous equation L[y] = 0. We will the

particular solution of

L[y] = y′′ + p(t)y′ + q(t)y = g(t)

in the following cases


When the right hand side is an exponential i.e. g(t) = eαt.

The second order differential equation becomes

L[y] = y′′ + p(t)y′ + q(t)y = eαt (3.119)

Let the specific solution of (3.119) be y = Aeat where A is a constant. Then

y′ = Aaeat, y′′ = Aa2eat

Therefore, the equation (3.119) becomes

Aa2eat + pAaeat + qeat = eat

or, (a2 + pa+ q)A = 1

When Aa2 + pa+ q 6= 0. Then

A =1

a2 + pa+ q

Thus, the required particular solution is

y =1

a2 + pa+ qeat

When a2 + pa+ q = 0 let y = Ateat be solution. Then

y′ = aAteat +Aeat, y′′ = a2Ateat +Aeat = a2Ateat + 2aAeat

Now the equation (3.119), becomes

Aa2teat + 2aAeat + p(aAteat +Aeat) + qAteat = eat

or, (a2 + pa+ q)At+ (2a+ p)Aeat = eat

or, (2a+ p)A = 1

or, A =1

2a+ p

Hence, the particular solution is

y =t

2a+ peat

Example 106.

Find a particular solution of

y′′ − 3y′ − 4y = 3e2t


y′′ − 3y′ − 4y = 3e2t (3.120)


Let the specific solution of (3.120) be Y = Ae2t, where A is a constant. Then

Y ′ = 2Ae2t, Y ′′ = 4Ae2t


Y ′′ − 3Y ′ − 4Y = 3e2t

or, 4Ae2t − 6Ae2t − 4Ae2t = 3e2t

or, − 6A = 3

or, A = −1

2

Hence the paricular solution is

Y (t) = −1

2e2t

Example 107.


y′′ − 2y′ − 3y = 3e2t


y′′ − 2y′ − 3y = 3e2t (3.121)

Let the specific solution of (3.121) be Y = Ae2t, where A is a constant. Then

Y ′ = 2Ae2t, Y ′′ = 4Ae2t


Y ′′ − 2Y ′ − 3Y = 3e2t

or, 4Ae2t − 4Ae2t − 3Ae2t = 3e2t

or, − 3A = 3

or, A = −1

Hence the paricular solution is

Y (t) = −e2t

Example 108.


y′′ − 3y′ − 4y = 2e−t

Solution: The given diffential equation

y′′ − 3y′ − 4y = 3e2t (3.122)


Note

L[a] = L[1] = (−1)2 − 3(−1)− 4 = 0

Let the specific solution of (3.122) be Y = Ate−t where A is a constant. Then

Y ′ = −Ate−t +Ae−t, Y ′′ = Ate−t −Ae−t −Ae−t = Ate−t − 2Ae−t


Y ′′ − 2Y ′ − 3Y = 3e2t

or, Ate−t − 2Ae−t − 3(−Ate−t +Ae−t)− 4Ate−t = 2e−t

or, − 5Ae−t = 2e−t

or, A = −2

5

Hence the particular solution

Y (t) = −2

5te−t

When the R .H.S. g(t) = sin at

In this case the second order non-homogeous equation becomes

L[y] = y′′ + py′ + qy = sin at (3.123)

Let

Y = A sin at+B cos at

be a particular solution of (3.123). Then

Y ′ = aA cos at− aB sin at, Y ′′ = −a2A sin at− a2B cos at

Now the equation (3.123) becomes

−a2(A sin at+B cos at) + p a(A cos at−B sin at) + q(A sin at+B cos at) = sin at

or, (−Aa2 − pBa+ qA) sin at+ (−a2B + apA+ qB) cos at = sin at

Equating the coefficients of cos at and sin at, we get

−Aa2 − pBa+ qA = 1

or,A(−a2 + q)− apB − 1 = 0 (3.124)

and apA+ (−a2 + q)B + 0 = 0 (3.125)

Solving (3.124) and (3.125) by method of cross multiplication

A

0 + (−a2 + b)=

B

−ap− 0=

1

(−a2 + q)2 + a2p2

or,A

−a2 + b=

B

−ap=

1

a4 − 2a2q + q2 + a2p2

or, A =−a2 + b

a4 − 2a2q + q2 + a2p2, B =

ap

a4 − 2a2q + q2 + a2p2


Hence the particular solution is

Y (t) =−a2 + b

a4 − 2a2q + q2 + a2p2sin at− ap

a4 − 2a2q + q2 + a2p2cos at

Similar method for g(t) = cos at.

Note :

1. If g(t) = c1 sin at+ c2 sin at, where c1 and c2 are constants, then we must form

the particular solution of the form

Y (t) = A sin at+B cos at

where A and B are undetermined coefficients to be determined by the given

equation.

2. If ai is not root of the auxiliary equation r2 + pr + q = 0 then the particular

solution as

Y (t) = A cos at+B sin at

3. If ai is a root of the auxiliary equation r2+pr+q = 0, then we let the particular

solution as

Y (t) = t(A cos at+B sin at)

Example 109.

Find the particular solution of y′′ − 3y′ − 4y = 2 sin t

Solution: Let

L[y] = y′′ − 3y′ − 4y = 2 sin t (3.126)

Let

Y (t) = A sin t+B cos t


Y ′(t) = A cos t−B sin t, Y ′′ = −A sin t−B cos t

Now the equations (3.126) becomes

−A sin t−B cos t− 3(A cos t−B sin t)− 4(A sin t+B cos t) = 2 sin t

or, (−2A+ 3B) sin t+ (−3A− 5B) cos t = 2 sin t+ 0 cos t

Equating coefficients of cos t and sin t, we get

−2A+ 3B = 1 (3.127)

−3A− 5B = 0 (3.128)



A = − 5

17, B =

3

17

Putting the values of A and B in Y (t), the particular solution is

Y (t) = − 5

17sin t+

3

17cos t

Example 110.

Find the particular solution of y′′ + 2y′ + 5y = 3 cos 2t

Solution: Let

L[y] = y′′ + 2y′ + 5y = 3 cos 2t (3.129)

Let

Y (t) = A sin 2t+B cos 2t


Y ′(t) = 2A cos 2t− 2B sin 2t, Y ′′ = −4A sin 2t− 4B cos 2t

Now the equations (3.126) becomes

−4A sin 2t− 4B cos 2t− 3(2A cos 2t− 2B sin 2t)− 4(A sin 2t+B cos 2t) = 3 cos 2t

or, (A− 4B) sin 2t+ (4A−B) cos 2t = 0 sin 2t+ 3 cos 2t


A− 4B = 0 (3.130)

4A−B = 3 (3.131)


A = −12

17, B = − 3

17

Putting the values of A and B in Y (t), the particular solution is

Y (t) = −12

17sin 2t− 3

17cos 2t

Example 111.

Find the solution of initial value problem of

y′′ + 4y = 2 sin 2t, y(0) = 2, y′(0) = −1

Solution: Let

L[y] = y′′ + 4y = 2 sin 2t, y(0) = 2, y′(0) = −1 (3.132)



r2 + 4 = 0

Solving, we get

r = ±2i

Thus, the complementry function is

yc = c1 cos 2t+ c2 sin 2t (3.133)

Since g(t) = 2 sin 2t and ia = 2i is a solution of r2 + 4 = 0 i.e. auxiliary equation,

so we let the particular solution as

Y (t) = t(A sin 2t+B cos 2t)

Then

Y ′(t) = t(2A cos 2t− 2B sin 2t) +A sin 2t+B cos 2t

Y ′′(t) = −t(4A sin 2t+ 4B cos 2t) + (2A cos 2t− 2B sin 2t) + 2A cos 2t− 2B sin 2t

= −4t(A sin 2t+B cos 2t) + 4A cos 2t− 4B sin 2t

Putting the values in the equation (3.132)

−4t(A sin 2t+B cos 2t) + 4A cos 2t− 4B sin 2t+ 4t(A sin 2t+B cos 2t) = 2 sin 2t

or, 4A cos 2t− 4B sin 2t2t = 3 sin 2t


4A = 0, −4B = 2

or, A = 0, B = −1

2

Hence the particular solutionn is

Y (t) = −1

2t cos 2t

y(t) = yc(t) + Y (t)

= c1 cos 2t+ c2 sin 2t− 1

2t cos 2t (3.134)

Also,

y′(t) = −2c1 sin 2t+ 2c2 cos 2t− 1

2cos 2t+ t sin 2t (3.135)


c1 = 2


and the initial condition y′(0) = −1 in (3.135), we get

0 + 2c2 −1

2+ 0 = −1

or, c2 = −1

4

Putting the values of c1 and c2 in (3.134), the initial value problem is

y(t) = 2 cos 2t− 1

4sin 2t− 1

2t cos 2t

When R.H.S. is a polynomial function

Let

g(t) = a0tn + a1t

n−1 + a2tn−2 + · · ·+ an

In this caes, the nonhomogeneous equation becomes

ay′′ + by′ + cy = a0tn + a1t

n−1 + a2tn−2 + · · ·+ an (3.136)

To obtained the a particular solution we assume that

Y (t) = A0tn +A1t

n−1 +A2tn−2 + · · ·+An (3.137)

Differentiating, we get

Y ′(t) = nA0tn−1 + (n− 1)A1t

n−2 + (n− 2)A2tn−3 + · · ·+An−1

Y ′′(t) = n(n− 1)A0tn−2 + (n− 1)(n− 2)A1t

n−3 + (n− 2)(n− 3)A2tn−4 + · · ·+An−2

Then the equation (3.136) becomes

an(n− 1)A0t

n−2 + (n− 1)(n− 2)A1tn−3 + (n− 2)(n− 3)A2t

n−4 + · · ·+An−2

+bnA0t

n−1 + (n− 1)A1tn−2 + (n− 2)A2t

n−3 + · · ·+An−1

+cA0t

n +A1tn−1 +A2t

n−2 + · · ·+An

= a0tn + a1t

n−1 + a2tn−2 + · · ·+ an (3.138)

Equating the coefficients of like term, we get

cA0 = a0

cA1 + nbA0 = a1

cAn + bAn−1 + 2aAn−2 = an

Now we have the following cases

Case I: If c 6= 0, then from above we get

A0 =a0

c


and the remaining equations give the values of A1, A2, A3, · · · , An successively.

Case II: If c = 0 and b = 0 then the polynomial on the left (3.138) is of degree

n− 1 and we can not satsify. In this case, we choose

Y (t) = t(A0t

2 +A1tn−1 +A2t

n−2 + · · ·+An−1t+An)

Case III If c = 0 and b = 0 then the polynomial on the left (3.138) is of degree

n− 2 and we can not satisfy. In this case we choose

Y (t) = t2(A0t

2 +A1tn−1 +A2t

n−2 + · · ·+An−1t+An)

Example 112.


y′′ + y′ − 2y = 2t, y(0) = 0, y′(0) = 1

Solution: Let

L[y] = y′′ + y′ − 2y = 2t, y(0) = 0, y′(0) = 1 (3.139)


r2 + r − 2 = 0

or, r2 + 2r − r − 2 = 0

or, r(r + 2)− 1(r + 2) = 0

or, (r − 1)(r + 2) = 0

or, r = 1, r = −2

Thus, the complementry function is

yc = c1et + c2e

−2t (3.140)

Since g(t) = 2t, so we let the particular solutions as

Y (t) = At+B

∴ Y ′ = A, Y ′′ = 0

Putting the values of Y ′′, Y ′ and Y in the given equation

0 +A− 2At−B = 2t

or, (A−B)− 2At = 2t

Equating the coefficients of like terms, we get

A−B = 0 (3.141)

2A = 2 (3.142)


Solving (3.141) and (3.142), we get

A = −1, B = −1

2

Hence particular solution is

Y (t) = −t− 1

2

Hence, the general solution is

y(t) = yc(t) + Y (t)

or, y(t) = c1et + c2e

−2t − t− 1

2(3.143)

Differentiating,

y(t) = c1et − 2c2e

−2t − 1 (3.144)


c1 + c2 =1

2or, 2c1 + 2c2 = 1 (3.145)

and using the initial condition y′(0) = 1 in (3.145), we get

c1 − 2c2 − 1 = 1

or, c1 − 2c2 = 2 (3.146)


c1 = 1, c2 = −1

2

Putting the values of c1 and c2, the solution of initial value problem is

y(t) = et − 1

2e−2t − t− 1

2

IV When the right side is of the form g(t) = eαtPn(t)

The second order nonhomogeous equation is

ay′′ + by′ + cy = eαtPn(t) (3.147)

Let Y (t) = eαtu(t) be a solution. Then

Y ′(t) = eαu′(t) + αeαtu(t) = eαt(αu(t) + u′(t))

Y ′′(t) = αeαt(αu(t) + u′(t)) + eαt(αu′(t) + u′′(t)) = eαt(u′′(t) + 2αu′(t) + α2u(t))

The equation (3.147)

aαeαt[u′′(t) + 2αu′(t) + α2u(t)

]+ beαt(αu(t) + u′(t)) + eαtαu(t) = eαtPn(t)

or, au′′(t) + (2aα+ b)u′(t) + (aα2 + bα+ c)u(t) = Pn(t)


If aα2 + bα+ c 6= 0 then, we consider

u(t) = A0tn +A1t

n−1 +A2tn−2 + · · ·+An

and hence the particular solution is

Y (t) = eαt(A0tn +A1t

n−1 +A2tn−2 + · · ·+An)

If aα2 + bα+ c = 0, 2aα+ b 6= 0 then, we consider

u(t) = t(A0tn +A1t

n−1 +A2tn−2 + · · ·+An)


Y (t) = teαt(A0tn +A1t

n−1 +A2tn−2 + · · ·+An)

If aα2 + bα+ c = 0, 2aα+ b = 0 then, we consider

u(t) = t2(A0tn +A1t

n−1 +A2tn−2 + · · ·+An)


Y (t) = t3eαt(A0tn +A1t

n−1 +A2tn−2 + · · ·+An)

V. When the right side g(t) is product of a trigonometric and exponential

function

Let g(t) = eαt cosβt. The second order nonhomogeous equation is

ay′′ + by′ + cy = eαt cosβt (3.148)

Let Y (t) = eαt(A cosβt+B sinβt) be a solution. Then

Y ′(t) = αeαt(A cosβt+B sinβt) + βeαt(−A sinβt+B cosβt)

= eαt [(αA+ βB) cosβt+ (αB − βA) sinβt]

Y ′′ = αeαt [(αA+ βB) cosβt+ (αB − βA) sinβt]

+eαt [−β(αA+ βB) sinβt+ β(αB − βA) cosβt]

= eαt[A(α2 − β2) + 2αβB cosβt+ B(α2 − β2)− 2αβA sinβt

]Putting the values of Y, Y ′, Y ” in (3.148), we get

aeαt[A(α2 − β2) + 2αβB cosβt+ B(α2 − β2)− 2αβA sinβt

]+beαt [(αA+ βB) cosβt+ (αB − βA) sinβt] + ceαt(A cosβt+B sinβt)

= eαt(cosβt+ 0 sinβt)

[(α2 − β2 + αa+ b)A+ (2αβ + aβ)B] cosβt+

[(α2 − β2 + αa+ b)B − (2αβ + aβ)A] sinβt cosβt+ 0 sinβt



(α2 − β2 + αa+ b)A+ (2αβ + aβ)B = 1

(α2 − β2 + αa+ b)B − (2αβ + aβ)A = 0

Solving the equations we get the values of A and B, and hence we get a particular

solution.

Example 113.

Find the particular solution of

y′′ − 3y′ − 4y = −8et cos 2t


y′′ − 3y′ − 4y = −8et cos 2t (3.149)

In this case, we assume that the prticular solution Y (t) as product of et and the

linear combination of sin 2t and cos 2t.

Y (t) = et(A cos 2t+B sin 2t)

Differentiating,

Y ′(t) = et(A cos 2t+B sin 2t) + et(−2A sin 2t+B cos 2t)

= et[(A+ 2B) cos 2t+ (−2A+B) sin 2t]

Y ′′(t) = et[(A+ 2B) cos 2t+ (−2A+B) sin 2t] + et[−2(A+ 2B) cos 2t+ 2(−2A+B) cos 2t]

= et[(−3A+ 4B) cos 2t+ (−4A− 3B) sin 2t]

Substituting the values of y, y′ and y′′ in the equation (3.149)

et[(−3A+ 4B) cos 2t+ (−4A− 3B) sin 2t]− 3et[(A+ 2B) cos 2t+ (−2A+B) sin 2t]

−4et(A cos 2t+B sin 2t) = −8et cos 2t

or, − (10A+ 2B) cos 2t+ (2A− 10B) sin 2t = −8 cos 2t+ 0 sin 2t

Equating the coefficient of like terms, we get

10A+ 2B = 8

2A− 10B = 0

Solving the equations we get the values of A and B,

A =10

13, B =

2

13


the particular solution as

Y (t) = et(

10

13sin 2t+

2

13sin 2t

)VI When g(t) = eαtPn cosβt or eαtPn cosβt

The second order nonhomogeous equation is

ay′′ + by′ + cy = eαtPn cosβt or ay′′ + by′ + cy = eαtPn sinβt

From Euler’s Formula

cosβt =eiβt + e−iβt

2, sinβt =

eiβt − e−iβt

2i

Now

g(t) = eαtPn(t) sinβt

= Pn(t)eαt+iβt − eαt−iβt

2i

In this case, we choose a particular solution as

Y (t) = eαt+iβt(C0tn + C1t

n−1 + Cn−1tn−2 + · · ·+ Cn)

+eαt−iβt(D0tn +D1t

n−1 +Dn−1tn−2 + · · ·+Dn)

= eαteiβt(C0tn + C1t

n−1 + Cn−1tn−2 + · · ·+ Cn)

+eαte−iβt(D0tn +D1t

n−1 +Dn−1tn−2 + · · ·+Dn)

= eαt(cosβt+ i sinβt)(C0tn + C1t

n−1 + Cn−1tn−2 + · · ·+ Cn)

+eαt(cosβt− i sinβt)(D0tn +D1t

n−1 +Dn−1tn−2 + · · ·+Dn)

= eαt(A0tn +A1t

n−1 +An−1tn−2 + · · ·+An) cosβt+

eαt(B0tn +B1t

n−1 +Bn−1tn−2 + · · ·+Bn) sinβt

Same result for g(t) = eαtPn(t) cosβt.

When g(t) = g1(t) + g2(t)

In this case the given differential equation becomes

ay′′ + by′ + cy = g1(t) + g2(t) (3.150)

Let Y1 and Y2 be two solutions of the equations

ay′′ + by′ + cy = g1(t)

ay′′ + by′ + cy = g2(t)

Then

aY ′′1 + bY ′1 + cY1 = g1(t) (3.151)

aY ′′2 + bY ′2 + cY2 = g1(t) (3.152)


Putting y = Y1 + Y2 in (3.150), we get

a(Y1 + Y2)′′ + b(Y1 + Y2)′ + c(Y1 + Y2) = g(t) + g2(t)

or,a(Y ′′1 + Y ′′2 ) + b(Y ′1 + Y ′2) + c(Y1 + Y2) = g(t) + g2(t)

or, aY ′′1 + bY ′1 + cY1 + aY ′′2 + bY ′2 + cY2 = g1(t) + g2(t)

or, g1 + g2 = g1 + g2

This shows that y = Y1 +Y2 is a solution of (3.150). Similar conclusion holds if g(t)

is the sum of any finite number of functions.

Example 114.


y′′ + 4y = t2 + 3et, y(0) = 0, y′(0) = 2

Solution: Let

y′′ + 4y = t2 + 3et, y(0) = 0, y′(0) = 2 (3.153)


r2 + 4 = 0

Solving, we get r = ±2i Thus, the complementry function is

yc = c1 cos 2t+ c2 sin 2t (3.154)

Since g(t) = t2 + 3et, so we let the particular solutions as

Y (t) = At2 +Bt+ C +Det

∴ Y ′ = 2At+B +Det, Y ′′ = 2A+Det

Putting the values of Y ′′, Y ′ and Y in the given equation

2A+Det + 4(At2 +Bt+ C +Det) = t2 + 3et

or, 4At2 + 4Bt+ 5Det + (2A+ 4C) = t2 + 3et

(3.155)


4A = 1, 4B = 0, 5D = 3, 2A+ 4C = 0

or, A =1

4, B = 0, D =

3

5, C = −1

8

Hence particular solution is

Y (t) =1

4t2 − 1

8+

3

8et



y(t) = yc(t) + Y (t)

or, y(t) = c1 cos 2t+ c2 sin 2t+1

4t2 − 1

8+

3

8et (3.156)

Differentiating,

y′(t) = −2c1 sin 2t+ 2c2 cos 2t+1

2t+

3

8et (3.157)


c1 + 0 + 0− 1

8+

3

8= 0

or, c1 = −19

40(3.158)

and using the initial condition y′(0) = 2 in (3.157), we get

0 + 2c2 + 0 +3

5= 0

or, c2 =7

10

Putting the values of c1 and c2, the solution of initial value problem is

y(t) = −19

40cos 2t+

7

10sin 2t+

1

4t2 − 1

8+

3

5et

3.6 Variation of Parameters

Let us consider a differential equation

y′′ + p(t)y′ + q(t)y = q(t) (3.159)

where p(t) and q(t) are given continuous functions. Let

yc(t) = c1y1 + c2y2 (3.160)

is general solution of the homogeneous equation.

y′′ + p(t)y′ + q(t)y = 0 (3.161)

Let us replace c1 and c2 by the functions u1(t) and u2(t), in (3.161), we get

y(t) = u1(t)y1(t) + u2(t)y2(t) (3.162)

We try to determine u1(t) and u2(t), so that the expression (3.162) is a solution of

the nonhomogeous equation (3.159). Now differentiating (3.162) with respect to t,

y′ = u′1(t)y1(t) + u1(t)y′1(t) + u′2(t)y2(t) + u2(t)y′2(t)

= u1(t)y′1(t) + u2(t)y′2(t) + u′1(t)y1(t) + u′2(t)y2(t) (3.163)

3.6. VARIATION OF PARAMETERS 193

We choose u1 and u2 such that

u′1(t)y1(t) + u′2(t)y2(t) = 0 (3.164)

Now the equation (3.163), becomes

y′ = u1(t)y′1(t) + u2(t)y′2(t) (3.165)

Further, by differentiating again, we obtain

y′′ = u′1(t)y′1(t) + u1(t)y′′1(t) + u′2(t)y′2(t) + u2(t)y′′2(t) (3.166)

Putting values of y, y′, y” in (3.159) from

u′1(t)y′1(t) + u1(t)y′′1(t) + u′2(t)y′2(t) + u2(t)y′′2(t)

+p(t)(u1(t)y′1(t) + u2(t)y′2(t))

+q(t)(u1(t)y1(t) + u2(t)y2(t)) = g(t)

or, u1(t)(y′′1(t) + p(t)y′1(t) + q(t)y1(t))

+u2(t)(y′′2(t) + p(t)y′2(t) + q(t)y2(t))

u′1(t)y′1(t) + u′1(t)y′2(t) = g(t) (3.167)

Since y1 and y2 are solutions of (3.161), so

y′′1(t) + p(t)y′1(t) + q(t)y1(t) = 0

and

y′′2(t) + p(t)y′2(t) + q(t)y2(t) = 0

Using these results in (3.167), we get

u′1(t)y′1(t) + u′2(t)y′2(t) = g(t) (3.168)

Solving the equations (3.164) and (3.168) by Cramer’s rule,

u′1 =

∣∣∣∣∣ 0 y2(t)

g(t) y′2(t)

∣∣∣∣∣∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣, u′2 =

∣∣∣∣∣ y1(t) 0

y′1(t) g(t)

∣∣∣∣∣∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣or, u′1(t) =

−y2(t)g(t)

W (y1, y2)(t), or, u′2(t) =

y1(t)g(t)

W (y1, y2)(t),

where W (y1, y2) is the Wronskian of y1 and y2 which is non-zero. Integrating,

u1(t) =

∫−y2(t)g(t)

W (y1, y2)(t)dt, u2(t) =

∫y1(t)g(t)

W (y1, y2)(t)dt

After substiting the values of u1 and u2 in

y = u1(t)y1(t) + u2(t)y2(t)

we get the required general solution of the given differential equation.


Theorem 12.

If the function p(t), q(t) and g(t) continuous on an open open interval I, and if

the functions y1 and y2 are a fundamental set of solution of the homogeous equation

y′′ + p(t)y′ + q(t)y = 0

Then a particular solution of

y′′ + p(t)y′ + q(t)y = g(t)

is

Y (t) = −y1(t)

∫ t

0

y2(s)g(t)

W (y1, y2)(s)ds+ y2(t)

∫ t

0

y1(s)g(s)

W (y1, y2)(s)ds

where t0 is any conveniently chosen point in I0. Then general solution is

y = c1y1(t) + c2y2(t) + Y (t)

Example 115.

Find the general solution of y′′−y′−2y = 2e−t by using the method of variation

of parameter.


y′′ − y′ − 2y = 2e−t (3.169)


r2 − r − 2 = 0

or, r2 − 2r + r − 2 = 0

or, r(r − 2) + 1(r − 2) = 0

or,(r + 1)(r − 2) = 0

∴ r = −1, r = 2

Thus, the complemetry solution is

yc = c1e−t + c2e

2t (3.170)

Let y1 = e−t and y2(t) = e2t. Then

y′1 = −e−t y2 = 2e2t

Also

W =

∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣=

∣∣∣∣∣ e−t e2t

−e−t 2e2t

∣∣∣∣∣ = 2et + et = 3et 6= 0


Hence y1 and y2 are fundamental solutions of the given diffential equation. Also,

g(t) = 2e−t

Nowy2(t)g(t)

W (y1, y2)=e2t 2e−t

3et=

2

3

andy1(t)g(t)

W (y1, y2)=e−t 2e−t

3et=

2

3e−3t

The particular solution is

Y (t) = −y1(t)

∫y2(t)g(t)

W (y1, y2)+ y2(t)

∫y1(t)g(t)

W (y1, y2)dt

= − e−t∫

2

3dt+ e2t

∫2

3e−3tdt

= −2

3te−t − 2

9e−t

Hence the general solution

y(t) = yc(t) + Y (t)

or, y(t) = c1e−t + c2e

2t − 2

3te−t − 2

9e−t

Example 116.

Find the general solution of y′′ + n2y = secnt by using the method of variation

of parameter.


y′′ + n2y = secnt (3.171)


r2 + n2 = 0

or, n = ±ni


yc = c1cosnt+ c2 sinnt (3.172)

Let y1 = cosnt and y2(t) = sinnt. Then

y′1 = −n sinnt y′2 = n cosnt


Also

W =

∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣=

∣∣∣∣∣ cosnt sinnt

−n sinnt n cosnt

∣∣∣∣∣ = n(sin2 nt+ cos2 nt) = n 6= 0


g(t) = secnt

Nowy2(t)g(t)

W (y1, y2)=

sinnt secnt

n=

tannt

n

andy1(t)g(t)

W (y1, y2)=

cosnt secnt

n=

1

n


Y (t) = −y1(t)

∫y2(t)g(t)

W (y1, y2)+ y2(t)

∫y1(t)g(t)

W (y1, y2)dt

= − cosnt

∫tannt

ndt+ sinnt

∫1

ndt

= −cosnt ln(secnt)

n2+t

nsinnt


y(t) = yc(t) + Y (t)

or, y(t) = c1 cosnt+ c2 sinnt− cosnt ln(secnt)

n2+t

nsinnt

Example 117.

Find the general solution of y′′ + y = tan t by using the method of variation of

parameter.


y′′ + y = tan t (3.173)


r2 + 1 = 0

or, n = ±i


yc = c1cost+ c2 sin t (3.174)


Let y1 = cos t and y2(t) = sin t. Then

y′1 = − sin t y′2 = cos t

Also

W =

∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣=


− sin t cos t

∣∣∣∣∣ = (sin2 t+ cos2 t) = 1 6= 0


g(t) = tan t

Nowy2(t)g(t)

W (y1, y2)=

sin t tan t

1= sin2 t sec t

andy1(t)g(t)

W (y1, y2)=

cos t tan t

n= sin t


Y (t) = −y1(t)

∫y2(t)g(t)

W (y1, y2)+ y2(t)

∫y1(t)g(t)

W (y1, y2)dt

= − cos t

∫sin2 t sec tdt+ sin t

∫sin tdt

= − cos t

∫(1− cos2 t) sec t− sin t cos t

= − cos t

∫sec tdt+ cos t

∫cos tdt− sin t cos t

= − cos t ln(sec t+ tan t) + sin t cos t− sin t cos t

= − cos t ln(sec t+ tan t)


y(t) = yc(t) + Y (t)

or, y(t) = c1 cos t+ c2 sin t− cos t ln(sec t+ tan t)

Example 118.

Find the general solution of y′′ + 4y′ + 4y = t−2e−2t by using the method of

variation of parameter.


y′′ + 4y′ + 4y = t−2e−2t (3.175)



r2 + 4r + 4 = 0

or, (r + 2)2 = 0

or, n = −2


yc = (c1 + c2t)e−2t (3.176)

Let y1 = e−2t and y2(t) = te−2t. Then

y′1 = −2e−2t, y′2 = e−2t − 2te−2t

Also

W =

∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣=

∣∣∣∣∣ e−2t te−2t

−2e−2t e−2t − 2te−2t

∣∣∣∣∣ = e−2t − 2te−4t + 2te−4t = e−4t 6= 0


g(t) = t−2e−2t

Nowy2(t)g(t)

W (y1, y2)=te−2tt−2e−2t

e−4t=

1

t

andy1(t)g(t)

W (y1, y2)=e−2tt−2e−2t

e−4t= t


Y (t) = −y1(t)

∫y2(t)g(t)

W (y1, y2)+ y2(t)

∫y1(t)g(t)

W (y1, y2)dt

= −e2t

∫1

tdt+ te−2t

∫t−2dt

= −e−2t ln t− te−2t 1

t= −e−2t ln t− e−2t


y(t) = yc(t) + Y (t)

or, y(t) = (c1 + c2t)e−2t − e−2t ln t− e−2t


Example 119.

Find the general solution of 4y′′+ y = 2 sec t/2 by using the method of variation

of parameter.


4y′′ + y = 2 sec t/2 (3.177)


4r2 + 1 = 0

or, n = ±1

2i


yc = c1cost/2 + c2 sin t/2 (3.178)

Let y1 = cos t/2 and y2(t) = sin t/2. Then

y′1 = −1

2sin t/2 y′2 =

1

2cos t/2

Also

W =

∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣=

∣∣∣∣∣ cos t/2 sin t/2

−12 sin t/2 1

2 cos t/2

∣∣∣∣∣ =1

2(sin2 t/2 + cos2 t/2) =

1

26= 0


g(t) = tan t

Nowy2(t)g(t)

W (y1, y2)=

cos t/2 · 2 sec t/2

1/2= 4 tan t/2

andy1(t)g(t)

W (y1, y2)=

cos t/2 · 2 sec t/2

1/2= 2


Y (t) = −y1(t)

∫y2(t)g(t)

W (y1, y2)+ y2(t)

∫y1(t)g(t)

W (y1, y2)dt

= − cos t/2

∫4 tan t/2dt+ sin t/2

∫2dt

= − cos t/2(− ln(cos t/2))

1/2+ 2t sin t/2

= 8 cos t/2 ln(cos t/2) + 2t sin t/2



y(t) = yc(t) + Y (t)

or, y(t) = c1 cos t/2 + c2 sin t/2 + 8 cos t/2 ln(cos t/2) + 2t sin t/2

Example 120.

Find the general solution of y′′ + 4y = 3 csc t by using the method of variation

of parameter.


y′′ + 4y = 3 csc t (3.179)


r2 + 4 = 0

or, n = ±2i


yc = c1cos2t+ c2 sin 2t (3.180)

Let y1 = cos 2t and y2(t) = sin 2t. Then

y′1 = −2 sin 2t y′2 = 2 cos 2t

Also

W =

∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣=

∣∣∣∣∣ cos 2t sin 2t

−2 sin 2t 2 cos 2t

∣∣∣∣∣ = 2(sin2 2t+ cos2 2t) = 2 6= 0


g(t) = tan t

Nowy2(t)g(t)

W (y1, y2)=

sin 2t · 3 csc t

2=

6 sin t cos t csc t

2= 3 cos t

andy1(t)g(t)

W (y1, y2)=

cos 2t3 csc t

2=

3

2(1− 2 sin2 t) csc t =

3

2(csc t− 2 sin t)


Y (t) = −y1(t)

∫y2(t)g(t)

W (y1, y2)+ y2(t)

∫y1(t)g(t)

W (y1, y2)dt

= − cos 2t

∫3 cos tdt+ sin 2t

∫3

2(csc t− 2 sin t)dt

= −3 cos 2t sin t+3

2sin 2t ln | csc t− cot s|+ 3 sin 2t cos t



y(t) = yc(t) + Y (t)

or, y(t) = c1cos2t+ c2 sin 2t− 3 cos 2t sin t+3

2sin 2t ln | csc t− cot s|+ 3 sin 2t cos t

Example 121.

Verify that y1 = t2, y2(t) = t−1 satisfy the corresponding homogeneous equa-

tion of the differential equation and find a particular solution of the non homogeneous

equation

t2y′′ − 2y = 3t2 − 1 t > 0


t2y′′ − 2y = 3t2 − 1 t > 0

or, y′′ − 2

t2y =

3t2 − 1

t2(3.181)

Comparing the equation (3.181), with

y′′ + p(t)y′ + q(t)y = g(t)

we get

g(t) =3t2 − 1

t2

The corresponding he homogeneous equation is

y′′ − 2

t2y = 0 (3.182)

We take, y1(t) = t2, we have

y′1 = 2t, y′′1 = 2

Now, from (3.182), we have

y′′1 −2

t2y1 = 0

or, 2− 2

t2t2 = 0

or, 0 = 0

This shows that y1 = t2 is a solution of (3.182).

Again, we take, y2(t) = t−1, we have

y′2 = −t−2, y′′2 = 2t−3


Now, from (3.182), we have

y′′1 −2

t2y1 = 0

or,2

t3− 2

t2t−1 = 0

or,2

t3− 2

t3= 0

or, 0 = 0

This shows that y2 = t−1 is a solution of (3.182).

Also,

W =

∣∣∣∣∣ y1(t) y2(t)

y′1(t) y′2(t)

∣∣∣∣∣=

∣∣∣∣∣ t2 t−1

2t −t−2

∣∣∣∣∣ = −1− 2 = −3 6= 0

Hence y1 and y2 form a fundamental set of solution of (3.181) Now,

y2(t)g(t)

W (y1, y2)=

1t

(2t2−1t2

)−3

=3t2 − 1

−3t3

y1(t)g(t)

W (y1, y2)=t2(

2t2−1t2

)−3

=3t2 − 1

−3


Y (t) = −y1(t)

∫y2(t)g(t)

W (y1, y2)+ y2(t)

∫y1(t)g(t)

W (y1, y2)dt

= −t2∫ (

3t2 − 1

−3t2

)dt+

1

t

∫ (3t2 − 1

(−3)

)dt

=t2

3

∫ (3

t− t−3

)dt− 1

3t

∫(3t2 − 1)dt

=t2

3

(3 ln t+

1

2t2

)− 1

3t(t3 − t)

= t2 ln t+1

6− 1

3t2+

1

3

= t2 ln t− 1

3t2+

1

2

3.7 Mechanical and Electrical Vibration

Liear homogeous differential equations with constant coefficients have important

application in physics and enginearing. We will discuss two important areas of

application in the field of Mechanical and Electric ocillations. Let us consider a

3.7. MECHANICAL AND ELECTRICAL VIBRATION 203

mass m hanging on the end of a spring of original length l. The mass causes an

enolgation L of spring the downward direction. In this case there are two force L

of spring in the downward direction. In this case there are two forces acting to the

spring. They are

1. The gravitational force or weight of the mass, act on downward direction and

has the magnitude mg, where g is accelaration due to gravity.

2. A force Fs due to spring, which acts upward direction. From Hook’s Law, the

upward spring force is proportional to the enolgation L, of the spring

Fs = −kL (3.183)

where the constant of proportinality k is called the spring constant, and the

minus sign is due to the fact that the spring force act in upward direction.

Since the mass is in equilibrium, the two forces balance each other, i.e.

mg − kL = 0 (3.184)

Let an external force F (t) act be applied on the mass m and u(t) be displcement

of the mass from equilibrium position at time t. Then u(t) is related to the forces

acting on the mass. From Newton’s Law of motion

mu′′(t) = f(t) (3.185)

where u′′ is acceleration of the mass and f(t) is net force acting on the mass. Here

both u(t) and f(t) are functions of t. In determing f there are four forces that must

be considered:

1. The weight w = mg of mass always act downward.


2. Spring force

Fs = −k(L+ u)

where L + u is total elongation of the spring. If L + u > 0, the spring is

extended and spring force act upward. On the other hand L + u < 0, the

spring is compressed by the length |L+ u| and the spring acts downward.

3. The damping or resistive force Fd always acts in the direction opposite to that

motion of the mass. This force may arise from several sources: resistance

from to the air or other medium in which the mass moves, internal energy

dissipiation due to the extension or compression of the spring, friction between

the mass and the guides that constrains its motion. It is assume that the

resistive force is proportional to the speed∣∣∣∣dudt∣∣∣∣ = |u′(t)|

of the mass. This is usually reffered to as viscous damping. Then Fd is directed

upward and is given by

Fd(t) = −γu′(t) (3.186)

where γ is proportionality constant, which is positive, is known as the damping

constant.

4. An applied force F (t) is directed upward or downward as F (t) is negative or

positive. The force may vary with time t, but independent of displacement u

and velocity u′ of the mass. The motion is called free if F = 0 and is forced if

F 6= 0.

Hence the net act force is

f(t) = mg + Fs(t) + Fd(t) + F (t)

From Newton second law of motion, i.e. from Eq.(3.186), we have

mu′′(t) = mg + Fs(t) + Fd(t) + F (t)

or, mu′′(t) = mg − k[L+ u(t)]− γu′(t) + F (t)

or, mu′′(t) = mg − kL− ku(t)]− γu′(t) + F (t)

or, mu′′(t) = −ku(t)]− γu′(t) + F (t) as mg − kL = 0

or, mu′′(t) + γu′(t) + ku′(t) = F (t) (3.187)

where the coefficients m, γ and k are positive constants. It is ordinary differential

equation of second order. to complete formulation of the vibration problem required

two initial conditions, the initial position u0 and the initial velocity v0 of the mass

u(0) = u0, u′(0) = u0


Example 122.

A mass weight 4lb streches a spring 2 in. Suppose that the mass is displaced an

additional 6 in. in the positive direction and the realsed. The mass is in a medium

that exerts a viscous force of 6 lb when velocity of 3 ft/s. Under the assumption,

formulate the initial value problem that govern the motion of the mass.

Solution: Let us measure the displacement u in feet. Since nothing about the

statement of external force, we assume that F (t) = 0. To determine m, note that

m =w

g=

4lb

32lb/ft2=

1

8lbs2/ft

For damping force Fd = 6lb and velocity u′(t) = 3ft/s

γ =Fdu′(t)

=6 lb

32ft2/s2= 2lbs/ft

The spring constant k is found from the statement that the mass stretches the spring

by l = 2in = 212ft = 1

6ft.

k =w

l=

4lb

1/6ft= 24lb/ft

Therefore, the equation of motion

mu′′(t) + γu′(t) + ku′(t) = F (t)

becomes

1

8u′′ + 2u′ + 24u = 0 (3.188)

For u(0) and u′(0): We have, the mass is displaced an additional 6in. in positive

direction when the mass is released. Therefore,

u(0) = 6in. =6

12ft =

1

2ft

and

u′(0) = 0ft/s

The second intial condition is implies by realesed in the statement of problem, when

we interpret to mean that the mass is set in motion with no initial velocity.

3.7.1 Spring Problem: Undamped Free Vibrations

We discuss a particular case when there is no damping and external force. Then

γ = 0, F (t) = 0


The equation of motion

mu′′(t) + γu′(t) + ku′(t) = F (t)

becomes

mu′′ + ku = 0

or, u′′ +k

mu = 0

or, u′′ + w20u = 0 (3.189)

where w20 = k

m is a positive constant. The auxiliary equation of (3.189) is

r2 + w20 = 0

or, r = ±w0i

Thus, the general solution of (3.189) is

u(t) = A cosw0t+B sinw0t (3.190)

Let

A = R cos δ and B = R sin δ

Then

R =√A2 +B2, tan δ =

B

A

Thus, the general solution (3.190) becomes

u(t) = R cos δ cosw0t+R sin δ sinw0t = R cos(w0t− δ) (3.191)

The corresponding motion of (3.189) is harmonic oscillation. Since the trigonometric

function cos(w0t− δ) is periodic and has period

T =2π

w0=

2π√k/m

= 2π

√m

k

1. The circular frquency

w0 =

√k

m

measure in radian per unit time, is called the naural frquency of the vibration.

2. A maximum value of cos(w0t− δ) is 1, so the maximum displacement of mass

from equlibrium is R, which is called the amplitude of the motion.

3. The dimensionless parameter δ is called the phase or phase angle and measure

the displacement of the wave from its normal position δ = 0.

From (3.191), we have the following obervation


1. As there is no damping, the motion has a constant amplitude R, thereby

dissipating no energy.

2. For a given mass m and the spring constant k, the frequency of the motion is

w =

√k

m

is always constant.

3. The time period T of the oscillation increase with increasing mass m and

decrease with increase of spring constant k. It means that the spring vibrates

more rapidly for small mass and stiffer string.

Example 123.

Express the following in the form of u = R cos(w0t− δ).

1. u = − cos t+√

3 sin t

3. u = 3 cos 2t+ 4 sin 2t

Solution: 1. Here u = − cos t+√

3 sin t. Let

− cos t+√

3 sin t = R(cos δ cos t+ sin δ sin t) = R cos(t− δ) (3.192)

Thus R cos δ = 1 and R sin δ =√

3. Then sqiaring and adding, we get

R2(cos2 δ + sin2 δ) = 1 + 3 = 4

∴ R = 2

Also,

tan δ =B

A=

√3

−1

∴ δ = π − π

3=

2π

3

Thus from (3.192), we get

u(t) = 2 cos

(t− 2π

3

)1. Here u = − cos t+

√3 sin t. Let

− cos t+√

3 sin t = R(cos δ cos t+ sin δ sin t) = R cos(t− δ) (3.193)

Thus R cos δ = 1 and R sin δ =√

3. Then sqiaring and adding, we get

R2(cos2 δ + sin2 δ) = 1 + 3 = 4


∴ R = 2

Also,

tan δ =B

A=

√3

−1

∴ δ = π − π

3=

2π

3


u(t) = 2 cos

(t− 2π

3

)where w0 = 1.

2. Here u = − cos t+√

3 sin t. Let

3 cos 2t+ 4 sin 2t = R(cos δ cos 2t+ sin δ sin 2t) = R cos(2t− δ) (3.194)

Thus R cos δ = 3 and R sin δ = 4. Then sqiaring and adding, we get

R2(cos2 δ + sin2 δ) = 9 + 16 = 25

∴ R = 5

Also,

tan δ =B

A=

√4

3

∴ δ = tan−1

(4

3

)Thus from (3.194), we get

u(t) = 5 cos

(2t− tan−1

(4

3

))where w0 = 2.

Example 124.

A mass weighting 3lb stretches a spring 3in. if the mass is pushed upward,

contracting the spring 1in., and then set in motion with downward velocity of 2ft/s

and if there is no damping, find the expression for the position u of the mass at any

time t. Also determine the frequency, period, amplitude and phase of the motion.

Solution: Let us measure the displacement u in feet. Since nothing about the

statement of external force, we assume that F (t) = 0. To determine m, note that

m =w

g=

3lb

32ft2/s2=

3

32lbs2/ft


and

w20 =

k

m=

12

3/32= 128/s2

and the equation of motion

u′′(t) + w20u(t) = 0

becomes

u′′ + 128u = 0 (3.195)

For u(0) and u′(0):

We have, the mass is pushed in 1in. upward direction. Therefore,

u(0) = −1in = − 1

12ft, u′(0) = 2ft/s


r2 + 128 = 0

or, r = ±8√

2i


u(t) = A cos 8√

2t+B sin 8√

2t (3.196)

Using the initial condition

u(0) = − 1

12

=⇒ A cos 0 +B sin 0 = − 1

12

or, A = − 1

12

Again, differentiating (3.196), we get

u′(t) = −8√

2A sin 8√

2t+√

2B cos 8√

2t

But

u′(0) = 2

or, − 8√

2A sin 8√

2 0 + 8√

2B cos√

2 0 = 2

or, B =1

4√

2

Putting the values of A and B in (3.196), the displacement is given by

u(t) = − 1

12cos 8

√2t+

1

4√

2sin 8√

2t (3.197)


Now the natural frequency

w0 = 8√

2

The time period

T =2π

w0=

2π

8√

2=

π

4√

2s

The amplitude

A =√A2 +B2 =

√1

144+

1

32=

√11

288

The phase angle

tan δ =B

A= −

14√

2

− 112

= − 3√2

Since A < 0 and B > 0, so phase angle δ lies on the second quadrant

δ = π − tan−1

(3√2

)Example 125.

A mass of 100 g stretches a spring 5cm. If the mass is set in motion from

equilibrium position from equlibrium position when a downward velocity 10cm/s

and there is no damping, determine the position u of the mass at any time t. When

does the mass first return to its equlibrium position?

Solution: Let us measure the displacement u in cm. Since nothing about the

statement of external force, we assume that F (t) = 0 and mass m = 100g. Since

there is nothing about damping , so γ = 0. To determine m, note that

k =w

l=

100× 980

5= 19600g/s2

and

w20 =

k

m=

19600

100= 196


u′′(t) + w20u(t) = 0

becomes

u′′ + 196u = 0 (3.198)



u(0) = 0cm, u′(0) = 10cm/s


r2 + 196 = 0

or, r = ±14i



u(t) = A cos 14t+B sin 14t (3.199)


u(0) = 0

=⇒ A cos 0 +B sin 0 = 0

or, A = 0


u′(t) = −14A sin 14t+ 14B cos 14t

But

u′(0) = 10

or, 14B = 10

or, B =5

7


u(t) =5

7sin 14t (3.200)

For the equilibrium position,

u(t) = 0

or,5

7sin 14t

or, 14t = nπ

or, t =nπ

14

For the first equilibrium n = 1,

t =π

14s

Example 126.

Suppose that a mass of weighting 100lb stretches a spring 2 in. If the mass is

displaced an additional 2 in. and set in motion with an initial velocity of 1 ft/s.

Determine the position of the mass at any time t. Also, determine the period, ampli-

tude and phase angle of the motion. Solution: Let us measure the displacement

u in feet. Since nothing about the statement of external force, we assume that

F (t) = 0. To determine m, note that

m =w

g=

10lb

32ft2/s2=

10

32lb s2/ft2


Since there is no damping, so γ = 0. The spring constant k is found from the

statement that the mass stretches the spring by l = 2in. = 212ft = 1

6ft.

k =w

l=

10lb

1/6ft= 60lb/ft

Also

w20 =

k

m=

60

10/32= 192/s2


u′′(t) + w20u(t) = 0

becomes

u′′ + 192u = 0 (3.201)



u(0) = 2in =2

12ft =

1

6ft, u′(0) = −1ft/s


r2 + 192 = 0

or, r = ±8√

3i


u(t) = A cos 8√

3t+B sin 8√

3t (3.202)


u(0) =1

6

=⇒ A cos 0 +B sin 0 =1

6

or, A =1

6


u′(t) = −8√

3A sin 8√

3t+ 8√

3B cos 8√

3t

But

u′(0) = −1

or, − 8√

3A sin 8√

3 0 + 8√

3B cos√

3 0 = −1

or, B = − 1

8√

3



u(t) =1

6cos 8

√3t− 1

8√

3sin 8√

3t (3.203)

Now the natural frequency

w0 = 8√

3

The time period

T =2π

w0=

2π

8√

3=

π

4√

3s

The amplitude

A =√A2 +B2 =

√1

144+

1

32=

√19

756

The phase angle

tan δ =B

A= −− 1

8√

216

= −√

3

4

Since A > 0 and B < 0, so phase angle δ lies on the fourth quadrant

δ = − tan−1

(√3

4

)

3.7.2 Spring Problem: Free Vibration with Damping

If we include the effect of damping γ 6= 0, with no external force (F (t) = 0, the

differential equation of governing the motion of the mass becomes

mu′′ + γu′ + ku = 0 (3.204)

The characterstics equation is

mr2 + γr + k = 0

This equation has two roots. Let r1 and r2 be roots of the equation. Then

r1, r2 = −−γ ±√γ2 − 4km

2m(3.205)

Taking poisitive sign we get one root and negative sign we get another root. Now

we have the following cases

Case I If γ2 − 4km > 0 then the solution is in the form

u(t) = er1t + c2er2t (3.206)

where r1 and r2 are real and distinct. Since γ, k,m are positive,so

γ2 − 4km < γ2

or, ± γ2 − 4km < γ

or, − γ ± γ2 − 4km < 0

=⇒ r1, r2 =−γ ± γ2 − 4km

2m< 0


This implies that the both roots are negative. Now

limt→∞

er1t = 0, limt→∞

er2t = 0 as r1 < 0, r2 < 0

=⇒ limt→∞

u(t) = limt→∞

(c1er1t + c2e

r2t) = 0

Hence, the motion is non-oscillatory and dies out increasing time.

Case II If γ2 − 4km = 0, then solution is

u(t) = (A+Bt)e−γt2m

Again, u(t) → ∞ as t → ∞. hence the motion is non-oscillatory and die out with

increasing time.

Case III If γ2 − 4km < 0, then roots of the characterstics equation are

r1 =−γ + i

√(4km− γ2)

2m, r2 =

−γ − i√

(4km− γ2)

2m

or, r1 = − γ

2m+ iµ, r2 = − γ

2m− iµ, where µ =

√(4km− γ2)

2mIn this case, the solution of the given equation is

u(t) = e−γt2m (A cosµt+B sinµt) (3.207)

Let A = R cos δ and B = R sin δ. Then the equation (3.207) becomes

u(t) = e−γt2m (cos δ cosµt+ sin δ sinµt)

= e−γt2m cos(µt− δ)

Also,

−1 ≤ cos(µt− δ) ≤ 1

=⇒ −Re−γt2m ≤ Re−

γt2m cos(µt− δ) ≤ Re−

γt2m

=⇒ −Re−γt2m ≤ u(t) ≤ Re−

γt2m

Hence u(t) lies between −Re−γt2m and Re−

γt2m and the amplitude of the wave decrease

as t increase. Also due to factor cos(µt− δ) the distance u(t) = 0 for infinitely many

value of t. Thus, the motion is damped oscilliation. Although, the motion is not

periodic, the parameter µ determine the frequency with which the mass oscilliates

back and forth; cosequently µ is called the quasi frequency. The quantity

Td =2π

µ

is called the quasi period. If w0 is initial frequency of the undamped motion then

µ

w0=

√(4km−γ2)

2m√km

=

(1− γ2

4km

)1/2

= 1− γ2

8kmapproximately

µ

w0< 1 as

γ2

8km> 0 (3.208)


for small value of γ2

4km . Thus, the quasi-frequency is slightly less than the natural

frequency w0.

The relation between Td and T is given by

TdT

=

2πµ2πw0

=w0

µ

=

(w0

µ

)−1

=

(1− γ2

4km

)−1/2

= 1 +γ2

8kmappro. (3.209)

for small value of γ2

4km . Therefore, Td > T , this shows that a small damping increase

the quasi-period.

Example 127.

The motion of a certain spring mass system is generated by the differential

equation

u′′ + 0.125u′ + u = 0

where u is measure in feet and t is in second. If u(0) = 2 and u′(0) = 0, determine

the position of the mass at any time t. Find the quasi-frequency and quasi-period,

as well as the time at which the mass pass through the equilibrium position.

Solution: We have ,

u′′ + 0.125u′ + u = 0 (3.210)

with the initial conditions

u(0) = 2, u′(0) = 0

The characterstic equation of (3.210)

r2 + 0.125r + 1 = 0

or, r2 +1

8r + 1 = 0

or, r =−1

8

2±

√18 − 4

2

or, r = − 1

16±√

225

16

or, r = − 1

16± i√

255

16


Hence the general solution is

u(t) = e−t16

(A cos

√255t

16+B sin

√255t

16

)(3.211)

Applying the initial condition

u(0) = 0 =⇒ A = 2

Again differentiating u(t) with respest to t, we get

u′(t) = − 1

16e−

t16

(A cos

√255t

16+B sin

√255t

16

)

+e−t16

(−√

255

16A sin

√255t

16+

√255

16B cos

√255t

16

)Using

u′(0) = 0 =⇒√

255

16B − 1

16A = 0

Putting A = 2, we get

B =2√255

Putting the values of A and B in (3.211), we get

u(t) = e−t16

(2 cos

√255t

16+

2√255

sin

√255t

16

)(3.212)

If we take

2 = R cos δ,2√255

= R sin δ

Then

R =

√22 +

(2√255

)2

=32√255

(3.213)

and

tan δ =

2√255

2=

1√255

∴ δ = 0.0625


u(t) = e−t16R

(cos δ cos

√255t

16+ sin δ sin

√255t

16

)

= e−t16

32√255

cos

(√255t

16− δ

)


where δ = 0.0625. Which gives the displacement of mass as function of time.

The quassi-frquency

µ =

√255

16= 0.998

The quasi-period

Td =2π

µ= 6.295s

The mass return to equilibrum, when

u(t) = 0

or, cos

(√255t

16− δ

)= 0

=⇒√

255t

16− δ =

π

2

or, t =16√255

(π2

+ δ)

= 1.637s

Example 128.

A spring is stretched 20cm by a force 6N . A mass of 2kg is hung from the spring

and is attached to a viscous damper that exerts a force of 3N when the velocity of

the mass 5m/s. If the mass is pulled down 5cm below its equilibrium position and

given an initial velocity of 10cm/s, determine its position u at any time t. Find the

quassi-frequency µ and the ratio of µ to the natural frequency of the corresponding

undamped motion.

Solution: Here the spring constant

k =FsL

=6

0.2= 30N/m

Damping constant

γ =Fdu′

=3

5Ns/m

Since there is no external force, F (t) = 0. The equation of motion is

mu′′ + γu′ + ku = 0

or, 2u′′ +3

5u′ + 30u = 0

or, 10u′′ + 3u′ + 150u = 0 (3.214)


u(0) = 5cm = 0.05m, u′(0) = 10cm/s = 0.1m/s


10r2 + 3r + 150 = 0

or, r =−3±

√9− 6000

20

or, r = − 3

20±√

5991

20i



u(t) = e−3t20

(A cos

√5991t

20+B sin

√5991t

20

)(3.215)

Applying the initial condition

u(0) = 0.05 =⇒ A = 0.05

Again differentiating u(t) with respest to t, we get

u′(t) = − 3

20e−

3t20

(A cos

√5991t

20+B sin

√5991t

20

)

+e−3t20

(−√

5991

20A sin

√5991t

20+

√5991

20B cos

√5991t

20

)

Using

u′(0) = 0.1 =⇒ −3

20A+

√5991

20B = 0.1

Putting A = 0.05, we get

B = 0.028

Putting the values of A and B in (3.211), we get

u(t) = e−3t16

(0.05 cos

√5991t

20+ 0.028 sin

√5991t

20

)(3.216)

If we take

0.05 = R cos δ, 0.028 = R sin δ

Then

R =√

(0.05)2 + (0.028)2 = 0.0571 (3.217)

and

tan δ =0.028

0.05∴ δ = 0.50709


u(t) = e−t16R

(cos δ cos

√255t

16+ sin δ sin

√255t

16

)

= 0.0571e−t16 cos

(√5991t

20− 0.50709

)


Which gives the displacement of mass as function of time.

The quassi-frquency

µ =

√5991

20= 3.87008

But the natural frequency

w0 =

√k

m=

√30

2= 3.8729

Therefore,µ

w0=

3.87008

3.8729= 0.9993

3.7.3 Electric Vibration

We shall considerflow of electric current in a simple RLC-circuit. Now we have the

following terms.

1. The curent I, measure in amperes (A), which is function of time t.

2. The resistance R, measure in ohms (Ω).

3. The capacitance C, measure in farads (F ).

4. The inductance L in henerys (H).

5. The impress voltage E in volts (V).

6. The total charge Q in coulombs (C) on the capacitor.


The resistance R, capacitance C, and inductance L, all are assumed to be positive

constants and voltage E is function of time and charge Q is function of time.

The relation between charge Q and current I is

I =dQ

dt

We have, from elementry law of electricity

1. The voltage drop across the resister is IR.

2. The voltage drop across the capacitor is QC .

3. The drop a cross the inductance is LdIdt .

From Krichhoff’s second law: In a closed circuit the impressed voltage is equal to

sum of voltage drop in the rest of the circuit. Hence

LdI

dt+RI +

1

CQ = E(t)

Substituting the values of I = dQdt , we get the second order differential equation

Ld2Q

dt2+R

dQ

dt+

1

CQ = E(t) (3.218)

for charge. We consider following initial conditions

Q(t0) = Q0, Q′(t0) = I(t0) = I0

Example 129.

A series circuit has a capacitor of 10−5F , a resistor of 3× 102Ω, and the charge

an inductor of 0.2H. The initial charge is 10−6 and there is no initial current. Find

the charge Q on the capacitor at any time t.

Solution: Here

C = 10−5F, R = 3× 102Ω, L = 0.2H

Since there is no external impressed voltage, so E(t) = 0.

Initial condition are

Q(0) = 10−6, Q′(0) = I0 = 0

Now the differential equation

Ld2Q

dt2+R

dQ

dt+

1

CQ = E(t)

becomes

0.2d2Q

dt2+ 300

dQ

dt+

1

10−5Q = 0

or,d2Q

dt2+ 1500

dQ

dt+ 5× 105Q = 0 (3.219)



r2 + 1500r + 5× 105 = 0

or, r =−1500±

√225× 104 − 20× 105

2or, r = −750± 250


Q(t) = Ae−1000t +Be−500t (3.220)

Using initial condition

Q(0) = 10−6 =⇒ A+B = 10−6 (3.221)

Differentiating, (3.220) with respect to t,

Q′(t) = −1000Ae−1000t − 500Be−500t (3.222)

Again

Q′(0) = 0 =⇒ −100A− 500B = 0

or, B = −2A (3.223)


A = −10−6, B = 2× 10−6

Putting the values of A and B in (3.220)

Q(t) = 10−6(e−1000t + e−500t

)Example 130.

A series circuit has a capacitor of 0.25 × 10−6Fand inductance 1H, the initial

charge is 10−6C and there is no initial current. Find the charge Q on the capacitor

at any time t.

Solution: Here

C = 0.25× 10−5F, R = 0Ω, L = 1H


Initial condition are

Q(0) = 10−6, Q′(0) = I0 = 0


Ld2Q

dt2+R

dQ

dt+

1

CQ = E(t)


becomes

d2Q

dt2+ 4× 106Q = (3.224)


r2 + 4× 106 = 0

or, r = ±2× 103i


Q(t) = A cos 200t+B sin 200t (3.225)

Using initial condition

Q(0) = 10−6 =⇒ A = 10−6 (3.226)

Differentiating, (3.225) with respect to t,

Q′(t) = −200A sin 2000t+ 2000B cos 2000t (3.227)

Again

Q′(0) = 0 =⇒ 2000B = 0

or, B = 0 (3.228)

Q(t) = 10−6cos2000tC

Example 131.

If a series current has a capacitor of C = 1.6 × 10−6F and an inductance of

L = 0.4H, find the resistance so that the current is critically damped.

Solution: Here

C = 1.6× 10−6F, R = 0Ω, L = 0.4H



Ld2Q

dt2+R

dQ

dt+

1

CQ = E(t)

becomes

0.4d2Q

dt2+R

dQ

dt+ 4× 106Q = (3.229)


r2 +Rr + 1.6× 10−6 = 0

(3.230)

The circuit will be critically damped if the discriminat of the equation is zero i.e

b2 − 4ac = 0

or, R2 − 4× 0.4× 106

1.6= 0

or, R = 1000

3.8. FORCE VIBRATION 223

3.8 Force Vibration

3.8.1 Forced Vibration with Damping

We have the general equation of spring -mass system subject to an extenal force

F (t) is

mu′′(t) + γu′(t) + ku(t) = F (t) (3.231)

where m, γ and k are the mass, damping coefficient and spring constant.

Let the external force be given by

F (t) = F0 coswt

where F0 and w are positive constants, representing, the amplitude and frequency,

respectively of the force. Then the equation (3.231)

mu′′(t) + γu′(t) + ku(t) = F0 coswt (3.232)

Let

uc(t) = c1u1 + c2u2

be complementary function of (3.234) and

U(t) = A coswt+B sinwt

be a particular solution of (3.234), where A and B are to be determine. Then the

general solution is

uc(t) = c1u1 + c2u2 +A coswt+B sinwt (3.233)


mr2 + γr + k = 0

This equation has two roots. Let r1 and r2 be roots of the equation. Then

r1, r2 = −−γ ±√γ2 − 4km

2m(3.234)

Taking poisitive sign we get one root and negative sign we get another root. Now

we have the following cases

Case I If γ2 − 4km > 0 then the complementary solution is in the form

uc(t) = er1t + c2er2t (3.235)

where r1 and r2 are real and distinct. Since γ, k,m are positive,so

γ2 − 4km < γ2

or, ± γ2 − 4km < γ

or, − γ ± γ2 − 4km < 0

=⇒ r1, r2 =−γ ± γ2 − 4km

2m< 0


This implies that the both roots are negative. Now

limt→∞

er1t = 0, limt→∞

er2t = 0 as r1 < 0, r2 < 0

=⇒ limt→∞

uc(t) = limt→∞

(c1er1t + c2e

r2t) = 0

Hence, the motion is non-oscillatory and dies out increasing time.

Case II If γ2 − 4km = 0, then solution is

uc(t) = (A+Bt)e−γt2m

Again, uc(t)→∞ as t→∞.

Case III If γ2 − 4km < 0, then roots of the characterstics equation are

r1 =−γ + i

√(4km− γ2)

2m, r2 =

−γ − i√

(4km− γ2)

2m

or, r1 = − γ

2m+ iµ, r2 = − γ

2m− iµ, where µ =

√(4km− γ2)

2m

In this case, the solution of the given equation is

uc(t) = e−γt2m (A cosµt+B sinµt) (3.236)

Let A = R cos δ and B = R sin δ. Then the equation (3.236) becomes

uc(t) = e−γt2m (cos δ cosµt+ sin δ sinµt)

= e−γt2m cos(µt− δ)

Also,

−1 ≤ cos(µt− δ) ≤ 1

=⇒ −Re−γt2m ≤ Re−

γt2m cos(µt− δ) ≤ Re−

γt2m

=⇒ −Re−γt2m ≤ uc(t) ≤ Re−

γt2m

Hence uc(t) lies between−Re−γt2m andRe−

γt2m and the amplitude of the wave decrease

as t increase. Thus uc(t) → 0 as t → ∞. Hence in all cases uc(t) is transient

solution. In many application it is of little importance and depending on γ, may

well undetactable after only a few seconds.

The particular solution


does not die out as t increase but persist indefinitely as the external force applied.

They represent a steady oscillation with the same frequency as th external force and

are called the steady state or forced response. Now



Differentiating,

U ′(t) = −Aw sinwt+Bw coswt

U ′′(t) = −Aw2 coswt−Bw2 sinwt

But U(t) satisfy the equation (3.232), so

−mw2(A coswt+B sinwt) + γw(−Aw sinwt+Bw coswt) + k(A coswt+B sinwt) = F0 coswt

or, (kA−mw2A+ γwB − F0) coswt+ (−γwA− w2mB + kB) sinwt = F0 coswt

Equatinf the both sides, we get

(k −mw2)A+ γwB − F0 = 0 (3.237)

and − γwA+ (k − w2m)B − 0 · F0 = 0 (3.238)

Solving the equations (3.237) and (3.238), by method of cross-multiplication

A

k −mw2=

B

γw=

F0

(k −mw2)2 + γ2w2

or,A

k −mw2=

B

γw=

F0

(k −mw2)2 + γ2w2=F0

∆2=R

∆

(3.239)

where

∆ =√

(k −mw2)2 + γ2w2, R =F0

∆

Thus

A =R

∆(k −mw2), B =

R

∆γw (3.240)

If A = R sin δ and B = R sin δ, then from (3.240), we get

sin δ =γw

∆

cos δ =k −mw2

δ= m

km − w

2

∆=m(w2

0 − w2)

∆

where

w0 =

√k

m

is natural frequency of the spring mass system without damping and forcing force

is zero. Also,

∆ =√

(k −mw2)2 + γ2m2

=

√m2

(k

m− w2

)+ γ2m2

=√m2(w2

0 − w2) + γ2w2


Substituting the values of A and B in U(t)


= R coswt cos δ +R sinwt sin δ

= R cos(wt− δ)

Now we investigate how the amplitude R of the steady oscillation depends on the

frequency w of the external force. We have

R =F0

∆(3.241)

Now,

∆ =√m2(w2

0 − w2)2 + γ2w2

= mw20

√(1− w2

w20

)2

+γ2w2

m2w20w

20

= mk

m

√(1− w2

w20

)2

+γ2w2

m2 kmw

20

where w20 =

k

m

= k

√(1− w2

w20

)2

+γ2

km

w2

w20

(3.242)

= k

√(1− w2

w20

)2

+ Γw2

w20

where Γ =γ2

km

From (3.241)

R =F0

k

√(1− w2

w20

)2+ Γw2

w20

or,RF0k

=1√(

1− w2

w20

)2+ Γw2

w20

(3.243)

which gives the ratio of the amplitude R of forced response to F0k , the static dis-

placement of the spring produced by a force F0.

From (3.243), for low frequency excitation (w → 0), we have

RF0k

→ 1

=⇒ R→ F0

k

and for high frequency excitation w →∞,

RF0k

→ 0


=⇒ R→ 0

For maximum value of R: We have

R =F0

∆

=F0√

m2(w20 − w2)2 + γ2w2

R will be maximum, when

g(w) = m2(w20 − w2)2 + γ2w2

is minimum. For minimum value of g(w), differentiating g(w) with respect to w,

g′(w) = −4m2w(w20 − w2) + 2γ2w

g′′(w) = (8m2w2 + 2γ2) > 0 (3.244)

For minimum value of g(w):

g′(w) = 0

or, − 4m2w(w20 − w2) + 2γ2w = 0

or, w2 = w20 −

γ2

2m2

At this value of w, R has maximum value, let this value be denoted by wmax. Then

w2max = w2

0 −γ2

2m2

= w20

(1− γ2

2m2w20

)= w2

0

(1− γ2

2mk

)w2max

w20

= 1− γ2

2mk(3.245)


Since γ2

2mk > 0, so wmax < w0. For small value of γ, wmax ' w0

The maximum value of R is

Rmax =F0

∆

=F0

k

√(1− w2

max

w20

)2+ γ2

kmw2max

w20

from (3.241) (3.246)

=F0

k

√(1−

(1− γ2

2mk

))2+ γ2

km

(1− γ2

2mk

) from (3.245)

=F0

k√

γ4

4m2k2+ γ2

km

(1− γ2

2mk

) from (3.245)

=F0

k√

γ2

km −γ2

4k2m2

=F0

k γ√km

√1− γ

4km

=F0

γ√

km

√1− γ2

4km

=F0

γw0

(1− γ2

4km

)−1/2

≡ F0

γw0

(1 +

γ2

8km

)for small value of γ.

Example 132.

A mass of 6kg stretches a spring 12cm. The mass 5kg is acted on by an external

force of 10 sin t/2N and moves in a medium that impart a viscous force 2N when

the speed of the mass is 4cm/s. If the mass is set in motion from its equilibrium

position with an initial velocity of 3cm/s.

1. Formulate the initial value problem describing the motion of the mass.

2. Find the solution of the initial value problem.

3. Identify the transient and steady state solution.


k =FsL

=6.8

0.12= 490N/m


Damping constant

γ =Fdu′

=2

0.0= 50s/M

Extenal force F (t) = 10 sin t/2N and mass m = 5kg. The equation of motion is

mu′′ + γu′ + ku = F (t)

or, 5u′′ + 50u′ + 490u = 10 sin t/2

or, u′′ + 10u′ + 98u = 2 sin t/2 (3.247)

Since the motion is set from equilibrium from its equilibrium position with an initial

velocity 3cm/s, so

u(0) = 2, u′(0) = 3cm/s = 0.03m/s


r2 + 10r + 98 = 0

or, r =−10

2±√

102 − 4 · 1 · 98

2

or, r = −5±√

73i

Hence the complementrary solution is

u(t) = e−5t(c1 cos

√73t+ c2 sin

√73t)

(3.248)

Let U(t) = A cos t/2 +B sin t/2 be a particular solution of (3.247). Then differenti-

ating,

U ′(t) = −A2

cos t/2 +B

2sin t/2

U ′′(t) = −A4

cos t/2− B

4sin t/2

Putting the values of U(t), U ′(t) and U ′′(t) in (3.247)

A

4cos t/2− B

4sin t/2 + 10

(−A

2cos t/2 +

B

2sin t/2

)+

98 (A cos t/2 +B sin t/2) = 0 · cos t/2 + 2 sin t/2

or,

(−A

4+ 5B + 98A

)cos t/2 +

(−B

2− 5A+ 98B

)sin t/2 = 0 · cos t/2 + 2 sin t/2

Equating the coefficients of like term

−A2

+ 5B + 98A = 0

or, B = −391

20A (3.249)

and − B

4− 5A+ 98B = 2

or, − 20A+ 391B = 8 (3.250)


Solving Eq. (3.249) and Eq. (3.250), we get

A = − 160

153281, B =

3138

153281

Substituting the values of A and B in particular solution, we get

U(t) = − 160

153281cos t/2 +

3138

153281sin t/2


u(t) = C.F.+ P.I.

= e−5t(c1 cos

√73t+ c2 sin

√73t)− 160

153281cos t/2 +

3138

153281sin t/2(3.251)


u(0) = 0 =⇒ c1 −160

153281= 0

or, c1 =160

153281

Differentiating (3.251),

u′(t) = −5e−5t(c1 cos

√73t+ c2 sin

√73t)

+ e−5t(−√

73c1 sin√

73t+√

73c2 cos√

73t)

+160

2× 153281sin t/2 +

3138

2× 153281cos t/2

Using the initial conditions

u′(0) =2

100

or, − 5c1 +√

73c2 +3138

2× 153281=

3

100

or, c2 =383443

153281× 100√

73(3.252)

Putting the values of c1 and c2 in (3.251),

u(t) = e−5t

(160

153281cos√

73t+383443

153281× 100√

73sin√

73t

)− 160

153281cos t/2 +

3138

153281sin t/2 (3.253)

Hence the transient part is

e−5t

(160

153281cos√

73t+383443

153281× 100√

73sin√

73t

)and steady part is

− 160

153281cos t/2 +

3138

153281sin t/2


Example 133.

A spring is stretched 6 in. by a mass that weight 8lb. The mass is attached to a

dashpot mechanism that has a damping of 0.25lb.s/ft and acted on by an external

force of 4 cos 2tlb. Determine the steady state response of the system.


k =FsL

=8

1/2= 16lb/ft

Damping constant

γ = 0.25 =1

4lbs2/ft

Extenal force F (t) = 2 sin 2tN and mass m = Wg = 8

32 = 14 lbs

2/ft. The equation of

motion is

mu′′ + γu′ + ku = F (t)

or,1

4u′′ +

1

4u′ + 16u = 4 cos 2t

or, u′′ + u′ + 64u = 16 cos 2t (3.254)

Let U(t) = A cos 2t+B sin 2t be a particular solution of (3.254). Then differentiating,

U ′(t) = −2A sin 2t+ 2B cos 2t

U ′′(t) = −4A cos 2t− 4B sin 2t


(−4A cos 2t− 4B sin 2t)− 2A sin 2t+ 2B cos 2t+ 16(A cos 2t+B sin 2t) = 16 sin 2t

or, (2B + 60A) cos 2t+ (−2A+ 60B) sin 2t = 16 sin 2t+ 0 sin 2t


2B + 60A = 8

or, B + 30A = 4 (3.255)

and − 2A+ 60B = 0

or, A = 30B (3.256)


A =240

901, B =

8

901


U(t) =240

901cos 2t+

8

901sin 2t


Example 134.

A spring mass system has spring constants 3N/s. A mass of 2kg attached to

the string and the motion takes place in a viscous fluid that offers a resistance

numerically equal to the magnitude of the instaneous velocity. If the system is driven

by an external force of 3 cos 3t − 2 sin 3t N. Determine the steady state response of

this system. Express your answer in the form of R cos(wt− δ).Solution: Here the spring constant

k = 3N/s

Damping constant

γ = 1Ns/m

Extenal force F (t) = 3 cos 3t−2 sin 3tN and mass m = 2kg. The equation of motion

is

mu′′ + γu′ + ku = F (t)

or, 2u′′ + u′ + 3u = 3 cos 3t− 2 sin 3t (3.257)

Let U(t) = A cos 3t+B sin 3t be a particular solution of (3.257). Then differentiating,


U ′′(t) = −9A cos 3t− 9B sin 3t


−2(9A cos 3t+ 9B sin 3t)− 3A sin 3t+ 3B cos 3t+ 3(A cos 3t+B sin 3t) = 3 cos 3t− 2 sin 3t

or, (−15A+ 3B) cos 3t+ (−15B − 3A) sin 3t = 3 cos 3t− 2 sin 3t


−15A+ 3B = 3

or, − 5A+B = 1 (3.258)

and 15B + 3A = 2 (3.259)


A = −13

78= −1

6, B =

1

6


U(t) = −1

6cos 3t+

1

6sin 3t

Let

R cos δ, R sin δ =1

6


Then

R =

√1

36+

1

36=

√2

6

and

tan δ = −1 =⇒ δ =3π

4

Thus, U(t) becomes

U(t) =

√2

6

(cos 3t sin

3π

4+ cos 3t sin

3π

4

)=

√2

6cos

(3t− 3π

4

)

3.8.2 Forced Vibration with out Damping

Let us assume that the mass m be attached to a spring with spring constant k,

without damping γ = 0 and a forcing function be F0 coswt. Then the equation of

motion becomes

mu′′ + ku = F0 coswt

or, u′′ +k

mu =

F0

mcoswt

or, u′′ + w20u =

F0

mcoswt

where w20 = k

m is natural frequency. The solution of the equation on the forcing

frequency w is different from the w0 or is equal to the natural frequency w0 of

unforced frequency .

Case I Let w 6= w0. The auxiliary equation is

r2 + w20 = 0

or, r = ±i

The complementry function is

C.F. = c1 cosw0t+ c2 sinw0t

Let U(t) = A coswt+B sinwt be a paricular solution. Then

U ′(t) = −wA sinwt+ w0B coswt

U ′′(t) = −w2A coswt− w2B sinwt

Putting the values of ′(t), U ′(t) and U ′′(t) in (3.260)

−w2(A coswt+B sinwt) + w20(A coswt+B sinwt) =

F0

mcoswt

or, (w20 − w2)(A coswt+B sinwt) =

F0

mcoswt



A =F0

m(w20 − w2)

, B = 0

Putting the values of A and B in U(t)

U(t) =F0

m(w20 − w2)

coswt


u(t) = C.F.+ P.I.

or, u(t) = c1 cosw0t+ c2 sinw0t+F0

m(w20 − w2)

coswt (3.260)

The constants c1 and c2 are determined by the initial conditions. The resulting

motion is, in general, the sum of two periodic motions of different frequencies (w0

and w) and different amplitudes as well.

In particular case, let us assume that the mass initially at rest, so that the initial

conditions

u(0) = 0, u′(0) = 0

Now

u(0) = 0 =⇒ c1 +F0

m(w20 − w2)

= 0

c1 = − F0

m(w20 − w2)

Again differentiating both sides with respect to t,

u′(t) = −w0c1 sinw0t+ w0c2 cosw0t−wF0

m(w20 − w2)

sinwt

Using

u′(0) = 0 =⇒ c2 = 0


u(t) = − F0

m(w20 − w2)

cosw0t+ +F0

m(w20 − w2)

coswt

=F0

m(w20 − w2)

(coswt− cosw0t)

=

[2F0

m(w20 − w2)

sin(w0 − w)t

2

]sin

(w0 + w)t

2

(3.261)

If |w − w0| is small, then w0 + w is much greater than |w0 − w|. Consequeently,

sin (w+w0)t2 is rapidly oscilating function compare to sin (w−w0)t

2 . The motion is


a rapid oscillation with the frequency w+w02 but with a slowly varying sinusoidal

amplitude.

R =2F0

m[w20 − w2]

∣∣∣∣sin (w + w0)t

2

∣∣∣∣Case II Let w = w0. As previous case the

C.F. = c1 cosw0t+ c2 sinw0t

Since the frequency of forcing function is same as the natural frequency of the system.

Let a particular solution is of the form

U(t) = t(A cosw0t+B sinw0t)

Then

U ′(t) = A cosw0t+B sinw0t+ w0t(−A sinw0t+B cosw0t)

U ′′(t) = −Aw0 sinw0t+Bw0 cosw0t+ w0(−A cosw0t+B sinw0t)

−w20t(A cosw0t+B sinw0t)

Putting the value of U(t), U ′(t) and U ′′(t) in (3.260)

−Aw0 sinw0t+Bw0 cosw0t+ w0(−A cosw0t+B sinw0t)

−w20t(A cosw0t+B sinw0t) + w2

0t(A cosw0t+B sinw0t)

=F0

mcosw0t

or, − 2w0A sinw0t+ 2w0B sinw0t =F0

mcosw0t

Therefore,

A = 0, B =F0

2mw0


U(t) =F0

2mw0sinw0t


u(t) = C.F.+ P.I.

or, u(t) = c1 cosw0t+ c2 sinw0t+F0

2mw0t sinw0t

Because of te present of the term t sinw0t, the solution u(t) will be unbounded as

t → ∞. In reality, the unbounded oscilliation do not occur. As soon as u becomes

large, the mathematical model is not valid.

Example 135.


Express the following expression as a product of two trigonometric functions of

the different frequencies (1). cos 9t− cos 7t (2). cosπt+ cos 2πt.

Solution (1).

cos 9t− cos 7t = −2 sin9t+ 7t

2sin

9t− 7t

2= −2 sin 8t sin t

Hence it is product of two sine functions with frequencies 8 and 1.

(2).

cosπt+ cos 2πt = 2 cosπt+ 2πt

2cos

π + 2πt

2= 2 cos

3πt

2cos

πt

2

Hence it is product of two cosine functions with frequencies 3π2 and π

2 .

Example 136.

A mass weight 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in

positive direction from its equlibrium position and released with no initial velocity.

Assume that there is no damping and is acted on by an external force of 2 cos 3t lb,

formulate the initial value problem describing the motion of the mass and solve it.

Solution: Let us measure the displacement u in feet. To determine m, note that

m =w

g=

4lb

32ft2/s2=

1

8lb s2/ft2

Since there is no damping, so γ = 0 and external fforce F (t) = 2 cos 2tlb The

spring constant k is found from the statement that mass stretches the spring by

l = 1.5in. = 212ft = 1

6ft by weight 4lb.

k =w

l=

4lb

1/8ft= 32lb/ft

The equation of motion

mu′′(t) + γu′ + ku = F (t)

becomes

1

8u′′ + 32u = 4 cos 3t

or, u′′ + 256u = 16 cos 3t (3.262)



u(0) = 2in =2

12ft =

1

6ft, u′(0) = 0


r2 + 256 = 0

or, r = ±16i



u(t) = c1 cos 16t+ c2 sin 16t (3.263)

Let U(t) = A cos 3t+B sin 3t be a particular solution of (3.262) Then


U ′′(t) = −9A cos 3t− 9B sin 3t

Putting the values of U(t), U ′(t) and U ′′(t) in (3.262), we get

−9A cos 3t− 9B sin 3t+ 256(A cos 3t+B sin 3t) = 16 cos 3t

or, 247A cos 3t+ 247B sin 3t = 16 cos 3t

Equating the coefficients

A =16

247, B = 0


U(t) =16

247cos 3t

hence the general solution is

u(t) = C.F.+ P.I.

= c1 cos 16t+ c2 sin 16t+16

247cos 3t (3.264)


u(0) =1

6

=⇒ c1 +16

247=

1

6

or, c1 =151

1482


u′(t) = −16c1 sin 16t+ 16c2 cos 16t− 48

247sin 3t

u′(0) = 0 =⇒ c2 = 0

Putting the values of c1 and c2 in (3.264), the displacement is given by

u(t) =151

1482cos 16t+

16

247cos 3t


Chapter 4

Higher Order Linear Equations

4.1 Higher Order Linear Equation

The nth order ordinary linear differential equation has form

P0(t)dny

dtn+ P1(t)

dn−1y

dtn−1+ · · ·+ Pn−1(t)

dy

dt+ Pn(t)y = G(t) (4.1)

where P0(t), P1(t), P2(t), · · · , Pn(t) and G(t) are continuous real-valued functions on

an open interval I = (α, β) and P0(t) 6= 0 for all t ∈ I.

Dividing by P0(t)

L[y] =dny

dtn+ p1(t)

dn−1y

dtn−1+ · · ·+ pn−1(t)

dy

dt+ pn(t)y = g(t) (4.2)

The eqution (4.2), is consider with the following initial values

y(t0) = y0, y′(t0) = y′0, · · · , y(n−1)(t0) = y(n−1)0 (4.3)

where t0 ∈ I = (α, β) and y0, y′0, · · · , y

(n−1)0 are real numbers, is higher order IVPs.

Theorem 13. Existence and Uniqueness


dny

dtn+ p1(t)

dn−1y

dtn−1+ · · ·+ pn−1(t)

dy

dt+ pn(t)y = g(t)

with initial values: y(t0) = y0, y′(t0) = y′0, · · · , y(n−1)(t0) = y(n−1)0

where p1(t), p2(t), p3(t) · · · pn(t) and g(t) are continous on an open interval I that

contains the point t0. Then there is exactly one solution y = φ(t) of this problem,

and the solution exists throughout the interval I.

Example 137.

239

240 CHAPTER 4. HIGHER ORDER LINEAR EQUATIONS

Find the interval in which the solution of the following equation is certain to

exists

1. ty′′′ + (sin t)y′ + 3y = cos t

2. y′′′ + ty′′ + t2y′ + t3y = ln t

3. t(t− 1)y(4) + ety′′ + 4t2y = 0


ty′′′ + (sin t)y′ + 3y = cos t

or,y′′′ +sin t

ty′ +

3

ty =

cos t

t

Since the functions sin tt ,

33 ,

cos tt all are continuous for all real t excect t = 0, so

the solution of the given differential equation lies on the interval −∞ < t < 0 and

0 < t <∞.


y′′′ + ty′′ + t2y′ + t3y = ln t

Since the functions t, t2 all are continuous for all real t but ln t continuous all real

except x = 0, so the solution of the given differential equation lies on the interval

−∞ < t < 0 and 0 < t <∞. Solution: 3. The given differential equation

t(t− 1)y(4) + ety′′ + 4t2y = 0

or, y(4) +et

t(t− 1)y′′ +

4t2

t(t− 1)y =

cos t

t

Since the functions et

t(t−1) ,4t2

t(t−1) all are continuous for all real t except t = 0, t = 1,

so the solution of the given differential equation lies on the interval −∞ < t < 0, 0 <

t < 1 and 0 < t <∞.

4.2 Solution of Linear Homogeneous equations; the Wron-

skian

Let us consider the homogeneous linear nth order differential equation of the form

L[y] =dny

dtn+ p1(t)

dn−1y

dtn−1+ · · ·+ pn−1(t)

dy

dt+ pn(t)y = 0 (4.4)

where p1, p2, · · · , pn are continuous real valued functions on an open interval I =

(α, β) where α and β are finite or α = ∞ or β = ∞ or both are infinite. Let


y1, y2, · · · , yn be solutions of (4.4). Then

dny1

dtn+ p1(t)

dn−1y1

dtn−1+ · · ·+ pn−1(t)

dy1

dt+ pn(t)y1 = 0 (4.5)

dny2

dtn+ p1(t)

dn−1y2

dtn−1+ · · ·+ pn−1(t)

dy2

dt+ pn(t)y2 = 0 (4.6)

...dnyndtn

+ p1(t)dn−1yndtn−1

+ · · ·+ pn−1(t)dyndt

+ pn(t)yn = 0 (4.7)

Let us consider the linear combination

y = c1y1 + c2y2 + · · ·+ cnyn (4.8)

Putiing the value of y in L. H. S. of (4.4), we get

dn

dtn(c1y1 + c2y2 + · · ·+ cnyn) + p1(t)

dn−1

dtn−1(c1y1 + c2y2 + · · ·+ cnyn)

+ · · ·+ pn−1(t)d

dt(c1y1 + c2y2 + · · ·+ cnyn) + pn(t)(c1y1 + c2y2 + · · ·+ cnyn)

= c1

(dny1

dtn+ p1(t)

dn−1y1

dtn−1+ · · ·+ pn−1(t)

dy1

dt+ pn(t)y1

)+c2

(dny1

dtn+ p1(t)

dn−1y1

dtn−1+ · · ·+ pn−1(t)

dy1

dt+ pn(t)y1

)+ · · ·

+cn

(dnyndtn

+ p1(t)dn−1yndtn−1

+ · · ·+ pn−1(t)dyndt

+ pn(t)yn

)= c1 · 0 + c2 · 0 + · · ·+ cn · cn = 0 using (4.5), (4.6), · · · , (4.7)

Thus y = c1y1 + c2y2 + · · · + cnyn is solution of the Eq. (4.4). The values of

c1, c2, · · · , cn are determined by the initial conditions

y(t0) = y0, y′(t0) = y′0, · · · , y(n−1)(t0) = y(n−1)0 (4.9)

Using, the initial conditions (4.9) to (4.8), we get the following system of linear

equations

c1y1(t0) + c2y2(t0) + · · ·+ cnyn(t0) = y0

c1y′1(t0) + c2y

′2(t0) + · · ·+ cny

′n(t0) = y′0

... (4.10)

c1y(n−1)1 (t0) + c2y

(n−1)2 (t0) + · · ·+ cny

(n−1)n (t0) = y

(n−1)0

Equations (4.10) can be solved uniquely for the constants c1, c2 · · · , cn provided that

the determinant of the coefficients is not zero. If the determinant of the coefficients is

zero, then it is always possible to chhose values of y0, y′0, · · · , y

(n−1)0 so that equations

(4.10) do not have a solution. Hence a neccessary and sufficient condition for the


existance of a solution of the equations (4.10) for arbitrary values of y0, y′0, · · · , y

(n−1)0

is that the Wronskian

W (y1, y2, · · · , yn)(t0) =

∣∣∣∣∣∣∣∣∣∣y1(t0) y2(t0) · · · yn(t0)

y′1(t0) y′2(t0) · · · y′n(t0)...

......

...

y(n−1)1 (t0) y

(n−1)2 (t0) · · · y

(n−1)n (t0)

∣∣∣∣∣∣∣∣∣∣6= 0 (4.11)

If W (y1, y2, · · · , yn)(t0) 6= 0, then the solution of the equations (4.10) givenby

c1 =1

W

∣∣∣∣∣∣∣∣∣∣y0(t0) y2(t0) · · · yn(t0)

y′0(t0) y′2(t0) · · · y′n(t0)...

......

...

y(n−1)0 (t0) y

(n−1)2 (t0) · · · y

(n−1)n (t0)

∣∣∣∣∣∣∣∣∣∣

c2 =1

W

∣∣∣∣∣∣∣∣∣∣y1(t0) y0(t0) · · · yn(t0)

y′1(t0) y′0(t0) · · · y′n(t0)...

......

...

y(n−1)1 (t0) y

(n−1)0 (t0) · · · y

(n−1)n (t0)

∣∣∣∣∣∣∣∣∣∣...

cn =1

W

∣∣∣∣∣∣∣∣∣∣y1(t0) y2(t0) · · · y0(t0)

y′1(t0) y′2(t0) · · · y′0(t0)...

......

...

y(n−1)1 (t0) y

(n−1)2 (t0) · · · y

(n−1)0 (t0)

∣∣∣∣∣∣∣∣∣∣Example 138.

Verify that the functions 1, cos(2t), sin(2t) are solutions of the differential equa-

tion y′′′ + 4y′ = 0 and determine their Wronskian.


y′′′ + 4y′ = 0 (4.12)

Let y1 = 1, y2 = cos 2t, y3 = sin 2t. Then substituting, y1 = 1 in (4.12)

0 + 4 · 0 = 0

or, 0 = 0

Hence y1 = 1 is a solution of (4.12).

Again, y2 = cos 2t, y′2 = −2 sin 2t, y′′2 = −4 cos 2t, y′′′2 = 8 sin 2t. Then substituting,

y1 = 1 in (4.12)

8 sin 2t− 8 sin 2t = 0

or, 0 = 0


Hence y2 = cos 2t is a solution of (4.12).

Again, y3 = sin 2t, y′3 = 2 cos 2t, y′′3 = −4 sin 2t, y′′′3 = −8 cos 2t. Then substitut-

ing in (4.12)

−8 cos 2t+ 8 cos 2t = 0

or, 0 = 0

Hence y1 = sin 2t is a solution of (4.12). Now the Wronskian of y1, y2, and y3, is

W (y1, y2, y3)(t) =

∣∣∣∣∣∣∣y1(t) y2(t) y3(t)

y′1(t) y′2(t) y′3(t)

y′′1(t) y′′2(t) y′′3(t)

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1 cos 2t sin 2t

0 −2 sin 2t 2 cos 2t

0 −4 cos 2t −4 sin 2t

∣∣∣∣∣∣∣= 1

∣∣∣∣∣ −2 sin 2t 2 cos 2t

−4 cos 2t −4 sin 2t

∣∣∣∣∣= 8(sin2 2t+ cos2 2t) = 8 6= 0

Example 139.

Verify that the functions et, e−t, e−2t are solutions of the differential equation

y′′′ + 2y′′ − y′ − 2y = 0 and determine their Wronskian.


y′′′ + 2y′′ + y′ − 2y = 0 (4.13)

Let y1 = et, y2 = e−t, y3 = e−2t. Then substituting, y1 = et, y′1 = et, y′′1 = et in

(4.13)

et + 2et − et − 2et = 0

or, 0 = 0

Hence y1 = et is a solution of (4.13).

Again, y2 = e−t, y′2 = −e−t, y′′2 = e−t, y′′′2 = −e−t. Then substituting, in (4.13)

−e−t + 2e−t + e−t − 2e−t = 0

or, 0 = 0

Hence y2 = e−t is a solution of (4.13).

Again, y3 = e−2t, y′2 = −2e−2t, y′′2 = 4e−2t, y′′′2 = −8e−2t. Then substituting, in

(4.13)

−8e−2t + 8e−2t + 2e−2t − 2e−2t = 0

or, 0 = 0


Hence y1 = e−2t is a solution of (4.13).

W (y1, y2, y3)(t) =

∣∣∣∣∣∣∣y1(t) y2(t) y3(t)

y′1(t) y′2(t) y′3(t)

y′′1(t) y′′2(t) y′′3(t)

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣et e−t e−2t

et −e−t −2e−2t

et e−t −4e−2t

∣∣∣∣∣∣∣= et e−t e−2t

∣∣∣∣∣∣∣1 1 1

1 −1 −2

1 1 −4

∣∣∣∣∣∣∣= e−2t

∣∣∣∣∣∣∣1 0 0

1 −2 −3

1 0 3

∣∣∣∣∣∣∣ = e−2t(−6) = −6e−2t

Example 140.

Verify that the functions x, x2, x−1 are solutions of the differential equation

x3y′′′ + x2y′′ − 2xy′ + 2y = 0 and determine their Wronskian.


x3y′′′ + x2y′′ − 2xy′ + 2y = 0 (4.14)

Let y1 = x, y2 = x2, y3 = x−1. Then substituting, y1 = x, y′1 = 1, y′′1 = 0, y′′′1 = 0

in (4.14)

x3 · 0 + x2 · 0− 2x+ 2x = 0

or, 0 = 0

Hence y1 = x is a solution of (4.14).

Again, y2 = x2, y′2 = 2x, y′′2 = 2, y′′′2 = 0. Then substituting, in (4.14)

x3 · 0 + 2x2 − 4x2 + 2x2 = 0 = 0

or, 0 = 0

Hence y2 = x2 is a solution of (4.14).

Again, y3 = x−1, y′3 = −x−2, y′′2 = 2x−3, y′′′2 = −6x−4. Then substituting, in

(4.14)

−6x3x−4 + 2x2x−3 + 2x x−2 + 2x−1 = 0

or, − 6

x+

6

x= 0

or, 0 = 0


Hence y1 = x−1 is a solution of (4.14).

W (y1, y2, · · · , yn)(x) =

∣∣∣∣∣∣∣y1(x) y2(x) y3(x)

y′1(x) y′2(x) y′3(x)

y′′1(x) y′′2(x) y′′3(x)

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣x x2 x−1

1 2x −x−2

0 2 2x−3

∣∣∣∣∣∣∣= x

∣∣∣∣∣ 2x −x−2

2 2x−3

∣∣∣∣∣− 1

∣∣∣∣∣ x2 −x−1

2 2x−3

∣∣∣∣∣= x

(4

x2+

2

x2

)− 2

x+

2

x

=6

x

4.2.1 Linear Dependence and Independence

The set of functions f1, f2, · · · , fn are said to be linearly dependent on the interval

I if there exists a set of constants k1, k2, · · · kn not all zero such that

k1f1 + k2f2 + · · ·+ knfn = 0 (4.15)

for all t ∈ I.

The set of functions f1, f2, · · · , fn are said to be linearly independent on the

interval I if

k1f1 + k2f2 + · · ·+ knfn = 0 =⇒ k1 = k2 = k3 = · · · kn = 0

for all t ∈ I.

Example 141.

Determine wheather the functions f1(t) = 1, f2(t) = t and f3(t) = t2 are linearly

independent or dependent on the interval I :∞ < t <∞.

Solution: From the linear combination

k1f(t) + f2(t) + k3f(t) = k1 + k2t+ k3t2

Let set this linear combination with zero

k1 + k2t+ k3t2 = 0 (4.16)


The equation (4.16) is hold for all t. Thus we take t = 0, t = 1, and t = −1 in (4.16)

and obtained the following equations

k1 = 0

k1 + k2 + k3 = 0 (4.17)

k1 − k2 + k3 = 0

Solving the equations in (4.17), we get k1 = 0, k2 = 0, k3 = 0. Thus the given

functions are linearly independent on I.

Example 142.

Determine wheather the functions f1(t) = 1, f2(t) = 2 + t and f3(t) = 3− t2 are

linearly independent or dependent on the interval I : −∞ < t <∞.


k1f(t) + f2(t) + k3f(t) = k1 + k2(2 + t) + k3(3− t2)


k1 + k2(2 + t) + k3(3− t2) = 0 (4.18)

The equation (4.18) is hold for all t. Thus we take t = 0, t = −2, and t = −1 in

(4.16) and obtained the following equations

k1 + 2k2 + 3k3 = 0

k1 − k3 = 0 (4.19)

k1 + 3k2 + 2k3 = 0

Solving the equations in (4.17), we get k1 = 1, k2 = −4, k3 = 5. Thus the given

functions are linearly dependent on every I.

Example 143.

Determine wheather the functions f1(t) = 2t−3, f2(t) = t2+1 and f3(t) = 2t2−3t

are linearly independent or dependent on the interval I : −∞ < t <∞.


k1f(t) + f2(t) + k3f(t) = k1(2t− 3) + k2(t2 + 1) + k3(2t2 − 3t)

= (k2 − 3k1) + (2k1 − 3k3)t+ (k2 + 2k3)t2


(k2 − 3k1) + (2k1 − 3k3)t+ (k2 + 2k3)t2 = 0

or, (k2 − 3k1) + (2k1 − 3k3)t+ (k2 + 2k3)t2 = 0 + 0 · t+ 0 · t2

4.3. HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS 247

The equation (4.18) is hold for all t. Equating the coefficients of the like terms, we

get the following equations

k2 − 3k1 = 0

2k1 − 3k3 = 0 (4.20)

k2 + 2k3 = 0

Solving the equations in (4.17), we get k1 = 0, k2 = 0, k3 = 0. Thus the given

functions are linearly independent on every I.α

4.3 Homogeneous Equations with Constant Coefficients

Let us consider nth order linear homogeneous differential equation of constant coef-

ficients

L[y] = a0y(n) + a1y

(n−1) + · · ·+ an−1y′ + any = 0 (4.21)

where a0, a1, a2, · · · an are real numbers. Let y = ert a solution of (4.21). Then

y′ = rert, y′′ = r2ert, y′′ = r3ert, yn = rnert

From the equation (4.21), we get

ert(a0rn + a1r

n−1 + · · ·+ an−1r + an) = 0

Since ert 6= 0, so

Z(r) = a0rn + a1r

n−1 + · · ·+ an−1r + an = 0 (4.22)

The polynomial Z(r) is called characterstics polynomial and Z(r) = 0 is called

the characterstics equation of the differential equation. The equation (4.22) is a

polynomial equation of the degree n, has n roots r1, r2, · · · , rn, some of them are

equal. We can write the polynomial equation as

Z(r) = a0(r − r1)(r − r2)(r − r3) · · · (r − rn) (4.23)

We have the following cases

Case I : Real and unequal If the roots of the characterstics equation are real and

no two are equal, then we have n distinct solutions er1t, er2t, er3t, · · · , ernt of (4.21).

If these function are linearly independent i.e. Wronskian W 6= 0, then the general

solution of (4.21) is

y = c1er1t + c2e

r2t + · · ·+ cnernt (4.24)

Case II Complex roots If the characterstics equation has complex roots, they

must occur in conjugate pairs λ ± iµ since the coefficients a0, a1, · · · , an are real


number. Provided that none of the roots is repeated, the general solution (4.24).

However as the second order, we can replace the complex-valued solutions e(λ+iµ)t

and e(λ−iµ)t by real-valued solutions

eλt cosµt and eλt sinµt (4.25)

obtained as the real and imaginary part of e(λ+iµ)t.

Case III: Repeated roots If the roots of the characterstics equation are not

distinct, then the solution of (4.21) is not in the form (4.24). Suppose the real root

m, be repeated in s times, where s ≤ n. Then

emt, temt, t2emt, · · · , ts−1emt

If a complex root λ + iµ is repeated s times, then the comple conjugate λ − iµ is

also repeated s times. Thus, 2s complex valued solutions can be replaced by real

solutions

eλt cosµt, teλt cosµt, t2eλt cosµt, · · · , ts−1eλt cosµt

eλt sinµt, teλt sinµt, t2eλt sinµt, · · · , ts−1eλt sinµt

Example 144.

Find the general solution of 2y′′′ − 4y′′ − 2y′ + 4y = 0.

Solution: The fiven differential equation

2y′′′ − 4y′′ − 2y′ + 4y = 0

Its characterstic equation is

2r3 − 4r2 − 2r + 4 = 0

or, 2r2(r − 2)− 2(r − 2) = 0

or, (r2 − 1)(r − 2) = 0

or, (r − 1)(r + 1)(r − 2) = 0

or, r = −1, 1, 2

Hence, required general solution is

y = c1e−t + c2e

t + c3e2t

Example 145.

Find the general solution of y(4) + y′′′− 7y′′− y′+ 6y = 0. Also find the solution

that satisfies the initial conditions

y(0) = 1, y′(0) = 0, y′′(0) = −2, y′′′(0) = −1



y(4) + y′′′ − 7y′′ − y′ + 6y = 0

with initial conditions

y(0) = 1, y′(0) = 0, y′′(0) = −2, y′′′(0) = −1


r4 + r3 − 7r2 − r + 6 = 0

or, r4 + r3 − 7r2 − 7r + 6r + 6 = 0

or, r3(r + 1)− 7r(r + 1) + 6(r + 1) = 0

or, (r + 1)(r3 − 7r + 6) = 0

or, (r + 1)(r3 − r2 + r2 − r − 6r + 6) = 0

or, (r + 1)[r2(r − 1) + r(r − 1)− 6(r − 1)] = 0

or, (r + 1)(r − 1)(r2 + r − 6) = 0

or, (r + 1)(r − 1)(r2 + 3r − 2r − 6) = 0

or, (r + 1)(r − 1) [r(r + 3)− 2(r + 3)] = 0

or, (r + 1)(r − 1)(r − 2)(r + 3) = 0

or, r = −1, 1, 2, −3

Therefore the general solution is

y = c1e−t + c2e

t + c3e2t + c4e

−3t

Differentiating both sides with respect t

y′ = −c1e−t + c2e

t + 2c3e2t − 3c4e

−3t

y′′ = c1e−t + c2e

t + 4c3e2t + 9c4e

−3t

y′′′ = −c1e−t + c2e

t + 8c3e2t − 27c4e

−3t


y(0) = 1, y′(0) = 0, y′′(0) = −2, y′′′(0) = −1

we get

c1 + c2 + c3 + c4 = 1

c1 − c2 + 2c3 − 3c4 = 0 (4.26)

c1 + c2 + 4c3 + 9c4 = −2

c1 − c2 + 8c2 − 27c4 = −1


By solving this system of algebraic equations, we find

c1 =11

8, c2 =

5

12, c3 = −2

3, c4 = −1

8

Therefore the solution of the initial value problem is

y =11

8et +

5

12e−t − 2

3e2t − 1

8e−3t

Example 146.

Find the general solution of 6y′′′ + 5y′′ + y′ = 0. Also find the solution that

satisfies the initial conditions

y(0) = −2, y′(0) = 2, y′′(0) = −1


6y′′′ + 5y′′ + y′ = 0


y(0) = 1, y′(0) = 0, y′′(0) = −2, y′′′(0) = −1


6r3 + 5r2 + r = 0

or, r(6r2 + 5r + 1) = 0

or, r(6r2 + 3r + 2r + 1) = 0

or, r(3r(2r+1) + 1(2r + 1)) = 0

or, r(2r + 1)(3r + 1) = 0

or, r = 0, −1

2,−1

3


y = c1e0 t + c2e

− 12t + c3e

− 13t = c1 + c2e

− 12t + c3e

− 13t


y′ = −1

2c2e− 1

2t − 1

3c3e− 1

3t

y′′ =1

4c2e− 1

2t +

1

9c3e− 1

3t

Using the initial condition, we get

y(0) = 2 =⇒ c1 + c2 + c3 = −2

y′(0) = 2 =⇒ −1

2c2 −

1

3c3 = 2 =⇒ 3c2 + 2c2 = −12

y′′(0) = 0 =⇒ 1

4c2 +

1

9c3 = 0 =⇒ 9c2 + 4c3 = 0



c1 = 8, c2 = 8, c3 = −18


y = 8 + 8e−12t − 18e−

13t

Example 147.

Find the general solution of 2y(4)−y′′′−9y′′+4y′+4y = 0. Also find the solution

that satisfies the initial conditions

y(0) = −2, y′(0) = 0, y′′(0) = −2, y′′′(0) = 0


2y(4) − y′′′ − 9y′′ + 4y′ + 4y = 0


y(0) = −2, y′(0) = 0, y′′(0) = −2, y′′′(0) = 0


2r4 − r3 − 9r2 + 4r + 4 = 0

or, 2r4 − 2r3 + r3 − r2 − 8r2 + 8r − 4r + 4 = 0

or, 2r3(r − 1) + r2(r − 1)− 8r(r − 1)− 4(r − 1) = 0

or, (r − 1)(2r3 + r2 − 8r − 4) = 0

or, (r − 1)(r2(2r + 1)− 4(2r + 1)) = 0

or, (r − 1)(2r + 1)(r2 − 4) = 0

or, r = 1, −1

2, 2, −2


y = c1et + c2e

2t + c3e−2t + c4e

− 12t


y′ = c1et + 2c2e

2t − 2c3e−2t − 1

2c4e− 1

2t

y′′ = c1et + 4c2e

2t + 4c3e−2t +

1

4c4e− 1

2t

y′′′ = c1et + 8c2e

2t − 8c3e−2t − 1

8c4e− 1

2t


Using the initial condition, we get

y(0) = −2 =⇒ c1 + c2 + c3 + c4 = −2

y′(0) = 0 =⇒ c1 + 2c2 − 2c3 −1

2c4 = 0 (4.27)

y′′(0) = −2 =⇒ c1 + 4c2 + 4c3 +1

4c4 = −2

y′′′(0) = 0 =⇒ c1 + 8c2 − 8c3 −1

4c4 = 0

(4.28)


c1 = −2

3, c2 = − 1

10, c3 = −1

6, c4 =

16

15


y = −2

3e−t +

1

10e2t − 1

6e−2t +

16

18e−1/2t

Example 148.

Find the general solution of y(6) + y = 0.


y(6) + y = 0

It’s characterstic equation is

r6 + 1 = 0

or, r6 = −1

or, r6 = cosπ + i sinπ Polar form

or, r6 = cos(π + 2kπ) + i sin(π + 2kπ) General Polar form

or, r = [cos(π + 2kπ) + i sin(π + 2kπ)]1/6 De-Moivre’s Theorem

or, r = cos

(π + 2kπ

6

)+ i sin

(π + 2kπ

6

)k = 0, 1, 2, 3, 4, 5

Thus, the solution are

r =

√3

2− i

2,

√3

2+i

2, i, −i, −

√3

2+i

2, −

√3

2− i

2


y = e√32t

(c1 cos

t

2+ c2 sin

t

2

)+ c3 cos t+ c4 sin t+ e

−√3

2t

(c1 cos

t

2+ c2 sin

t

2

)Example 149.


Find the general solution of y(4) + y = 0.


y(6) + y = 0


r4 + 1 = 0

or, r4 = −1

or, r4 = cosπ + i sinπ Polar form

or, r4 = cos(π + 2kπ) + i sin(π + 2kπ) General Polar form

or, r = [cos(π + 2kπ) + i sin(π + 2kπ)]1/4De-Moivre’s Theorem

or, r = cos

(π + 2kπ

4

)+ i sin

(π + 2k

4

)k = 0, 1, 2, 3

Thus, the solution are

r =1√2− i√

2, − 1√

2+

i√2, − 1√

2− i√

2, − 1√

2+

i√2


y = et√2

(c1 cos

t√2

+ c2 sint√2

)+ e− t√

2

(c1 cos

t√2

+ c2 sint√2

)Example 150.

Find the general solution of y(4) + 2y′′ + y = 0.


y(6) + 2y′′ + y = 0


r4 + 2r2 + 1 = 0

or, (r2 + 1)(r2 + 1) = 0

or, r = i, i,−i,−i


y = c1 cos t+ +c2 sin t+ c3t cos t+ c4t sin t

Example 151.

Find the general solution of y′′′ + 3y′′ − 4y = 0.


y′′′ + 3y′′ − 4y = 0



r3 + 3r2 − 4 = 0

or, r3 − r2 + 4r2 − 4 = 0

or, r2(r − 1) + 4(r2 − 1) = 0

or, r2(r − 1) + 4(r − 1)(r + 1) = 0

or, (r − 1)(r2 + 4r + 4) = 0

or, (r − 1)(r + 2)2 = 0

∴ r = 1, −2, −2


y = c1et + c2e

−2t + c2t e−2t

Example 152.

Find the general solution of y(6) − y′′ = 0.


y(6) − y′′ = 0


r6 − r2 = 0

or, r2(r4 − 1) = 0

or, r2(r2 − 1)(r2 + 1) = 0

or, r = 0, 0, 1, −1, i, −i


y = (c1 + tc2)e0t + c3et + c4e

−t + c5 cos t+ c6 sin t

or, y = c1 + c2t+ c3et + c4e


Example 153.

Find the general solution of y(4) − 4y′′′ + 4y′′ = 0.


y(4) − 4y′′′ + 4y′′ = 0


r4 − 4r3 + 4r2 = 0

or, r2(r2 − 4r + 4) = 0

or, r2(r − 2)2 = 0

or, r = 0, 0, −2, −2



y = (c1 + c2t)e0t + (c3 + tc4)e−2t

or, y = c1 + c2t+ (c3 + tc4)e−2t

Example 154.

Find the general solution of y(4)− y = 0. Also find the solution that satisfies the

initial conditions

y(0) =7

2, y′(0) = −4, y′′(0) =

5

2, y′′′(0) = −2


y(4) − y = 0


y(0) =7

2, y′(0) = −4, y′′(0) =

5

2, y′′′(0) = −2


r4 − 1 = 0

or, (r2 − 1)(r2 + 1) = 0

or, r = 1, −1, i, −i


y = c1et + c2e

−t + c3 cos t+ c4 sin t (4.29)

Differentiating

y = c1et + c2e


or, y′ = c1et − c2e

−t − c3 sin t+ c4 cos t

or, y′′ = c1et + c2e

−t − c3 cos t− c4 sin t

or, y′′′ = c1et − c2e

−t + c3 sin t− c4 cos t

Using the initial conditions

y(0) =7

2=⇒ c1 + c2 + c3 =

7

2y′(0) = −4 =⇒ c1 − c2 − c4 = −4 (4.30)

y′′(0) =5

2=⇒ c1 + c2 − c3 =

5

2y′′′(0) = −2 =⇒ c1 − c2 − c4 = −2

Solving the equations (4.30) we get

c1 = 0, c2 = 3, c3 =1

2, c4 = −1

Thus, the solution of the given initial value problem from (4.29), we get

y = 3e−t +1

2cos t− sin t


4.4 Nonhomogeneous Equations; Method of Undeter-

mined Coefficients

We consider the second order nonhomogeneous equation

L[y] = a0yn + a1y

n−1 + · · ·+ any = g(t) (4.31)

We know that the Yc(t) is solution of homogeneous equation L[y] = 0. Let Z(r) =

a0rn + a1r

n−1 + · · · + an = 0 be auxiliary equation of (4.31). We will find the

particular solution of

L[y] = g(t)

in the follwing cases

Case I: When the right hand side is an exponential i.e. g(t) = eat where α is a

constant, the particular solution must be of the form

(a) Y (t) = Aeat if α is not a root of auxiliary equation Z(r) = 0.

(b) Y (t) = Atseαt if α is a root of multiplicity s the auxiliary equation Z(r) = 0.

where A is undetermied coefficient.

Case II: When the right hand side is an exponential i.e. g(t) = b0tn+b1t

n−1+· · ·+bnwhere b0, b1, · · · , bn are constants, the particular solution must be of the form

(a) Y (t) = A0tn+A1t

n−1 + · · ·+An if 0 is not a root of auxiliary equation Z(r) = 0.

(b) Y (t) = ts(A0tn + A1t

n−1 + · · · + An) is a root of multiplicity sthe auxiliary

equation Z(r) = 0. where A0, A1, · · · , An are undetermied coefficients.

Case III: When the right hand side is an exponential i.e. g(t) = A1 cosβt+A2 sinβt

where A1, A2 are constants, the particular solution must be of the form

(a) Y (t) = B1 cosβt+B2 sinβt if iβ is not a root of auxiliary equation Z(r) = 0.

(b) Y (t) = ts(B1 cosβt + B2 sinβt) if iβ is a root of multiplicity s the auxiliary

equation Z(r) = 0. where B1, B1 are undetermied coefficients.

Case IV: When the right hand side is an exponential i.e. g(t) = eαt(A1 cosβt +

A2 sinβt) where A1, A2 are constants, the particular solution must be of the form

(a) Y (t) = eαt(B1 cosβt + B2 sinβt) if α ± iβ is not a root of auxiliary equation

Z(r) = 0.

(b) Y (t) = tseαt(B1 cosβt+B2 sinβt) if α±iβ is a root of multiplicity sthe auxiliary

equation Z(r) = 0. where B1, B1 are undetermied coefficients.

Case V: When the right hand side is an exponential i.e. g(t) = eαt(b0tn + b1t

n−1 +

· · · + bn) where b0, b1, · · · , bn are constants, the particular solution must be of the

form

(a) Y (t) = eαt(A0tn + A1t

n−1 + · · · + An) if α is not a root of auxiliary equation

Z(r) = 0.

(b) Y (t) = tseαt(A0tn+A1t

n−1+· · ·+An) if α is a root of multiplicity s the auxiliary

equation Z(r) = 0. where A0, A1, · · · , An are undetermied coefficients.

Case VI: When the right hand side is an exponential i.e. g(t) = (b0tn + b1t

n−1 +

· · ·+ bn)(d1 cosβ+d2 sinβt) where b0, b1, · · · , bn, d1, d2 are constants, the particular

4.4. NONHOMOGENEOUS EQUATIONS; METHODOF UNDETERMINED COEFFICIENTS257

solution must be of the form

(a) Y (t) = (A0tn + A1t

n−1 + · · ·+ An) cosβt+ (A0tn + A1t

n−1 + · · ·+ An) sinβt if

±iβ are not a root of auxiliary equation Z(r) = 0.

(b) Y (t) = ts

(A0tn +A1t

n−1 + · · ·+An) cosβt+ (A0tn +A1t

n−1 + · · ·+An) sinβt

if±iβ is a root of multiplicity sthe auxiliary equation Z(r) = 0. whereA0, A1, · · · , Anare undetermied coefficients.

Case VII: When the right hand side is an exponential i.e. g(t) = eαt(b0tn+b1t

n−1 +

· · ·+ bn)(d1 cosβ+d2 sinβt) where b0, b1, · · · , bn, d1, d2 are constants, the particular

solution must be of the form

(a) Y (t) = eαt(A0tn +A1t

n−1 + · · ·+An) cosβt+ (A0tn +A1t

n−1 + · · ·+An) sinβt

if α± iβ are not a root of auxiliary equation Z(r) = 0.

(b) Y (t) = ts

(A0tn +A1t

n−1 + · · ·+An) cosβt+ (A0tn +A1t

n−1 + · · ·+An) sinβt

if α ± iβ is a root of multiplicity s the auxiliary equation Z(r) = 0. where

A0, A1, · · · , An are undetermied coefficients.

Case VIII If g(t) = g1(t) + g1(t) + g1(t) + · · ·+ gn(t) then find the particular Yi(t)

solution for

L[y] = gi(t) i = 1, 2, · · · , n

Then the sum Y (t) = Y1 + Y2 + · · · + Yn is the particular solution of the given

equation.

Example 155.

Find a general solution of

y′′′ − 3y′′ + 3y′ − y = 4et


y′′′ − 3y′′ + 3y′ − y = 4et (4.32)

The characterstics solution is

r3 − 3r2 + 3r − 1 = 0

or, (r − 1)3 = 0

or, r = 1, 1 , 1

Thus the roots is repeated three times, hence the complemetry function is

yc(t) = c1et + c2te

t + c3t2et (4.33)

Since α = 1 is repeated root of the characterstic equation three times, so let the

specific solution (4.32) be Y = At3e2t where A is a constant. Then

Y ′ = A(t3et + 3t2et) = Aet(t3 + 3t2)

Y ′′ = A(t3et + 3t2et + 3t2et + 6tet) = Aet(t3 + 6t2 + 6t)

Y ′′′ = Aet(t3 + 6t2 + 6t) +Aet(3t2 + 12t+ 6) = Aet(t3 + 9t2 + 18t+ 6)



Aet(t3 + 9t2 + 18t+ 6)− 3Aet(t3 + 6t2 + 6t) + 3Aet(t3 + 3t2)−At3e2t = 4e2t

or, 6Aet = 4et

or, A =2

3


Y (t) =2

3t3et

Required general solution is

y = yc(t) + Y (t)

or, y = c1et + c2te

t + c3t2et +

2

3t3et

Example 156.

Find the general solution of y(4) + 2y′′ + y = 3 sin t− 5 cos t.


y(4) + 2y′′ + y = 3 sin t− 5 cos t (4.34)


r4 + 2r2 + 1 = 0

or, (r2 + 1)(r2 + 1) = 0

or, r = i, i,−i,−i


y = c1 cos t+ +c2 sin t+ c3t cos t+ c4t sin t

For particular solution, here g(t) = 3 sin t− 5 cos t, and β = 1. Also, ±iβ = ±i, are

solutions of the characterstic equation repeated two times, the particular solution

Y (t) = t2A cos t+Bt2 sin t. Differentiating both sides

Y ′ = 2At sin t+At2 cos t+ 2Bt cos t−Bt2 sin t

Y ′′ = 2A sin t+ 2At cos t+ 2At cos t−At2 sin t+ 2B cos t− 2B sin t− 2Bt sin t−Bt2 cos t

= 2A sin t+ 4At cos t−At2 sin t+ 2B cos t− 3Bt sin t−Bt2 cos t

Y ′′′ = 2A cos t+ 4A cos t− 4At sin t− 2At sin t−At2 cos t− 2B sin t− 3B sin t

−3Bt cos t− 2Bt cos t+Bt2 cos t

= 6A cos t− 6At sin t−At2 cos t− 5B sin t− 5Bt sin t+Bt2 sin t

y(4) = −6A sin t− 6A sin t+ 6At cos t− 2At cos t+At2 sin t− 5B cos t

−5B cos t+ 5Bt sin t+ 2Bt sin t+Bt2 cos t

= −12A sin t+ 4At cos t+At2 sin t− 10B cos t+ 5B sin t+ 7Bt sin t+ bt2 cos t


Putting the values in (4.34)

−12A sin t+ 4At cos t+At2 sin t− 10B cos t+ 5B sin t+ 7Bt sin t+ bt2 cos t+

2(2A sin t+ 4At cos t−At2 sin t+ 2B cos t− 3Bt sin t−Bt2 cos t) +

t2A cos t+Bt2 sin t = 3 sin t− 5 cos t

or,− 8A sin t− 8B cos t = 3 sin t− 5 sin t

Equating coefficients of sin t and cos t on both sides, we get

−8A = 3, −8B = −5

A = −3

8, B =

5

8

Thus the specific solution is

Y (t) = −3

8t2 sin t+

5

8t2 cos t

Hence, required general solution of given differential equation (4.34)

y = yc(t) + Y (t)

= c1 cos t+ +c2 sin t+ c3t cos t+ c4t sin t− 3

8t2 sin t+

5

8t2 cos t

Example 157.

Find the general solution of y(4) + y′′ = sin 2t.


y(4) + y′′′ = sin 2t (4.35)


r4 + r3 = 0

or, r3(r + 1) = 0

or, r = 0, 0, 0,−1


y = c1 + c2t+ c3t2 + c4e

−t

For particular solution, here g(t) = sin 2t, and β = 2. Also, ±iβ = ±2i, are

not solutions of the characterstic equation ,so let the particular solution Y (t) =

A cos 2t+B sin 2t. Differentiating both sides

Y ′ = −2A sin 2t+ 2B cos 2t

Y ′′ = −4A cos 2t− 4B sin 2t

Y ′′′ = 8A sin 2t− 8B cos 2t

Y (4) = 16A cos 2t+ 16B sin 2t


Putting the values in (4.35)

16A cos 2t+ 16B sin 2t+ 8A sin 2t− 8B cos 2t = sin 2t

or,(16A− 8B) cos 2t+ (16B + 8A) sin 2t = 0 cos 2t+ sin 2t

Equating coefficients of sin t and cos t on both sides, we get

16A− 8B = 0, 16B + 8A = 1

A =1

40, B =

1

20

Thus the specific solution is

Y (t) =1

40cos 2t+

1

20sin 2t

Hence, required general solution of given differential equation (4.34) is

y = yc(t) + Y (t)

= c1 + c2t+ c3t2 + c4e

−t +1

40cos 2t+

1

20sin 2t

Example 158.


y′′′ + y′′ + y′ + y = 4t+ et


y′′′ + y′′ + y′ + y = 4t+ et (4.36)


r3 + r2 + r + 1 = 0

or, r2(r + 1) + 1(r + 1) = 0

or,(r + 1)(r2 + 1) = 0

or, r = −1, i ,−i

Thus the complemetry function is

yc(t) = c1et + c2 cos t+ c3 sin t (4.37)

Since g(t) = e−t + 4t sum of the two functions, so the particular solution is sum of

particular solutions of following equations

y′′′ + y′′ + y′ + y = e−t (4.38)

y′′′ + y′′ + y′ + y = 4t (4.39)


For the particular solution of (4.38): Here α = −1 is solution of the characterstics

equation, so Y1(t) = A te−t particular solution but not Ae−t.

Y ′ = A(e−t − te−t) = A(1− t)e−t

Y ′′ = −Ae−t −A(1− t)e−t = A(t− 2)e−t

Y ′′′ = Ae−t −A(t− 2)e−t = A(3− t)e−t


A(3− t)e−t +A(t− 2)e−t +A(1− t)e−t +A te−t = e−t

or, A(3− t+ t− 2 + 1− t+ t) = 1

or, 2A = 1

or, A =1

2


Y1(t) =1

2t e−t

Again, for the particular solution of (4.39), let Y2 = Bt+ C

Y ′ = B, , Y ′′ = 0, Y ′′′ = 0

(4.40)


0 + 0 +B +Bt+ C = 4t

or, Bt+ (B + C) = 4t+ 0

∴ B = 4, B + C = 0

or, B = 4, C = −4


Y2(t) = 4t− 4

Hence the required particular solution is

Y (t) = Y1(t) + Y2(t) =1

2t e−t + 4t− 4


y = yc(t) + Y (t)

or, y = c1et + c2 cos t+ c3 sin t+

1

2t e−t + 4t− 4

Example 159.


Find a general solution of

y(4) − y = 3t+ sin t


y(4) − y = 3t+ sin t (4.41)


r4 − 1 = 0

or, (r2 − 1)(r2 + 1) = 0

or, r = −1,−1, i ,−i


yc(t) = c1e−t + c2e

t + c2 cos t+ c3 sin t (4.42)

Since g(t) = 3t+ sin t sum of the two functions, so the particular solution is sum of

particular solutions of following differential equations

y(4) − y = 3t (4.43)

y(4) − y = sin t (4.44)

Again, for the particular solution of (4.39), let Y2 = Bt+ C

Y ′ = B, , Y ′′ = 0, Y ′′′ = 0, y(4) = 0

(4.45)


−(Bt+ C) = 3t

∴ B = −3, C = 0


Y2(t) = −3t

For the particular solution of (4.44), here ±iβ = ±i is solution of the characterstics

equation, so Y1(t) = t(A cos t+B sin t) solution but not A cos t+B sin t.

Y ′ = (A cos t+B sin t) + t(−A sin t+B cos t)

Y ′′ = (−A sin t+B cos t) + (−A sin t+B cos t) + t(−A cos t−B sin t)

= −2A sin t+ 2B cos t+ t(−A cos t−B sin t)

Y ′′′ = −2A cos t− 2B sin t−A cos t−B sin t+ t(A sin t−B cos t)

= −3A cos t− 3B sin t+ t(A sin t−B cos t)

y(4) = −3A sin t− 3B cos t+ (A sin t−B cos t) + t(A cos t+B sin t)

= −2A sin t− 4B cos t+ t(A cos t+B sin t)

(4.46)



−2A sin t− 4B cos t+ t(A cos t+B sin t)

or, 2A = 1

or, A =1

2


Y1(t) =1

2t e−t


Y (t) = Y1(t) + Y2(t) =1

2t e−t + 4t− 4


y = yc(t) + Y (t)

or, y = c1et + c2 cos t+ c3 sin t+

1

2t e−t + 4t− 4

Example 160.


y′′′ − 4y′ = t+ 3 cos t+ e−2t


y′′′ − 4y′ = t+ 3 cos t+ e−2t (4.47)


r3 − 4r = 0

or, r = 0, −2, 2


yc(t) = c1 + c2e2t + c2e

−2t (4.48)

Since g(t) = t+ 3 cos t+ e−2t sum of the two functions, so the particular solution is

sum of particular solutions of following equations

y′′′ − 4y′ = t (4.49)

y′′′ − 4y′ = 3 cos t (4.50)

y′′′ − 4y′ = e−2t (4.51)


Again, for the particular solution of (4.49), here t = 0 is a root so let particular

solution be Y1 = t(At+B) = At2 +Bt, not (At+B).

Y ′1 = 2At+B, , Y ′′1 = 2A, Y ′′′1 = 0


0− 4(2At+B) = t

or, − 8At− 4B = t+ 0

or, − 8B = 1, −4B = 0

or, B = −1

8, B = 0


Y1(t) = −1

8t2

For the particular solution of (4.50), here ±iβ = ±i are not solutions of the charac-

terstics equation, so Y2(t) = (C cos t+D sin t) solution.

Y ′2 = −C sin t+D cos t

Y ′′2 = −C cos t−D sin t

Y ′′′2 = C sin t−D cos t

(4.52)


C sin t−D cos t− 4(−C sin t+D cos t) = 3 cos t

or, − 5D cos t+ 5C sin t = 3 cos t+ 0 sin t

or, D = −3

5, C = 0


Y2(t) = −3

5sin t

Finally for the equation (4.51) here α = −2 is single root of the characterstic equa-

tion, so we assume that the particular solution as

Y3 = Ete−2t

Differentiating,

Y ′3 = E(e−2t − 2te−2t

)Y ′′3 = E

(−2e−2t − 2e−2t + 4te−2t

)= E

(−4e−2t + 4te−2t

)Y ′′′3 = E

(8e−2t + 4e−2t − 8te−2t

)= E

(12e−2t − 8te−2t

)(4.53)



E(12e−2t − 8te−2t

)− 4E

(e−2t − 2te−2t

)= e−2t

or, 8E = 1

or, E =1

8

Y3 =1

8e−2t


Y (t) = Y1(t) + Y2(t) = −1

8t2 − 3

5sin t+

1

8e−2t


y = yc(t) + Y (t)

or, y = c1 + c2e2t + c2e

−2t − 1

8t2 − 3

5sin t+

1

8e−2t

Example 161.

Find the solution of the following initial value problem

y′′′ − 3y′′ + 2y′ = t+ et, y(0) = 1, y′(0) = −1

4, y′′(0) = −3

2


y′′′ − 3y′′ + 2y′ = t+ et (4.54)

with the initial value

y(0) = 1, y′(0) = −1

4, y′′(0) = −3

2


r3 − 3r2 + 2r = 0

or,r(r − 1)(r − 2) = 0

or, r = 0, 1, 2


yc(t) = c1 + c2et + c2e

2t (4.55)

Since g(t) = t + et sum of the two functions, so the particular solution is sum of

particular solutions of following equations

y′′′ − 3y′′ + 2y′ = t (4.56)

y′′′ − 3y′′ + 2y′ = et (4.57)


Again, for the particular solution of (4.54), here t = 0 is a root so let particular

solution be Y1 = t(At+B) = At2 +Bt, not (At+B).

Y ′1 = 2At+B, , Y ′′1 = 2A, Y ′′′1 = 0


0− 6A+ 2(2At+B) = t

or, 4At− 6A+ 2B = t+ 0

or, 4A = 1, −6A+ 2B = 0

or, A =1

4, B =

3

4


Y1(t) =1

4t2 +

3

4t

Finally for the equation (4.57) here α = 1 is single root of the characterstic equation,

so we assume that the particular solution as

Y3 = Ctet

Differentiating,

Y ′3 = C(et + tet

)Y ′′3 = C

(et + et + tet

)= E

(2et + tet

)Y ′′′3 = C

(2et + et + tet

)= E

(3et + tet

)(4.58)


C(3et + tet

)− 3C

(2et + tet

)+ 2C(et + tet) = et

or, C = −1

Y3 = −tet


Y (t) = Y1(t) + Y2(t) =1

4t2 +

3

4t− tet


y = yc(t) + Y (t)

or, y = c1 + c2et + c2e

2t +1

4t2 +

3

4t− tet


Differentiating, both sides with respect to t,

y′ = c2et + 2c3e

2t +1

2t+

3

4− et − tet

y′′ = c2et + 4c3e

2t +1

2− et − et − tet

Using initial conditions

y(0) = 1 =⇒ c1 + c2 + c3 = 1

y′(0) = −1

4=⇒ c2 + 2c3 +

3

4− 1 = −1

4(4.59)

y′′(0) = −3

2=⇒ c2 + 4c3 +

1

2− 2 = −3

2

Solving the equations (4.59), we get c1 = 1, c2 = c3 = 0. The solution of the initial

value problem is

y(t) = 1− 1

4t2 +

3

4t− tet

4.5 Variation of Parameters

Let us consider a differential equation

L[y] = a0yn + a1y

n−1 + · · ·+ any = g(t) (4.60)

Let

yc(t) = c1y1 + c2y2 + c3y3 + · · ·+ cnyn (4.61)

is general solution of the homogeneous equation

L[y] = a0yn + a1y

n−1 + · · ·+ any = (4.62)

Let us replace c1, c2, · · · , cn by the functions u1(t), u2(t), · · · , un(t), in (4.61) we get

Y (t) = u1y1 + u2y2 + u3y3 + · · ·+ unyn (4.63)

We try to determine u1(t), u2(t), · · · , un(t) so that the expression (4.63) is a solution

of the nonhomogeneuos equation (4.60). Now differentiating (4.63) with respect to

t

Y ′ = u1y′1 + u2y

′2 + u3y

′3 + · · ·+ uny

′n + u′1y1 + u′2y2 + u′3y3 + · · ·+ u′nyn (4.64)

We choose u1(t), u2(t), · · · , un(t) such that

u′1y1 + u′2y2 + u′3y3 + · · ·+ u′nyn = 0 (4.65)

. Now the equation (4.64) becomes

Y ′ = u1y′1 + u2y

′2 + u3y

′3 + · · ·+ uny

′n (4.66)


Further, by differentiating (4.66) again, we obtain

Y ′′ = u1y′′1 + u2y

′′2 + u3y

′3 + · · ·+ uny

′′n + u′1y

′1 + u′2y

′2 + u′3y

′3 + · · ·+ u′ny

′n (4.67)

We choose u1(t), u2(t), · · · , un(t) such that

u′1y′1 + u′2y

′2 + u′3y

′3 + · · ·+ u′ny

′n = 0 (4.68)

Continuning this process in a similar mannar through n− 1 derivatives of Y gives

Y (m) = u1y(m)1 + u2y

(m)2 + u3y

(m)3 + · · ·+ uny

(m)n , m = 0, 1, 2, · · · , n− 1 (4.69)

and the following n− 1 conditions on the fuctions u1, u2, · · · , un

u′1y(m−1)1 + u′2y

(m−1)2 + u′3y

(m−1)3 + · · ·+ u′ny

(m−1)n = 0, m = 0, 1, 2, · · · , n− 1(4.70)

The nth derivative of Y is

Y (n) = u1y(n)1 + u2y

(n)2 + u3y

(n)3 + · · ·+ uny

(n)n + u′1y

(n)1 + u′2y

(n)2 + · · ·+ u′ny

(n)n (4.71)

Finally we impose the condition that Y must be a solution of (4.60). On substituting

for the derivatives of Y i.e. Y, Y ′, Y ′′ · · ·Y (n), collecting terms, and making use of

the fact taht L[yi] = 0, i = 1, 2, · · · , n, we obtain

u′1y(n−1)1 + u′2y

(n−1)2 + u′3y

(n−1)3 + · · ·+ u′ny

(n−1)n = g(t) (4.72)

Equation (4.72), coupled with the n−1 equations (4.70), gives n simultaneous linear

nonhomogeneous algebraic equations for u′1, u′2 · · · , u′n:

u′1y′1 + u′2y

′2 + u′3y

′3 + · · ·+ u′ny

′n = 0

u′1y′′1 + u′2y

′′2 + u′3y

′3 + · · ·+ u′ny

′′n = 0

u′1y′′′1 + u2y

′′′2 + u′3y

′′′3 + · · ·+ u′ny

′′′n = 0 (4.73)

...

u′1y(n−1)1 + u′2y

(n−1)2 + u′3y

(n−1)3 + · · ·+ u′ny

(n−1)n = g(t)

Above syetem is a linear algebraic system for the unknowns u′1, u′2 · · · , un. Solving

the equations by Cramer’s rule

u′m(t) =g(t)Wm(t)

W (t), m = 1, 2, 3 · · · ,m (4.74)

where

W (t) = W (y1, y2, · · · , yn)(t)

=

∣∣∣∣∣∣∣∣∣∣y1 y2 · · · yny′1 y′2 · · · y′n...

y(n−1)1 y

(n−2)2 · · · y

(n−1)n

∣∣∣∣∣∣∣∣∣∣


Wm(t) is the determinat obtained from W (t) by replacing mth column by the column0

0...

1

(4.75)

From (4.74),

um(t) =

∫ t

0

g(t)Wm(t)

W (t)ds m = 1, 2, · · · , n for some t0 ∈ I.

Hence,

Y (t) = u1y1 + u2y2 + u3y3 + · · ·+ unyn

=n∑

m=1

ym

∫ t

0

g(t)Wm(t)

W (t)ds

Aslo, from Abel’s theorem

W (t) = W (y1, y2, · · · , yn)(t) = ce−∫p1(t)dt

Example 162.

Find the general solution of y′′′ − y′ = t by using the method of variation of

parameters.


y′′′ − y′ = t (4.76)

The auxiliary equation of (??) is

r3 − r = 0

or, r(r2 − 1) = 0

or, r = 0, 1 − 1

Thus, the complementrary solution is

yc = c1e0t + c1e

t + c2e−t = c1 + c2e

t + c2e−t

Let y1 = 1, y2 = et and y3 = e−t.Then

y′1 = 0, y′′1 = 0, y′2 = et, y′′2 = et, y′3 = −e−t, y′′3 = e−t,

Also

W (y1, y2, y3) =

∣∣∣∣∣∣∣y1(t) y2(t) y3(t)

y′1(t) y′2(t) y3(t)′

y′′1(t) y′′2(t) y′′3(t)

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣1 et e−t

0 et −e−t

0 et e−t

∣∣∣∣∣∣∣


= 1

∣∣∣∣∣ et −e−tet e−t

∣∣∣∣∣ = ete−t + ete−t = 2 6= 0

Hence y1, y2, y3 are fundamental solutions of the given differential equation. Also,

g(t) = t

W1(t) =

∣∣∣∣∣∣∣0 y2(t) y3

0 y′2(t) y′31 y′′2(t) y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣0 et e−t

0 et −e−t

1 et e−t

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ et e−t

et −e−t

∣∣∣∣∣= −ete−t − ete−t = −2

W2(t) =

∣∣∣∣∣∣∣y1 0 y3

y′1 0 y′3y′′1 1 y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1 0 e−t

0 0 −e−t

0 1 e−t

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ 0 e−t

1 −e−t

∣∣∣∣∣ = e−t

W3(t) =

∣∣∣∣∣∣∣y1 y2 0

y′1 y′2 1

y′′1 y′′2 1

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1 et 0

0 et 0

0 et 1

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ et 0

et 1

∣∣∣∣∣ = et


u1(t) =

∫g(t)W1(t)

W (t)dt =

∫−2t

2dt = −1

2t2

u2(t) =

∫g(t)W2(t)

W (t)dt

=

∫te−t

2dt

=1

2

∫te−tdt

=1

2

t

∫e−tdt−

∫ (dt

dt

∫e−tdt

)dt

=

1

2

−te−t +

∫e−tdt

=

1

2

−te−t − e−t

= −1

2(t+ 1)e−t

u3(t) =

∫g(t)W3(t)

W (t)dt

=

∫te−t

2dt

=1

2

∫tetdt

=1

2

t

∫etdt−

∫ (dt

dt

∫etdt

)dt

=

1

2

tet −

∫etdt

=

1

2

te−t − et

=

1

2(t− 1)et


Y (t) = y1(t)u1(t) + y2u2(t) + y3(t)u3(t)

= −1

2t2 · 1− 1

2(t+ 1)e−tet +

1

2e−tet(t− 1)

=1

2t2 − 1

which is a particular solution of (4.76). Hence the general solution

y(t) = yc(t) + Y (t)

or, y(t) = c1 + c2et + c2e

−t +1

2t2 − 1 = c1 + c2e

t + c2e−t +

1

2t2

we may assume c1 − 1 as c1.

Example 163.

Find the general solution of y′′′ − y′ = t by using the method of variation of

parameters.



y′′′ + y′ = sec t − π

2< t <

π

2(4.77)


r3 + r = 0

or, r(r2 + 1) = 0

or, r = 0, i − i


yc = c1e0t + c2 cos t+ c2 sin t = c1 + c2 cos t+ c2 sin t

Let y1 = 1, y2 = cos t and y3 = sin t. Then

y′1 = 0, y′′1 = 0, y′2 = − sin t, y′′2 = −cost y′3 = cos t, y′′3 = − sin t,

Also

W (y1, y2, y3) =

∣∣∣∣∣∣∣y1(t) y2(t) y3(t)

y′1(t) y′2(t) y3(t)′

y′′1(t) y′′2(t) y′′3(t)

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣1 cos t sin t

0 − sin t cos t

0 − cos t − sin t

∣∣∣∣∣∣∣= 1

∣∣∣∣∣ − sin t cos t

−cost − sin t

∣∣∣∣∣ = sin2 t+ cos2 t = 1 6= 0


g(t) = sec t

W1(t) =

∣∣∣∣∣∣∣0 y2(t) y3

0 y′2(t) y′31 y′′2(t) y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣0 cos t sin t

0 − sin t cos t

1 − cos t − sin t

∣∣∣∣∣∣∣ = 1


− sin t cos t

∣∣∣∣∣= cos2 t+ sin2 t = 1

W2(t) =

∣∣∣∣∣∣∣1 0 y3

y′1 0 y′3y′′1 1 y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1 0 sin t

0 0 cos t

0 1 − sin t

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ 0 cos t

1 − sin t

∣∣∣∣∣ = − cos t


W3(t) =

∣∣∣∣∣∣∣y1 y2 0

y′1 y′2 1

y′′1 y′′2 1

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣1 cos t 0

0 − sin t 0

0 − cos t 1

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ − sin t 0

− cos t 1

∣∣∣∣∣ = − sin t

u1(t) =

∫g(t)W1(t)

W (t)dt =

∫sec tdt = ln(sec t+ tan t)

u2(t) =

∫g(t)W2(t)

W (t)dt

=

∫sec t(− cos t)

1dt

= −∫dt = −t

u3(t) =

∫g(t)W3(t)

W (t)dt

=

∫sec t (− sin t)

1dt

= −∫t tan tdt

= ln(cos t)


Y (t) = y1(t)u1(t) + y2u2(t) + y3(t)u3(t)

= 1 · ln(sec t+ tan t)− t cos t+ sin t ln(cos t)

= ln(sec t+ tan t)− t cos t+ sin t ln(cos t)


y(t) = yc(t) + Y (t)

or, y(t) = c1 + c2 cos t+ c2 sin t+ ln(sec t+ tan t)− t cos t+ sin t ln(cos t)

.

Example 164.

Find the general solution of y′′′ − 2y′′ − y′ + 2y = e4t by using the method of

variation of parameters.


y′′′ − 2y′′ − y′ + 2y = e4t (4.78)



r3 − 2r2 − r + 2 = 0

or, r2(r − 2)− 1(r − 2) = 0

or, (r − 2)(r + 1)(r − 1) = 0

or, r = 2, 1 − 1


yc = c1e−t + c2e

t + c2e2t

Let y1 = e−t, y2 = et and y3 = e2t. Then

y′1 = −e−t, y′′1 = e−t, y′2 = et, y′′2 = et y′3 = 2e2t, y′′3 = 4e2t,

Also

W (y1, y2, y3) =

∣∣∣∣∣∣∣y1(t) y2(t) y3(t)

y′1(t) y′2(t) y3(t)′

y′′1(t) y′′2(t) y′′3(t)

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣e−t et e2t

−e−t et 2e2t

e−t et 4e2t

∣∣∣∣∣∣∣= e−tete2t

∣∣∣∣∣∣∣1 1 1

−1 1 2

1 1 4

∣∣∣∣∣∣∣= e2t

∣∣∣∣∣∣∣1 0 0

−1 2 3

1 0 3

∣∣∣∣∣∣∣ = 6e2t 6= 0


g(t) = sec t

W1(t) =

∣∣∣∣∣∣∣0 y2(t) y3

0 y′2(t) y′31 y′′2(t) y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣0 et e2t

0 et 2e2t

1 et 4e2t

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ et e2t

et 2e2t

∣∣∣∣∣= e3t

W2(t) =

∣∣∣∣∣∣∣1 0 y3

y′1 0 y′3y′′1 1 y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣e−t 0 e2t

−e−t 0 2e2t

e−t 1 4e2t

∣∣∣∣∣∣∣ = ete2t

∣∣∣∣∣∣∣1 0 1

−1 0 2

1 1 4

∣∣∣∣∣∣∣ = −3et


W3(t) =

∣∣∣∣∣∣∣y1 y2 0

y′1 y′2 0

y′′1 y′′2 1

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣e−t et 0

−e−t et 0

e−t et 1

∣∣∣∣∣∣∣ =

∣∣∣∣∣ e−t et

−e−t et

∣∣∣∣∣ = 1 + 1 = 2

u1(t) =

∫g(t)W1(t)

W (t)dt =

∫e4te3t

6e2tdt =

1

6

∫e5tdt =

1

30e5t

u2(t) =

∫g(t)W2(t)

W (t)dt =

∫e4t(−3)et

6e2tdt

= −1

2

∫e3tdt = −1

6e3t

u3(t) =

∫g(t)W3(t)

W (t)dt

=

∫e4t 2

6e2tdt

=1

3

∫e2tdt

=1

6e2t


Y (t) = y1(t)u1(t) + y2u2(t) + y3(t)u3(t)

= e−t · 1

30e5t + et

(−1

6e3t

)+ e2t 1

6e2t

=1

30e4t − 1

4e4t +

1

6e4t =

1

30e4t


y(t) = yc(t) + Y (t)

or, y(t) = c1e−t + c2e

t + c2e2t +

1

30e4t

.

Example 165.

Given that y1(t) = et, y2(t) = tet and y3 = e−t are solutions of the homoge-

neous equation corresponding to

y′′′ − y′′ − y′ + y = g(t)


detemine a paricular solution in term of an integral.


y′′′ − y′′ − y′ + y = g(t) (4.79)

Here y1 = et, y2 = tet and y3 = e−t. Then

y′1 = et, y′′1 = et, y′2 = tet + et, y′′2 = tet + 2et y′3 = −e−t, y′′3 = e−t,

Also

W (y1, y2, y3) =

∣∣∣∣∣∣∣y1(t) y2(t) y3(t)

y′1(t) y′2(t) y3(t)′

y′′1(t) y′′2(t) y′′3(t)

∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣et tet e−t

et (t+ 1)et −e−t

et (t+ 2)et e−t

∣∣∣∣∣∣∣= etete−t

∣∣∣∣∣∣∣1 t 1

1 t+ 1 −1

1 t+ 2 1

∣∣∣∣∣∣∣= et

∣∣∣∣∣∣∣0 t 1

2 t+ 1 −1

0 t+ 2 1

∣∣∣∣∣∣∣ Operating C1 → C1 − C2

= −2et

∣∣∣∣∣ t 1

t+ 2 1

∣∣∣∣∣= 4et

Hence y1, y2, y3 are fundamental solutions of the given differential equation.

W1(t) =

∣∣∣∣∣∣∣0 y2(t) y3

0 y′2(t) y′31 y′′2(t) y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣0 tet e−t

0 (t+ 1)et −e−t

1 (t+ 2)et e−t

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ tet e−t

(t+ 1)et −e−t

∣∣∣∣∣= −2t− 1

W2(t) =

∣∣∣∣∣∣∣y1 0 y3

y′1 0 y′3y′′1 1 y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣et 0 e−t

et 0 −e−t

et 1 e−t

∣∣∣∣∣∣∣ = ete−t

∣∣∣∣∣∣∣1 0 1

1 0 −1

1 1 1

∣∣∣∣∣∣∣ = 2


W3(t) =

∣∣∣∣∣∣∣y1 y2 0

y′1 y′2 0

y′′1 y′′2 1

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣et tet 0

et (t+ 1)et 0

et (t+ 2)et 1

∣∣∣∣∣∣∣ =

∣∣∣∣∣ et tet

et (t+ 1)et

∣∣∣∣∣ = e2t

u1(t) =

∫ t

0

g(t)W1(s)

W (s)ds =

∫ t

0

g(s)(−1− 2s)

4esds

u2(t) =

∫g(t)W2(t)

W (t)dt =

∫g(s)(2)

4esds

=1

2

∫ t

0g(s)e−sds

u3(t) =

∫ t

0

g(s)W3(s)

W (s)ds

=

∫ t

0

g(s)e2s

4esds

=1

4

∫ t

0g(s)esds


Y (t) = y1(t)u1(t) + y2u2(t) + y3(t)u3(t)

= et∫ s

0

g(s)(−1− 2s)

4esds+

1

2tet∫ t

0g(s)e−sds+

1

4e−t∫ t

0g(s)esds

Example 166.

Giventhat x, x2 and 1x are solutions of the homogeneous equation corresponding

to

x3y′′′ + x2y′′ − 2xy′ + 2y = 2x4, x > 0

determine a particular solution.


x3y′′′ + x2y′′ − 2xy′ + 2y = 2x4

y′′′ +1

xy′′ − 2

x2y′ +

2

x3y = 2x2 (4.80)

Here y1 = x , y2 = x2 and y3 = 1x . Then

y′1 = 1, y′′1 = 0, y′2 = 2x, y′′2 = 2, y′3 = − 1

x2, y′′3 =

2

x3,


Also

W (y1, y2, y3) =

∣∣∣∣∣∣∣y1(t) y2(t) y3(t)

y′1(t) y′2(t) y3(t)′

y′′1(t) y′′2(t) y′′3(t)

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣x x2 1/x

1 2x − 1x2

0 2 2x3

∣∣∣∣∣∣∣= x

∣∣∣∣∣ 2x − 1x2

2 2x2

∣∣∣∣∣− 1

∣∣∣∣∣ x2 1/x

2 2/x2

∣∣∣∣∣= x

(4x

x3+

2

x2

)− 1

(2x2

x3− 2

x

)=

6

x

Hence y1, y2, y3 are fundamental solutions of the given differential equation.

W1(t) =

∣∣∣∣∣∣∣0 y2(t) y3

0 y′2(t) y′31 y′′2(t) y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣0 x2 1

x

0 2x − 1x2

1 2 2x3

∣∣∣∣∣∣∣ = 1

∣∣∣∣∣ x2 1x

2x − 1x2

∣∣∣∣∣= −1− 2 = −3

W2(t) =

∣∣∣∣∣∣∣y1 0 y3

y′1 0 y′3y′′1 1 y′′3

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣x 0 1

x

1 0 − 1x2

0 1 2x3

∣∣∣∣∣∣∣ = −

∣∣∣∣∣ x 1/x

1 −1/x2

∣∣∣∣∣ = 2/x

W3(t) =

∣∣∣∣∣∣∣y1 y2 0

y′1 y′2 0

y′′1 y′′2 1

∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣x x2 0

1 2x 0

0 2 1

∣∣∣∣∣∣∣ =

∣∣∣∣∣ x x2

1 2x

∣∣∣∣∣ = 2x2 − x2 = x2

g(x) = 2x (4.81)


u1(x) =

∫g(x)W1(x)

W (x)dx =

∫2x(−3)

6/xdx = −x

3

3

u2(x) =

∫g(x)W2(x)

W (x)dx =

∫2x · 2/x

6/xdx

=4

6

∫xdx =

x2

3

u3(x) =

∫g(x)W3(x)

W (x)dx

=

∫ t

0

2x · x2

6/xdx

=x5

15


Y (t) = y1(t)u1(t) + y2u2(t) + y3(t)u3(t)

= −x4

3+x4

3+x4

15=x4

15

Chapter 5

System Of First Order Linear

Equations

5.1 Introduction

Systems of simultaneous ordinary differential equations arise naturally in prob-

lems involving several dependent variables, each of which is a function of the same

single independent variable.. We will denote the independent variable by t and

x1, x2, x3, · · · , xn dependent variables that are functions of t. Differentiation with

respect to t will be denoted by a prime.

Let us consider an arbitrary nth order equation

y(n) = F (t, y, y′, y′′, · · · , y(n−1)) (5.1)

This equation can be transformed into a system of n first order equations by intro-

ducing the variables x1, x2, · · · , xn defined by

x1 = y, x2 = y′, x3 = y′′, · · · , xn = y(n−1). (5.2)

Example 167.

Transform the following equations into a system of first orderequations

1. u′′ + 0.125u′ + u = 0 2. mu′′ + γu′ + ku = F (t)

3. u′′ + 2u′ + 2u = 0 4. t2u′′ + t2u′ + (t2 − 1)u = 0

5. u′′ + 0.5u′ + 2u = 3 sin t 6. u(4) − u = 0


u′′ + 0.125u′ + u = 0 (5.3)

Let x1 = u and x2 = u′. Then it follows that x′1 = u′ = x2 and u′′ = x′2. Then, by

substituting for u, u′ and u′′ in (5.3),

x′2 + 0.125x2 + x1 = 0

x′2 = −0.125x2 − x1

281

282 CHAPTER 5. SYSTEM OF FIRST ORDER LINEAR EQUATIONS

Thus, x1 and x2 satisfy the following system of two first order differential equations

x′1 = x2

x′2 = −x1 − 0.125x2

Solution 2. The given differential equation

mu′′ + γu′ + ku = 0 (5.4)



mx′2 + γx2 + kx1 = F (t)

x′2 = − kmx1 −

γ

mx2 +

1

mF (t)


x′1 = x2

x′2 = − kmx1 −

γ

mx2 +

1

mF (t)


u′′ + 2u′ + 2u = 0 (5.5)



x′2 + 2x2 + 2x1 = 0

x′2 = −2x2 − 2x1


x′1 = x2

x′2 = −2x1 − 2x2


t2u′′ + tu′ + (t2 − 1)u = 0 (5.6)



t2x′2 + tx2 + (t2 − 1)x1 = 0

x′2 = − t2 − 1

t2x1 −

1

tx2

5.1. INTRODUCTION 283


x′1 = x2

x′2 = − t2 − 1

t2x1 −

1

tx2


u′′ + 0.5u′ + 2u = 2 sin t (5.7)



x′2 + 0.5x2 + 2x1 = 2 sin t

x′2 = −0.5x2 − 2x1 + 2 sin t


x′1 = x2

x′2 = −2x1 − 0.5x2 + 2 sin t


u(4) − u = 0 (5.8)

Let x1 = u, x2 = u′, x3 = u′′ and x4 = u′′′. Then it follows that

x′1 = u′ = x+ 2

x′2 = u′′ = x3

x′3 = u′′′ = x4

From (5.8)

x′4 = u(4) = u = x1


x′1 = x2

x′2 = x3

x′3 = x4

x′4 = x1

(5.9)

Example 168.


Transform the given initial value problem into an initial value problem for two

first order equation, u′′ + u′ + 4u = 2 sin 4t, u(0) = 1, u′(0) = 3.

Solution: . The given initial value problem is

u′′ + u′ + 4u = 2 sin t u(0) = 1, u′(0) = 3 (5.10)



x′2 + x2 + 4x1 = 2 sin 4t

x′2 = −x2 − 4x1 + 2 sin 4t


x′1 = x2

x′2 = −4x1 − x2 + 2 sin 4t

Also,

u(0) = 1 =⇒ x1(0) = 1

u′(0) = 3 =⇒ x2(0) = 3

Hence initial value problem of first order

x′1 = x2, x1(0) = 1

x′2 = −4x1 − x2 + 2 sin 4t x2(0) = 3

Generalization: Let us consider an arbitrary nth order equation

y(n) = F (t, y, y′, y′′, · · · , y(n−1)) (5.11)

This equation can be transformed into a system of n first order equations by intro-

ducing the variables x1, x2, · · · , xn defined by

x1 = y, x2 = y′, x3 = y′′, · · · , xn = y(n−1). (5.12)

It follows that

x′1 = x2

x′2 = x3

... (5.13)

x′n−1 = xn

From equation (5.13)

xn = F (t, x1, x2, · · · , xn) (5.14)


From the equations (5.13) and (5.14) we get the following system

x′1 = Fn(t, x1, x2, · · · , xn)

x′2 = Fn(t, x1, x2, · · · , xn)

... (5.15)

x′n = Fn(t, x1, x2, · · · , xn)

and is a system of n first order equations.

A solution of the system (5.15) on the interval I = (α, β) is a set of n functions

x1 = φ1(t), x2 = φ2(t), · · · , xn = φn(t),

that are differentiable at all points in the interval I and that satisfy the system of

equations (5.15) at all points in this interval.

In addition to the differential equations, there may also given initial conditions of

the form

x1(t0) = x01, x2(t0) = x0

2 · · · , xn(t0) = x0n (5.16)

that t0 is a specified value of t in I, and x01, x

02, · · · , x0

2 are prescribed numbers. The

differential equations (5.15) with the initial conditions (5.16) form an initial value

problem.

The following conditions on F1, F2, · · · , Fn the initial value problem (5.15),(5.16)

has a unique solutions:

Theorem 14.

Let each functions F1, F2, · · · , Fn and the partial derivatives ∂F1∂x1

, · · · ∂F1∂xn

, · · · ∂Fn∂x1, · · · ∂Fn∂xn

becontinuous in a region R of tx1x2 · · ·xn−space defined by α < tβ, α1 < x1 <

β1, · · ·αn < tβn and the pont (t0, x01, x

02, · · · , x0

n) in R. Then there is an interval

|t− t0| < h in which there exists a unique solution x1 = φ1(t), · · ·xn = φn(t) of the

system of differential equations

x′1 = Fn(t, x1, x2, · · · , xn)

x′2 = Fn(t, x1, x2, · · · , xn)

...

x′n = Fn(t, x1, x2, · · · , xn)

that also satisifies the initial conditions

x1(t0) = x01, x2(t0) = x0

2 · · · , xn(t0) = x0n

Linear and Nonlinear system: If each the functions F1, F2, · · · , Fn in the

equation (5.15) is a linearfunction of dependent variables x1, x2, · · · , xn, then the


system of the equations is said to be linear, otherwise, it is non-linear. The most

general system of n first order linear equations has the form

x′1 = p11(t)x1 + · · ·+ p1n(t)xn + g1(t)

x′2 = p21(t)x1 + · · ·+ p2n(t)xn + g2(t)

... (5.17)

x′n = pn1(t)x1 + · · ·+ pnn(t)xn + gn(t)

If each of the functions g1(t), g2(t), · · · , gn(t) is zero for all t in the interval I, then

the system (5.17) is said to be homogeneous; otherwise, it is nonhomogeneous.

Theorem 15.

If p11(t), p12(t), · · · , p1n(t), · · · , pn1(t), · · · pn1(t) and g1(t), · · · , gn(t) are continu-

ous on an open interval I = (α, β) then there exists unique solution x1 = φ1(t), x2 =

φ2(t), · · · , xn = φn(t) of the system

x′1 = p11(t)x1 + · · ·+ p1n(t)xn + g1(t)

x′2 = p21(t)x1 + · · ·+ p2n(t)xn + g2(t)

... (5.18)

x′n = pn1(t)x1 + · · ·+ pnn(t)xn + gn(t)


x1(t0) = x01, x2(t0) = x0

2 · · · , xn(t0) = x0n (5.19)

Moreover, the solution exists throughout the interval I.

Example 169.

System of first order equations can be transformed into a single equation of

higher order. Consider the system

x′1 = −2x1 + x2, x′2 = x1 − 2x2

(a) Solve the first equation for x2 and substitute into the second equation, thereby

obtaining a second order equation for x1. Solve this equation for x1 and then deter-

mine x2

(b) Find the solution of the given system that also satisfies the initial conditions

x1(0) = 2 and x2(0) = 3. T.U. (2073)

Solution: The given system of equations is

x′1 = −2x1 + x2 (5.20)

x′2 = x1 − 2x2 (5.21)


Solving for x1

Differentiating (5.20) , we get

x′′1 = −2x′1 + x′2

Putting the value of x′2, from (5.21)

x′′1 = −2x′1 + x1 − 2x2

(5.22)

Again, putting x2 = x′1 + 2x1 from (5.20)

x′′1 = −2x′1 + x1 − 2(x′1 + 2x1)

or, x′′1 + 4x′1 + 3x1 = 0 (5.23)

which is second order linear equations in x1 with constant coefficients. Its charac-

terstic equation is

r2 + 4r + 3 = 0

or,r2 + 3r + r + 3 = 0

or, r(r + 3) + 1(r + 3) = 0

or,(r + 1)(r + 3) = 0

∴ r = −1,−3.

Thus the solution of (5.23) is

x1(t) = c1e−t + c2e

−3t (5.24)

Aslo from (5.20),

x2 = 2x1 + x′1

or, x2 = 2(c1e−t + c2e

−3t)− c1e−t − 3c2e

−3t

or, x2 = c1e−t − c2e

−3t (5.25)

Using the initial conditions x1(0) = 2, in (5.24), we get

x1(0) = c1 + c2

or,c1 + c2 = 2 (5.26)


x2(0) = c1 − c2

or,c1 − c2 = 3 (5.27)



c1 =5

2, c2 = −1

2

Hence requaired solutions from (5.24) and (5.25)

x1(t) =5

2e−t − 1

2e−3t

x2(t) =5

2e−t +

1

2e−3t

Example 170.

Consider the system

x′1 = 2x2, x1(0) = 3

x′2 = −2x1, x2(0) = 4

(a) Transform the above system into a single equation of second order.

(b) Find the solution of the given system that satisfies the initial conditions

(c) Sketch the graph of the solution in the x1x2−plane.


x′1 = 2x2 (5.28)

x′2 = −2x1 (5.29)

Solving for x1


x′′1 = x′2


x′′1 = −4x1

or,x′′1 + 4x1 = 0 (5.30)


terstic equation is

r2 + 4 = 0

or,r2 = −4

∴ r = ±2i.


x1(t) = c1 cos 2t+ c2 sin 2t (5.31)


Aslo from (5.28),

x2 =1

2x′1

or, x2 =1

2(−2c1 sin 2t+ 2c2 cos 2t)

or, x2 = −c1 sin 2t+ c2 cos 2t (5.32)


x1(0) = c1 + 0

or,c1 = 3 (5.33)


x2(0) = 0 + c2

or,c2 = 4 (5.34)

Hence requaired solutions from (5.31) and (5.32) are

x1(t) = 3 cos 2t+ 4 sin 2t

x2(t) = −3 sin 2t+ 4 cos 2t

Squaring and adding,

x21 + x2

2 = (3 cos 2t+ 4 sin 2t)2 + (−3 sin 2t+ 4 cos 2t)2

or, x21 + x2

2 = 9 cos2 2t+ 24 cos 2t sin 2t+ 16 sin2 2t+ 9 sin2 2t− 24 cos 2t+ 16 cos2 2t

or, x21 + x2

2 = 9(cos2 2t+ sin2 2t) + 16(sin2 2t+ cos2 2t)

or, x21 + x2

2 = 16

Example 171.

Consider the system

x′1 = x1 − 2x2, x1(0) = −1

x′2 = 3x1 − 4x2, x2(0) = 2

(a) Transform the above system into a single equation of second order.

(b) Find the solution of the given system that satisfies the initial conditions


x′1 = x1 − 2x2 (5.35)

x′2 = 3x1 − 4x2 (5.36)

Solving for x1


x′′1 = x′1 − 2x′2



x′′1 = x′1 − 2(3x1 − 4x2)

or, x′′1 = x′1 − 6x1 + 8x2

Again, putting x2 = 12(−x′1 + x1) from (5.35)

x′′1 = x′1 − 6x1 + 4(−x′1 + x1)

or, x′′1 + 3x′1 + 2x1 = 0 (5.37)


terstic equation is

r2 + 3r + 2 = 0

or, r2 + 2r + r + 2 = 0

or, r(r + 2) + 1(r + 2) = 0

or, (r + 1)(r + 2) = 0

∴ r = −1,−2.


x1(t) = c1e−t + c2e

−2t (5.38)

Aslo from (5.35),

x2 =1

2

(x1 − x′1

)or, x2 =

1

2

(c1e−t + c2e

−2t + c1e−t + 2c2e

−2t)

or, x2 = c1e−t +

3

2c2e−3t (5.39)

Using the initial conditions x1(0) = −1, in (5.38), we get

x1(0) = c1 + c2

or, c1 + c2 = −1 (5.40)


x2(0) = c1 +3

2c2

or,2c1 + 3c2 = 4 (5.41)


c1 = −7, c2 = 6

Hence requaired solutions from (5.24) and (5.25)

x1(t) = −7e−t + 6e−2t

x2(t) = −7e−t + 9e−2t

5.2. LINEAR INDEPENDENCE AND LINEAR DEPENDENCE 291

5.2 Linear Independence and Linear Dependence

A set of k vectors x(1),x(2), · · ·x(k) is said to be linearly dependent if there exists a

set of complex or real numbers c1, c2, · · · , ck, at least one of which is nonzero such

that

c1x(1) + c2x

(2) + · · ·+ x(n) = 0 (5.42)

On the other hand x(1),x(2), · · ·x(k) is said to be linearly independent if

c1x(1) + c2x

(2) + · · ·+ x(n) = 0 =⇒ c1 = 0, c2 = 0, · · · , cn = 0

Example 172.

Determine whether the vectors

x(1) =

1

2

−1

,x(2) =

2

1

3

,x(3) =

−4

1

−11

are linearly independent or dependent. If they are linearly dependent, find a linear

relation among them.

Solution: The given vectors are

x(1) =

1

2

−1

,x(2) =

2

1

3

,x(3) =

−4

1

−11

Consider

c1x(1) + c2x

(2) + x(3) = 0 (5.43)

or, c1

1

2

−1

+ c2

2

1

3

+ c3

−4

1

−11

=

0

0

0

or,

1 · c1 + 2 · c2 − 4 · c3

2 · c1 + 1 · c2 + 1 · c3

−1 · c1 + 3 · c2 − 11 · c3

=

0

0

0

or,

1 2 −4

2 1 1

−1 3 −11

c1

c2

c3

=

0

0

0

ı.e.Ax = b

where

A =

1 2 −4

2 1 1

−1 3 −11

, x =

c1

c2

c3

, b =

0

0

0


The augmented matrix is

[A : b] =

1 2 −4 : 0

2 1 1 : 0

−1 3 −11 : 0

∼

1 2 −4 : 0

0 −3 9 : 0

0 5 −15 : 0

, R2 → R2 − 2R1, R3 → R1 +R2

∼

1 2 −4 : 0

0 1 −3 : 0

0 0 0 : 0

, R2 → 2R2 + 3R1, R3 → 5R1 −R3

Thus obtained the equivalent system

c1 + 2c2 − 4c3 = 0

c2 − 3c3 = 0 (5.44)

From the second equation of (5.44), we get c2 = 3c3, and from the first we obtaine

c1 = 4c3 − 2c2 = −2c3. Thus we solved for c1 and c2 in term of c3, with the latter

remaining arbitrary. Hence the there are infinitly many solutions. Hence the system

is linearly dependent. Now from (5.43)

−2c3x(1) + 3c3x

(2) + c3x(3) = 0

or, 2x(1) − 3x(2) − x(3) = 0

Example 173.


x(1) =

2

1

0

,x(2) =

1

1

0

,x(3) =

−1

2

0




x(1) =

2

1

0

,x(2) =

1

1

0

,x(3) =

−1

2

0


Consider

c1x(1) + c2x

(2) + x(3) = 0 (5.45)

or, c1

2

1

0

+ c2

1

1

0

+ c3

−1

2

0

=

0

0

0

or,

2 · c1 + 1 · c2 − 1 · c3

1 · c1 + 1 · c2 + 2 · c3

0 · c1 + 0 · c2 + 0 · c3

=

0

0

0

or,

2 1 −1

1 1 2

0 0 0

c1

c2

c3

=

0

0

0

ı.e.Ax = b

where

A =

2 1 −1

1 1 1

0 0 0

, x =

c1

c2

c3

, b =

0

0

0


[A : b] =

2 1 −1 : 0

1 1 1 : 0

0 0 0 : 0

∼

1 1 2 : 0

2 1 −1 : 0

0 0 0 : 0

, Interchanging R1, R2

∼

1 1 2 : 0

0 −1 −5 : 0

0 0 0 : 0

, R2 → R2 − 2R1, R3 → 5R1 −R3


c1 + c2 + 2c3 = 0

−c2 − 5c3 = 0 (5.46)

From the second equation of (5.44), we get c2 = −5c3, and from the first we obtaine

c1 = 5c3 − 2c2 = 3c3. Thus we solved for c1 and c2 in term of c3, with the latter



3c3x(1) − 5c3x

(2) + c3x(3) = 0

or, 3x(1) − 5x(2) + x(3) = 0


Example 174.


x(1) =

1

2

−1,

0

,x(2) =

2

3

1

−1

,x(3) =

−1

0

2

2

x(3) =

3

−1

1

3




x(1) =

1

2

−1,

0

,x(2) =

2

3

1

−1

,x(3) =

−1

0

2

2

x(3) =

3

−1

1

3

Consider

c1x(1) + c2x

(2) + x(3) = 0 (5.47)

or, c1

1

2

−1

0

+ c2

2

3

1

−1

+ c3

−1

0

2

2

+ c4

3

−1

1

3

=

0

0

0

or,

1 · c1 + 2 · c2 − 1 · c3 + 3 · c4

2 · c1 + 3 · c2 + 0 · c3 − 1 · c4

−1 · c1 + 1 · c2 + 2 · c3 + 1 · c4

0 · c1 − 1 · c2 + 2 · c3 + 3 · c4

=

0

0

0

0

or,

1 2 −1 3

2 3 0 −1

−1 1 2 1

0 −1 2 3

c1

c2

c3

c4

=

0

0

0

0

ı.e.Ax = b

where

A =

1 2 −1 3

2 3 0 −1

−1 1 2 1

0 −1 2 3

, x =

c1

c2

c3

c4

, b =

0

0

0

0



[A : b] =

1 2 −1 3 : 0

2 3 0 −1 : 0

−1 1 2 1 : 0

0 −1 2 3 : 0

∼

1 2 −1 3 : 0

0 −1 2 −7 : 0

0 3 1 4 : 0

0 −1 2 3 : 0

, Interchanging R2 → R2 − 2R1, R3 → R1 +R3

∼

1 2 −1 3 : 0

0 −1 −5 −7 : 0

0 0 7 17 : 0

0 0 0 10 : 0

, R3 → 3R2 +R3, R4 → R4 −R2


c1 + 2c2 − c3 + 3 c4 = 0

−c2 + 2c3 − 7 c4 = 0

7c3 − 17 c4 = 0

10c4 = 0 (5.48)

Solving these equations, we get c4 = 0, c3 = 0, c2 = 0, c1 = 0 is only solution. Hence

the given vectors are linearly independent.

Example 175.


x(1) =

(et

2e−t

),x(2) =

(e−t

e−t

),x(3) =

(3e−t

0

)




x(1) =

(et

2e−t

),x(2) =

(e−t

e−t

),x(3) =

(3e−t

0

)


Consider

c1x(1) + c2x

(2) + x(3) = 0 (5.49)

or, c1

(et

2e−t

)+ c2

(3e−t

0

)+ c3

−4

1

−11

=

0

0

0

or,

e−t · c1 + e−t · c2 + 3e−t · c3

2e−t · c1 + e−t · c2 + 0 · c3

0 · c1 + 0 · c2 + 0 · c3

=

0

0

0

or,

e−t e−t 3e−t

2e−t e−t 0

0 0 0

c1

c2

c3

=

0

0

0

ı.e.Ax = b

where

A =

e−t e−t 3e−t

2e−t e−t 0

0 0 0

, x =

c1

c2

c3

, b =

0

0

0


[A : b] =

e−t e−t 3e−t : 0

2e−t e−t 0 : 0

0 0 0 : 0

∼

1 1 3 : 0

2 1 0 : 0

0 0 0 : 0

, R1 → etR1, R2 → etR1

∼

1 1 2 : 0

0 −1 0 : 0

0 0 0 : 0

, R2 → R2 − 2R1


c1 + 2c2 + 3c3 = 0

−c2 − 6c3 = 0 (5.50)

From the second equation of (5.44), we get c2 = 6c3, and from the first we obtaine

c1 = −4c3 − 3c3 = −7c3. Thus we solved for c1 and c2 in term of c3, with the latter



−7c3x(1) + 6c3x

(2) + c3x(3) = 0

or, 7x(1) − 6x(2) − x(3) = 0


Example 176.


x(1) =

(2 sin t

sin t

),x(2) =

(sin t

2 sin t

)are linearly independent or dependent. If they are linearly dependent, find a linear



x(1) =

(2 sin t

sin t

),x(2) =

(sin t

2 sin t

)Consider

c1x(1) + c2x

(2) = 0 (5.51)

or, c1

(2 sin t

sin t

)+ c2

(sin t

2 sin t

)=

(0

0

)

or,

(2 sin t · c1 + sin t · c2

sin t · c1 + 2 sin t · c2

)=

(0

0

)

or,

(2 sin t sin t

sin t 2 sin t

)(c1

c2

)=

0

0

0

ı.e.Ax = b

where

A =

(2 sin t sin t

sin t 2 sin t

), x =

(c1

c2

), b =

(0

0

)The augmented matrix is

[A : b] =

(2 sin t sin t : 0

sin t 2 sin t : 0

)

∼

(2 1 : 0

1 2 : 0

), R1 →

1

sin tR1, R2 →

1

sin tR2

∼

(1 2 : 0

2 1 : 0

), R1 ↔ R2

∼

(1 2 : 0

0 −3 : 0

), R2 → R2 − 2R1


c1 + 2c2 = 0

−3c3 = 0 (5.52)


Solving these equations, we get c2 = 0, c1 = 0.Hence the given vectors are linearly

independent.

5.3 Eigenvalues and Eigenvectors

The equation

Ax = y (5.53)

is a linear transformation that maps a given vector x into a new vector y. Vector that

are transformed into multiples of themselves are important. To find such vectors,

we set y = λx, where λ is a scalar proportionality factor. Now the equation (5.53)

becomes

Ax = λx (5.54)

or, (A− λI)x = 0

where I is the identity matrix of order n. The equation (5.54) has nonzero solution

if and only if det(A− λI) = 0.

The eigen value of a square matrix A = (ajk) is a number λ such that the equation

(5ei1) has a solution x 6= 0. The x is called an eigenvector of A corresponding

eigenvalue λ.

The equation det(A − λI) = 0 is called characterstics equations. It polynomial

equation of degree n in λ. The roots of the charecterstics equation are eighenvalue of

the matrix A. Substituting the value of λ in the equation (5ei1), we get eigenvectors

of A corresponding to the eigenvalues.

Example 177.

Find the eigenvalue and eigenvectors of the matrix

A =

(3 −1

1 1

)Solution: The given matrix is

A =

(3 −1

1 1

)The characterstic matrix is

= A− λI

=

(3 −1

1 1

)− λ

(1 0

0 1

)

=

(3− λ −1

1 1− λ

)(5.55)

5.3. EIGENVALUES AND EIGENVECTORS 299

The charetersics equation is

|A− λI| = 0

or,

∣∣∣∣∣ 3− λ −1

1 1− λ

∣∣∣∣∣ = 0

or, (3− λ)(1− λ) + 1 = 0

or, 3− λ− 3λ+ λ2 + 1 = 0

or, λ2 − 4λ+ 4 = 0

or, (λ− 2)2 = 0

or, λ = 2

Eigen value of the matrix A is λ = 2.

Let

x =

(x1

x2

)be eigenvector of the matrix A corresponding to eigenvalue λ = 2. Then the marix

equation is

(A− λI)x = 0

or,

(3− λ −1

1 1− λ

)(x1

x2

)=

(0

0

)But λ = 2

or,

(3− 2 −1

1 1− 2

)(x1

x2

)=

(0

0

)

or,

(1 −1

1 −1

)(x1

x2

)=

(0

0

)

or,

(x1 − x2

x1 − x2

)=

(0

0

)Thus each row of this vector equation leads to the condition x1 = x2. Let x1 = x1 =

c 6= 0. Hence, eigenvectors corrosponding to eigenvalue λ = 2 are

x =

(x1

x2

)=

(c

c

)= c

(1

1

), c 6= 0

Example 178.


A =

(3 −1

4 −2

)


Solution: The given matrix is

A =

(3 −1

4 2

)The characterstic matrix is

= A− λI

=

(3 −1

4 2

)− λ

(1 0

0 1

)

=

(3− λ −1

4 −2− λ

)(5.56)


|A− λI| = 0

or,

∣∣∣∣∣ 3− λ −1

4 −2− λ

∣∣∣∣∣ = 0

or, − (3− λ)(2 + λ) + 4 = 0

or, 6− 2λ+ 3λ− λ2 − 4 = 0

or, λ2 − λ− 2 = 0

or, λ2 − 2λ+ λ− 2

or, (λ− 2)(λ+ 1) = 0

or, λ = 2, −1

Eigen value of the matrix A are λ1 = 2 and λ2 = −1.

Let

x =

(x1

x2

)be eigenvector of the matrix A corresponding to eigenvalue λ = 2. Then the marix

equation is

(A− λI)x = 0

or,

(3− λ −1

4 −2− λ

)(x1

x2

)=

(0

0

)(5.57)

Taking λ = 2 in (5.57)

or,

(3− 2 −1

4 −2− 2

)(x1

x2

)=

(0

0

)

or,

(1 −1

4 −4

)(x1

x2

)=

(0

0

)

or,

(x1 − x2

4x1 − 4x2

)=

(0

0

)


Thus each row of this vector equation leads to the condition x1 = x2. Let x1 = x1 =

c 6= 0. Hence, eigenvectors corrosponding to eigenvalue λ = 2 are

x =

(x1

x2

)=

(c

c

)= c

(1

1

), c 6= 0

Taking λ = −1 in (5.57)

or,

(3 + 1 −1

4 −2 + 1

)(x1

x2

)=

(0

0

)

or,

(4 −1

4 −1

)(x1

x2

)=

(0

0

)

or,

(4x1 − x2

4x1 − 4x2

)=

(0

0

)

Thus each row of this vector equation leads to the condition 4x1 = x2. Let x1 =

c 6= 0. Then x2 = 4c Hence, eigenvectors corrosponding to eigenvalue λ = 2 are

x =

(x1

x2

)=

(c

4c

)= c

(1

4

), c 6= 0

Example 179.


A =

(3 −2

4 −1

)


A =

(3 −2

4 −1

)

The characterstic matrix is

= A− λI

=

(3 −2

4 −1

)− λ

(1 0

0 1

)

=

(3− λ −2

4 −1− λ

)(5.58)



|A− λI| = 0

or,

∣∣∣∣∣ 3− λ −2

4 −1− λ

∣∣∣∣∣ = 0

or, − (3− λ)(1 + λ) + 4× 2 = 0

or, − 3− 3λ− λ+ λ2 + 8 = 0

or, λ2 − 2λ+ 5 = 0

or, λ =−(−2)±

√(−2)2 − 4× 1× 5

2× 1

or, λ =2±√

4− 20

2=

2± 4i

2= 1± 2i

or, λ = 1 + 2i, 1− 2i

Eigenvalue of the matrix A are λ1 = 1 + 2i and λ2 = 1− 2i.

Let

x =

(x1

x2

)

be eigenvector of the matrix A corresponding to eigenvalue λ. Then the marix

equation is

(A− λI)x = 0

or,

(3− λ −2

4 −1− λ

)(x1

x2

)=

(0

0

)(5.59)


Taking λ = 1 + 2i in (5.59)

or,

(3− 1− 2i −2

4 −1− 1− 2i

)(x1

x2

)=

(0

0

)

or,

(2− 2i −1

4 −2− 2i

)(x1

x2

)=

(0

0

)

or,

((2− 2i)x1 − 2x2

4x1 − (2 + 2i)x2

)=

(0

0

)

or,

((1− i)x1 − x2

2x1 − (1 + i)x2

)=

(0

0

)

or,

((1− i)x1 − x2

2(1− i)x1 − (1 + i)(1− i)x2

)=

(0

0

)Multiplying second row by 1− i

or,

((1− i)x1 − x2

2(1− i)x1 − (12 − i2)x2

)=

(0

0

)

or,

((1− i)x1 − x2

2(1− i)x1 − 2x2

)=

(0

0

)

or,

((1− i)x1 − x2

(1− i)x1 − x2

)=

(0

0

)

Thus each row of this vector equation leads to the condition (1 − i)x1 = x2. Let

x1 = c 6= 0 Then x2 = (1 − i)c. Hence, eigenvectors corrosponding to eigenvalue

λ = 2 are

x =

(x1

x2

)=

(c

c(1− i)

)= c

(1

1− i

), c 6= 0


Taking λ = 1− 2i in (5.59)

or,

(3− 1 + 2i −2

4 −1− 1 + 2i

)(x1

x2

)=

(0

0

)

or,

(2 + 2i −2

4 −2 + 2i

)(x1

x2

)=

(0

0

)

or,

((2 + 2i)x1 − 2x2

4x1 + (−2 + 2i)x2

)=

(0

0

)

or,

((1 + i)x1 − x2

2x1 + (−1 + i)x2

)=

(0

0

)

or,

((1 + i)x1 − x2

2(1 + i)x1 − (1 + i)(1− i)x2

)=

(0

0

)Multiplying second row by 1− i

or,

((1 + i)x1 − x2

2(1 + i)x1 − (12 − i2)x2

)=

(0

0

)

or,

((1 + i)x1 − x2

2(1 + i)x1 − 2x2

)=

(0

0

)

or,

((1 + i)x1 − x2

(1 + i)x1 − x2

)=

(0

0

)

Thus each row of this vector equation leads to the condition (1 + i)x1 = x2. Let

x1 = c 6= 0 Then x2 = (1 + i)c. Hence, eigenvectors corrosponding to eigenvalue

λ = 1− 2i are

x =

(x1

x2

)=

(c

c(1 + i)

)= c

(1

1 + i

), c 6= 0

Example 180.


A =

119 −2

989

−29

29

109

89

109

59


A =

119 −2

989

−29

29

109

89

109

59

=1

9

11 −2 8

−2 2 10

8 10 5



= A− λI

=1

9

11 −2 8

−2 2 10

8 10 5

− λ 1 0 0

0 1 0

0 0 1

=

1

9

11− 9λ −2 8

−2 2− 9λ 10

8 10 5− 9λ


|A− λI| = 0

or,1

9

∣∣∣∣∣∣∣11− 9λ −2 8

−2 2− 9λ 10

8 10 5− 9λ

∣∣∣∣∣∣∣ = 0

or, (11− 9λ)

∣∣∣∣∣ 2− 9λ 10

10 5− 9λ

∣∣∣∣∣+ 2

∣∣∣∣∣ 2 10

8 5− 9λ

∣∣∣∣∣+ 8

∣∣∣∣∣ −2 2− 9λ

8 10λ

∣∣∣∣∣ = 0

or, (11− 9λ) [(2− 9λ)(5− 9λ)− 100] + 2(10− 18λ− 80) + 8(−20λ− 16 + 72λ) = 0

or, − 729(λ3 − 2λ2 − λ+ 2 = 0

or, λ2(λ− 2)− 1(λ− 2) = 0

or, (λ− 2)(λ2 − 1) = 0

or, (λ− 2)(λ− 1)(λ+ 1) = 0

or, λ = 2, 1, −1

Eigenvalue of the matrix A are λ1 = 2, λ2 = 1 and λ3 = −1.

Let

x =

x1

x2

x3


equation is

(A− λI)x = 0

or,1

9

11− 9λ −2 8

−2 2− 9λ 10

8 10 5− 9λ

x1

x2

x3

=

0

0

0

(5.60)

or,

11− 9λ −2 8

−2 2− 9λ 10

8 10 5− 9λ

x1

x2

x3

=

0

0

0

(5.61)


Taking λ = 1 in (5.59) 11− 9 −2 8

−2 2− 9 10

8 10 5− 9

x1

x2

x3

=

0

0

0

or,

2 −2 8

−2 −7 10

8 10 4

x1

x2

x3

=

0

0

0

Applying R2 → R2 +R1, R3 → R3 − 4R1

or,

2 −2 8

0 −9 18

0 18 −36

x1

x2

x3

=

0

0

0

or,

2 −2 8

0 −9 18

0 0 0

x1

x2

x3

=

0

0

0

Using R3 → R2 +R3

The correponding system is

2x1 − 2x2 + 8x3 = 0

−9x2 + 18x3 = 0

If we take x2 = c then from second equation x2 = 2c and from first 2x1 = 4c − 8c

or, x1 = −2c.

x =

x1

x2

x1

=

−2c

2c

c

= c

−2

2

1

, c 6= 0

Similarly eigenvectors can be computed for the eigenvalues λ = −1, 2.

Example 181.


A =

0 1 1

1 0 1

1 1 0


A =

0 1 1

1 0 1

1 1 0



= A− λI

=

0 1 1

1 0 1

1 1 0

− λ 1 0 0

0 1 0

0 0 1

=

−λ 1 1

1 −λ 1

1 1 −λ


|A− λI| = 0

or,

∣∣∣∣∣∣∣−λ 1 1

1 −λ 1

1 1 −λ

∣∣∣∣∣∣∣ = 0

or, − λ

∣∣∣∣∣ λ 1

1 −λ

∣∣∣∣∣− 1

∣∣∣∣∣ 1 1

1 −λ

∣∣∣∣∣+ 1

∣∣∣∣∣ 1 −λ1 1

∣∣∣∣∣ = 0

or, − λ(−λ2 − 1)− (−λ− 1) + 1(1 + λ) = 0

or, − (λ3 − 2λ2 − λ+ 2 = 0

or, − λ3 + 3λ+ 2 = 0

or, λ3 − 3λ− 2

or,λ3 − 2λ2 + 2λ2 − 4λ+ λ− 2 = 0

or, λ2(λ− 2) + 2λ(λ− 2) + 1(λ− 2) = 0

or, (λ− 2)(λ2 + 2λ+ 1) = 0

or, (λ− 2)(λ+ 1)2 = 0

or, λ = 2, −1, −1

Eigenvalue of the matrix A are λ1 = 2, λ2 = 1 and λ3 = −1.

Let

x =

x1

x2

x3


equation is

(A− λI)x = 0

or,

−λ 1 1

1 −λ 1

1 1 −λ

x1

x2

x3

=

0

0

0

(5.62)


Taking λ = 2 in (5.62) −2 1 1

1 −2 1

1 1 −2

x1

x2

x3

=

0

0

0

or,

1 −2 1

−2 1 1

1 1 −2

x1

x2

x3

=

0

0

0

Applying R↔ R2

Applying R2 → R2 + 2R1, R3 → R3 −R1

or,

1 −2 1

0 −3 3

0 3 −3

x1

x2

x3

=

0

0

0

or,

1 −2 1

0 −3 3

0 3 −3

x1

x2

x3

=

0

0

0

or,

1 −2 1

0 −3 3

0 0 0

x1

x2

x3

=

0

0

0

Using R3 → R2 +R3

The correponding system is

x1 − 2x2 + x3 = 0

−3x2 + 3x3 = 0

If we take x3 = c then from second equation x2 = c and from first x1 = 2c − c or,

x1 = c.

x(1) =

x1

x2

x1

=

c

c

c

= c

1

1

1

, c 6= 0

Taking λ = 1 in (5.62) 1 1 1

1 1 1

1 1 1

x1

x2

x3

=

0

0

0

The correponding system is a single equation

x1 + x2 + x3 = 0 (5.63)

Thus values for two of the quantities x1, x2, x3 can be choosen arbitrarily, and the

third is determined from the equation (5.63). For example, if x1 = c1, x2 = c2. In

vector notation

x =

x1

x2

x1

=

c1

c2

−c1 − c2

= c1

1

0

−1

+ c2

0

1

−1

5.4. BASIC THEORYOF SYSTEMOF FIRST ORDER LINEAR EQUATIONS309

Thus, by choosing c1 = 1 and c2 = 0 we obtaine the eigenvector

x(2) =

1

0

−1

Again, by choosing c1 = 0 and c2 = 1 we obtaine the eigenvector

x(3) =

0

1

−1

x(2) and x(3) are linealy independent vectors. Therefore, in this example, two linearly

independent eigenvector are associated with the double eigenvalue.

5.4 Basic Theory of System of First Order Linear Equa-

tions

Let us considera system of n first order linear differential equations

x′1 = p11(t)x1 + p12(t)x2 + · · ·+ p1n(t)xn + g1(t)

x′2 = p21(t)x1 + p22(t)x2 + · · ·+ p2n(t)xn + g2(t)

... (5.64)

x′n = pn1(t)x1 + pn2(t)x2 + · · ·+ pnn(t)xn + gn(t)

The system (??) can be written asx′1x′2...

x′n

=

p11(t) p12(t) · · · p1n(t)

p21(t) p22(t) · · · p2n(t)...

......

...

pn1(t) pn2(t) · · · pnn(t)

x1

x2

...

xn

+

g1(t)

g2(t)...

gn(t)

We can write the system as

x′ = P(t)x + g(t) (5.65)

where

x′ =

x1

x2

...

xn

, P (t) =

p11(t) p12(t) · · · p1n(t)

p21(t) p22(t) · · · p2n(t)...

......

...

pn1(t) pn2(t) · · · pnn(t)

x =

x1

x2

...

xn

, g(t) =

g1(t)

g2(t)...

gn(t)


P (t) is called the coefficient matrix. The matrices A(t) and g(t) are said to be contin-

ious on some interval I = (α, β) if each of the scalar functions p11, p12, · · · , p1n, g1(t), · · · , gn(t)

are continous on that interval.

A vector x = φ(t) is said to be a solution of (5.65) if its components satisfy the

system of equations (5.64).

If g(t) = 0, then the system

x′ = P(t)x (5.66)

is called homogeneous equation, obtained from (5.65). Once the homogeneous equa-

tion has been solved, there are several methods that can be used to solve the non-

homogeneous equation (5.65)

We first use the following notations for a specific solutions of (5.66)

x(1) =

x11(t)

x21(t)...

xn1(t)

, x(2)(t) =

x12(t)

x22(t)...

xn2(t)

, etc · · · (5.67)

We state following some theorems without proof.

Theorem 16.

If the vector functions x(1) and x(2) are solutions of the system

x′ = P(t)x

then the linear combination c1x(1) + c2x

(2) is also solution for for any constants c1

and c2.

Thus, the theorem state that if x(1),x(2), · · · ,x(n) are solutions of the equation

x′ = P(t)x

Then

x = c1x(1) + c2x

(2) + · · ·+ cnx(n)

is also a solution for any contansts c1, c2, · · · , cn.

Let x(1),x(2), · · · ,x(n) be solutions of the equation

x′ = P(t)x

Let us consider the matrix X(t) whose columns are the vectors x(1),x(2), · · · ,x(n)

X(t) =

x11(t) x12(t) · · · x1n(t)

x21(t) x22(t) · · · x2n(t)...

......

...

xn1(t) xn2(t) · · · xnn(t)

(5.68)


The columns of X(t) are linearly independent for a given value of t if and only if

det X 6= 0 for all value of t ∈ (α, β). This determinat is called the Wronskian of the

n solutions x(1),x(2), · · · ,x(n) and is denoted by W [x(1),x(2), · · · ,x(n)]. i.e.

W [x(1),x(2), · · · ,x(n)] = det X(t) =

∣∣∣∣∣∣∣∣∣∣x11(t) x12(t) · · · x1n(t)

x21(t) x22(t) · · · x2n(t)...

......

...

xn1(t) xn2(t) · · · xnn(t)

∣∣∣∣∣∣∣∣∣∣(5.69)

The solutions x(1),x(2), · · · ,x(n) are linearly independent if and only ifW [x(1),x(2), · · · ,x(n)] 6=0.

Example 182.

If x(1),x(2), · · · ,x(n) are linearly independent solutions of the homogeneous sys-

tem

x′ = P(t)x

for each points in the interval (α, β) then each solution x = φ(t) of the system can

be expressed as a linear combination of x(1),x(2), · · · ,x(n) i.e

φ(t) = c1x(1) + c2x

(2) + · · ·+ cnx(n)

in exactly one way.

Theorem 17.

If x(1),x(2), · · · ,x(n) are solutions of

x′ = P(t)x

on the interval (α, β), then in this interval W [x(1),x(2), · · · ,x(n)] either is identically

zeros or else never vanishes.

Trace of a square matrix: Let

P (t) =

p11(t) p12(t) · · · p1n(t)

p21(t) p22(t) · · · p2n(t)...

......

...

pn1(t) pn2(t) · · · pnn(t)

be a square matrix. Then trace of matrix P (t) is denoted by trP (t) and is

tr(P (t)) = p11(t) + p22(t) + p33(t) + · · ·+ pnn(t)


The Wronskian of x(1),x(2), · · · ,x(n) satisfies the following relation

dW

dt= tr(P (t))W

or,dW

W= tr(P (t))dt

or,

∫dW

W=

∫tr(P (t))dt

or, lnW =

∫tr(P (t))dt+ ln c

or, W = ce∫

tr(P (t))dt

or, W = ce∫

[p11(t)+p22(t)+p33(t)+···+pnn(t)]dt

where c is an arbitrary constant.

Example 183.

Show that the Wronskians of two fundamental set of of solutions of the system

x′ = P(t)x

where

P (t) =

p11(t) p12(t) · · · p1n(t)

p21(t) p22(t) · · · p2n(t)...

......

...

pn1(t) pn2(t) · · · pnn(t)

can differe at most by a multiplicative constant.

Solution: Let W1(t) and W2(t) be two fundamental sets of solutions of

x′ = P(t)x

where

P (t) =

p11(t) p12(t) · · · p1n(t)

p21(t) p22(t) · · · p2n(t)...

......

...

pn1(t) pn2(t) · · · pnn(t)

Then

W1(t) = c1e∫

[p11(t)+p22(t)+p33(t)+···+pnn(t)]dt (5.70)

W2(t) = c2e∫

[p11(t)+p22(t)+p33(t)+···+pnn(t)]dt (5.71)

Then dividing (5.70) by (5.71), we get

W1 =c1

c2W2

Hence W1(t) and W2(t) are differe by only in the multiplicative constants.


Theorem 18.

Let

e(1) =

1

0...

0

, e(2) =

0

1...

0

, · · · , e(n) =

0

0...

1

further let x(1),x(2), · · · ,x(n) are solutions of

x′ = P(t)x

that satisfy the initial conditions

x(1)(t0) = e(1),x(2)(t0) = e(2), · · · ,x(n)(t0) = e(n)

respectively, where t0 is any point in (α, β). Then x(1),x(2), · · · ,x(n) form a funda-

mental set of solution of the

x′ = P(t)x

Example 184.

Show that the general solution of x′ = P(t)x + g(t) is sum of particular solution

xp this equation and the general solution xc of the corresponding homogeneous

equation.


x′ = P(t)x + g(t) (5.72)

Let xg be general solution of (5.72). Then

dx(g)

dt= P(t)x(g) + g(t) (5.73)

For xc to be any solution of the homogeneous system

x′ = P(t)x

Then

dx(c)

dt= P(t)x(c) (5.74)

Also, x(p) is any particular solution of (5.72). Then

dx(p)

dt= P(t)x(p) + g(t) (5.75)


Now, we shall show that x(g) = x(c) + x(p) is a solution of (5.72). For this adding

(5.74) and (5.74), and using linearity of derivative and matrix multiplication

dx(c)

dt+dx(p)

dt= P(t)x(c) + P(t)x(p) + g(t)

or,d

dt(x(c) + x(p)) = P(t)

(x(c) + x(p)

)+ g(t)

Thus, x(g) = x(c) + x(p) is a solution of (5.72).

Example 185.

Consider the vectors

x(1)(t) =

(t

1

), x(2)(t) =

(t2

2t

)

(a) Compute the Wronskian of x(1) and x(2).

(b) In what interval are x(1) and x(2) linearly independent?

(c) What conclusion can be drawn about the coefficients in the system of homoge-

neous differential equations satisfied by x(1) and x(2)?

(d)Find the system of lequations and verify the conclusions of the part (c).


x(1)(t) =

(t

1

), x(2)(t) =

(t2

2t

)

(a) The Wronkian is

W (x(1)(t),x(2)(t)) =∣∣∣ x(1) x(2)

∣∣∣ =

∣∣∣∣∣ t t2

1 2t

∣∣∣∣∣ = 2t2 − t2 = t2

(b) x(1) and x(2) are linearly independent at each point where W = t2 6= 0 for t 6= 0.

So the vectors are independent on every interval not containing t = 0.

(c) Since W (t) should either identically be zero or nowhere vanishes on any interval

α < t < β on which the matrix is continuous. Also, as W (t) = 0 for t = 0 and

W (t) 6= 0 for t 6= 0, there should be at least one coefficcient in the matrix P (t) of the

homogeous system x′ = P(t)x satisfied by x(1) and x(2). Which are discontinuous

at t = 0.

(d) Let

P (t) =

(a11 a12

a21 a22

)Let the system of differential equations be

x′ = P(t)x (5.76)


Taking x =

(t

1

), x′ =

(1

0

)and P (t) =

(a11 a12

a21 a22

)in (5.76),

(1

0

)=

(a11 a12

a21 a22

)(t

1

)a11t+ a12 = 1 (5.77)

a21t+ a22 = 0 (5.78)

Again, taking x =

(t2

2t

), x′ =

(2t

2

)and P (t) =

(a11 a12

a21 a22

)in (5.76),

(2t

2

)=

(a11 a12

a21 a22

)(t2

2t

)a11t

2 + 2a12t = t2 =⇒ a11t+ 2a12 = t (5.79)

a21t2 + a22t = 2t =⇒ a21t+ a22 = 2 (5.80)

Solving (5.77), (5.78), (5.79) and (5.80) we get a11 = 0, a12 = 1, a21 = − 2t2

and

a22 = 2t . So the corresponding system

x′ =

(0 1

− 2t2

2t

)x

Both the coefficients in the second row are discontinuous at t = 0. So, the conclusion

of (c) is verified.

Example 186.

Consider the vectors

x(1)(t) =

(t

1

), x(2)(t) =

(et

et

)

(a) Compute the Wronskian of x(1) and x(2).

(b) In what interval are x(1) and x(2) linearly independent?

(c) What conclusion can be drawn about the coefficients in the system of homoge-

neous differential equations satisfied by x(1) and x(2)?

(d)Find the system of lequations and verify the conclusions of the part (c).


x(1)(t) =

(t

1

), x(2)(t) =

(et

et

)(a) The Wronkian is

W (x(1)(t),x(2)(t)) =∣∣∣ x(1) x(2)

∣∣∣ =

∣∣∣∣∣ t et

1 et

∣∣∣∣∣ = tet − et = et(t− 1)


(b) x(1) and x(2) are linearly independent at each point where W = et(t− 1) 6= 0 for

t 6= 1. So the vectors are independent on every interval not containing t = 1.

(c) Since W (t) should either identically be zero or nowhere vanishes on any interval

α < t < β on which the matrix is continuous. Also, as W (t) = 0 for t = 1 and

W (t) 6= 0 for t 6= 1, there should be at least one coefficcient in the matrix P (t) of the

homogeous system x′ = P(t)x satisfied by x(1) and x(2). Which are discontinuous

at t = 1.

(d) Let

P (t) =

(a11 a12

a21 a22

)Let the system of differential equations be

x′ = P(t)x (5.81)

Taking x =

(t

1

), x′ =

(1

0

)and P (t) =

(a11 a12

a21 a22

)in (5.81),

(1

0

)=

(a11 a12

a21 a22

)(t

1

)a11t+ a12 = 1 (5.82)

a21t+ a22 = 0 (5.83)

Again, taking x =

(et

et

), x′ =

(et

et

)and P (t) =

(a11 a12

a21 a22

)in (5.81),

(et

et

)=

(a11 a12

a21 a22

)(et

et

)a11e

t + a12et = et =⇒ a11 + a12 = 1 (5.84)

a21et + a22e

t = et =⇒ a21t+ a22 = 1 (5.85)

Solving (5.82), (5.83), (5.84) and (5.85) we get a11 = 0, a12 = 1, a21 = 11−t and

a22 = − t1−t . So the corresponding system

x′ =

(0 11

1−t −t

1−t

)x

Both the coefficients in the second row are discontinuous at t = 1. So, the conclusion

of (c) is verified.

Chapter 6

Differential Equation of First

Order but not of the First

degree

Introduction :

In this unit, we will consider the differential equations which are of first order but

are of degree higher than one.The most general form of a differential equation of

first order not of first degree is i.e. of n degree is(dy

dx

)n+ P1

(dy

dx

)n−1

+ P2

(dy

dx

)n−2

+ · · ·Pn−1

(dy

dx

)+ Pn = 0 (6.1)

where P1, P2, · · · , Pn−1, Pn are functions of x and y. Let p = dydx . Then the equation

(6.1) becomes

pn + P1pn−1 + P2p

n−2 + · · ·+ Pn−1p+ Pn = 0 (6.2)

This equation can be written as

f(x, y, p) = 0

The above equation can not be solved in general form. Here we discuss five special

types of such equations

1. Solvable for p

2. Solvable for y

3. Solvable for x

4. Homogeous in x and y.

5. Clairaut’s Equation.

6.1 Solvable for p

We have the first order differential equation higher order is

f(x, y, p) = 0 (6.3)

317

318CHAPTER 6. DIFFERENTIAL EQUATION OF FIRST ORDER BUT NOTOF THE FIRST DEGREE

Let the differential equation (6.3) can be solved for p and is of the equation can be

written as

[p− f1(x, y)][p− f2(x, y)] · · · [p− fn(x, y)] = 0 (6.4)

Equating each factor to zeros we get equations of the first order and the first degree.

Solving each equations we get solutions as

φ1(x, y, c1) = 0, φ2(x, y, c2) = 0, · · · , φn(x, y, cn) = 0

Without any loss of generality, we can write

c1 = c2 = · · · cn = c

Thus the solutions becomes

φ1(x, y, c) = 0, φ2(x, y, c) = 0, · · · , φn(x, y, c) = 0

Therefore, the solution of the equation (6.3) can be put in the form

φ1(x, y, c)φ2(x, y, c)φ3(x, y, c) · · ·φn(x, y, c) = 0

Example 187.

Solve (p− xy)(p− x2)(p− y2) = 0

Solution: The given equation is

(p− xy)(p− x2)(p− y2) = 0

Equating each factor with zero

p− xy = 0 (6.5)

or, p− x2 = 0 (6.6)

or, p− y2 = 0 (6.7)

From (6.5), we get

p− xy = 0

or, p = xy

or,dy

dx= xy

or,dy

y= xdx

or,

∫dy

y=

∫xdx

or, ln y =x2

2+ c

or, ln y − x2

2− c = 0 (6.8)

6.1. SOLVABLE FOR P 319

From (6.6), we get

p− x2 = 0

or, p = x2

or,dy

dx= x2

or, dy = x2dx

or,

∫dy =

∫x2dx

or, y =x3

3+ c

or, y − x3

3− c = 0 (6.9)

From (6.7), we get

p− y2 = 0

or, p = y2

or,dy

dx= y2

or,dy

y2= dx

or,

∫dy

y2=

∫dx

or, − 1

y= x+ c

or,1

y+ x+ c = 0 (6.10)

From (6.8), (6.9) and (6.10) the combined solution is(ln y − x2

2− c)(

y − x3

3− c)(

1

y+ x+ c

)= 0

Example 188.

Solve (p+ y + x)(xp+ y + x)(p+ 2x) = 0


(p+ y + x)(xp+ y + x)(p+ 2x) = 0


p+ y + x = 0 (6.11)

or, xp+ y + x = 0 (6.12)

or, p+ 2x = 0 (6.13)


From (6.11), we get

p+ y + x = 0

or,dy

dx+ y = −x (6.14)

(6.15)

which is linear. I.F.= e∫dx = ex. Multiplying the equation (6.14) by I.F., we get

exdy

dx+ exy = −xex

or,d

dx(exy) = −xex

or, d(exy) = −xexdx

Integrating, ∫d(exy) = −

∫xexdx+ c

or, exy = −[x

∫exdx−

∫ (dx

dx

∫exdx

)dx

]+ c

or, exy = −[xex −

∫exddx

]+ c

or, exy = −xex + ex + c

or, exy + xex − ex − c = 0

or, x+ y − 1− ce−x = 0 (6.16)

From (6.12), we get

xp+ y + x = 0

or, xdy

dx+ y = −x

or,d

dx(xy) = −x

or,d(xy) = −xdx

Integrating,

xy = −x2

2+ c

or, x2 + 2xy − c = 0 (6.17)

From (6.13), we get

p+ 2x = 0

or,dy

dx+ 2x = 0

or, dy + 2xdx = 0

or,

∫dy + 2

∫xdx = c

or, y + x2 − c = 0 (6.18)


From (6.16), (6.17) and (6.18), the combined solution is(x+ y − 1− ce−x

) (x2 + 2xy − c

) (y + x2 − c

)= 0

Example 189.

Solve p2 − 7p+ 12 = 0


p2 − 7p+ 12 = 0

or, p2 − 3p− 4p− 12 = 0

or, p(p− 3)− 4(p− 3) = 0

or, (p− 3)(p− 4) = 0


p− 3 = 0 (6.19)

or, p− 4 = 0 (6.20)

From (6.19), we get

dy

dx− 3 = 0

or, dy − 3dx = 0

∴ y − 3x− c = 0 (6.21)

From (6.20), we get

dy

dx− 4 =

or, dy − 4dx = 0

∴ y − 4x− c = 0 (6.22)

Thus, the combined solution is

(y − 3x− c)(y − 4x− c) = 0

Example 190.

Solve p2 + 2py cotx = y2


p2 + 2py cotx− y2 = 0


Solving for p

p =−2y cotx±

√4y2 cot2 +4y2

2

=−2y cotx± 2y cscx

2

= y

(−cosx

sinx± 1

sinx

)= y

(−cosx

sinx+

1

sinx

)or y

(−cosx

sinx− 1

sinx

)= y

(1− cosx

sinx

)or − y

(1 + cosx

sinx

)= y

(2 sin2 x/2

2 sinx/2 cosx/2

)or − y

(2 cos2 x/2

2 sinx/2 cosx/2

)dy

dx= y tan

x

2or − y cot

x

2

Thus, we get the following equations

dy

dx= y tan

x

2(6.23)

dy

dx= −y cot

x

2(6.24)


dy

y= tan

x

2dx

or,

∫dy

y=

∫tan

x

2dx

or, ln y = 2 ln secx

2+ lnB

or, ln y = ln(B sec2 x

2

)or, y = B sec2 x

2

or, y cos2 x

2= B

or, y(1 + cosx) = 2B = c

or, y(1 + cosx)− c = 0 (6.25)



dy

y= − cot

x

2dx

or,

∫dy

y= −

∫cot

x

2dx

or, ln y = −2 ln sinx

2+ lnA

or, ln y = − ln(

sin2 x

2

)+ lnA

or, y =A

sin2 x/2

or, y sin2 x

2= A

or, y(1− cosx) = 2A = c

or, y(1− cosx)− c = 0 (6.26)

From (6.25) and (6.26), the combined solution is

[y(1 + cosx)− c][(1− cosx)− c] = 0

Example 191.

Solve x2p2 + xyp− 6y2 = 0


x2p2 + xyp− 6y2 = 0

or, x2p2 + 3xyp− 2xyp− 6y2 = 0

or, xp(xp+ 3y)− 2y(xp+ 3y) = 0

or, (xp− 2y)(xp+ 3y) = 0

or, p =2y

x(6.27)

p = −3y

x(6.28)

From (6.27)

dy

dx=

2y

x

or,dy

y=

2dx

x

or,

∫dy

y= 2

∫dx

x+ lnc

or, ln y = 2 lnx+ ln c

or, ln y = ln(cx2)

or, y = cx2

or, y − cx2 = 0 (6.29)


From (6.28)

dy

dx= −3y

x

or,dy

y= −3dx

x

or,

∫dy

y= −3

∫dx

x+ lnc

or, ln y = −3 lnx+ ln c

or, ln y = ln(cx−3)

or, y = cx−3

or, yx3 = c

or, yx3 − c = 0 (6.30)

From (6.29) and (6.30), the combined solution is

(y − cx2)(yx3 − c) = 0

Example 192.

Solve xy2(p2 + 2) = 2py3 + x3


xy2(p2 + 2)− 2py3 − x3

or, xy2p2 − x3 − 2py3 + 2xy2 = 0

or, x(y2p2 − x2)− 2y2(yp− x) = 0

or, x(yp+ x)(yp− x)− 2y2(yp− x) = 0

or,(yp− x)(xyp− x2 − 2y2) = 0

Thus

yp− x = 0 (6.31)

Or,

xyp− x2 − 2y2 = 0 (6.32)

From (6.31),

ydy

dx− x = 0

or, ydy − xdx = 0

or,y2

2− x2

2=c

2or, y2 − x2 − c = 0 (6.33)


Again, from (6.32)

xydy

dx= 2y2 − x2

or,dy

dx=

2y2 − x2

xy(6.34)

Which is homogeous equation. Let us put y = vx. Then

dy

dx= x

dv

dx+ v

Now from (6.34)

xdv

dx+ v =

2v2x2 − x2

xvx=

2v2 − 1

v

or, xdv

dx=

2v2 − 1

v− v =

v2 − 1

v

or,2v

v2 − 1dv = 2

dx

x

∴∫

2v

v2 − 1dv = 2

∫dx

x+ ln c

or, ln(v2 − 1) = 2 lnx+ ln c

or, ln(v2 − 1) = ln(cx2)

or, v2 − 1 = cx2

or,y2

x2− 1 = cx2

or, y2 − x2 − cx4 = 0 (6.35)

Combined the solutions (6.34) and (6.35)

(y2 − x2 − c)(y2 − x2 − cx4) = 0

Example 193.

Solve p3 + 2xp2 − y2p2 − 2xy2p = 0.


p3 + 2xp2 − y2p2 − 2xy2p = 0

or, p2(p− y2) + 2xp(p− y2) = 0

or, (p− y2)(p2 + 2xp) = 0

or, p(p− y2)(p+ 2x) = 0

Therefore,

p = 0 (6.36)


or,

p+ 2x = 0 (6.37)

p− y2 = 0 (6.38)

From (6.36)

dy

dx= 0

or, dy = 0

or, y = c

or, (y − c) = 0 (6.39)

From (6.37), we get

p+ 2x = 0

or,dy

dx+ 2x = 0

or, dy + 2xdx = 0

or,

∫dy + 2

∫xdx = c

or, y + x2 − c = 0 (6.40)

From (6.38), we get

p− y2 = 0

or, p = y2

or,dy

dx= y2

or,dy

y2= dx

or,

∫dy

y2=

∫dx

or, − 1

y= x+ c

or,1

y+ x+ c = 0 (6.41)

Thus, from (6.39), (6.40) and (6.41) the combined solution is

(y − c)(y + x2 − c)(1

y+ x+ c) = 0

Example 194.


Slove x2p3 + y(1 + x2)p2 + y3p = 0.


x2p3 + y(1 + x2y)p2 + y3p = 0

or, p(x2p2 + y(1 + x2y)p+ y3) = 0

or, pp(x2p+ y) + y2(x2p+ y) = 0

or, p(p+ y2)(x2p+ y) = 0

Therefore,

p = 0 (6.42)

or,

p+ y2 = 0 (6.43)

x2p+ y = 0 (6.44)

From (6.42)

dy

dx= 0

or, dy = 0

or, y = c

or, (y − c) = 0 (6.45)

From (6.43), we get

p+ y2 = 0

or, p = −y2

or,dy

dx= −y2

or,dy

y2= −dx

or,

∫dy

y2= −

∫dx

or, − 1

y= −x+ c

or, x− 1

y− c = 0 (6.46)

From (6.44), we get

x2p+ y = 0

or, x2 dy

dx+ y = 0

or,dy

dx+

1

x2y = 0


which is linear and multiplying by integrating factor

I.F. = e∫

1x2dx = e−

1x

e−1xdy

dx+ e−

1x

1

x2y = 0

or,d

dx

(ye−

1x

)= 0

or, ye−1x = c

or, ye−1x − c = 0

Thus, from (6.39), (6.40) and (6.41) the combined solution is

(y − c)(x− 1

y− c)

(ye−

1x − c

)= 0

Example 195.

Solve yp2 + (x− y)p− x = 0


yp2 + (x− y)p− x = 0

or, yp2 + xp− yp− x = 0

or, p(yp+ x)− 1(yp+ x) = 0

or, (p− 1)(yp+ x) = 0

Therefore,

p = 1 (6.47)

or,

yp+ x = 0 (6.48)

From (6.47), we get

dy

dx− 1 = 0

or, dy − dx = 0

∴ y − x− c = 0 (6.49)

From (6.48), we get

ydy

dx+ x = 0

or, ydy + xdx = 0

∴y2

2+x2

2=c

2or,(x2 + y2 − c) = 0 (6.50)

Required solution is , form (6.49) and (6.50)

(y − x− c)((x2 + y2 − c) = 0

6.2. EQUATION SOLVABLE FOR Y 329

6.2 Equation Solvable for y

If the differential equation g(x, y, p) = 0 is solvable for y, then

y = f(x, p) (6.51)

Differentiating with respect to x, gives

p =dy

dx= φ

(x, p,

dp

dx

)(6.52)

which is an equation in two variables x and p, and it will give rise to a solution of

the form

F (x, p, c) = 0 (6.53)

The elimination of p between equations (6.51) and (6.53) gives a relation between

x, y and c, which is required solution. When the elimination of p between these

equation is not easily done, the values of x and y in term of p can be found, and

these together will constitute the required solution.

Example 196.

Solve y + px = x4p2


y = −px+ x4p2 (6.54)


p =dy

dx= −p− xdp

dx+ 4x3p2 + 2x4p

dp

dx

or, 2p− 4x3p2 = −xdpdx

+ 2x4pdp

dx

or, 2p(1− 2x3p) = −x(1− 2x3p)dp

dx

or, 2p = −xdpdx

or,dp

p+ 2

dx

x= 0

Integrating,

ln p+ 2 lnx = ln c

or, ln(x2p) = ln c

or, x2p = c

or, p =c

x2(6.55)

Substituting the value of p from (6.55) to (6.54)

y = − cx

+ c2

or, xy + c = c2x


Example 197.

Solve y = sin p− p cos p


y = sin p− p cos p (6.56)


p =dy

dx= cos p

dp

dx− cos p

dp

dx+ p sin p

dp

dx

or, p = p sin pdp

dxor, sin pdp = dx

Integrating,

− cos p+ c = x

or cos p = c− x (6.57)

From (6.56)

p cos p = sin p− y

or, p =sin p− y

cos p=

√1− cos2 p− y

cos p

or, p =

√1− (c− x)2 − y

c− x

or, p =

√1− c2 + 2cx− x2 − y

c− xPutting the value of p in (6.57)

c− x = cos

(√1− c2 + 2cx− x2 − y

c− x

)Example 198.

Solve y = yp2 + 2px


y = yp2 + 2px (6.58)

or, y =2px

1− p2(6.59)


dy

dx= p2 dy

dx+ 2yp

dp

dx+ 2p+ 2x

dp

dx

or, p = p3 + 2ypdp

dx+ 2p+ 2x

dp

dx

or, 0 = p3 + p+ 2ypdp

dx+ 2x

dp

dx

6.2. EQUATION SOLVABLE FOR Y 331

Putting the value of y from (6.59)

0 = p3 + p+ 2p2px

1− p2

dp

dx+ 2x

dp

dx

or, 0 = p3(1− p2) + p(1− p2) + 4p2xdp

dx+ 2x(1− p2)

dp

dx

or, 0 = p3 − p5 + p− p3 + (4p2x+ 2x− 2xp2)dp

dx

or, 0 = p(1− p4) + (2p2x+ 2x)dp

dx

or, 0 = p(1− p2)(1 + p2) + 2x(p2 + 1)dp

dx

or, 0 = p(1− p2) + 2xdp

dx

or,2dp

p(p− 1)(p+ 1)=dx

x(6.60)

Let

2

p(p− 1)(p+ 1)=A

p+

B

p− 1+

C

p+ 1(6.61)

or, 2 = (p− 1)(p+ 1)A+Bp(p+ 1) + Cp(p− 1) (6.62)

Putting p = 0 in (6.62) we get A = −2.

Putting p = 1 in (6.62) we get B = 1.

Putting p = −1 in (6.62) we get C = 1. Substituting the values of A,B and C in

(6.61)

2

p(p− 1)(p+ 1)= −2

p+

1

p− 1− 1

p+ 1(6.63)

Again, from (6.60)

−2dp

p+

dp

p− 1+

dp

p+ 1=dx

x

Integrating,

−2 ln p+ ln(p− 1) + ln(p+ 1) = lnx+ ln c

or, lnp2 − 1

p2= ln cx

or,p2 − 1

p2= cx

or, p2 − 1 = cxp2

or, p =1√

1− cx


From (6.58)

y =y

1− cx+

2√1− cx

x

or, y − cxy = y + 2x√

1− cxor, cxy + +2x

√1− cx = 0

6.3 Equation Solvable for x

If the differential equation g(x, y, p) = 0 is solvable for y, then

x = f(y, p) (6.64)

Differentiating with respect to y, gives

1

p=dx

dy= φ

(y, p,

dp

dx

)(6.65)

which is an equation in two variables y and p, and it will give rise to a solution of

the form

F (y, p, c) = 0 (6.66)

The elimination of p between equations (6.64) and (6.66) gives a relation between

x, y and c, which is required solution. When the elimination of p between these

equation is not easily done, the values of x and y in term of p can be found, and

these together will constitute the required solution.

Example 199.

Solve y2 ln y = xyp+ p2


y2 ln y = xyp+ p2 (6.67)

or, x =1

py ln y − p

y(6.68)

6.3. EQUATION SOLVABLE FOR X 333


dx

dy=

d

dy

(1

py ln y

)− d

dy

(p

y

)or,

1

p=p ddy (y ln y)− y ln y dpdy

p2−y dpdy − py2

or,1

p=p(1 + ln y)− y ln y dpdy

p2−y dpdy − py2

or,1

p=py2(1 + ln y)− y3 ln y dpdy − y

2pdpdy + p3

p2y2

or, py2 = py2(1 + ln y)− y3 ln ydp

dy− p2y

dp

dy+ p3

or, py2 = py2 + py2 ln y − y3 ln ydp

dy− p2y

dp

dy+ p3

or, 0 = p3 + py2 ln y − y3 ln ydp

dy− p2y

dp

dy

or, 0 = p(p2 + y2 ln y)− y(y2 ln y + p2

) dpdy

or, 0 = (p2y2 ln y)

(p− y dp

dy

)or,0 = p− y dp

dy

or,dp

p=dy

y

(6.69)

Integrating,

ln p = lnx+ ln c

or, ln p = ln(cx)

or, p = cx (6.70)

Substituting the value of p from (6.70) to (6.68)

ln y = cx+ c2

Example 200.

Solve xp3 = a+ bp


xp3 = a+ bp (6.71)

or, x =a

p3+

b

p2= ap−3 + bp−2 (6.72)



1

p= −3a

p4

dp

dy− 2b

p3

dp

dy

or, dy +3a

p3dp+

2b

p2dp = 0

Integrating,

y − 3a

2p2− 2b

p= c

y =3a

2p2+

2b

p+ c (6.73)

Hence eliminating, p from

x =a

p3+

b

p2

y =3a

2p2+

2b

p+ c

we get required solution.

Example 201.

Solve x = y + a ln p


x = y + a ln p (6.74)


1

p= 1 +

a

p

dp

dy

or,1− pp

=a

p

dp

dy

or, dy =a

1− pdp

Integrating,

y = c− a ln(1− p)

Also from (6.74)

x = c− a ln(1− p) + a ln p (6.75)

Hence eliminating, p from

x = c− a ln(1− p) + a ln p

y = c− a ln(1− p)


6.4. EQUATION HOMOGENEOUS IN X AND Y 335

6.4 Equation Homogeneous in x and y

When the equation is homogeneous in x and y, it can be written as

F

(dy

dx,y

x

)= 0

Then it is possible to solve it for dydx and solve it for y

x and obtained

y = xφ(p) (6.76)

Differentiating, (6.76) with respect to x, we get

p = φ(p) + xφ′(p)dp

dx,

dx

x=

φ′(p)

p− φ(p),

which is variable separated form.

Example 202.

Solve x2p2 − 2xyp+ 2y2 − x2 = 0.


x2p2 − 2xyp+ (2y2 − x2) = 0

Solving for p

p =2xy ±

√4x2y2 − 4x2(2y2 − x2)

2x2

or, p =2xy ± 2x

√y2 − (2y2 − x2)

2x2

or, p =y ±

√x2 − y2

x

or,dy

dx=y

x±√

1−(yx

)2(6.77)

which is homogeous equaation. Let y = vx. Then

dy

dx= x

dv

dx+ v


xdv

dx+ v = v ±

√1− v2

or,dv√

1− v2= ±dx

x


Integrating, we get ∫dv√

1− v2= ±

∫dx

x+ c

or, sin−1(v) = ± lnx+ ln c

Taking positive sign,

sin−1(v) = lnx+ ln c

or, sin−1 v = ln(cx)

or, v = sin ln cx

or, y = x sin ln cx

or, y − x sin ln cx = 0

Taking negative sign,

sin−1(v) = − lnx+ ln c

or, sin−1 v = ln(c/x)

or, v = sin ln c/x

or, y = x sin ln c/x

or, y − x sin ln c/x = 0


(y − x sin ln cx)(y − x sin ln

c

x

)= 0

6.5 Clairaut’s Equation

A first order differential equation of the form

y = px+ f(p) (6.78)

where p = dydx is called Clairaunt’s equation.

To find the solution of (6.78), differentiating equation (6.78) with respect to x,

dy

dx= p+ x

dp

dx+ f ′(p)

dp

dx

or,p = p+ xdp

dx+ f ′(p)

dp

dx

or,0 = xdp

dx+ f ′(p)

dp

dx

or,(x+ f ′(p))dp

dx= 0

Thus,

dp

dx= 0 (6.79)

6.5. CLAIRAUT’S EQUATION 337

Or,

x+ f ′(p) = 0 (6.80)

From (6.79), integrating

p = c

Putting the value of p = c in (6.78), we get

y = px+ f(p) (6.81)

is the complete solution.

If we eliminate p from

y = px+ f(p)

and

x+ f ′(p) = 0

we get a solution, which ia does not contains any arbitrary constant . This solution

is called a singular solution.

Example 203.

Find the general solution of (y − px)(p− 1) = p.


(y − px)(p− 1) = p

or, y = px+p

p− 1(6.82)

Differentiating both sides with respect to x,

p = p+ xdp

dx+

(p− 1) dpdx − pdpdx

(p− 1)2

or,

(x− 1

(p− 1)2

)dp

dx= 0

Therefore,dp

dx= 0

Integrating,

p = c

Putting the value of p in (6.82), we get

(y − cx)(c− 1) = c

Example 204.


Find the general solution of p = ln(px− y).


p = ln(px− y) (6.83)

or, ep = px− yor, y = px− ep


p = p+ xdp

dx− ep dp

dx

or, (x− ep) dpdx

= 0

Therefore,dp

dx= 0

Integrating,

p = c


c = ln(cx− y)

Example 205.

Find the general solution of sin px cos y = cos px sin y + p.


sin px cos y = cos px sin y + p (6.84)

or, sin px cos y − cos px sin y = p

or, sin(px− y) = p

or, px− y = sin−1 p

or, y = px− sin−1 p


p = p+ xdp

dx− 1√

1− p2

dp

dx

or,

(x− 1√

1− p2

)dp

dx= 0

Therefore,dp

dx= 0

Integrating,

p = c


sin cx cos y = cos cx sin y + c


Example 206.

Find the general and singular solution of y = px+ ap .


y = px+a

p(6.85)


p = p+ xdp

dx− a

p2

dp

dx

or,

(x− a

p2

)dp

dx= 0

Thus,

dp

dx= 0 (6.86)

Or,

x− a

p2= 0 (6.87)

Therefore, from (6.86)dp

dx= 0

Integrating,

p = c


y = cx+a

c

which is general soution.

Again, from (6.87)

x− a

p2= 0

or, x =a

p2

or, p =

√a

x

Again, from (6.85)

y =

√a

xx+

a√ax

or, y =√ax+

√ax

or, y2 = 4ax

which is singular solution.


Example 207.

Find the general and singular solution of y = px+√a2p2 + b2.


y = px+√a2p2 + b2 (6.88)


p = p+ xdp

dx+

2a2p

2√

1− p2

dp

dx

or,

(x+

a2p√a2p2 + b2

)dp

dx= 0

Thus,

dp

dx= 0 (6.89)

Or,

x+a2p√

a2p2 + b2= 0 (6.90)

Therefore, from (6.89)dp

dx= 0

Integrating,

p = c


y = cx+√a2c2 + b2

which is general soution.

Again, from (6.90)

x+a2p√

a2p2 + b2= 0

or, x = − a2p√a2p2 + b2

(6.91)

or,x

a= − ap√

a2p2 + b2

Again, from (6.88)

y = − a2p2√a2p2 + b2

+√a2p2 + b2 =

b2√a2p2 + b2

y

b=

b√a2p2 + b2

(6.92)


Squaring and adding (6.92) and (6.92)

x2

a2+y2

b2=

a2p2

a2p2 + b2+

b2

a2p2 + b2

or,x2

a2+y2

b2= 1

which is singular solution.

6.5.1 Reducible to Clairaut’s Form

By using appropriate substitutions, some differential equations can be reduced to

Clairaunt’s form.

Example 208.

Reduce xyp2 − (x2 + y2 − 1)p+ xy = 0 to Clairaunt’s form and find its singular

solution.


xyp2 − (x2 + y2 − 1)p+ xy = 0 (6.93)

Let x2 = u and y2 = v. Then

du = 2xdx, dv = 2ydy

∴dv

du=y

x

dy

dx=

√v√up

or, p =

√u√v

dv

du(6.94)

Now, from (6.93)

√u√vu

v

(dv

du

)2

− (u+ v − 1)

√u√v

dv

du+√u√v = 0

or, u

(dv

du

)2

− (u+ v − 1)dv

du+ v = 0

or, uP 2 − (u+ v − 1)P + v = 0 where P =dv

duor, uP 2 − uP − vP + P + v = 0

or, v(1− P )− uP (1− P ) + P = 0

or, v(1− P ) = uP (1− P )− P

or, v = uP +P

P − 1(6.95)


which is of Clairaut’s form. Differentiating it with respect to u

dv

du=

d

du(uP ) +

d

du

(P

P − 1

)or, P = P + u

dP

du+

(P − 1)dPdu − Pd(P−1)du

(P − 1)2

or, 0 = udP

du+

(1− P )dPdu − PdPdu

(P − 1)2

or, udP

du+

(P − 1− P(P − 1)2

)dP

du= 0

or, udP

du− 1

(P − 1)2

dP

du= 0

or,

(u− 1

(P − 1)2

)dP

du= 0

(6.96)

To get singular solution, we consider

u− 1

(P − 1)2= 0

or, P − 1 =1√u

or, P = 1 +1√u

(6.97)

Putting the value of P in (6.95)

v = u

(1 +

1√u

)+

(1 + 1√

u

)1 + 1√

u− 1

or, v = u+√u+√u+ 1

or, v = u+ 2√u+ 1

or, y2 = x2 + 2x+ 1

or, y2 = (x+ 1)2

which is required singular solution.

Example 209.

Solve (px− y)(py + x) = h2p.


(px− y)(py + x) = h2p = 0

or, xyp2 + (x2 − y2 − h2)p− xy = 0 (6.98)




∴dv

du=y

x

dy

dx=

√v√up

or, p =

√u√v

dv

du(6.99)

Now, from (6.93)

√u√vu

v

(dv

du

)2

+ (u− v − h2)

√u√v

dv

du−√u√v = 0

or, u

(dv

du

)2

+ (u− v − h2)dv

du− v = 0

or, uP 2 + (u− v − h2)P − v = 0 where P =dv

duor, uP 2 + uP − vP − h2P − v = 0

or, − v(1 + P ) + uP (1 + P )− h2P = 0

or, v(1 + P ) = uP (1 + P )− h2P

or, v = uP − h2P

P + 1(6.100)

which is of Clairaut’s form. Differentiating it with respect to u

dv

du=

d

du(uP )− h2 d

du

(P

P + 1

)or, P = P + u

dP

du− h2 (P + 1)dPdu − P

d(P+1)du

(P + 1)2

or, 0 = udP

du− h2 (P + 1)dPdu − P

dPdu

(P + 1)2

or, udP

du− h2

(P + 1− P(P + 1)2

)dP

du= 0

or, udP

du− h2

(P + 1)2

dP

du= 0

or,

(u− h2 1

(P + 1)2

)dP

du= 0

(6.101)

For general solution, we consider

dP

du= 0 =⇒ P = c

Hence, from (6.100)

v = cu− h2c

c+ 1

or, y2 = cx2 − h2c

c+ 1


which is general solution. To get singular solution, we consider

u− h2 1

(P + 1)2= 0

or, P + 1 =h√u

or, P =h√u− 1

(6.102)

Putting the value of P in (6.100)

v = u

(h√u− 1

)− h2

(h√u− 1)

h√u− 1 + 1

or, v = h√u− u− h(h−

√u)

or, v = 2h√u− u− h2

or, y2 = 2hx− x2 − h2


Example 210.

Solve y = 2px+ y2p3.


y = 2px+ y2p3

or, y2 = 2pxy + y3p3 = 0 (6.103)

Let y2 = v. Then differentiating both sides with respect to x

dv

dx= 2y

dy

dx= 2√vp

or, p =1

2√v

dv

dx=

1

2√vP where P =

dv

dx

Now from (6.103)

v = 2√vx

1

2√vP + v

√v

(1

2√vP

)3

or, v = Px+1

8P 3

which Clairaut’s form and the solution is

v = cx+1

8c3

or, y2 = cx+1

8c3


Example 211.

Solve y = 2px+ yp2.


y = 2px+ yp2

or, y2 = 2pxy + y2p2 = 0 (6.104)

Let y2 = v. Then differentiating both sides with respect to x

dv

dx= 2y

dy

dx= 2√vp

or, p =1

2√v

dv

dx=

1

2√vP where P =

dv

dx

Now from (6.103)

v = 2√vx

1

2√vP + v

(1

2√vP

)2

or, v = Px+1

4P 2

which Clairaut’s form and the solution is

v = cx+1

4c2

or, y2 = cx+1

4c2

Example 212.

Reduce axyp2 + (x2 − ay2 − b)p − xy = 0 to Clairaunt’s form and solve the

equation.


axyp2 + (x2 − ay2 − b)p− xy = 0 (6.105)



∴dv

du=y

x

dy

dx=

√v√up

or, p =

√u√v

dv

du(6.106)


Now, from (6.93)

a√u√vu

v

(dv

du

)2

+ (u− av − b)√u√v

dv

du−√u√v = 0

or, au

(dv

du

)2

+ (u− av − b)dvdu− v = 0

or, auP 2 + (u− av − b)P − v = 0 where P =dv

duor, auP 2 + uP − avP − bP − v = 0

or, − v(aP + 1) + uP (aP + 1)− bP = 0

or, v(aP + 1) = uP (aP + 1)− bP

or, v = uP − bP

aP + 1(6.107)

which is of Clairaut’s form. The general solution is

v = uc− bc

ac+ 1

or, y2 = cx2 − bc

ac+ 1

Chapter 7

Partial Differential Equations of

the first order

7.1 Definition and Examples

Equations which contain one or more partial derivatives are called Partial differen-

tial equations.

Such equations arise in geometry, engineering and physics when the number of in-

dependent variables in the problem under consideration are two or more. Whenever

we consider case of two independent variables, x and y usually be taken as the inde-

pendent variables and z as the dependent variable. We adopt the following notations

through out the text.

p =∂z

∂x, q =

∂z

∂y, r =

∂2z

∂x2, s =

∂2z

∂x∂y, t =

∂2z

∂y2

Examples

1.∂z

∂x+∂z

∂y= xy2 + z

2. (∂z

∂x

)3

+∂2z

∂y2= xy

∂z

∂y

3.

∂3z

∂x3=

(1 +

∂z

∂y

)1/2

4.

z∂z

∂x+∂z

∂y= x2

347

348CHAPTER 7. PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

7.1.1 Order of a Partial Differential Equation

The order of a partial differential equation is defined as the order of the highest

partial derivative occurring in the partial differential equation.

The examples 1 and 4 are first order, example 2 is second order and 3 is third order

partial differential equations.

7.1.2 Degree of a Partial Differential Equation

The degree of a partial differential equation is defined as the degree of the highest

order of partial derivative involved in the PDE after it has been made free from

radicals and fraction in partial derivatives and dependent variable.

For examples, the equation 1,2,3 have degree 1 and the equation 3 has degree 2.

The mathematical definition of degree of a PDE as stated above is unimpressive.

The concept of degree cannot be attributed to all PDE. For example, the PDE

sin p+ eq = 1

has no degree.

7.1.3 Linear and Non-linear Partial Differential Equations

A partial differential equation is said to be linear if the dependent variable and

its partial derivatives occurs in the first power only and are not multiplied. A

partial differential equation which is not linear is called non-linear partial differential

equation. Example 1 is linear and 2,3,4 are non-linear.

7.2 Origin Of First Order Partial Differential Equation

Partial differential equations can be derived in two ways:

1. By the elimination of the arbitrary constants from a relation between x, y and

z.

2. By the elimination of arbitrary functions of these variables.

7.2.1 By Elimination of Arbitrary Constants

Let z be a function of x and y such that

f(x, y.z, a, b) = 0 (7.1)

where a and b are arbitrary constants.

Differentiating (7.1) partially with respect to x and y, we get the relations

∂f

∂x+∂f

∂z

∂z

∂x= 0 i.e.,

∂f

∂x+∂f

∂zp = 0 (7.2)

7.2. ORIGIN OF FIRST ORDER PARTIAL DIFFERENTIAL EQUATION 349

and

∂f

∂y+∂f

∂z

∂z

∂y= 0 i.e.,

∂f

∂x+∂f

∂zq = 0 (7.3)

By help of the equations (7.1), (7.2) and (7.3) two constants a and b can be eliminated

and we obtained a relation of the form

φ(x, y, z, p, q) = 0 (7.4)

This shows that the system of surfaces (7.1) gives to a partial differential equation

of the first order given by (7.4)

In similar method it can be shown that if there are more arbitrary constants than

the number of independent variables, the above method of elimination will give rise

to a partial differential equations of higher order.

Eliminating the constants from above method we may have following cases:

1. Case I. When the number of arbitrary constants is less than the number of

independent variables, then the elimination of arbitrary constants usually give

rise to more than one PDE of order one.

2. Case II.When the number of arbitrary constants is equal to the number of

independent variables, then the elimination of arbitrary constants usually give

rise to one PDE of order one.

3. Case III. When the number of arbitrary constants is greater than the number

of independent variables, then the elimination of arbitrary constants usually

give rise to a PDE of order greater than one.

Example 213. Form a partial differential equation by elimination of the arbitrary

constant a from z = ax+ y

Note in this case there is only one arbitrary constant a and two independent vari-

ables x and y.

Solution: Given equation of the plane is

z = ax+ y (7.5)

Differentiating (7.5) partially w.r.t x and y

∂z

∂x= a (7.6)

and

∂z

∂y= 1 (7.7)

Eliminating a from (7.5) and (7.6) gives

z = x∂z

∂x+ y (7.8)

Since (7.7) does not contains arbitrary constant, so (7.7) is a PDE under consider-

ation. Thus we get two PDEs (7.7) and (7.8).


Example 214. The equation of all spheres of radius r having its center in XOY

plane is

(x− a)2 + (y − b)2 + z2 = r2

Form a partial differential equation by eliminating the constant a and b, assume r

as fixed.

Solution: Given equation of sphere is

(x− a)2 + (y − b)2 + z2 = r2 (7.9)

Differentiating both sides, (7.9) partially w.r.t x and y

2(x− a) + 2z∂z

∂x= 0

Or, x− a = −zp (7.10)

and

2(y − b) + 2z∂z

∂x= 0

Or, y − b = −zq (7.11)

Substituting the values from (7.10) and (7.11) in the given equation (7.9)

z2(p2 + q2 + 1) = r2

which is required PDE.

Example 215. Find the partial differential equation by the elimination of a and c

from

x2 + y2 + (z − c)2 = a2

Solution: The equation of sphere is

x2 + y2 + (z − c)2 = a2 (7.12)

Differentiating (7.12), partially w.r.t x and y

2x+ 2(z − c)∂z∂x

= 0

Or, z − c = −xp

(7.13)

and

2y + 2(z − c)∂z∂y

= 0

Or, z − c = −yq

(7.14)

From (7.13) and (7.14)−xp

=−yq

Or, xq − yp = 0



Example 216. Find the PDE of the set of all right circular cones whose axes

coincide with z-axis.

Solution: The set of all right circular cones whose axes coincide with z-axis, having

semi-vertical angle α and vertex (0, 0, c) is given by

x2 + y2 = (z − c)2 tan2 α (7.15)

where c and α both are arbitrary constants. Differentiating both sides of (7.15)

w.r.t. x and y we get

2x = 2(z − c)∂z∂x

tan2 α

Or, x = p(z − c) tan2 α (7.16)

and

2y = 2(z − c)∂z∂y

tan2 α

Or, y = q(z − c) tan2 α (7.17)

From (7.16) and (7.17)x

y=p

q

Or, xq − yp = 0

Example 217. Find the PDE by eliminating a and b from

z = (x+ a)(y + b)


z = (x+ a)(y + b) (7.18)

Differentiating both sides of (7.18) w.r.t x and y we get

∂z

∂x= (y + b)

Or, p = y + b (7.19)

and

∂z

∂y= x+ a

Or, q = x+ a (7.20)

From (7.19) and (7.20)

z = pq


Example 218. Find the PDE by eliminating a and b from

2z = (ax+ y)2 + b


2z = (ax+ y)2 + b (7.21)

Differentiating both sides of (7.18) w.r.t x and y we get

2∂z

∂x= 2a(ax+ y)

Or, p = a(ax+ y) (7.22)

and

2∂z

∂y= 2(ax+ y)

Or, q = ax+ y (7.23)

From (7.22) and (7.23)

p = aq

a =p

q(7.24)

Putting values of a in (7.23) we get desired PDE as

q =p

qx+ y

Or, q2 = px+ qy

7.2.2 By the Elimination of Arbitrary Functions

Let us suppose we have relation between x, y and z of type

f(u, v) = 0 (7.25)

where u and v are functions of x, y and z and f is arbitrary function of u and v.

Differentiating (7.25) partially with respect to each of the independent variables x

and y, regarding z as dependent variable

∂f

∂u

(∂u

∂x

∂x

∂x+∂u

∂y

∂y

∂x+∂u

∂z

∂z

∂x

)+∂f

∂v

(∂v

∂x

∂x

∂x+∂v

∂y

∂y

∂x+∂v

∂z

∂z

∂x

)= 0

As∂x

∂x= 1,

∂y

∂x= 0

∂z

∂x= p

∂z

∂y= q


we get∂f

∂u

(∂u

∂x+∂u

∂zp

)+∂f

∂v

(∂v

∂x+∂v

∂zp

)= 0

Or,∂f

∂u

(∂u

∂x+∂u

∂zp

)= −∂f

∂v

(∂v

∂x+∂v

∂zp

)(7.26)

and similarly

Or,∂f

∂u

(∂u

∂y+∂u

∂zq

)= −∂f

∂v

(∂v

∂y+∂v

∂zq

)(7.27)

Eliminating ∂f∂u and ∂f

∂v from (7.26) and (7.27) i.e. dividing

∂u∂x + ∂u

∂z p∂u∂y + ∂u

∂z q=

∂v∂x + ∂v

∂zp∂v∂y + ∂v

∂z q

Or,

(∂u

∂x+ p

∂u

∂z

)(∂v

∂y+ q

∂v

∂z

)=

(∂u

∂y+ q

∂u

∂z

)(∂v

∂x+ p

∂v

∂z

)(∂u

∂y

∂v

∂z− ∂v

∂y

∂u

∂z

)p+

(∂v

∂x

∂u

∂z− ∂u

∂x

∂v

∂z

)q =

∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

Pp+ qQ = R (7.28)

where

P =∂u

∂y

∂v

∂z− ∂v

∂y

∂u

∂z=∂(u, v)

∂(y, z),

Q =∂v

∂x

∂u

∂z− ∂u

∂x

∂v

∂z=∂(u, v)

∂(z, x)

and

R =∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x=∂(u, v)

∂(x, y)

The equation (7.28) is a partial equation of the first order.

Example 219. Eliminate the arbitrary function from the function

z = xy + f(x2 + y2)

Solution: The given function is

z = xy + f(x2 + y2) (7.29)

Differentiating (7.29) partially with respect to x and y

p = y + 2xf ′(x2 + y2)

Or,p− y

2x= f ′(x2 + y2) (7.30)


and

q = x+ 2yf ′(x2 + y2)q − x

2y= f ′(x2 + y2) (7.31)

From (7.30) and (7.31)p− y

2x=q − x

2y

Or py − y2 = qx− x2


Example 220. Find a PDE,by eliminating the arbitrary function from the function

z = x+ y + f(xy)


z = x+ y + f(xy) (7.32)


p = 1 + yf ′(xy)

Or,p− 1

y= f ′(xy) (7.33)

and

q = 1 + xf ′(xy)

Or,q − 1

x= f ′(xy) (7.34)

From (7.33) and (7.34)p− 1

y=q − 1

x

Or px− x = qy − y



z = f(xyz

)Solution: The given function is

z = f(xyz

)(7.35)



p = yf ′(xyz

) ∂

∂x

(xz

)Or, p = yf ′

(xyz

)(z · 1− x · pz2

)Or,

pz2

y(z − xp)= f ′

(xyz

)(7.36)

and

q = xf ′(xyz

) ∂

∂y

(yz

)Or, p = xf ′

(xyz

)(z · 1− y · qz2

)Or,

qz2

x(z − yq)= f ′

(xyz

)(7.37)

From (7.33) and (7.34)pz2

y(z − xp)=

qz2

x(z − yq)Or, xp(z − yq) = qy(z − xp)

Or, xpz − pqxy = qyz − pqxy

Or, px = qy



f(x2 + y2 + z2, z2 − 2xy

)= 0


f(u, v) = 0

where

u = x2 + y2 + z2, v = z2 − 2xy

The required PDE is

∂(u, v)

∂(y, z)p+

∂(u, v)

∂(z, x)q =

∂(u, v)

∂(x, y)(7.38)

Now∂(u, v)

∂(y, z)=

∂u∂y

∂u∂z

∂v∂y

∂v∂z

=2y 2z

−2x 2z


= 4(y + x)z

∂(u, v)

∂(z, x)=

∂u∂z

∂u∂x

∂v∂z

∂v∂x

=2z 2x

2z −2y

= −4(y + x)z

∂(u, v)

∂(x, y)=

∂u∂x

∂u∂y

∂v∂x

∂v∂y

=2x 2y

−2y −2x

= −4(x2 − y2)

From equation (7.38), the required equation is

4(x+ y)zp− 4z(y + x)q = −4(x2 − y2)

z(p− q) + x− y = 0

Example 223. (T.U.2070) Eliminate f from

z = y2 + 2f

(1

x+ loge x

)Solution: Given function

z = y2 + 2f

(1

x+ loge y

)Differentiating both sides partially w.r.t x and y

p = − 2

x2f ′(

1

x+ loge y

)

Or, − x2p = 2f ′(

1

x+ loge y

)(7.39)

and

q = 2y +1

yf ′(

1

x+ loge y

)

Or, y(q − 2y) = f ′(

1

x+ loge y

)(7.40)

From (7.39) and (7.40)

−x2p = y(q − 2y)

Or, x2p+ yq = 2y2

which is desired PDE.

7.3. CAUCHY’S PROBLEM FOR FIRST-ORDER EQUATIONS 357

7.3 Cauchy’s Problem for First-Order Equations

All Partial differential equation do not have solution and if solution exist, it may

not be unique. In this section we discuss about the existence and uniqueness of first

order PDE

Theorem 19. Cauchy’s Problem for the First Order

If

(a) x0(µ), y0(µ) and z0(µ) are functions which, together with their first derivatives,

are continuous in the interval M defined by µ1 < µ < µ2;

(b) F (x, y, z, p, q) is a continuous function of x, y, z, p and q in a certain region U of

the xyzpq space, then we find the existence of a function φ(x, y) with the following

properties:

(1) φ(x, y) and its partial derivatives with respect to x and y are continuous functions

of x and y in a region R of the xy space.

(2) For all values of x and y lying in R, the point x, y, φ(x, y), φx(x, y), φy(x, y)lies in U and

F (x, y, φ(x, y), φx(x, y), φy(x, y)) = 0

(3) For all µ ∈M , the point x0(µ), y0(µ) ∈ R, and

φx0(µ), y0(µ) = z0(µ)

Geometrically, it tells us that there exists a surface z = f(x, y) that passes through

the curve Γ with the parametric equations

x = x0(µ), y = y0(µ) and z = z0(µ) (7.41)

and at every point of which the direction (p, q,−1) of normal is such that

F (x, y, z, p, q) = 0 (7.42)

Theorem 20.

If

(a) g(y) and all its derivatives are continuous for |y − y0| < δ i.e. immediate neigh-

borhood of y0

(b) z0 = g(y0), q0 = g′(y0) where x0 is given number and

(c) f(x, y, z, q) and all its partial derivatives are continuous in a region S given by

|x− x0| < δ, |y − y0| < δ, |q − q0| < δ

then there exists a unique function φ(x, y) such that:

(a) φ(x, y) and all its partial derivatives are continuous in a region R defined by

|x− x0| < δ1, |y − y0| < δ2


(b) For all (x, y) ∈ R, z = φ(x, y) is a solution of the equation

∂z

∂x= f

(x, y, z,

∂z

∂y

)(c) For all values of y in the interval |y − y0| < δ1, φ(x0, y) = g(y).

7.4 Linear Equation of the First Order

The partial differential equation of the type

Pp+Qq = R

where P, Q, R are functions of x, y, z, is called a linear partial equation of the first

order or Lagrange’s linear equation.

Theorem 21.

The general solution of the linear partial differential equation

Pp+Qq = R

is

F (u, v) = 0

where F is arbitrary function and u(x, y, z) = c1 and v(x, y, z) = c2 form solution of

the equationsdx

P=dy

Q=dz

R

Proof

We have Let u and v be two function of x, y, z connected by the relation

F (u, v) = 0 (7.43)

where F is arbitrary function. It is now to be shown that, on the elimination of the

arbitrary function F from (7.43), a partial differential equation will be formed and

moreover it will be linear. Regarding z as dependent variable and differentiating

partially with respect to x and y, we get

∂F

∂u

(∂u

∂x+∂u

∂z

∂z

∂x

)+∂F

∂v

(∂v

∂x+∂v

∂z

∂z

∂x

)= 0

Or,∂F

∂u

(∂u

∂x+ p

∂u

∂z

)+∂F

∂v

(∂v

∂x+ p

∂v

∂z

)= 0 (7.44)

7.4. LINEAR EQUATION OF THE FIRST ORDER 359

and

∂F

∂u

(∂u

∂y+∂u

∂z

∂z

∂y

)+∂F

∂v

(∂v

∂y+∂v

∂z

∂z

∂y

)= 0

Or,∂F

∂u

(∂u

∂y+ q

∂u

∂z

)+∂F

∂v

(∂v

∂y+ q

∂v

∂z

)= 0 (7.45)

Eliminating ∂F∂u ,

∂F∂v between (7.44) and (7.45) by determinant method, we get∣∣∣∣∣∣∣

∂u∂x + p∂u∂z

∂v∂x + p∂v∂z

∂u∂y + q ∂u∂z

∂v∂y + q ∂v∂z

∣∣∣∣∣∣∣ = 0

Or

(∂u

∂x+ p

∂u

∂z

)(∂v

∂y+ q

∂v

∂z

)−(∂u

∂y+ q

∂u

∂z

)(∂v

∂x+ p

∂v

∂z

)= 0

Or,

(∂u

∂y

∂v

∂z− ∂u

∂z

∂v

∂y

)p+

(∂u

∂z

∂v

∂x− ∂u

∂x

∂v

∂z

)q =

∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

Or,∂(u, v)

∂(y, z)p+

∂(u, v)

∂(z, x)q =

∂(u, v)

∂(x, y)(7.46)

which is in the form of

Pp+Qq = R

where P is the Jacobian

P =∂(u, v)

∂(y, z)=

∣∣∣∣∣∣∣∂u∂y

∂u∂z

∂v∂y

∂v∂z

∣∣∣∣∣∣∣and so on for Q and R.

Consider u(x, y, z) = c1 and v(x, y, z) = c2 where c1 and c2 are arbitrary constants.

Taking differentials, du = 0 and dv = 0 we get(∂u

∂x

)dx+

(∂u

∂y

)dy +

(∂u

∂z

)dz = 0 (7.47)(

∂v

∂x

)dx+

(∂v

∂y

)dy +

(∂v

∂z

)dz = 0 (7.48)

Solving (7.47) and (7.48), by method of cross multiplication for dx, dy and dz

dx∂u∂y

∂v∂z −

∂u∂z

∂v∂y

=dy

∂u∂z

∂v∂x −

∂u∂x

∂v∂z

=dz

∂u∂x

∂v∂y −

∂u∂y

∂v∂x

Using P,Q,R as Jacobian, we get

dx

P=dy

Q=dz

R(7.49)


Now (7.49) represent the differential equation whose solutions are u(x, y, z) = c1

and v(x, y, z) = c2. Having found u and v, we can say F (u, v) = 0 is the solution of

Pp+Qq = R. Note The differential equations

dx

P=dy

Q=dz

R

are called Lagrange’s auxiliary equations or subsidiary equations or characteristic

equation for

Pp+Qq = R

Steps for solving Pp+Qq = R

1. Put the given PDE in the standard form Pp+Qq = R

2. Write down Lagrange’s auxiliary equations

dx

P=dy

Q=dz

R

3. Solve these equations by well known method. Let u(x, y, z) = c1 and v(x, y, z) =

c2 are two independent solutions.

4. The general solution is then written one of the equivalent form

F (u, v) = 0, u = F (v) or v = F (u)

where F is an arbitrary function.

Solved examples

Rule for Solving dxP = dy

Q = dzR

Suppose that one the variables either cancels or absent from two fraction of the

auxiliary equation. Then an integral can be obtained by usual methods. The same

procedure can be repeated with the another set of two fractions of given auxiliary

equation.

Example 224. Find the general integral of the differential equation

y2z

xp+ zxq = y2

Solution: The given PDE is in the form

Pp+Qq = R


The Lagrange’s auxiliary equations are

dx

P=dy

Q=dz

ROr,

dx

y2z/x=dy

zx=dz

y2

From first two fractionsxdx

y2z=dy

zx

Or, x2dx = y2dy

Integrating,

x3 − y3 = c1

Also from first and third fractions

xdx

y2z=dz

y2

Or, xdx = zdz

Integrating,

x2 − z2 = c2

Hence general integral is given by

F (x3 − y3, x2 − y2) = 0 Or x3 − y3 = F (x2 − y2)


y2p− xyq = x(z − 2y)

Solution: The Lagrange’s auxiliary equations are

dx

y2=

dy

−xy=

dz

x(z − 2y)

From first two ratios

xdx+ ydy = 0

Integrating,

x2 + y2 = c1

From second and third ratiosdy

−y=

dz

z − 2y

Or, ydz + zdy = 2y

Or, d(yz) = 2y

Integrating,

zy − y2 = c2


Therefore, the required solution is

F (x2 + y2, zy − y2) = 0

Rule II for solving dxP = dy

Q = dzR

To solve, such equations some time we choose multipliers P1, Q1, R1 so that P1dx+

Q1dy +R1dz is exact differential of P1P +Q1Q+R1R. Then

P1dx+Q1dy +R1dz

P1P +Q1Q+R1R

is combined with a suitable ratio of dxP = dy

Q = dzR . This combination will give a

solution.

Example 226. Find the general integral of the linear partial differential equation

px(x+ y) = qy(x+ y)− (x− y)(2x+ 2y + z))

Solution: The given PDE is

px(x+ y)− qy(x+ y) = −(x− y)(2x+ 2y + z))


dx

x(x+ y)=

dy

−y(x+ y)=

dz

−(x− y)(2x+ 2y + z)(7.50)

From first two ratiosdx

x= −dy

y

Integrating,

loge x = − loge y + loge c1

Or, loge(xy) = loge c1

∴ xy = c1 (7.51)

Again the auxiliary eq (7.50) can be written as

dx+ dy

x(x+ y)− y(x+ y)= − dz

(x− y)(2x+ 2y + z)

Or,dx+ dy

(x− y)(x+ y)= − dz

(x− y)(2x+ 2y + z)

Or,dx+ dy

x+ y= − dz

2x+ 2y + z

Or,dx+ dy

x+ y=

−dz − dx− dy2x+ 2y + z − x− y


Or,dx+ dy

x+ y+dx+ dy + dz

x+ y + z= 0

Integrating,

loge(x+ y) + loge(x+ y + z) = loge c2

Or, loge(x+ y)(x+ y + z) = loge c2

∴ (x+ y)(x+ y + z) = c2 (7.52)

Hence the required general solution from (7.51) and (7.52), is

(x+ y)(x+ y + z) = F (xy)

Example 227. Find the general solution of the differential equation

x2 ∂z

∂x+ y2 ∂z

∂y= (x+ y)z

Solution: The Lagrange’s auxiliary equations for the given equation are

dx

x2=dy

y2=

dz

(x+ y)z(7.53)

From first two ratiosdx

x2=dy

y2

Integrating,

1

x− 1

y= c1

x− y = C1xy (7.54)

Again from (7.53)

dx− dyx2 − y2

=dz

(x+ y)z

Or,dx− dyx− y

=dz

z

Integrating,

loge(x− y) = loge z + loge c2

Or, logex− yz

= loge c2

x− yz

= c2 (7.55)

From (7.54) and (7.55) we get

xy

z= c3 (7.56)

From (7.55) and (7.56), the general solution is

F

(xy

z,x− yz

)= 0


Example 228. Find the general solution of the linear partial differential equation

px(z − 2y2) = (z − qy)(z − y2 − 2x3)

Solution: Rewriting the given differential equation as

px(z − 2y2) + qy(z − y2 − 2x3) = z(z − y2 − 2x3)


dx

x(z − 2y2)=

dy

y(z − y2 − 2x3)=

dz

z(z − y2 − 2x3)(7.57)

From last two ratiosdy

y=dz

z

Integrating,

loge y = loge z + loge c1

Or, logey

z= loge c1

y

x= c1 (7.58)

Choosing 0,−2y, 1 as multipliers, for each fraction of (7.57) and equating with first

ratiodx

x(z − 2y2)=

−2ydy + dz

−2y2(z − y2 − 2x3) + z(z − y2 − 2x3)

Or,dx

x(z − 2y2)=

dz − 2ydy

(z − 2y2)(z − y2 − 2x3)

Or,dx

x=

dz − 2ydy

z − y2 − 2x3

Let t = z − y2. Then dt = dz − 2ydy,and above equation becomes

dx

x=

dt

t− 2x3

Or,dt

dx=t− 2x3

x

Or,dt

dx− 1

xt = −2x2 (7.59)

which is ordinary linear differential equation whose I.F. is

I.F. = e−∫

1xdx = e− loge x = eloge x

−1=

1

x


Multiplying equation (7.59) by I.F.

1

x

dt

dx− 1

x2t = −2x

Or,d

dx

(t

x

)= −2x

Or, d

(t

x

)= −2xdx

Integrating,t

x= −x2 + c2

Or,z − y2

x+ x2 = c2

Or,z − y2 + x3

x= c2 (7.60)

From (7.58) and (7.60), the general solution of the given equation is

F

(y

x,z − y2 + x3

x

)= 0

Example 229. Solve

p cos(x+ y) + q sin(x+ y) = z

Solution: The Lagrange’s auxiliary equations are

dx

cos(x+ y)=

dy

sin(x+ y)=dz

z

Or,dx+ dy

cos(x+ y) + sin(x+ y)=

dx− dycos(x+ y)− sin(x+ y)

=dz

z

Taking first two members

cos(x+ y)− sin(x+ y)

cos(x+ y) + sin(x+ y)(dx+ dy) = dx− dy

Integrating,

loge(cos(x+ y) + sin(x+ y)) = x− y + loge c1

cos(x+ y) + sin(x+ y) = c1ex−y

(cos(x+ y) + sin(x+ y))ey−x = c1 (7.61)

Again taking first and the third members, we get

dx+ dy

cos(x+ y) + sin(x+ y)=dz

z


Or,dx+ dy√

2(sin(x+ y) cos π4 + cos(x+ y) sin π

4

) =dz

z

Or,√

2dz

z= cosec

(x+ y +

π

4

)(dx+ dy)

Integrating√

2 loge z = loge tan1

2

(x+ y +

π

4

)+ loge c2

Or, z√

2 = c2 tan1

2

(x+ y +

π

4

)

c2 = z√

2 cot1

2

(x+ y +

π

4

)= z

√2 cot

(x2

+y

2+π

8

)(7.62)

Hence, from (7.61) and (7.62), the general solution of given equation is given by

(cos(x+ y) + sin(x+ y))ey−x = F(z√

2 cot(x

2+y

2+π

8

))Rule III for solving dx

P = dyQ = dz

R

The given Lagrange’s can be written as

dx

P=dy

Q=dz

R=P1dx+Q1dy +R1dz

P1P +Q1Q+R1R

This is true for any multipliers P1, Q1 and R1. But we choose the multipliers as

P1P +Q1Q+R1R = 0

Then we get an equation of the form

P1dx+Q1dy +R1dz = 0

Integrating, we can get an integral of the form u(x, y, z) = 0 . With help of another

set of suitable chosen of multipliers, or other method we may obtain another integral

v(x, y, z) = c2. Then required general solution is F (u, v) = 0.


z(xp− yq) = y2 − x2


zxp− zyq = y2 − x2

The Lagrange’s auxiliary equation are

dx

zx=

dy

−zy=

dz

y2 − x2(7.63)


From first two ratios,dx

zx=

dy

−zy

Or,dx

x+dy

y= 0

Integrating

loge x+ loge y = loge c1

Or, loge(xy) = loge c1

∴ xy = c1 (7.64)

Again choosing multiplier x, y, z to the fraction of (7.63) we get

dx

zx=

dy

−zy=

dz

y2 − x2=

xdx+ ydy + zdz

zx2 − zy2 + zy2 − zx2=xdx+ ydy + zdz

0

∴ xdx+ ydy + zdz = 0

Integrating,x2

2+y2

2+z2

2=c2

2

Or, x2 + y2 + z2 = c2 (7.65)

Here (7.64) and (7.65) are the integral of the auxiliary equations. The general

solution is

x2 + y2 + z2 = F (xy)

Or we may write the general solution as

F (x2 + y2 + z2, xy) = 0

Example 231. Find the general solution of the linear partial differential equation

(y + zx)p− (x+ yz)q = x2 − y2

Solution: The auxiliary equation are

dx

y + zx=

dy

−(x+ yz)=

dz

x2 − y2(7.66)

Taking x, y,−z as multipliers on (7.66),

dx

y + zx=

dy

−(x+ yz)=

dz

x2 − y2=

xdx+ ydy − zdzxy + zx2 − xy − y2z − zx2 + zy2

=xdx+ ydy − zdz

0

∴ xdx+ ydy − zdz = 0


Integrating,

x2 + y2 − z2 = c1 (7.67)

Again taking y, x, 1 as multiplier on (7.66)

dx

y + zx=

dy

−(x+ yz)=

dz

x2 − y2=

ydx+ xdy + dz

y2 + xyz − x2 − xyz + x2 − y2=ydx+ xdy + dz

0

∴ ydx+ xdy + dz = 0

d(xy) + dz = 0

Integrating,

xy + z = c2 (7.68)

From Eq. (7.67) and (7.68), the general solution is

F (x2 + y2 − z2, xy + z) = 0

Example 232. (TU 2059,072) Solve

(mz − ny)p+ (nx− lz)q = ly −mx

Solution The Lagrange’s auxiliary equations for the given equation is

dx

mx− ny=

dy

nx− lz=

dz

ly −mx(7.69)

Choosing x, y, z as multipliers, each fraction

dx

mz − ny=

dy

nx− lz=

dz

ly −mx=xdx+ ydy + zdz

0

∴ xdx+ ydy + zdz = 0

Integrating

x2 + y2 + z2 = c1 (7.70)

Again taking l,m, n as multipliers, each fraction

dx

mz − ny=

dy

nx− lz=

dz

ly −mx=ldx+mdy + ndz

0

∴ ldx+mdy + ndz = 0

Integrating

lx+my + nz = c2 (7.71)

Here (7.70) and (7.71) are the integrals of the auxiliary equations. The general

solution is

F (x2 + y2 + z2, lx+my + nz) = 0

where F is arbitrary function.

7.5. LAGRANGE’S METHOD FOR MORE THAN TWO VARIABLES 369

7.5 Lagrange’s Method for More than Two Variables

The Lagrange’s method for two variables can be extended to the case of n inde-

pendent variables. Let the n independent variables be x1, x2, · · · , xn. Let z be the

dependent variable and

pi =∂z

∂xi: i = 1, 2, · · · , n.

Lagrange’s theorem can be written as

Theorem 22.

If ui(x1, x2, · · · , xn, z) = ci, i = 1, 2, 3 · · · , n are n independent solution of the

equationsdx1

P1=dx2

P2= · · · = dxn

Pn=dz

R

then the relation φ(u1, u2, · · · , un) = 0, in which φ is arbitrary function, is a general

solution of the linear partial differential equation

P1∂z

∂x1+ P2

∂z

∂x2+ · · ·+ Pn

∂z

∂xn= R ///

Example 233. Solve

x∂u

∂x+ y

∂u

∂y+ z

∂u

∂z= xyz

Solution: The auxiliary equations of given PDE

dx

x=dy

y=dz

z=

du

xyz

Taking first and second members

dx

x=dy

y

Integrating,

loge x = loge y + loge c1

Or, logex

y= loge c1

Similarly, taking first and third members, we get

x

z= c2

Again,dx

x=dy

y=dz

z=

du

xyz=yzdx+ zxdy + xydz

xyz + xyz + xyz

∴du

xyz=yzdx+ zxdy + xydz

3xyz

Or, 3du = d(xyz)


Integrating,

3u = xyz + c3

Or, 3u− xyz = c3

Hence the general solution of given PDE is

φ

(x

y,x

z, 3u− xyz

)= 0

where φ is an arbitrary function.

Example 234. Find the general solution of the partial differential equation

(y − z)∂u∂x

+ (z − x)∂u

∂y+ (x− y)

∂u

∂z= 0

Solution: The auxiliary equation of the given PDE are

dx

y − z=

dy

z − x=

dz

x− y=du

0

The equations are equivalent to the three relations

du = 0

dx+ dy + dz = 0

xdx+ ydy + zdz = 0

Integrating above three equations, we get following integrals

u = c1

x+ y + z = c2

x2 + y2 + z2 = c3


u = φ(x+ y + z, x2 + y2 + z2)

7.6 Integral Surface Passing through a Given Curve

We know that the general solution of the PDE Pp + Qq = R is φ(u, v) = 0 where

φ is arbitrary function. Now we will discuss two methods of using such a general

solution to get the integral surface which passes through given curve.

Method I

Let Pp+Qq = R be a linear equation of first oder and its auxiliary equations are

dx

P=dy

Q=dz

R

7.6. INTEGRAL SURFACE PASSING THROUGH A GIVEN CURVE 371

Let

u(x, y, z) = c1 and v(x, y, z) = c2 (7.72)

be two independent solutions of the auxiliary equations. Then we know that any

solution of the corresponding linear equation is of the form

F (u, v) = 0 (7.73)

which arising from a relation

F (c1, c2) = 0 (7.74)

We now determine the function F in special circumstances. Let it be the integral

surface pass through the parametric curve

x = x(t), y = y(t), z = z(t)

where t is a parameter. Then the particular solution (7.72) must be such that

u (x(t), y(t), z(t)) = c1 and v (x(t), y(t), z(t)) = c2

We eliminate single parameter t from this and we will get a relation containing c1

and c2. Finally we eliminate the constants c1 and c2 with help of (7.72) and we

obtained the required integral surface.

Method II Let

u(x, y, z) = c1 and v(x, y, z) = c2 (7.75)

be two independent solutions of auxiliary equations of the PDE

Pp+Qq = R.

Suppose we wish to find the integral surface passing through the curve which deter-

mined by the equations

f(x, y, z) = 0 and g(x, y, z) = c2 (7.76)

We eliminate x, y, z from four equations of (7.75) and (7.76) and obtain a relation

between c1 and c2. Finally putting c1 = u and c2 = v in that relation we get required

integral surface.

Example 235. Find the integral surface of the linear partial equation

x(y2 + z)p− y(x2 + z)q = (x2 − y2)z

containing the line x+ y = 0, z = 1.



dx

x(y2 + z)=

dy

−y(x2 + z)=

dz

(x2 − y2)z

Taking 1/x, 1/y, 1/z as multipliers

dx

x(y2 + z)=

dy

−y(x2 + z)=

dz

(x2 − y2)z=

1/xdx+ 1/ydy + 1/zdz

y2 + z − x2 − z + x2 − y2

∴dx

x+dy

y+dz

z= 0

Integrating,

loge x+ loge y + loge z = loge c1

Or, loge(xyz) = loge c1

∴ xyz = c1

Again taking x, y,−1 multipliers

dx

x(y2 + z)=

dy

−y(x2 + z)=

dz

(x2 − y2)z=

xdx+ ydy − dzx2y2 + x2z − x2y2 − y2z − x2z + y2z

∴ xdx+ ydy − dz = 0

Integrating,

x2 + y2 − 2z = c2

Hence solutions of auxiliary equations are

xyz = c2 and x2 + y2 − 2z = c2 (7.77)

The equation of straight line is

x+ y = 0, z = 1

From Method I

Parametric form of the straight line is

x = t, y = −t and z = 1

Putting values of x, y, z in equation (7.77), we get

−t2 = c1 2t2 − 2 = c2

Eliminating t from them, we find a relation

2c1 + c2 + 2 = 0


Substituting the values of c1 and c2

x2 + y2 + 2xyz − 2z + 2 = 0

which is desire integral surface.

From method II Using z = 1 in (7.77), we get

xy = c1, x2 + y2 − 2 = c2

=⇒ (x+ y)2 − 2xy = 2 + c2

Or, 02 − 2c1 = 2 + c2 as x+ y = 0

=⇒ c2 + 2c1 + 2 = 0

Putting values of c2 and c1 we get

x2 + y2 + 2xyz − 2z + 2 = 0

which is desire integral surface.

Example 236. Find the equation of the integral surface of the differential equation

2y(z − 3)p+ (2x− z)q = y(2x− 3)

which passes through the circle z = 0, x2 + y2 = 2x.


dx

2y(z − 3)=

dy

2x− z=

dz

y(2x− 3)(7.78)

Taking first and last fractions of (7.78),

(2x− 3)dx− 2(z − 3)dz = 0

Integrating

x2 − 3x− z2 + 6z = c1 (7.79)

Again choosing 1/2, y,−1 as multipliers, each fraction of (7.78)

dx

2y(z − 3)=

dy

2x− z=

dz

y(2x− 3)=

12dx+ ydy − dz

yz − 3y + 2xy − yz − 2xy + 3y

=⇒ 1

2dx+ ydy − dz = 0

Integrating

x+ y2 − 2z = c2 (7.80)


Again equation of circle is

z = 0, x2 + y2 = 2x

The parametric form of the circle is

x = t y = (2t− t2)1/2, z = 0

Substituting these value in (7.79) and (7.80), we have

c1 = t2 − 3t, c2 = 3t− t2

Eliminating t from these equations

c1 + c2 = 0

Putting values of c1 and c2 from (7.79) and (7.80),

x2 − 3x− z2 + 6z + x+ y2 − 2z = 0

Or, x2 + y2 − z2 − 2x+ 4z = 0

which is desired integral surface.

Example 237. Find the integral surface of PDE

(2xy − 1)p+ (z − 2x2)q = 2(x− yz)

which passes through the line x = 1, y = 0.

Solution: The Lagrange’s equations for the given equation are

dx

2xy − 1=

dy

z − 2x2=

dz

2(x− yz)

Taking z, 1, x as multipliers, each of fraction

dx

2xy − 1=

dy

z − x2=

dz

2(x− yz)=

zdx+ dy + xdz

2xyz − z + z − 2x2 + 2x2 − 2xyz

=⇒ zdx+ dy + xdz = 0

=⇒ d(xz) + dy = 0

Integrating

xz + y = c1 (7.81)

Again, taking x, y, 1/2 as multipliers, each fraction

dx

2xy − 1=

dy

z − 2x2=

dz

2(x− yz)=

xdx+ ydy + 12dz

2x2y − x+ yz − 2x2y + x− yz


=⇒ xdx+ ydy +1

2dz = 0

Integrating

x2 + y2 + z = c2 (7.82)

Also the solutions (7.81) and (7.82) passes through the line

x = 1 y = 0

Putting x = 1 and y = 0 in the equations (7.81) and (7.82) we get

z = c1 z + 1 = c2

Eliminating z from these two equations

c1 + 1 = c2

Putting values of c1 and c2 from (7.81) and (7.82) we get

xz + y + 1 = x2 + y2 + z

Or, x2 + y2 − xz − y + z = 1

which is desired integral surface.

Example 238. Find the integral surface of the equation

(x− y)y2p+ (y − x)x2q = (x2 + y2)z

passing through the curve xz = a3, y = 0

Solution: The Lagrange’s equations are

dx

(x− y)y2=

dy

(y − x)x2=

dz

(x2 + y2)z

From first two members, we get

x2dx+ y2dy = 0

Integrating

x3 + y3 = c1 (7.83)

Againdx− dy

y2(x− y) + x2(x− y)=

dz

(x2 + y2)z


Or,dz

z=dx− dyx− y

Integrating

loge z = loge(x− y) + loge c2

z

x− y= c2 (7.84)

Also the solutions (7.83) and (7.84) pass through the curve

xz = a3 y = 0

Putting y = 0 in the equations (7.83) and (7.84) we get

x3 = c1 and c2 =z

x

=⇒ x = c1/31 and z = c2 c

1/31

Putting values of x and y in xz = a3

c1/31 c2 c

1/31 = a3

Cubing

c21c

32 = a3

Again putting values of c1 and c2

(x3 + y3)2

(z

x− y

)3

= a3

(x3 + y3)2z3 = a2(x− y)3

which is required surface.

Example 239. Find the general integral of the equation

(x− y)p+ (y − x− z)q = z

and the particular solution through the circle z = 1, x2 + y2 = 1.

Solution: The Lagrange’s auxiliary equations corresponding to the given equa-

tion aredx

x− y=

dy

y − x− z=dz

z

Choosing 1, 1, 1 as multipliers,

dx

x− y=

dy

y − x− z=dz

z=dx+ dy + dz

0


=⇒ dx+ dy + dz = 0

Integrating

x+ y + z = c1 (7.85)

From last two members

dy

y − c1 − y=dz

zas x+ z = c1 − y

Or,2dy

2y − c1=

2dz

z

Integrating

loge(2y − c1) = 2 loge z + loge c2

Or, loge2y − c1

z2= loge c2

Or,2y − c1

z2= c2

Putting value of c1 from (7.85)

y − x− zz2

= c2 (7.86)

Hence the general integral equation is

F

(x+ y + z,

y − x− zz2

)= 0

Since the required curve is given by

z = 1, x2 + y2 = 1

so putting these values in (7.85) and (7.86), we get

x+ y = c1 − 1 y − x = c2 − 1

But we have

2(x2 + y2) = (x+ y)2 + (y − x)2

Or, 2 = (c1 − 1)2 + (c2 − 1)2

Or, 2 = c21 − 2c1 + 1 + c2

2 − 2c2 + 1

Or, c21 + c2

2 − 2c1 − 2c2 = 0

Putting values of c1 and c2 from (7.85) and (7.86), we get the general surface

z4(x+ y + z)2 + (y − x− z)2 − 2z4(x+ y + z) + 2z2(y − x− z) = 0


Example 240. Find the solution of xp+ yq = z representing a surface meeting the

parabola y2 = 4x, z = 1.TU 2056,2065

Solution: The auxiliary equations are

dx

x=dy

y=dz

z

From first and last membersdx

x=dz

z

Integrating,

logx = loge z + loge c1

x

z= c1 (7.87)

Similarly, taking the second and third members

y

z= c2 (7.88)

Since the required curve is given by

y2 = 4x, z = 1

so putting these values in (7.87) and (7.88), we get

x = c1 and y = c2

Putting values of c1 and c2 in y2 = 4x

c22 = 4c1

Again putting values of c1 and c2 from (7.87) and (7.88)

y2

z2= 4

x

z

Or, y2 = 4xz

7.7 Geometrical Interpretation of Pp+Qq = R

We have

P p+Q q + (−1)R = 0 (7.89)

We know that the direction ratios of the normal at a point on the surface f(x, y, z) =

0 are∂f

∂x,∂f

∂y,∂f

∂z

7.7. GEOMETRICAL INTERPRETATION OF PP +QQ = R 379

−∂f∂x/∂f

∂z,−∂f

∂y/∂f

∂z,−1

From the partial differential of implicit function

∂z

∂x= −∂f

∂x/∂f

∂zand

∂z

∂y= −∂f

∂y/∂f

∂z

Thus, the direction ratios of the normal at a point on the surface f(x, y, z) = 0 are

∂z

∂x,∂z

∂y,−1 Or, p, q, −1

Therefore, given equation (7.89) shows that the normal to the desired surface is

perpendicular to the line whose direction ratios are P,Q,R.

We also know that the simultaneous equations

dx

P=dy

q=dz

R

represent a family of curves such that the tangent at any point on those curves have

direction ratios P,Q,R. Therefore, these curves lie wholly on the surface. Thus, the

general solution of (7.89) is the surface φ(x, y) = 0, which passes through the curves

u = c1 and u = c2

Theorem 23. Show that the equation

Pdx+Qdy +Rdz = 0

(if it integrable) represents a family of surfaces orthogonal to the family represented

by

Pp+Qq = R.

Proof: We know that the direction ratios of the normal at a point (x, y, z) to

the surface represented by Pdx+Qdy +Rdz = 0 are P,Q,R.

Also the direction cosines of the normal at the point (x, y, z) to the surface Pp+Qq =

R are p, q,−1. But the equation Pp + Qq = R shows that the two normals are

perpendicular to each other.

Hence the two surfaces given by the differential equations

Pdx+Qdy +Rdz = 0 and Pp+Qq = R

are orthogonal to each other. Proved

Working rule:

Case I If given family has the differential equation

Pp+Qq = R

then the orthogonal surface is given by

Pdx+Qdy +Rdz = 0


Case II If the given family of surface is

f(x, y, z) = c

then is normal will have direction ratios are

∂f

∂x,∂f

∂y,∂f

∂z

Also if the desired orthogonal surface is

φ(x, y, z) = 0

then its normal will have direction ratios

∂φ

∂x,∂φ

∂y,∂φ

∂zi.e. p, q,−1

But these two surfaces φ(x, y, z) = 0 and f(x, y, z) = c are orthogonal, so

p∂f

∂x+ q

∂f

∂y− ∂f

∂z= 0

Or, p∂f

∂x+ q

∂f

∂y=∂f

∂z

This is the partial differential equation of orthogonal surface. Solving it, we can find

the desired surface.

Example 241. Find the surface which intersects the surface of the system

z(x+ y) = c(3z + 1)

orthogonally and which passes through the circle

x2 + y2 = 1 z = 1

Solution: The given surface is

f(x, y, z) =z(x+ y)

3z + 1

Differentiating f partially with respect to x, y and z

∂f

∂x=

z

3z + 1,

∂f

∂y=

z

3z + 1

and∂f

∂z= (x+ y)

(3z + 1) · 1− z · 3(3z + 1)2

=x+ y

(3z + 1)2

The differential equation of orthogonal surface is

p∂f

∂x+ q

∂f

∂y=∂f

∂z


z

3z + 1p+

z

3z + 1q =

x+ y

(3z + 1)2

Or, z(3z + 1)p+ z(3z + 1)q = (x+ y)

Its Lagrange’s auxiliary equations are

dx

z(3z + 1)=

dy

z(3z + 1)=

dz

x+ y(7.90)

From first two members of (7.90)

dx = dy

Integrating

x− y = c1 (7.91)

Using x, y,−z(3z + 1) as multipliers in (7.90)

dx

z(3z + 1)=

dy

z(3z + 1)=

dz

x+ y=

xdx+ ydy − z(3z + 1)dz

xz(3z + 1) + yz(3z + 1)− (x+ y)z(3z + 1)

=⇒ xdx+ ydy − z(3z + 1)dz = 0

Integratingx2

2+y2

2− 3z3

3− z2

2=c2

2

x2 + y2 − 2z3 − z2 = c2 (7.92)

From (7.91) and (7.92), the general surface orthogonal to given surface is

x2 + y2 − 2z3 − z2 = φ(x− y) (7.93)

Since the general surface contains the circle

x2 + y2 = 1, z = 1

so from (7.93)

−2 = φ(x− y)

Now from (7.93), required surface is

x2 + y2 − 2z3 − z2 + 2 = 0

Example 242. Find the surface which is orthogonal to the one parameter system

z = cxy(x2 + y2)

and which passes through the rectangular hyperbola x2 − y2 = a2, z = 0.


Solution: The given surface is

f(x, y, z) =z

xy(x2 + y2)= c

Differentiating f partially with respect to x, y and z

∂f

∂x= − z(3x2y + y3)

[xy(x2 + y2)]2= − z(3x2 + y2)

x2y(x2 + y2)2

∂f

∂y=−z(3y2 + x2)

xy2(x2 + y2)2

and∂f

∂z=

1

xy(x2 + y2)


p∂f

∂x+ q

∂f

∂y=∂f

∂z

− z(3x2 + y2)

x2y(x2 + y2)2p+−z(3y2 + x2)

xy2(x2 + y2)2q =

1

xy(x2 + y2)

Or,3x2 + y2

xp+

x2 + 3y2

yq = − z

x2 + y2

Its Lagrange’s auxiliary equations are

xdx

3x2 + y2=

ydy

x2 + 3y2= − zdz

x2 + y2(7.94)

From first two members of (7.90) and equating with third

xdx+ ydy

4(x2 + y2)=−zdzx2 + y2

Or,xdx+ ydy = −4zdz

Integrating

x2 + y2 + 4z2 = c1 (7.95)

From first two members of (7.90) and equating with third

xdx− ydy2(x2 − y2)

=−zdzx2 + y2

2xdx− 2ydy

(x2 − y2)= − 4zdz

−4z2 + c1

Integrating

loge(x2 − y2) =

1

2loge(c1 − 4z2) +

1

2c2


Or,(x2 − y2)2

x2 + y2= c2 (7.96)

From (7.95) and (7.96), the general surface orthogonal to given surface is

x2 + y2 + 4z2 = φ

((x2 − y2)2

x2 + y2

)(7.97)

Since the general surface contains the circle

x2 − y2 = a2, z = 0

so from (7.97)

x2 + y2 = φ

(a4

x2 + y2

)(7.98)

Let

v =a4

x2 + y2

Then

x2 + y2 =a4

v

Now from (7.98)

φ(v) =a4

v

φ

((x2 − y2)2

x2 + y2

)=

a4

(x2−y2)2

x2+y2

=a4(x2 + y2)

(x2 − y2)2

Now (7.97) becomes

x2 + y2 + 4z2 =a4(x2 + y2)

(x2 − y2)2

Or, (x2 + y2 + 4z2)(x2 − y2) = a4(x2 + y2)

which is required surface.

Example 243. Find the equation of the system of surfaces which cut orthogonally

the cones of the

x2 + y2 + z2 = cxy

Solution: The given equation of system of surface is

f(x, y, z) =x2 + y2 + z2

xy=x

y+y

x+z2

xy= c

Differentiating f partially w.r.t x, y and z we get

∂f

∂x=

1

y− y

x2− z2

x2y


∂f

∂y= − x

y2+

1

x− z2

xy2

∂f

∂z=

2z

xy


p∂f

∂x+ q

∂f

∂y=∂f

∂z

Or,

(1

y− y

x2− z2

x2y

)p+

(− x

y2+

1

x− z2

xy2

)q =

2z

xy

Or,x2 − y2 − z2

x2yp+

y2 − x2 − z2

xy2q =

2z

xy

Or,x2 − y2 − z2

xp+

y2 − x2 − z2

yq = 2z

The auxiliary equations are

xdx

x2 − y2 − z2=

ydy

y2 − x2 − z2=dz

2z(7.99)

Taking 1, 1, z as multipliers

xdx

x2 − y2 − z2=

ydy

y2 − x2 − z2=dz

2z=xdx+ ydy + zdz

0

=⇒ xdx+ ydy + zdz = 0

Integrating,

x2 + y2 + z2 = c1 (7.100)

Again from first and second and equating with third

xdx− ydyx2 − y2

=dz

z

Integrating,

loge(x2 − y2) = 2 loge z + loge c2

Or,x2 − y2

z2= c2 (7.101)

From (7.100) and (7.101) we get required surface

x2 + y2 + z2 = φ

(x2 − y2

z2

)

7.8. CHARPIT’S METHOD 385

7.8 Charpit’s Method

7.8.1 Non-Linear Partial Differential Equation

A differential equation containing p and q with higher power other than one and

product of p and q is called non-linear partial differential equation. For example

z2(p2 + q2 + 1) = 1

7.8.2 Types of Solutions

Solution

A solution of a differential equation is a relation between variables, which is free

from the derivatives, which satisfy the given differential equation.

Complete solution:

We have seen that the relation of type

φ(x, y, z, a, b) = 0 (7.102)

give rise to a partial differential equation of the first order of the form

f(x, y, z, p, q) = 0 (7.103)

on the elimination of the arbitrary constants a and b. Here x and y are independent

variables and z is dependent variable. The relation (7.102) is solution of (7.103). Any

such relation which contains as many arbitrary constants as there are independent

variables and is a solution of partial differential equation of the first order is called

a complete solution or a complete integral of that equation.

Particular solution:

A particular solution of a partial differential equation is obtained by giving particular

values to the arbitrary constants in the complete integral.

Singular solution:

The equation of envelope of the surfaces represented by the complete integral of a

given differential equation is called its singular solution.

If there is a differential equation

f(x, y, z, p, q) = 0

whose complete integral is

φ(x, y, z, a, b) = 0

then the equation is obtained by eliminating a and b from

φ = 0,∂φ

∂a= 0, and

∂φ

∂b= 0

is a singular solution of the PDE.

General solution:


Let φ(x, y, z, a, b) be a complete solution of f(x, y, z, p, q) = 0. Let the constant b be

a function of a i.e.

b = Ψ(a)

Then the solution becomes

φ(x, y, z, a,Ψ(a)) = 0 (7.104)

It is one parameter subfamily of the family (7.102). The equation of the envelope of

the family of surfaces represented by (7.104) is also a solution of the equation (7.103).

This solution is called the general solution of f(x, y, z, p, q) = 0 corresponding to the

complete integral φ(x, y, z, a, b) = 0

The general solution or equation of envelope of the surfaces represented by (7.104)

is obtained by eliminating a from

φ(x, y, z, a,Ψ(a)) = 0, and∂f

∂a= 0

7.8.3 General Method of Solution of a Non-linear Partial Differen-

tial Equation of Order One with Two Independent Variable

(Charpit’s method)

Let the given partial differential equation be

f(x, y, z, p, q) = 0 (7.105)

Since z is dependent variable of independent variables x and y, so

dz =∂z

∂xdx+

∂z

∂ydy = pdx+ qdy (7.106)

The fundamental idea in Charpit’s method is to introduce a second partial differen-

tial equation of first order

g(x, y, z, p, q, a) = 0 (7.107)

containing an arbitrary constant a such that

1. The equation (7.105) and (7.107) can be solved to find

p = p(x, y, z, a), q = q(x, y, z, a)

2. Substituting these values of p and q in (7.106), the equation

dz = p(x, y, z, a)dx+ q(x, y, z, a)dy (7.108)


is integrable.

If such a relation (7.107) has been founded, the solution of equation (7.108)

φ(x, y.z, a, b) = 0 (7.109)

containing two arbitrary constants a and b will be a solution of the equation (7.105).

Also a complete solution of the equation (7.105).

Now the main problem is to derive a method of finding the relation (7.107) . Let us

assume that (7.107) is the relation which when taken along with (7.105) gives those

values of p and q which make (7.106) integrable. Differentiating (7.105) and (7.107)

with respect to x, we get

∂f

∂x+∂f

∂z

∂z

∂x+∂f

∂p

∂p

∂x+∂f

∂q

∂q

∂x= 0

Or,∂f

∂x+∂f

∂zp+

∂f

∂p

∂p

∂x+∂f

∂q

∂q

∂x= 0 (7.110)

and similarly

∂g

∂x+∂g

∂zp+

∂g

∂p

∂p

∂x+∂g

∂q

∂q

∂x= 0 (7.111)

Again, differentiating (7.105) and (7.107) with respect to y, we get

∂f

∂y+∂f

∂zq +

∂f

∂p

∂p

∂y+∂f

∂q

∂q

∂y= 0 (7.112)

and

∂g

∂y+∂g

∂zq +

∂g

∂p

∂p

∂y+∂g

∂q

∂q

∂y= 0 (7.113)

For elimination of ∂p∂x from (7.110) and (7.111) ,multiplying (7.110) by ∂g∂p and (7.111)

by ∂f∂p and subtracting we get(∂f

∂x

∂g

∂p− ∂f

∂p

∂g

∂x

)+ p

(∂f

∂z

∂g

∂p− ∂f

∂p

∂g

∂z

)+∂q

∂x

(∂f

∂q

∂g

∂p− ∂f

∂p

∂g

∂q

)= 0 (7.114)

and eliminating ∂q∂y from (7.112) and (7.113)(∂f

∂y

∂g

∂q− ∂f

∂q

∂g

∂y

)+ q

(∂f

∂z

∂g

∂q− ∂f

∂q

∂g

∂z

)+

∂p

∂y

(∂f

∂p

∂g

∂q− ∂f

∂q

∂g

∂p

)= 0 (7.115)

Since∂q

∂x=

∂2z

∂x∂y=∂p

∂y


Hence adding (7.114) and (7.115) and rearranging, we get(∂f

∂x+ p

∂f

∂z

)∂g

∂p+

(∂f

∂y+ q

∂f

∂z

)∂g

∂q+(

−p∂f∂p− q∂f

∂p− q∂f

∂q

)∂g

∂z+

(−∂f∂p

)∂g

∂x+

(−∂f∂q

)∂g

∂y= 0 (7.116)

This is a linear partial differential equation of the first oder with x, y, z, p, q as

independent variables and g as dependent variable Lagrange’s auxiliary equations

are

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=

dy

−fq(7.117)

This equations are known as Charpit’s auxiliary equations, where

fx =∂f

∂xand so on.

Any integral of eq. (7.117) satisfy eq. (7.116). If such an equation contains p or

q, it can be taken as the required second relation. It should be noted that not all

of Charpit’s auxiliary equation (7.117) need be used, but that p or q must in the

solution obtained. Of course, the simpler the integral involving p or q or both that

derived from (7.117), the easier will be subsequent labour in finding the solution of

the given equation. Substituting the values of p and q in

dz = pdx+ qdy

and integrating we get desired solution.

Example 244. Find complete integral of the equation

p2x+ q2y = z (TU2059, 2073)


f : p2x+ q2y − z = 0 (7.118)

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=dy

fq

Ordp

p2 − p=

dq

q2 − q=

dx

−2px=

dy

−2qy= · · ·

It follows that

p2dx+ 2pxdp

−2p3x− 2p2x+ 2p3x=

q2dy + 2qydq

−2q3y − 2q2y + 2q3x

Or,p2dx+ 2pxdp

p2x=q2dy + 2qydq

q2y


Integrating,

loge p2x = loge q

2y + loge a

∴, p2x = aq2y (7.119)

Solving (7.118) and (7.119) for p and q we get

q =

z

(1 + a)y

1/2

and p =

az

(1 + a)x

1/2

Putting values of (p) and q in

dz = pdx+ qdy

we get

dz =

az

(1 + a)x

1/2

dx+

z

(1 + a)y

1/2

dy

√1 + a

dz√z

=√adx√x

+dy√y

Integrating, √(1 + a)z =

√ax+

√y + b

which is required complete integral of (7.118)


(p2 + q2)y = qz


f : (p2 + q2)y − qz = 0 (7.120)

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=dy

fq

Ordp

−pq=

dq

p2 + q2 − q2=

dx

−2py=

dy

−2qy + z= · · ·

From which it follows thatdp

−pq=dq

p2

Or, pdp+ qdq = 0

Integrating,

p2 + q2 = c

For simplicity set c = a2

∴ p2 + q2 = a2 (7.121)


Solving (7.120) and (7.121) for p and q we get

q =a2y

z, p =

a

z

√z2 − a2y2

Putting values of (p) and q in

dz = pdx+ qdy

we get

dz =a

z

√z2 − a2y2dx+

a2y

zdy

zdz − a2ydy = a√z2 − a2y2dx

zdz − a2ydy√z2 − a2y2

= adx

Integrating, √z2 − a2y2 = ax+ b

Or, z2 = a2y2 + (ax+ b)2

which is required complete integral of (7.120).

Example 246. Find a complete integral of

p = (qy + z)2.


f : p− (qy + z)2 = 0 (7.122)

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=dy

fq

Ordp

2p(qy + z)=

dq

4q(qy + z)=dx

−1=

dy

−2y(qy + z)= · · ·


p= −dy

y

Integrating,

py = a

Putting p = ay in (7.122)

(qy + z)2 =a

y

∴ q =1

y

√a

y− z

(7.123)


Putting the value of p and q in

dz = pdx+ qdy

we get

ydz + zdy = adx+

√a

ydy

Or, d(zx) = adx+

√a

ydy

Integrating,

yz = ax+ 2√ay + b



z2 = pqxy


f : z2 − pqxy = 0 (7.124)

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=dy

fq

Or,dp

−pqy + 2pz=

dq

−pqx+ 2qz=

dx

qxy=

dy

pxy= · · ·

From which it follows that

xdp+ pdx

2pxz=ydq + qdy

2qyz

d(xp)

xp=d(yq)

yq

Integrating,

loge px = loge qy + loge a2

px = a2qy (7.125)

Solving (7.124) and (7.124) simultaneously for p and q, we get

q =z

ayand p =

az

x

∴ q =1

y

√a

y− z

(7.126)



dz = pdx+ qdy

we get

dz =az

xdx+

z

aydy

Or,dz

z=a

xdx+

1

a

dy

y

Integrating,

loge z = a loge x+1

aloge y + loge b

z = bxay1/a



2(z + px+ qy) = yp2


f : 2(z + px+ qy)− yp2 = 0 (7.127)

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=dy

fq

Or,dp

2p+ 2p=

dq

2q − p2 + 2q=

dx

−2x+ 2py=

dy

pxy= · · ·


2p=dy

−yIntegrating,

loge p+ 2 loge y = loge a

py2 = a (7.128)


p =a

y2and q = −z

y− ax

y3+

a3

2y4


dz = pdx+ qdy

we get

dz =a

y2dx+

(−zy− ax

y3+

a3

2y4

)dy


Or, ydz + zdy = a

(ydx− xdy

y2

)+

a2

2y3dy

Or, d(yz) = ad(yx

)+

a2

2y3dy

Integrating,

yz =ax

y2− a2

4y3+ b

z =ax

y3− a2

4y4+b

y


Example 249. Solve by General method

z = pq TU2061, 2066, 2067


f : z − pq = 0 (7.129)

The Charpit’s auxiliary equations

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=dy

fq

Or,dp

0 + p=

dq

0 + q= · · ·


p=dq

q

Integrating,

loge p = loge q + loge a

p = aq (7.130)


q =

√z

aand q = a

√z

a=√az


dz = pdx+ qdy

we get

dz =√azdx+

√z

ady

Or,


Or,√adz√z

= adx+ dy

Integrating,

2√az = ax+ y + b



q = px+ p2 TU2071


f : q − px− p2 = 0 (7.131)

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=

dy

−fq

Or,dp

−p=dq

0=

dx

x+ 2p=dy

−1· · ·


p=dy

1

Integrating,

loge p = y + loge a

logep

a= y

p = aey (7.132)


p = aey and q = axey + a2e2y


dz = pdx+ qdy

we get

dz = aeydx+ axeydy + a2e2ydy

Or, dz = ad(xey) + a2e2ydy

Integrating,

z = axey +a2

2e2y + b


7.9. SPECIAL TYPES OF EQUATIONS 395


(p+ q)(px+ qy)− 1 = 0 TU2069


f : (p+ q)(px+ qy)− 1 = 0 (7.133)

dp

fx + pfz=

dq

fy + qfz=

dz

−pfp − qfq=

dx

−fp=

dy

−fq

Or,dp

p2 + pq=

dq

pq + q2· · ·


p=dq

q

Integrating,

loge p = loge q + loge a

loge p = loge aq

p = aq (7.134)


p = aq and q =1√

1 + a√ax+ y

Or, p =a√

1 + a√ax+ y

and q =1√

1 + a√ax+ y


dz = pdx+ qdy

we get

dz =1√

1 + a

adx+ dy√ax+ y

Integrating,

z =2√

1 + a

√ax+ y + b


7.9 Special Types of Equations

In this section we shall consider some shorter methods for special types of first oder

partial differential equations.


7.9.1 Type I: Equations That Tnly Involve p and q

Let the partial differential equation containing p and q be

f(p, q) = 0 (7.135)

The Charpit’s auxiliary equations are

dx

−fp=

dy

−fq=

dz

−pfp − qfq=

dp

fx + pfz=

dq

fy + qfz

reduce todp

0=dq

0= · · ·

=⇒ dp = 0, dq = 0

Integrating,

p = a, and q = b

where a and b arbitrary constants. In (7.135), replacing p and q by a and b respec-

tively, we get

f(a, b) = 0

Solving this for b, we get

b = φ(a)

Putting these values in

dz = pdx+ qdy

we get

dz = adx+ φ(a)dy

Integrating,

z = ax+ φ(a)y + c (7.136)

which is a complete solution of (7.135). It contains two arbitrary constants a and

c.

Singular Integral: The singular integral, if it exists, is obtained by eliminating a

and c between the complete integral (7.136) and the equations formed by differenti-

ating (7.136) partially w.r.t. a and c i.e. between the equations

z = ax+ φ(a)y + c

0 = x+ φ′(a)y

and 0 = 1

Since 1 = 0 is inconsistent, therefore, in this case there is no singular solution.


Example 252. Solve

p2 + q2 = m2

where m is a constant.


p2 + q2 = m2 (7.137)

where m is a constant, is of the form f(p, q) = 0. Its solution is of the form

z = ax+ by + c

Therefore, putting p = a and q = b in (7.137) we get

a2 + b2 = m2

b =√m2 − a2

Putting the value of b in above solution, the general solution is

z = ax+√m2 − a2 y + c (7.138)

Singular Solution: Differentiating (7.138) partially with respect to a and c sepa-

rately, we get

0 = x− ax√m2 − a2

and 0 = 1

However 0 = 1 is not possible. Hence the equation (7.137) has no singular solution.

General integral Put c = φ(a) in complete solution (7.138). Then we get

z = ax+ y√m2 − a2 + φ(a) (7.139)

Differentiating (7.139) partially with respect to a

0 = x− ay√m2 − a2

+ φ′(a) (7.140)

Eliminating a from (7.139) and (7.140) , we get general solution of (7.137).


p+ q = pq


p+ q − pq = 0 (7.141)

is of the form f(p, q) = 0. Its solution is of the form

z = ax+ by + c



a+ b− ab = 0

b =a

a− 1


z = ax+a

a− 1y + c


pq = 1


pq = 1 (7.142)

is of the form f(p, q) = 0. Its solution is of the form

z = ax+ by + c


ab = 1

b =1

a


z = ax+1

ay + c

7.9.2 Type II: Equations not Involving the Independent Variables

If the partial differential equation does not contain the independent variables x and

y, then it is of the form

f(z, p, q) = 0 (7.143)

In this case, Charpit’s auxiliary equations reduces to

dx

−fp=

dy

−fq=

dz

−pfp − qfq=

dp

pfz=

dy

qfz

Form which it follows thatdp

p=dq

q

Integrating,

loge q = loge p+ loge a


∴ q = ap

where a is arbitrary constant Solving

q = ap and f(z, p, q) = 0

we obtained expression for p and q, and integrating

dz = pdx+ qdy


Alternative method

From

dz = pdx+ qdy

we get

dz = p(dx+ ady) as q = ap

Or, dz = pdX, where X = x+ ay

∴ p =dz

dXand q = a

dz

dX

Putting these values of p and q in (7.143), we get

f

(z,dz

dX, adz

dX

)= 0

which is an ordinary differential equation of first oder and solving it a complete

integral can be obtained.


p2z2 + q2 = 1


f : p2z2 + q2 − 1 = 0 (7.144)

which is in the form f(z, p, q) = 0. Putting

p = aq

where a is arbitrary constant, in (7.144) we get

q =1√

1 + a2z2

and

p = a1√

1 + a2z2


Again putting values of p and q in dz = pdx+ qdy we get√(1 + (az)2)dz = adx+ dy

Integrating,

1

2aaz√

(1 + (az)2) +12

2alog(az +

√(1 + (az)2) = ax+ y + b

Or, z√

(1 + a2z2) + loge(az +√

(1 + a2z2) = 2a(ax+ y + b)

which is required general integral.


z = p2 − q2


f : z − (p2 − q2) = 0 (7.145)


p = aq


q =

√z√

a2 − 1

and

p = a

√z√

a2 − 1

Again putting values of p and q in dz = pdx+ qdy we get√a2 − 1

dz√z

= adx+ dy

Integrating,

2√a2 − 1

√z = ax+ y + b



zpq = p+ q



f : zpq − p− q = 0 (7.146)


p = aq


q =1 + a

az

and

p =1 + a

z

Again putting values of p and q in dz = pdx+ qdy we get

dz =(1 + a)(adx+ dy)

az

Or, zdz =(1 + a)(adx+ dy)

a

Integrating,z2

2=

(1 + a)

a(ax+ y) +

b

2

Or,z2 = 2(1 + a)(x+

y

a

)+ b



z2(p2 + q2 + 1) = c2 TU2057


f : z2(p2 + q2 + 1) = c2 (7.147)

which is in the form f(z, p, q) = 0. Let

u = x+ ay

where a is arbitrary constant. Now

p =dz

duand q = a

dz

du

Putting values of p and q in (7.147) we get

z2

((dz

du

)2

+ a2

(dz

du

)2

+ 1

)= c2


Or, z2

((dz

du

)2

+ a2

(dz

du

)2)

= c2 − z2

Or, zdz

du

√1 + a2 =

√a2 − z2

Or, − 1

2

√a2 + 1

(−2z)dz√c2 − z2

dz = du

Integrating,

−1

2

√a2 + 1 2

√a2 − z2 = u+ b as

∫f ′(x)√f(x)

dx = 2√f(x) + c

Or, −√

(a2 + 1)√a2 − z2 = (x+ ay + b)

Squaring

(a2 + 1)(a2 − z2) = (x+ ay + b)2

which is complete solution.

7.9.3 Type III: Separable Equation

We say that a first oder partial equation is separable if it can be written in the form

f(x, p) = g(y, q)

For such an equation Charpit’s equations becomes

dx

fp=

dy

−gq=

dp

−fx=

dq

−gq

from which it follows that

fxdx+ fpdp = 0 i.e.∂f

∂xdx+

∂f

∂pdp = 0

df = 0

=⇒ f(x, p) = a = constant

Thus given equation can be written as

f(x, p) = g(y, q) = a

Solving these equations for p and q , we determine

p = F1(x, a) and q = F2(y, a)

Putting the values of p and q in dz = pdx+ qdy we get

dz = F1(x, a)dx+ F2(y, a)dy


Integrating

z =

∫F1(x, a)dx+

∫F2(y, a)dy + b

where b is arbitrary constant. This is the complete solution, since it contains two

arbitrary constants.


p2 + q2 = x+ y

Solution: The given equation can be written as

p2 − x = y − q2 = a

∴ p =√x+ a and q =

√y − a

Putting values of p and q in

dz = pdx+ qdy

we get

dz =√x+ adx+

√y − ady

Integrating,

z =2

3(x+ a)3/2 +

2

3(y − a)3/2 + b

which is a complete integral of the given equation.


p2y(1 + x2) = qx2


p2(1 + x2)

x2=q

y= a

∴ p =ax√

1 + x2and q = a2y


dz = pdx+ qdy

we get

dz =ax√

1 + x2dx+ a2ydy

dz =a

2

2x√1 + x2

dx+ a2ydy

Integrating,

z =a

22√

1 + x2 +a2y2

2+ b

Or, z = a√

1 + x2 +a2y2

2+ b




p2q(x2 + y2) = p2 + q


x2 + y2 =1

q+

1

p2

Or, x2 − 1

p2=

1

q− y2 = a2

∴ p =1√

x2 − a2and q =

1√a2 + y2


dz = pdx+ qdy

we get

dz =1√

x2 − a2dx+

1√a2 + y2

dy

Integrating,

z = loge(x+√x2 − a2) + loge(y +

√a2 + y2) + b


Example 262. Solve the partial differential equation√p+√q = 2x. (TU 2060)


√p− 2x = −√q = a

∴ p = (a+ 2x)2 and q = a2


dz = pdx+ qdy

we get

dz = (a+ 2x)2dx+ a2dy

Integrating,

z =(a+ 2x)3

6+ a2y + b (7.148)


Singular solution Differentiating (7.148) with respect to a and b, separately

0 =1

2(a+ 2x)2 + 2ay and 0 = 1


Since 0 = 1 is not possible, singular solution does not exist.

General solution: Take b = φ(a), where φ denotes an arbitrary function. Then

equation (7.148) becomes

z =(a+ 2x)3

6+ a2y + φ(a) (7.149)

Differentiating (7.149) partially with respect to a we get

0 =1

2(a+ 2x)2 + 2ay + φ′(a) (7.150)

Eliminating a from (7.149) and (7.150), we get required general solution.

7.9.4 Type IV: Clairaut’s Equation

A first oder partial differential equation of the form

z = px+ qy + f(p, q) (7.151)

where f is a function a function of p and q, is said to be of Clairaunt’s partial

differential equation.

The Charpit’s auxiliary equations take the form

dx

p+ fp=

dy

y + fq=dp

0=dq

0

From which it follows that

dp = 0 dq = 0

=⇒ p = a q = b


Substituting the values of p and q in (7.151), we get

z = ax+ by + f(a, b) (7.152)

This is the complete integral, since it contains two arbitrary constants.

To find the general solution, put b = φ(a) in (7.152), where φ is an arbitrary function.

Then

z = ax+ φ(a)y + f(a, φ(a)) (7.153)

Differentiating (3) partially with respect to a, we get

0 = x+ yφ′(a) + f ′(a, φ(a)) (7.154)

Eliminating a from (7.153) and (7.154), the general solution is obtained.

To find the singular integral, eliminate a and b from the three equations

z = ax+ by + f(a, b)

0 = x+∂f

∂a

0 = y +∂f

∂b


Example 263. Find a complete integral of the equation

(p+ q)(z − xp− yq) = 1

Solution: Writing the given equation in the form

z = xp+ yq +1

p+ q

Putting p = a and q = b

z = ax+ by +1

a+ b

which is a complete integral.

Example 264. Find a complete integral of the equation

pqz = p2(xq + p2) + q2(yp+ q2)

Solution: Writing the given equation in the form

z = xp+ yq +p4 + q4

pq

Putting p = a and q = b

z = ax+ by +a4 + b4

ab

which is a complete integral.

Example 265. Find the singular integral of

z = px+ qy + loge(pq)

Solution: The complete integral of the given equation is

z = ax+ by + loge(ab) (7.155)

Differentiating (7.155) partially with respect to a and b, we get

0 = x+1

aand 0 = y +

1

b

∴ a = −1

xand b = −1

y

Eliminating a and b between (7.155), we get

z = x

(−1

x

)+ y

(−1

y

)+ loge

(1

xy

)Or, z = −2− loge(xy)

which is required general solution.


Example 266. Solve

z = px+ qy + pq (TU 2064)

Solution: The given differential equation is in the Clairaut’s form. Therefore,

a complete solution is

z = ax+ by + ab (7.156)


Singular Integral: Differentiating (7.156) partially with respect to a and b sepa-

rately, we get

0 = x+ b and 0 = y + a

x = −b and y = −a

Putting values of a and b in (7.156) we get

z = −xy − xy + xy

Or, z = xy


General Integral: Take b = φ(a), where φ is arbitrary function. Then (7.156)

becomes

z = ax+ φ(a)y + aφ(a) (7.157)

Differentiating (7.157) partially with respect to a, we get

0 = x+ φ′(a)y + φ(a) + aφ′(a) (7.158)

Eliminating a from (7.157) and (7.158), we get required general solution of given

PDE.

Example 267. Solve

z = px+ qy −√pq (TU 2064, 2060, 2065)

Solution: The given differential equation is in the Clairaut’s form. Therefore,

a complete solution is

z = ax+ by − 2√ab (7.159)


Singular Integral: Differentiating (7.159) partially with respect to a and b sepa-

rately, we get

0 = x− 2b

2√ab

and 0 = y − 2a

2√ab


Or, x =

√b

aand y =

√a

b

∴ xy = 1

is one singular solution. Which contains only x and y. For another singular solution

which contains x, y, z

Putting values of x and y in (7.159) we get

z = a

√b

a+ b

√a

b− 2√ab = 0

∴ (x− z) =

√b

aand y − z =

√b

a

Hence

(x− z)(y − z) = 1

is required singular solution.

General Integral: Take b = φ(a), where φ is arbitrary function. Then (7.159)

becomes

z = ax+ φ(a)y −√aφ(a) (7.160)

Differentiating (7.160) partially with respect to a, we get

0 = x+ φ′(a)y − 2φ(a) + aφ′(a)

2√aφ(a)

0 = x+ φ′(a)y − φ(a) + aφ′(a)√aφ(a)

(7.161)

Eliminating a from (7.160) and (7.161), we get required general solution of given

PDE.

Chapter 8

Partial Differential Equations of

Second Order

8.1 The Origin of Second Order Partial Differential Equa-

tion

Suppose that the function z is given by

z = f(u) + g(v) + w (8.1)

where f and g are arbitrary functions of u and v respectively and u, v and w are

functions of x and y. The writing

p =∂z

∂x, q =

∂z

∂y, r =

∂2z

∂x2, s =

∂2z

∂x∂y, t =

∂2z

∂y2(8.2)

Differentiating (8.1) partially with respect to x and y, we get

p = f ′(u)ux + g′(v) vx + wx

q = f ′(u) uy + f ′(v) vy + wy

Again, differentiating p and q

r = f ′′(u) u2x + g′′(v) v2

x + f ′(u) uxx + g′(v) vxx + wxx

s = f ′′(u) ux uy + g′′(v) vx vy + f ′(u) uxy + g′(v) vxy + wxy

t = f ′′(u) u2y + g′′(v) v2

y + f ′(u) uyy + g′(v) vyy + wyy

We now have five equations involving the four arbitrary quantities f ′, f ′′, g′, g′′. If

we eliminate these four quantities from the equations, we obtained the relation

p− wx ux vx 0 0

q − wy uy vy 0 0

r − wxx uxx vxx u2x v2

x

s− wxy uxy vxy uxuy vxvyt− wyy uyy vyy u2

y v2y

= 0 (8.3)

409

410CHAPTER 8. PARTIAL DIFFERENTIAL EQUATIONS OF SECONDORDER

which involves only the derivatives p, q, r, s, t and known functions of x and y. It is

second order partial differential equation.

The determinant of (8.3), when we expanded with respect to the first column gives

a partial differential equation of the form

Rr + Ss+ Tt+ Pp+Qq = W

where R,S, T, P,Q and R are known functions of x and y. So equation (8.1) can be

considered as a solution of the second order linear PDE (8.3).

Example 268. Find a differential equation from z = f(x + ay) + g(x − ay) by

eliminating the arbitrary functions f and g, where a is a constant.


z = f(x+ ay) + g(x− ay) (8.4)

Differentiating (8.4) partially w.r.t. x and y

p =∂z

∂x= f ′(x+ ay) + g′(x− ay) (8.5)

q =∂z

∂y= af ′(x+ ay)− ag′(x− ay) (8.6)

Again, differentiating (8.5) partially with respect to x, we get

r =∂2z

∂x2= f ′′(x+ ay) + g′′(x− ay) (8.7)

and differentiating (8.6) partially with respect to y, we get

t =∂2z

∂y2= a2f ′(x+ ay) + a2g′(x− ay) (8.8)

From (8.7) and (8.8)

t = a2r


Example 269. Verify that the partial differential equation

∂2z

∂x2− ∂2z

∂y2=

2z

x2

is satisfied by

z =1

xφ(y − x) + φ′(y − x)

where φ is arbitrary function.

8.1. THE ORIGIN OF SECONDORDER PARTIAL DIFFERENTIAL EQUATION411


z =1

xφ(y − x) + φ′(y − x) (8.9)

Differentiating (8.9) partially w.r.t. x

∂z

∂x= −

1

xφ′(y − x) +

1

x2φ(y − x)

− φ′′(y − x) (8.10)

Again, differentiating (8.10) partially with respect to x, we get

∂2z

∂x2= −

−1

xφ′′(y − x)− 1

x2φ′(y − x)− 1

x2φ′(y − x)− 2

x3φ(y − x)

+ φ′′′(y − x)

=1

xφ′′(y − x) +

2

x2φ′(y − x) +

2

x3φ(y − x) + φ′′′(y − x) (8.11)

Differentiating (8.9) partially with respect to y, we get

∂z

∂y=

1

xφ′(y − x) + φ′′(y − x)

∂2z

∂y2=

1

xφ′′(y − x) + φ′′′(y − x) (8.12)

Subtracting (8.12) and (8.11) we get

∂2z

∂x2− ∂2z

∂y2=

2

x2φ′(y − x) +

2

x3φ(y − x)

=2

x2

1

xφ(y − x) + φ′(y − x)

=

2z

x2


Example 270. If u = f(x + iy) + g(x − iy), where the functions f and g are

arbitrary, show that∂2u

∂x2+∂2u

∂y2= 0


u = f(x+ iy) + g(x− iy) (8.13)


∂u

∂x= f ′(x+ iy) + g′(x− iy)

∂2u

∂x2= f ′′(x+ iy) + g′′(x− iy)

∂u

∂y= i f ′(x+ iy)− i g′(x− iy)

∂2u

∂y2= i2f ′′(x+ iy) + i2g′′(x− iy)

= −(f ′′(x+ iy) + g′′(x− iy)

)as i2 = −1


∴∂2u

∂x2+∂2u

∂y2= f ′′(x+ iy) + g′′(x− iy)− (f ′′(x+ iy) + g′′(x− iy)) = 0

Example 271. Show that, if f and g are arbitrary functions, then

u = f(x− vt+ iαy) + g(x− vt− iαy)

is solution of∂2u

∂x2+∂2u

∂y2=

1

c2

∂2u

∂t2

provided that

α2 = 1− v2

c2


u = f(x− vt+ iαy) + g(x− vt− iαy) (8.14)

Differentiating partially with respect to x

∂u

∂x= f ′(x− vt+ iαy) + g′(x− vt− iαy)

∂2u

∂x2= f ′′(x− vt+ iαy) + g′′(x− vt− iαy)

Differentiating partially with respect to y

∂u

∂y= (iα)f ′(x− vt+ iαy) + (−iα)g′(x− vt− iαy)

∂2u

∂y2= (iα)2f ′′(x− vt+ iαy) + (−iα)2g′′(x− vt− iαy)

∂2u

∂y2= −α2f ′′(x− vt+ iαy)− α2g′′(x− vt− iαy)

Differentiating partially with respect to t

∂u

∂t= (−v)f ′(x− vt+ iαy) + (−v)g′(x− vt− iαy)

∂2u

∂t2= v2f ′′(x− vt+ iαy) + v2g′′(x− vt− iαy)

1

v2

∂2u

∂t2= f ′′(x− vt+ iαy) + g′′(x− vt− iαy) (8.15)

Now,

∂2u

∂x2+∂2u

∂y2= (1− α2)f ′′(x− vt+ iαy) + (1− α2)g′′(x− vt− iαy)

or,∂2u

∂x2+∂2u

∂y2= (1− α2)(f ′′(x− vt+ iαy) + g′′(x− vt− iαy))

8.1. THE ORIGIN OF SECONDORDER PARTIAL DIFFERENTIAL EQUATION413

Using the equaation (8.15)

∂2u

∂x2+∂2u

∂y2=

(1− α2)

v2

∂2u

∂t2(8.16)

But we have

α2 = 1− v2

c2

or,v2

c2= 1− α2

or,1

c2=

1− α2

v2

Now, from (8.16), we get

∂2u

∂x2+∂2u

∂y2=

1

c2

∂2u

∂t2(8.17)

Hence (8.14) is solution of (8.17).

Example 272. If

z = f(x2 − y) + g(x2 + y)

where f, g are arbitrary functions, prove that

∂2z

∂x2− 1

x

∂z

∂x= 4x2 ∂

2z

∂y2

Solution: Here

z = f(x2 − y) + g(x2 + y) (8.18)

Differentiating (8.18) partially with respect to x,

∂z

∂x= 2x f ′(x2 − y) + 2x g′(x2 + y) (8.19)

or,1

x

∂z

∂x= 2

(f ′(x2 − y) + g′(x2 + y)

)(8.20)

Again, differentiating (8.19) with respect to x

∂2z

∂x2= 2 f ′(x2 − y) + 2 g′(x2 + y) + 4x2 f ′′(x2 − y) + 4x2 g′′(x2 + y) (8.21)

Differentiating (8.18) with respect to y

∂z

∂y= −f ′(x2 − y) + g′(x2 + y)

∂2z

∂y2= f ′′(x2 − y) + g′′(x2 + y)

∴ 4x2 ∂2z

∂y2= 4x2f ′′(x2 − y) + 4x2g′′(x2 + y) (8.22)


Now from (8.20) and (8.22)

∂2z

∂2x− 1

x

∂z

∂x= 2 f ′(x2 − y) + 2 g′(x2 + y) + 4x2 f ′′(x2 − y) + 4x2 g′′(x2 + y)

−2(f ′(x2 − y) + g′(x2 + y

)= 4x2f ′′(x2 − y) + 4x2g′′(x2 + y)

From (8.22)

∂2z

∂x2− 1

x

∂z

∂x= 4x2 ∂

2z

∂y2

Example 273. Verify that tha partial differential equation

2r + 5s+ 2t = 0

is satisfied by

z = f(2y − x) + g(y − 2x)

where f and g are arbitrary functions. TU 2066.

Solution: Here

p =∂z

∂x= −f ′(2y − x)− 2g′(y − 2x)

q =∂z

∂y= 2f ′(2y − x) + g′(y − 2x)

r =∂2z

∂x2= f ′′(2y − x) + 4g′′(y − 2x)

t =∂2z

∂y2= 4f ′′(2y − x) + g′′(y − 2x)

s =∂2z

∂x∂y= −2f ′′(2y − x)− 2g′′(y − 2x)

Now,

2r + 5s+ 2t = 2f ′′(2y − x) + 8g′′(y − 2x)− 10f ′′(2y − x)− 10g′′(y − 2x)

+8f ′′(2y − x) + 2g′′(y − 2x) = 0

Hence, the partial differential equation

2r + 5s+ 2t = 0

is satisfied by

z = f(2y − x) + g(y − 2x)

where f and g are arbitrary functions.

8.2. LINEAR PARTIAL DIFFERENTIAL EQUATIONWITH CONSTANT COEFFICIENTS 415

Example 274. Verify that tha partial differential equation

4r + 12s+ 9t = 0

is satisfied by

z = f(2y − 3x) + xg(2y − 3x)

where f and g are arbitrary functions. TU 2066.

Solution: Here

z = f(2y − 3x) + xg(2y − 3x)

Diferetiating partially with respect to x and y,

p =∂z

∂x= −3f ′(2y − 3x) + g(2y − 3x)− 3xg′(2y − 3x)

q =∂z

∂y= 2f ′(2y − 3x) + 2xg′(2y − 3x)

r =∂2z

∂x2= 9f ′′(2y − 3x)− 3g′(2y − 3x)− 3g′(2y − 3x) + 9xg′(2y − 3x)

t =∂2z

∂y2= 4f ′′(2y − 3x) + 4xg′′(2y − 3x)

s =∂2z

∂x∂y= −6f ′′(2y − 3x) + 2g′(2y − 3x)− 6xg′′(2y − 3x)

Now,

4r + 12s+ 9t = 36f ′′(2y − 3x)− 12g′(2y − 2x)− 12g′(2y − 2x) + 36xg′′(2y − 3x)

−72f ′′(2y − 3x) + 24g′(2y − 3x)− 72xg′′(2y − 3x) + 36f ′′(2y − 3x) +

36xg′′(2y − 3x) = 0

(8.23)

Hence, the partial differential equation

4r + 12s+ 9t = 0

is satisfied by

z = f(2y − 3x) + xg(2y − 3x)

where f and g are arbitrary functions.

8.2 Linear Partial Differential Equation with Constant

Coefficients

A partial differential equation is said to be linear if all the terms in the equations

involve the partial derivatives at most in first degree. A general linear partial dif-


ferential equation will be of the form(A0

∂nz

∂xn+A1

∂nz

∂xn−1∂y+ · · ·+An

∂nz

∂yn

)+

+

(B0

∂n−1z

∂xn−1+B1

∂n−1z

∂xn−2∂y+ · · ·+Bn−1

∂n−1z

∂yn−1

)+ · · ·+(

M0∂z

∂x+M1

∂z

∂y

)+N0z = f(x, y) (8.24)

where A0, A1, · · · , An, B0, · · · ,M0,M1, N0 are all functions of x, y or constants.

If A0, A1, · · · , An, B0, · · · ,M0,M1, N0 are all constants then the equation (8.24) is

called linear partial differential equation of first order with constant coefficients.

For simplicity, we denote the operators,

D =∂

∂x, and D′ =

∂

∂y, and Dn =

∂n

∂x2, D′n =

∂n

∂y2

In this notations the equation (8.24) becomes

F (D,D′)z = f(x, y) (8.25)

where

F (D,D′) = (A0Dn+A1D

n−1D′+· · ·+AnD′n)+(B0Dn−1+B0D

n−2D′+· · ·+Bn−1Dn−1)

+ · · ·+ (M0D +M1D′) +N0

8.2.1 Linear Homogeneous and Non-homogeneous Equation with

Constant Coefficients

A partial diffeential equation of the form

A0∂nz

∂xn+A1

∂nz

∂xn−1∂y+ · · ·+An

∂nz

∂yn= f(x, y) (8.26)

where A0, A1, · · · , An all are constants and f(x, y) is function of independent vari-

ables x and y, is called as linear homogenous partial differential equation of nth order

with constant coefficients. i.e.

F (D,D′)z = f(x, y)

is homogenous if F (D,D′) is a homogeous function in D,D′ of degree n.

When all the derivatives in F (D,D′) are not of the same order, then it is called a

non-homogenous linear partial differential equation with constant coefficients.


8.2.2 Linear Differential Operator

We classify linear differential operator F (D,D′) into two type

1. Reducible A linear operator F (D,D′) is said to be reducible if it can be

written as the product of linear factor of the form D + aD′ + b where a and

b are constants. For example, the oprator F (D,D′) = D2 −D′2 is reducible,

since it can be written as

F (D,D′) = (D −D′)(D −D′)

2. Irreducible A linear operator F (D,D′) is said to be irreducible if it can not

be written as the product of linear factor of the form D+aD′+ b where a and

b are constants. For example, the oprator F (D,D′) = D2 +D′2 is irreducible,

since it can not be written as product of two linear factors.

8.2.3 Solution

As in case of ordinary differential equations, the complete solution of partial differ-

ential equation

F (D,D′)z = f(x, y)

consists of two parts

1. Complementary function: The complementry function of the partial dif-

ferential equation

F (D,D′)z = f(x, y)

is the general solution of

F (D,D′)z = 0

2. Particular Integral: Let the differential equation can be

F (D,D′)z = f(x, y)

Operating the oprator1

F (D,D′)

on both sides, we get

1

F (D,D′)F (D,D′)z =

1

F (D,D′)f(x, y)

or, z =1

F (D,D′)f(x, y)

as,1

F (D,D′)and F (D,D′) inverse of each of others.


Thus the value of

z =1

F (D,D′)f(x, y)

is called particular integral.

Theorem 24. If u is the complementary function and z1 a particular integral inte-

gral of a linear partial differential equation, then u+ z1 is a general solution of the

equation.

Proof: The given differential equation be

F (D,D′)z = f(x, y) (8.27)

Since u is complementary function of (8.27), so it is general solution of F (D,D′)z =

0, so

F (D,D′)u = 0 (8.28)

and u contains as many arbitrary functions as in the order of (8.27). Also z1 is a

particular integral of (8.27) and it does not contains any arbitrary functions and

constants. Thus

F (D,D′)z1 = f(x, y) (8.29)

Adding, the equations (8.28) and (8.29)

F (D,D′)u+ F (D,D′)z1 = f(x, y)

F (D,D′)(u+ z1) = f(x, y)

This shows that u+ z1 is general solution of (8.27)

Theorem 25. If u1, u2, u3, · · · , un are solutions of homogeneous linear partial dif-

ferential equation F (D,D′)z = 0 then

n∑r=1

crur

where the cr are arbitrary constants, is a solution.

Proof: Here, u1, u2, u3 · · · , un are solutions of homogeneous linear partial differ-

ential equation

F (D,D′)z = 0 (8.30)

Then

F (D,D′)ur = 0 for all r = 1, 2, · · · , n (8.31)


Since, the differential operator F (D,D′) is linear, so

F (D,D′)(crur) = crF (D,D′)ur for any constant cr

and

F (D,D′)

n∑r=1

ur =

n∑r=1

F (D,D′)ur

Thus,

F (D,D′)

n∑r=1

crur =

n∑r=1

F (D,D′)crur = cr

n∑r=1

F (D,D′)ur = 0 using (8.31)

Hencen∑r=1

crur

is a solution of homogeneous linear partial differential equation.

8.2.4 Method of Finding the Complementary Function of PDE with

Constant Coefficients.

Let F (D,D′)z = f(x, y) be the given partial differential equation with constant

coefficients. The complementary function is solution of F (D,D′)z = 0.

Let

F (D,D′) = (D = m1D′)(D −m2D

′) · · · (D −mnD′)

where m1,m2,m3, · · · ,mn are constants. Consider

(D −mnD′)z = 0

or,∂z

∂x−mn

∂z

∂y= 0

or, p−mnq = 0

which is linear. The lagrange’s auxiliary equations are

dx

1=

dy

−mn=dz

0


dy +mndx = 0

Integrating,

y +mnx = c1 (8.32)

and from last two ratios

dz = 0


Integrating,

z = c2 (8.33)

From (8.32) and (8.33), we have

c2 = φn(c1)

or, z = φn(y +mnx)

is a solution of (D−mnD′)z = 0 and is of F (D,D′)z = 0. When m1,m2,m3, · · · ,mn

are all are distinct, we get n distinct solutions z = φr(y+mrx), where r = 1, 2, · · · , nof F (D,D′)z = 0 and from theorem (25),

z =

n∑r=1

φr(y +mrx)

is solution of F (D,D′)z = 0. Thus, the complementary function of F (D,D′)z =

f(x, y) is

z = φ1(y +m1x) + φ2(y +m2x) + φ3(y +m3x) + · · ·+ φn(y +mnx)

1. Let z = φ(y+mx) be a solution of the homogeneous partial differential equa-

tion

(Dn +A1Dn−1D′ +A2D

n−2D′ + · · ·+AnD′n)z = 0 (8.34)

Then

Dz =∂z

∂x= mφ′(y+mx), Drz = mrφ(r)(y+mx), Dnz = mnφ(n)(y+mx)

D′z =∂z

∂x= φ′(y +mx), D′rz = φ(r)(y +mx), D′nz = φ(n)(y +mx)

But r + s = n, so

DrD′sz = DrD′sφ(y+mx) = Drφ(s)(y+mx) = mrφ(r+s)(y+mx) = mrφ(n)(y+mx)

Putting z = φ(y +mx) in the equation (8.33) , we get

(mn +A1mn−1 +A2m

n−2 + · · ·+An)φ(n)(y +mx) = 0

=⇒ mn +A1mn−1 +A2m

n−2 + · · ·+An = 0

Thus, if y = φ(y +mx), where φ is arbitrary function, is soution of

(Dn +A1Dn−1D′) +A2D

n−2D′ + · · ·+AnD′n)z = 0

then m is a solution of

mn +A1mn−1 +A2m

n−2 + · · ·+An = 0 (8.35)

This equation (8.35), is called auxiliary equation of (8.34). Which is obtained

by putting D = m and D′ = 1 in F (D,D′) = 0.


2. 2 If the partial differential equation of second order, then

(D2 +A1DD′ +A2D

′2)z = 0


m2 +A1m+A2 = 0

which is quadratic in m and it has two solutions m1 and m2.

(a) If m1 and m2 are unequal, then

z = φ1(y +m1x) + φ2(y +m2x)

is required solution.

(b) If two roots equal say m, then the given equation can be written as

(D −mD′)(D −mD′)z = 0 (8.36)

Let u = (D −mD′)z. Then the equation (8.36) becomes

(D −mD′)u = 0

and the equation has solution

u = φ(y +mx)

Thus,

(D −mD′)z = φ1(y +mx)

or, p−mq = φ1(y +mx)

The lagrange’s auxiliary equations are

dx

1=

dy

−m=

dz

φ1(y +mx)

From, first two ratios

dy +mdx = 0

Integrating,

y +mx = c1 (8.37)

From first and last ratios,

dx

1=

dz

φ1(y +mx)

or, dx =dz

φ(c1)


or, dz = φ1(c1)dx

Integrating,

z = xφ1(c1) + c2

or, c2 = z − xφ1(y +mx)

Required solution is

c2 = φ2(c1)

or, z − xφ1(y +mx) = φ2(y +mx)

or, z = xφ1(y +mx) + φ2(y +mx) (8.38)


If a root m is repeated r times, then the general solution is

z = φ1(y +mx) + xφ2(y +mx) + x2φ3(y +mx) + · · ·+ xr−1φr(y +mx)

8.2.5 Determination of Particular Integral (P.I.)

We have, the particular integral of linear homogeneous partial differential equation

F (D,D′)z = f(x, y) is

P.I. =1

F (D,D′)f(x, y)


Case I

1

F (D,D′)φ(ax+by) =

1

F (a, b)×nth integral of φ(t) where t = ax+by if F (a, b) 6= 0

Case II1

F (D,D′)eax+by =

1

F (a, b)eax+by if F (a, b) 6= 0

Case III1

F (D,D′)sin(ax+ by) and

1

F (D,D′)sin(ax+ by)

is obtained by putting D2 = −a2, DD′ = −ab and D′2 = −b2, provided the denom-

inator is not zero.

Case IV1

F (D,D′)

(eax+by V

)= eax+by 1

F (D + a,D + b)V

Case V1

F (D,D′)xmyn = (F (D,D′))−1xmyn

and expand the expression [F ‘(D,D′)] as Dn[1 + φ(D,D′)]

Notes

r =∂2z

∂x2= D2z, s =

∂2z

∂x∂y= DD′z, and t =

∂2z

∂y2= D′2z


Example 275. Solve the equation

∂4z

∂x4− ∂4z

∂y4=

∂4z

∂x2∂y2

Solution: We have

D =∂

∂x, D′ =

∂

∂y

The given PDE is

(D4 +D′4 − 2D2D′2)z = 0

Put D = m, D′ = 1, the auxiliary equation is

m4 − 2m2 + 1 = 0

or, (m− 1)2(m+ 1)2 = 0

∴ m = 1, 1 and m = −1,−1

Thus, the solution is

z = φ1(y + x) + xφ2(y + x) + φ3(y − x) + xφ3(y − x)

where φ1, φ2, φ3 and φ4 are arbitrary functions.

Example 276. Solve

r = a2t


r = a2t

or,∂2x

∂x2= a2 ∂

2z

∂y2

or, (D2 − a2D′2)z = 0


m2 − a2 = 0

∴ m = a, m = −a

Therefore, the solution is

z = φ1(y + ax) + φ2(y − ax)

Example 277. Solve∂4z

∂x4− ∂4z

∂y4= 0


Solution: The given partial diffeential is

(D4 −D′4)z = 0


m4 − 1 = 0

or, (m2 + 1)(m2 − 1) = 0

∴ m = ±i, m = ±1

Thus, the solution is

z = φ1(y + x) + φ2(y − x) + φ3(y + ix) + φ4(y − ix)

Note:1

Df =

∫fdx here we integrate taking y as a constant

1

D′f =

∫fdy here we integrate taking x as a constant

Example 278. Find the solution of the equation

∂2z

∂x2− ∂2z

∂y2= x− y


(D2 −D′2)z = x− y


m2 − 1 = 0

Solving,

m = 1, −1

Thus, the complementary function is

C.F. = φ1(y + x) + φ2(y − x)


For particular integral, we have

P.I. =1

F (D,D′)(x− y)

=1

D2 −D′2(x− y)

=1

D2

(1− D′2

D2

)−1

(x− y)

=1

D2

(1 +

D′2

D2+D′4

D4+ · · ·

)(x− y)

as (1 + x)n = 1 + nx+n(n− 1)x2

2!+n(n− 1)(n− 2)x3

3!+ · · ·

=1

D2

((x− y) +

D′2

D2(x− y) +

D′4

D4(x− y) + · · ·

)=

1

D2((x− y) + 0)

as D′2(x− y) =∂2

∂y2(x− y) =

∂

∂y

(∂

∂y(x− y)

)=

∂

∂y(−1) = 0

D′4(x− y) = 0, and so on

=1

D

∫(x− y)dx

=1

D

(x2

2− xy

)=

1

2

∫x2 − y

∫xdx

=x3

6− x2y

2


z = C.F.+ P.I.

= φ1(y + x) + φ2(y − x) +x3

6− x2y

2


∂x2− ∂2z

∂x∂y− 6

∂2z

∂y2= xy


(D2 −DD′ − 6D′2)z = xy


m2 −m− 6 = 0

or, (m− 3)(m+ 2) = 0


∴,m = 3, m = −2

THus, the complementary function is

C.F. = φ1(y + 3x) + φ2(y − 2x)

and

P.I. =1

F (D,D′)f(x, y)

=1

D2 −DD′ − 6D′2xy

=1

D2

(1−

(D′

D+ 6

D′2

D2

))−1

xy

=1

D2

(1 +

D′

D+D′2

D2+ · · ·

)(xy)

=1

D2

(xy +

D′

D(xy) +

D′2

D2(xy) + · · ·

)=

1

D2

(xy +

1

Dx+ 0

)as D′(xy) =

∂

∂y(xy) = x

=1

D2

(xy +

∫xdx

)=

1

D2

(xy +

x2

2

)=

1

D

∫ (xy +

x2

2

)dx

=1

D

(x2y

2+x4

4

)=

y

2

∫x2dx+

1

4

∫x4dx

=x3y

6+x5

24


z = C.F.+ P.I.

or, z = φ1(y + 3x) + φ2(y − 2x) +x3y

6+x5

24

Example 280. Solve

r − 7s+ 12t = ex−y


(D2 − 7DD′ + 12D′2)z = ex−y


m2 − 7m+ 12 = 0


or, (m− 3)(m− 4) = 0

∴ m = 3, m = 4

The complementary function is

C.F. = φ1(y + 3x) + φ2(y + 4x)

Particular integral

P.I. =1

D2 − 7DD′ + 12D′2ex−y =

ex−y

12 − 7 · 1 · (−1) + 12 · (−1)2=ex−y

20

The complete solution is

z = C.F.+ P.I.

or,z = φ1(y + 3x) + φ2(y + 4x) +ex−y

20

Example 281. Solve

r + s− 2t = ex+y


(D2 +DD′ − 2D′2)z = ex+y


m2 +m− 2 = 0

or, (m+ 2)(m− 1) = 0

∴ m = −2, m = 1


C.F. = φ1(y − 2x) + φ2(y + x)

Particular integral

P.I. =1

D2 +DD′ − 2D′2ex+y (8.39)

If we substitute D = 1 and D′ = 1, the denominator vanishes, so we multiply the

expression by x and differentiate denominator with respect to D

P.I. = x1

2D +D′ex+y = x

ex+y

2 · 1 + 1=x

3ex+y


z = C.F.+ P.I.

or,z = φ1(y − 2x) + φ2(y + x) +xex+y

3


Theorem 26. If the given partial differential equation is F (D,D′)z = φ(ax + by),

where F (D,D′) is homogeous function of D and D′ of degree n, then the particular

integral of the PDE is

1

F (D,D′)φ(n)(ax+ by) =

1

F (a, b)φ(ax+ by)

where F (a, b) 6= 0, φ(n) being the nth derivative of φ(ax+ by) with respect to ax+ by.

Note:

P.I. =1

F (a, b)× nthderivative of φ(ax+ by) with respect to ax+ by.

i.e. P.I. =1

F (a, b)× nthderivative of φ(t) with respect to t = ax+ by.

(8.40)


∂x2+∂2z

∂y2= 24(x+ y)


(D2 +D′2)z = ex+y


m2 + 1 = 0

or, m2 = −1 = i2

∴ m = −i, m = i


C.F. = φ1(y − ix) + φ2(y + ix)

Particular integral

P.I. =1

D2 +D′224(x+ y)

=24

12 + 12

∫ ∫(tdt)dt

= 12 · 1

2

∫t2dt

=12

6t3 = 2(x+ y)3


z = C.F.+ P.I.

or, z = φ1(y − ix) + φ2(y + ix) + 2(x+ y)3



∂x2+ 3

∂2z

∂x∂y+ 2

∂2z

∂y2= x+ y


(D2 + 3DD” + 2D′2)z = x+ y


m2 + 3m+ 2 = 0

or, (m+ 2)(m+ 1) = 0

∴ m = −2, m = −1


C.F. = φ1(y − x) + φ2(y − 2x)

Particular integral

P.I. =1

D2 + 3DD′ + 2D′2(x+ y)

=1

12 + 3 · 1 · 1 + 2 · 12

∫ ∫(tdt)dt

=1

6· 1

2

∫t2dt

=1

36t3 =

(x+ y)3

36


z = C.F.+ P.I.

or, z = φ1(y − 2x) + φ2(y − x) +(x+ y)3

36

Example 284. Solve

∂3z

∂x3− 2

∂3z

∂x2∂y− ∂3z

∂x∂y2+ 2

∂3z

∂y3= ex+y

Solution: The given partial differential equation can be written as

(D3 − 2D2D′ −DD′2 + 2D′3)z = ex+y


m3 − 2m2 −m+ 2 = 0

or, m2(m− 2)− 1(m− 2) = 0

or, (m− 2)(m− 1)(m+ 1) = 0

∴ m = 1, m = −1, m = 2



C.F. = φ1(y + x) + φ2(y + 2x) + φ3(y − x)

Particular integral

P.I. =1

D3 − 2D2D′ −DD′2 + 2D′3ex+y

= x1

3D2 − 4DD′ −D′2ex+y as F (1, 1) = 0

=1

3 · 12 − 4 · 1 · 1− 12

∫ ∫etdt dt

= −x2et = −xe

x+y

2


z = C.F.+ P.I. = φ1(y + x) + φ2(y + 2x) + φ3(y − x)− xex+y

2

General Method for Particular Integral

Let us consider the equation

(D −mD′)z = f(x, y)

or,∂z

∂x−m∂z

∂y= f(x, y)

or, p−mq = f(x, y)

The Lagrange’s auxiliary equation are

dx

1=

dy

−m=

dz

f(x, y)


dy = −mdx

Integrating,

y +mx = c

From first and third ratios

dz = f(x, y) = f(x, c−mx)dx Putting value of y

Integrating,

z =

∫f(x, c−mx)dx

After integration c−mx is repleaced by y, it does not contains any arbitrary constant.

Now

(D −mD′)z = f(x, y)

=⇒ 1

D −mD′f(x, y) = z


Putting values of z

1

D −mD′f(x, y) =

∫f(x, c−mx)dx

Which gives particular integral of (D −mD′)z = f(x, y).

Again, let the given partial differential equation be

F (D,D′)z = f(x, y)

Let F (D,D′) = (D −m1D′)(D −m2D

′)(D −m3D′) · (D −mnD

′). Then

P.I. =1

F (D,D′)f(x, y) =

1

D −m1D′1

D −m2D′1

D −m3D′1

D −mnD′f(x, y)

can be determined by the repeated application of the method given above.

Example 285. Solve

r + s− 6t = y cosx

Solution: The given partial differential equation can be written as

(D2 +DD′ − 6D′2)z = y cosx


m2 +m− 6 = 0

or, (m+ 3)(m− 2) = 0

∴ m = −3, m = 2


C.F. = φ1(y − 3x) + φ2(y + 2x)


Also, particular integral

P.I. =1

D2 +DD′ − 6D′2y cosx

=1

(D − 2D′)(D + 3D′)y cosx

=1

D − 2D′1

D + 3D′y cosx

=1

D − 2D′

∫(c+ 3x) cosxdx

=1

D − 2D′

(c+ 3x)

∫cosxdx−

∫ (d

dx(c+ 3x)

∫cosxdx

)dx

=

1

D − 2D′

(c+ 3x) sinx− 3

∫sinx dx

=

1

D − 2D′(c+ 3x) sinx+ 3 cosx

=1

D − 2D′y sinx+ 3 cosx

=

∫(k − 2x) sinx+ 3 cosxdx

=

∫(k − 2x) sinx dx+ 3

∫cosxdx

= (k − 2x)

∫sinxdx−

∫ (d

dx(k − 2x)

∫sinxdx

)dx+ 3 sinx

= −(k − 2x) cosx− 2

∫cosxdx+ 3 sinx

= −(k − 2x) cosx− 2 sinx+ 3 sinx

= −y cosx+ sinx


z = C.F.+ P.I. = φ1(y − 3x) + φ2(y + 2x)− y cosx+ sinx

8.3 Non-homogeneous

The linear partial differential equation which is not homogeneous is known as non-

homogeneous linear partial differential equation. We classify the linear operator

F (D,D′) into two type

(i) Reducible: A linear operator F (D,D′) is said to be reducible, if it can be

written as the product of linear factors of the form D + aD′ + b where a and b are

constants. For example, the operator F (D,D′) = D2−D′2 is reducible, since it can

be written as

F (D,D′) = D2 −D′2 = (D −D′)(D +D′)

8.3. NON-HOMOGENEOUS 433

(i) Irreducible: A linear operator F (D,D′) is said to be irreducible if it can not

be written as the product of linear factors of the form D + aD′ + b where a and b

are constants. For example, the operator F (D,D′) = D2 +D′2 is irreducible, since

it can be written as product of linear factors.

8.3.1 Reducible Partial Differential Equations

We have following results

Theorem 27. If the operator F (D,D′) is reducible in linear form, then the order

in which the linear factors occurs is unimportant.

Theorem 28. If αrD + βrD′ + γr is factor of F (D,D′) and φr(ξ) is an arbitrary

function of single variable ξ; and if αr 6= 0, then

ur = e−γrxαr φr(βrx− αry)

is a solution of F (D,D′)z = 0.

Theorem 29. If βrD′+ γr is factor of F (D,D′) and φr(ξ) is an arbitrary function

of single variable ξ; and if βr 6= 0, then

ur = e−γrxαr φr(βrx)

is a solution of F (D,D′)z = 0.

Theorem 30. If (αrD+βrD′+γr)

n is factor of F (D,D′) and φr1(ξ), φr2(ξ), · · · , φr2(ξ)

are arbitrary functions of single variable ξ; and if αr 6= 0, then

ur = e−γrxαr

[φr1(βrx− αry) + xφr2(βrx− αry) + x2φr2(βrx− αry) + · · ·+ xn−1φrn(βrx− αry)

]is a solution of F (D,D′)z = 0.

Theorem 31. If (βrD′ + γr)

n is factor of F (D,D′) and φr1(ξ), φr2(ξ), · · · , φr2(ξ)

are arbitrary functions of single variable ξ; and if βr 6= 0, then

ur = e−γrxαr

[φr1(βrx) + xφr2(βrx) + x2φr2(βrx) + · · ·+ xn−1φrn(βrx)

]is a solution of F (D,D′)z = 0.

Example 286. Solve

r + s− 2t− p− 2q = 0


(D2 +DD′ − 2D′2 −D − 2D′)z = 0

or, [D2 + 2DD′ −DD′ − 2D′2 − 1(D + 2D′)]z = 0

or, [D(D + 2D′)−D′(D + 2D′)− 1(D + 2D′)]z = 0

or, [(−D +D′ + 1)(D + 2D′)]z = 0


Hence, the given partial differential equation is reducible. The solution correspond-

ing to

(D + 2D′)z = 0

is


or, u1 = e0φ1(2x− y) = φ1(2x− y)

and the solution corresponding to

(−D +D′ + 1)z = 0

is


or, u2 = e−x−1 φ2((1)x− (−1)y) = exφ2(x+ y)


z = u1 + u2

z = φ1(2x− y) + exφ2(x+ y)

Example 287. Solve

r + 2s+ t+ 2p+ 2q + z = 0


(D2 + 2DD′ +D′2 + 2D + 2D′ + 1)z = 0

or, [(D +D′)2 + 2(D +D′) + 12]z = 0

or, (D +D′ + 1)2z = 0


z = e−γrxαr [φ1(β1x− α1y) + xφ2(β1x− α1y)]

or, z = e−x[φ1(y − x) + xφ2(y − x)]

Example 288. Solve:

(D2 −DD′ +D′ − 1)z = cos(x+ 2y) + ey

Solution: We have

D2 −DD′ +D′ − 1 = D2 − 1−DD′ +D′

= (D − 1)(D + 1)−D′(D − 1)

= (D − 1)(D −D′ + 1)


Thus, the given partial differential equation is reducible to linear

(D − 1)(D −D′ + 1)z = cos(x+ 2y) + ey

C.F. = e−γ1xα1 φ1(β1x− α1y) + e

−γ2xα2 φ2(β2x− α2y)

= exφ1(y) + e−xφ2(y + x)

Now, particular integral

P.I. =1

D2 −DD′ +D′ − 1[cos(x+ 2y) + ey] = P1 + P2

where,

P1 =1

D2 −DD′ +D′ − 1cos(x+ 2y)

=1

−12 − (−1 · 2) +D′ − 1cos(x+ 2y)

=1

D′cos(x+ 2y)

=

∫cos(x+ 2y)dy

=sin(x+ 2y)

2

P2 =1

D2 −DD′ +D′ − 1ey

=1

(D − 1)(D −D′ + 1)ey

= ey1

(D − 1)(D − 1 + 1)1

= −ey 1

D(1−D)1

= −ey 1

D(1−D)−1 · 1

= −ey 1

D(1 +D +D2 + · · · )1

= −ey 1

D1

= −ey∫dx

= −xey

Hence, the complete solution is

z = C.F. + P.I. = C.F. + P1 + P2

z = exφ1(y) + e−xφ2(y + x) +1

2sin(x+ 2y)− xey

Example 289.

(D2 −DD′ − 2D)z = sin(3x+ 4y)



D(D −D′ − 2)z = sin(3x+ 4y)

or,C.F. = φ(y) + e2xφ2(y + x)

P.I. =1

D2 −DD′ − 2Dsin(3x+ 4y)

=1

−32 − (−3 · 4)− 4Dsin(3x+ 4y)

=1

3− 2Dsin(3x+ 4y)

=3 + 2D

9− 4D2sin(3x+ 4y)

=3 + 3D

9− 4(−9)sin(3x+ 4y)

=3 + 3D

45sin(3x+ 4y)

=1

45(3 sin(3x+ 4y) + 3D sin(3x+ 4y))

=1

45

(3 sin(3x+ 4y) + 3

∂

∂xsin(3x+ 4y)

)=

1

45(3 sin(3x+ 4y) + 9 cos(3x+ 4y))

=1

15(sin(3x+ 4y) + 3 cos(3x+ 4y))

8.3.2 Irreducible Partial Differential Equation

If the operator F (D,D′) is irreducible, it is not always possible to find a solution

which contains full arbitrary functions.

Example 290. Solve:∂2z

∂x2=

1

k

∂z

∂t

Solution: Let

D =∂

∂xand D′ =

∂

∂t

Then the given partial differential equation is

(kD2 −D′)z = 0 (8.41)

Here, kD2 −D′ is not reducible to linear factor in D and D′. Let

z = Aeax+by

be a trial solution of the given partial differential equation. Then

D2z = Aa2eax+by, D′z = Abeax+by


Now from (8.41), we get

kAa2eax+by = Abeax+by

=⇒ b = ka2

Thus,

z = Aeax+ka2y

and the most general solution is

z =∑i

Aieaix+ka2i y

Example 291. Solve

r − s+ 2q − z = x2y2

Solution: Let

D =∂

∂xand D′ =

∂

∂y

Then the given partial differential equation is

(D2 −DD′ + 2D − 1)z = x2y2 (8.42)

For C.F. let

z = Aeax+by (8.43)

be solution of

(D2 −DD′ + 2D − 1)z = 0 (8.44)

Then

D2z = Aa2eax+by, DD′z = Aabeax+by, D′z = Abeax+by

From (8.44)

(Aa2 −Aab+ 2Ab−A)eax+by = 0

or, A(a2 − ab+ 2b− 1) = 0

or, a2 − ab+ 2b− 1 = 0

or, b =1− a2

2− aThus, the required C.F. is

C.F. =∑

Aeax+by where b =1− a2

2− aFor particular integral

P.I. =1

D2 −DD′ + 2D′ − 1x2y2

= −[1− (D2 −DD′ + 2D′)

]−1x2y2

= −[1 + (D2 −DD′ + 2D′) + (D2 −DD′ + 2D′)2 + (D2 −DD′ + 2D′)3 + · · · ](x2y2

)


Expanding each term and taking the term involving up toD2 andD′2 a asD3(x2y2) =

0 and D′3(x2y2) = 0 and other higher order derivatives terms are zeros.

P.I. = −[1 +D2 −DD′ + 2D′ +D2D′2 + 4D′2 + 4D2D′ − 4DD′2 + 12D2D′2 · · · ](x2y2

)= −(x2y2 + 2y2 − 4xy + 4x2y + 4 + 8x2 + 16y − 16x+ 48)

= −x2y2 − 2y2 + 4xy − 4x2y − 8x2 − 16y + 16x− 52

Therefore, the general solution of the given equation is

z = C.F. + P.I.

=∑

Aeax+by − x2y2 − 2y2 + 4xy − 4x2y − 8x2 − 16y + 16x− 52

where b =1− a2

2− a

Example 292. Solve

(D2 −D′)z = 2y − x2


(D2 −D′)z = 2y − x2

Since D2 −D′ can not be resolved into linear factors in D and D′, so for C.F. let

z = Aeax+by

be solution of

(D2 −D′)z = 0 (8.45)

Then



(Aa2 −Ab)eax+by = 0

=⇒ b− a2 = 0 as Aeax+by 6= 0

=⇒ b = a2

The required C.F. is ∑Aeax+by where b = a2

i.e. C.F.=∑

Aeax+a2y


For particular integral

P.I. =1

D2 −D′(2y − x2)

= − 1

D′

(1− D2

D′

)−1

(2y − x2)

= − 1

D′

(1 +

D2

D′+D4

D′2+ · · ·

)(2y − x2)

=

(− 1

D′− D2

D′2− D4

D′3+ · · ·

)(2y − x2)

= − 1

D′(2y − x2)− D2

D′2(2y − x2) as

D4

D′3(2y − x2) = 0, · · ·

= −2

∫ydy + x2

∫dy +

1

D′2(2)

= −2y2

2+ x2y + 2

∫ (∫dy

)dy

= −y2 + x2y − 2y2

2= −y2 + x2y + y2 = x2y

Therefore, the general solution is

z = C.F.+ P.I. =∑

Aeax+a2y + x2y

Example 293. Solve

(D2 −D′)z = e2x+y


(D2 −D′)z = e2x+y


z = Aeax+by

be solution of

(D2 −D′)z = 0 (8.46)

Then




=⇒ b− a2 = 0 as Aeax+by 6= 0

=⇒ b = a2



i.e. C.F.=∑

Aeax+a2y


P.I. =1

D2 −D′eax+by

=1

22 − 1e2x+y =

eax+by

3


z = C.F.+ P.I. =∑

Aeax+a2y + x2y

Example 294. Solve

(D2 −D′)z = ex+y


(D2 −D′)z = ex+y


z = Aeax+by

be solution of

(D2 −D′)z = 0 (8.47)

Then




=⇒ b− a2 = 0 as Aeax+by 6= 0

=⇒ b = a2


i.e. C.F.=∑

Aeax+a2y



P.I. =1

D2 −D′eax+by

= ex+y 1

(D + 1)2 − (D′ + 1)1 since D2 −D′ = 12 − 1 = 0

= ex+y 1

D2 + 2D + 1−D′ − 1(1)

= ex+y 1

D2 + 2D −D′(1)

= −ex+y 1

D′

(1− D2 +D′

D

)−1

(1)

= −ex+y 1

D′

(1 +

D2 +D′

D′+

(D2 +D′)2

D′2· · ·)

(1)

= −ex+y 1

D′(1 + 0)

= −ex+y

∫dy

= −yex+y


z = C.F.+ P.I. =∑

Aeax+a2y − yex+y

Example 295. Solve

(D2 −D′)z = A cos(lx+my)


(D2 −D′)z = A cos(lx+my)


z = Aeax+by

be solution of

(D2 −D′)z = 0 (8.48)

Then

D2z = Ba2eax+by, D′z = Bbeax+by


(Ba2 −Bb)eax+by = 0

=⇒ b− a2 = 0 as Beax+by 6= 0

=⇒ b = a2


The required C.F. is ∑Beax+by where b = a2

i.e. C.F.=∑

Beax+a2y


P.I. =1

D2 −D′A cos(lx+my)

= A1

D2 −D′cos(lx+my)

= A1

−l2 −D′cos(lx+my)

= −A 1

l2 +D′cos(lx+my)

= −A l2 −D′

l4 −D′2cos(lx+my)

= −A l2 −D′

l4 − (−m2)cos(lx+my)

= −A l2 −D′

l4 +m2cos(lx+my)

=−A

l4 +m2

(l2 cos(lx+my)−D′ cos(lx+my)

)= − A

l4 +m2

(l2 cos(lx+my) +m sin(lx+my)

)Therefore, the general solution is

z = C.F.+ P.I. =∑

Beax+a2y − A

l4 +m2

(l2 cos(lx+my) +m sin(lx+my)

)8.4 Second Order Partial Differential Equation with Vari-

able Coefficients

8.4.1 Canonical Forms (Method of Transformations)

We shall consider the equation of form

Rr + Ss+ Tt+ F (x, y, z, p, q) = 0 (8.49)

where R,S, T are functions of x and y possessing continuous partial derivatives of

higher order as necessary.

We shall show that any equation of the type (8.49) can be reduced to one of the

three Canonical forms by a suitable change of independent variables.

Suppose we change the independent variable from x, y to u, v, where

u = u(x, y), v = v(x, y)

8.4. SECONDORDER PARTIAL DIFFERENTIAL EQUATIONWITH VARIABLE COEFFICIENTS443

Then, we have

p =∂z

∂x=∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x=∂u

∂x

∂z

∂u+∂v

∂x

∂z

∂v

q =∂z

∂y=∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y=∂u

∂y

∂z

∂u+∂v

∂y

∂z

∂v

∴∂

∂x=∂u

∂x

∂

∂u+∂v

∂x

∂

∂v

and∂

∂y=∂u

∂y

∂

∂u+∂v

∂y

∂

∂v

Now

r =∂2z

∂x2=

∂

∂x

(∂z

∂x

)=

(∂u

∂x

∂

∂u+∂v

∂x

∂

∂v

)(∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x

)=

∂u

∂x

∂

∂u

(∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x

)+∂v

∂x

∂

∂v

(∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x

)=

∂u

∂x

[∂

∂u

(∂z

∂u

∂u

∂x

)+

∂

∂u

(∂z

∂v

∂v

∂x

)]+∂v

∂x

[∂

∂v

(∂z

∂u

∂u

∂x

)+

∂

∂v

(∂z

∂v

∂v

∂x

)]=

∂u

∂x

[∂2z

∂u2

∂u

∂x+

∂

∂u

(∂u

∂x

)∂z

∂u+

∂2z

∂u∂v

∂v

∂x+

∂

∂u

(∂v

∂x

)∂z

∂v

]+∂v

∂x

[∂2z

∂u∂v

∂u

∂x+

∂

∂v

(∂u

∂x

)∂z

∂u+∂2z

∂v2

∂v

∂x+

∂

∂v

(∂v

∂x

)∂z

∂v

]Since u and v are independent variables, so

∂

∂u

(∂v

∂x

)= 0,

∂

∂v

(∂u

∂x

)= 0,

∂2z

∂u∂v=

∂2z

∂v∂u

r =∂u

∂x

[∂2z

∂u2

∂u

∂x+

∂

∂u

(∂u

∂x

)∂z

∂u+

∂2z

∂u∂v

∂v

∂x

]+∂v

∂x

[∂2z

∂u∂v

∂u

∂x+∂2z

∂v2

∂v

∂x+

∂

∂v

(∂v

∂x

)∂z

∂v

]=

∂2z

∂u2

(∂u

∂x

)2

+ 2∂2z

∂u∂v

∂u

∂x

∂v

∂x+∂2z

∂v2

(∂v

∂x

)2

+

+∂u

∂x

∂

∂u

(∂u

∂x

)∂z

∂u+∂v

∂x

∂

∂v

(∂v

∂x

)∂z

∂v

As∂u

∂x

∂

∂u

(∂u

∂x

)=∂2u

∂x2,

∂v

∂x

∂

∂v

(∂v

∂x

)=∂2v

∂2x

Therefore r becomes

r =∂2z

∂u2

(∂u

∂x

)2

+ 2∂2z

∂u∂v

∂u

∂x

∂v

∂x+∂2z

∂v2

(∂v

∂x

)2

+∂2u

∂x2

∂z

∂u+∂2v

∂x2

∂z

∂v


Similarly

t =∂2z

∂u2

(∂u

∂y

)2

+ 2∂2z

∂u∂v

∂u

∂y

∂v

∂y+∂2z

∂v2

(∂v

∂y

)2

+∂2u

∂y2

∂z

∂u+∂2v

∂y2

∂z

∂v

and

s =∂2z

∂x∂y=

∂

∂x

(∂z

∂y

)=

(∂u

∂x

∂

∂u+∂v

∂x

∂

∂v

)(∂u

∂y

∂z

∂u+∂v

∂y

∂z

∂v

)=

∂u

∂x

∂

∂u

(∂u

∂y

∂z

∂u+∂v

∂y

∂z

∂v

)+∂v

∂x

∂

∂v

(∂u

∂y

∂z

∂u+∂v

∂y

∂z

∂v

)=

∂2z

∂u2

∂u

∂x

∂u

∂y+

∂2u

∂y∂x

∂z

∂u+

∂2z

∂u∂v

∂u

∂x

∂v

∂y+

∂2u

∂x∂y

∂z

∂v+

∂2z

∂u∂v

∂u

∂y

∂v

∂x

+∂v

∂x

∂u

∂y

∂2z

∂v2

=∂2z

∂u2

∂u

∂x

∂u

∂y+

∂2u

∂y∂x

∂z

∂u+

∂2z

∂u∂v

(∂u

∂x

∂v

∂y+∂u

∂y

∂v

∂x

)+

∂2u

∂x∂y

∂z

∂v+

∂v

∂x

∂u

∂y

∂2z

∂v2

Putting the values of r, s and t in (8.49) we get

A∂2z

∂u2+ 2B

∂2z

∂u∂v+ C

∂2z

∂v2+ F

(u, v, z,

∂z

∂u,∂z

∂v

)= 0 (8.50)

where

A = R

(∂u

∂x

)2

+ S∂u

∂x

∂u

∂y+ T

(∂u

∂y

)2

(8.51)

B = R∂u

∂x

∂v

∂x+S

2

(∂u

∂x

∂v

∂y+∂u

∂y

∂v

∂x

)+ T

∂u

∂y

∂v

∂y(8.52)

C = R

(∂v

∂x

)2

+ S∂v

∂x

∂v

∂y+ T

(∂v

∂y

)2

(8.53)

and the F is transformed form of the function of f .

Now our aim is to determine u and v so that the equation (8.50) may reduce to

simpler form. The procedure is simple when the discriminant of the quadratic form

(8.51)

S2 − 4RT

is every where either positive, negative or zero.

Case I Let

S2 − 4RT > 0

In this case, two roots λ1, λ2 of the equation

Rλ2 + Sλ+ T = 0 (8.54)


will be real and distinct. Let u, v be such that

∂u

∂x= λ1

∂u

∂y(8.55)

∂v

∂x= λ2

∂v

∂y(8.56)

Now,

A = R

(∂u

∂x

)2

+ S∂u

∂x

∂u

∂y+ T

(∂u

∂y

)2

(8.57)

= (Rλ21 + Sλ1 + T )

(∂u

∂y

)2

= 0 ·(∂u

∂y

)2

= 0

Since λ1 is roots of (8.54), so

Rλ21 + Sλ1 + T = 0

Hence A = 0. Similarly C = 0.

Note that the equation (8.55) is linear. The lagrange’s auxiliary equations are

dx

1=

dy

−λ1=du

0=⇒ du = 0

∴ u = c1 (8.58)

Also,

dx

1=

dy

−λ1

or, dy + λ1dx = 0

Integrating

y + λ1x = c2 (8.59)

From (8.58) and (8.59), the solution of the equation (8.55) is

u = f1(y + λ1x)

We may suppose

u = f1(x, y)

Similarly, the genral solution of (8.56) is

u = f2(x, y)


Now

AC −B2 =

[R

(∂u

∂x

)2

+ S∂u

∂x

∂u

∂y+ T

(∂u

∂y

)2][

R

(∂v

∂x

)2

+ S∂v

∂x

∂v

∂y+ T

(∂v

∂y

)2]

−[R∂u

∂x

∂v

∂x+S

2

(∂u

∂x

∂v

∂y+∂u

∂y

∂v

∂x

)+ T

∂u

∂y

∂v

∂y

]2

= R2

(∂u

∂x

)2(∂v∂x

)2

+RS

(∂u

∂x

)2 ∂v

∂x

∂v

∂y+RT

(∂u

∂x

)2(∂v∂y

)2

+

RS

(∂v

∂x

)2 ∂u

∂x

∂u

∂y+ S2∂u

∂x

∂u

∂y

∂v

∂x

∂v

∂y+ ST

∂u

∂x

∂u

∂y

(∂v

∂y

)2

+

+RT

(∂u

∂y

)2(∂v∂x

)2

+ ST

(∂v

∂y

)2 ∂v

∂x

∂v

∂y+ T 2

(∂u

∂y

)2(∂v∂y

)2

−R2

(∂u

∂x

)2(∂v∂x

)2

− S2

4

(∂u

∂x

∂v

∂y+∂u

∂y

∂v

∂x

)2

− T 2

(∂u

∂y

)2(∂v∂y

)2

−RS∂u∂x

∂v

∂x

(∂u

∂x

∂v

∂y+∂u

∂y

∂v

∂x

)− ST ∂u

∂y

∂v

∂y

(∂u

∂x

∂v

∂y+∂u

∂y

∂v

∂x

)−2RT

∂u

∂x

∂v

∂x

∂u

∂y

∂v

∂y

= RT

[(∂u

∂x

)2(∂v∂y

)2

+

(∂u

∂y

)2(∂v∂x

)2

− 2∂u

∂x

∂v

∂x

∂u

∂y

∂v

∂y

]

−S2

4

(∂u

∂x

∂v

∂y+∂u

∂y

∂v

∂x

)2

+ S2∂u

∂x

∂u

∂y

∂v

∂x

∂v

∂y

=1

4(4RT − S2)

(∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

)2

(8.60)

As A = C = 0,so from (8.60)

−B2 =1

4(4RT − S2)

(∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

)2

or, B2 =1

4(S2 − 4RT )

(∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

)2

(8.61)

Since (S2− 4RT ) > 0, so from (8.61), B 6= 0. Dividing the equation (8.50) by B, as

setting A = 0 and B = 0, we get

∂2z

∂x∂y= φ

(u, v, z,

∂z

∂u,∂z

∂v

)(8.62)

which is reqiuired canonical form of (8.49).

Case II When S2 − 4RT = 0, in this case two roots of

Rλ2 + Sλ+ T = 0


are real and equal i.e. λ1 = λ2. In this case, we choose u such that

∂u

∂x= λ

∂u

∂y

As in case I, its general solution can be taken as

u = f(x, y)

Further, we take v to be any function of x and y and independent of u.

Similar to case I, A = 0. Also from (8.61) B = 0 as S2 − 4RT = 0. On the other

hand, in this case C 6= 0, otherwise v would be a function of u.

Putting A = 0, B = 0 and dividing both sides of (8.50) by C, we get desired

cannonical form∂2z

∂y2= φ

(u, v, z,

∂z

∂u,∂z

∂v

)Case III When S2 − 4RT < 0, in this case the roots of

Rλ2 + Sλ+ T = 0

will be complex and conjugate of each other. Proceeding as in Case I, in this case

also, we will get the same cannonical form as in Case I. But u and v are complex

conjugates. Further to obtained a real canonical form, we use

u = α+ iβ, v = α− iβ

where α, β are real.

∴, α =u+ v

2, β =

i(v − u)

2

=⇒ ∂z

∂u=∂z

∂α

∂α

∂u+∂z

∂β

∂β

∂u=

1

2

(∂z

∂α− i ∂z

∂β

)and

∂z

∂v=∂z

∂α

∂α

∂v+∂z

∂β

∂β

∂v=

1

2

(∂z

∂α+ i

∂z

∂β

)We also note that

∂

∂u=

1

2

(∂

∂α− i ∂

∂β

)∴

∂2z

∂u∂v=

∂

∂u

(∂z

∂x

)=

1

4

(∂

∂α− i ∂

∂β

)(∂z

∂α+ i

∂z

∂β

)=

1

4

(∂2z

∂α2+∂2z

∂β2

)which is real quantity. Using this in (8.62) we get(

∂2z

∂α2+∂2z

∂β2

)= ψ

(α, β, z,

∂z

∂α,∂z

∂β

)


Working Rules

Step I Compare the given equation with

Rr + Ss+ Tt+ f(x, y, z, p, q) = 0 (8.63)

and obtaine the values of

R,S, T, f

Step II Form the quadratic equation

Rλ2 + Sλ+ T = 0

and solve it, we will get two values λ1, λ2. Now we have the following cases

1. If λ1, λ2 are unequal, then form two ordinary differential equations

dy

dx+ λ1 = 0

dy

dx+ λ2 = 0

Solving the equations we get two solutions f(x, y) = c1 and g(x, y) = c2. Now

we use the following transformations

u = f(x, y), v = g(x, y)

2. 2 If λ1 = λ2 then we get a single ordinary differential equation

dy

dx+ λ1 = 0

Solving this equation, we will get a solution

f(x, y) = c1

and take u = f(x, y) .We choose the other any function of v = g(x, y) inde-

pendent of u. Thus, we have two transformations u, v .

Steps III Find the values of p, q interms of

∂z

∂u,∂z

∂v

and r, s, t interms of∂z

∂u,∂z

∂v,∂2z

∂u2,∂2z

∂v2,∂2z

∂u∂v

Step IV Putting the values of p, q, r, s, t in (8.63) , we get the required Canon-

ical form, in term of u, v and their derivatives.

Example 296. Reduce the equation

∂2z

∂x2= x2 ∂

2z

∂y2



r = x2t (8.64)

Comparing it to the stardard form

Rr + Ss+ Tt+ f(x, y, z, p, q) = 0

we get

R = 1, S = 0, T = −x2

Now the quadratic equation

Rλ2 + Sλ+ T = 0

becomes

λ2 − x2 = 0

λ = ±x

which are real and distinct. Let λ1 = x and λ2 = −x. Hence the equations

dy

dx+ λ1 = 0 and

dy

dx+ λ2 = 0

become

dy

dx+ x = 0 and

dy

dx− x = 0

or, dy + xdx = 0 and dy − xdx = 0

Integrating these two

y +1

2x2 = c1 and y − 1

2x2 = c2

where c1 and c2 are constants. Hence the relations transform x, y to u, v are

u = y +1

2x2 and v = y − 1

2x2

∂u

∂x= x,

∂u

∂y= 1,

∂v

∂x= −x, ∂v

∂y= 1


p =∂z

∂x=

∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x

= x∂z

∂u− x∂z

∂v

= x

(∂z

∂u− ∂z

∂v

)q =

∂z

∂y=

∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y

=∂z

∂u+∂z

∂v

r =∂2z

∂x2=

∂

∂x

(∂z

∂x

)=

∂

∂x

(x

(∂z

∂u− ∂z

∂v

))=

(∂z

∂u− ∂z

∂v

)∂x

∂x+ x

∂

∂x

(∂z

∂u− ∂z

∂v

)=

∂z

∂u− ∂z

∂v+ x

[∂

∂u

(∂z

∂u− ∂z

∂v

)∂u

∂x+

∂

∂v

(∂z

∂u− ∂z

∂v

)∂v

∂x

]=

∂z

∂u− ∂z

∂v+ x2

[∂

∂u

(∂z

∂u− ∂z

∂v

)− ∂

∂v

(∂z

∂u− ∂z

∂v

)]=

∂z

∂u− ∂z

∂v+ x2

[∂2z

∂u2− 2

∂2z

∂u∂v+∂2z

∂v2

]Again,

∂z

∂y=

∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y=∂z

∂u+∂z

∂v

∴∂

∂y=

∂

∂u+

∂

∂v

t =∂2z

∂y2=

(∂

∂u+

∂

∂v

)(∂z

∂u+∂z

∂v

)=

∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

Substituting the values in (8.64)

∂z

∂u− ∂z

∂v+ x2

[∂2z

∂u2− 2

∂2z

∂u∂v+∂2z

∂v2

]= x2

(∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

)or,

∂2z

∂u∂v=

1

4x2

(∂z

∂u− ∂z

∂v

)But

u = y +1

2x2 and v = y − 1

2x2 =⇒ u− v = x2

∴∂2z

∂u∂v=

1

4(u− v)

(∂z

∂u− ∂z

∂v

)which is required canonical form of the given equation.



(n− 1)2 ∂2z

∂x2− y2n ∂

2z

∂y2= nyn−1 ∂z

∂y

to canonical form and find the general solution.


(n− 1)2r − y2nt− nyn−1q = 0 (8.65)


Rr + Ss+ Tt+ f(x, y, z, p, q) = 0

we get

R = (n− 1)2, S = 0, T = −y2n


Rλ2 + Sλ+ T = 0

becomes

(n− 1)2λ2 − y2n = 0

λ = ± yn

n− 1

which are real and distinct. Let λ1 = yn

n−1 and λ2 = − yn

n−1 . Hence the equations

dy

dx+ λ1 = 0 and

dy

dx+ λ2 = 0

becomesdy

dx+

yn

n− 1= 0 and

dy

dx− yn

n− 1= 0

or, (n− 1)y−ndy + dx = 0 and dx− (n− 1)y−ndy = 0

Integrating these two

x+ y1−n = c1 and x− y1−n = c2

where c1 and c2 are constants. Hence for the relations transform x, y to u, v, we take

u = x− y1−n and v = x+ y1−n

∂u

∂x= 1,

∂u

∂y= (n− 1)y−n,

∂v

∂x= 1,

∂u

∂y= −(n− 1)y−n


p =∂z

∂x=

∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x=∂z

∂u+∂z

∂v

∴∂

∂x=

∂

∂u+

∂

∂v

q =∂z

∂y=

∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y

= (n− 1)y−n(∂z

∂u− ∂z

∂v

)r =

∂2z

∂x2=

(∂

∂u+

∂

∂v

)(∂z

∂u+∂z

∂v

)=

∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

t =∂2z

∂y2=

∂

∂y

((n− 1)y−n

(∂z

∂u− ∂z

∂v

))= −n(n− 1)y−n−1

(∂z

∂u− ∂z

∂v

)+ (n− 1)y−n

∂

∂y

(∂z

∂u− ∂z

∂v

)= −n(n− 1)y−n−1

(∂z

∂u− ∂z

∂v

)+(n− 1)y−n

[∂

∂u

(∂z

∂u− ∂z

∂v

)∂u

∂y+

∂

∂v

(∂z

∂u− ∂z

∂v

)∂v

∂y

]Using

∂u

∂y= (n− 1)y−n,

∂u

∂y= −(n− 1)y−n

t = −n(n− 1)y−n−1

(∂z

∂u− ∂z

∂v

)+ (n− 1)2y−2n

[∂

∂u

(∂z

∂u− ∂z

∂v

)+

∂

∂v

(∂z

∂u− ∂z

∂v

)]= −n(n− 1)y−n−1

(∂z

∂u− ∂z

∂v

)+ (n− 1)2y−2n

[∂2z

∂u2− ∂2z

∂u∂v+∂2z

∂v2

]Substituting the values of r, t and q in (8.65)

(n− 1)2

(∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

)+

y2nn(n− 1)yn−1

(∂z

∂u− ∂z

∂v

)− y2n(n− 1)2y−2n

[∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

]= ny2n−1(n− 1)y−n

(∂z

∂u− ∂z

∂v

)4∂2z

∂u∂v= 0

or,∂2z

∂u∂v= 0

which is required canonical form of the the given equation.

For solution, we have∂2z

∂v∂u= 0


Integrating with respect to v∂z

∂u= φ(u)

where φ(u) is arbitrary function of u. Again integrating with respect to u

z =

∫φ(u)du+ ψ2(v)

or, z = ψ1(u) + ψ2(v)

Putting the value of u and v,

z = ψ1(x− y1−n) + ψ2(x+ y1−n)

which is the general solution.


∂2z

∂x2+ 2

∂2z

∂x∂y+∂2z

∂y2= 0

to canonical form and hence solve it.


r + 2s+ t = 0 (8.66)


Rr + Ss+ Tt+ f(x, y, z, p, q) = 0

we get

R = 1, S = 2, T = 1


Rλ2 + Sλ+ T = 0

becomes

λ2 + 2λ+ 1 = 0

or, (λ+ 1)2 = 0

∴ λ = −1,−1

which are real and equal. Hence the equation

dy

dx+ λ = 0

becomesdy

dx− 1 = 0


or, dy − dx = 0

Integrating,

x− y = c

where c is arbitrary constant and this relation transform x, y to u,

u = x− y

In this case, we take v an arbitrary function of x and y, which is independent of u.

Let us take v = x+ y.

∂u

∂x= 1,

∂u

∂y= −1,

∂v

∂x= 1,

∂v

∂y= 1

p =∂z

∂x=∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x=∂z

∂u+∂z

∂v

∂

∂x=

∂

∂u+

∂

∂v

q =∂z

∂y=∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y= −∂z

∂u+∂z

∂v

∂

∂x= − ∂

∂u+

∂

∂v

r =∂

∂x

(∂z

∂x

)=

(∂

∂u+

∂

∂v

)(∂z

∂u+∂z

∂v

)=∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

s =∂

∂x

(∂z

∂y

)=

(∂

∂u+

∂

∂v

)(−∂z∂u

+∂z

∂v

)= −∂

2z

∂u2+∂2z

∂v2

t =∂

∂y

(∂z

∂y

)=

(− ∂

∂u+

∂

∂v

)(−∂z∂u

+∂z

∂v

)=

∂2z

∂u2− 2

∂2z

∂u∂v+∂2z

∂v2

Putting the values of r, s and t in (8.66), we get

∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2− 2

∂2z

∂u2+ 2

∂2z

∂v2+∂2z

∂u2− 2

∂2z

∂u∂v+∂2z

∂v2= 0

or, 4∂2z

∂v2= 0

or,∂2z

∂v2= 0


which is required canonical form of (8.66). Now to solve this equation, we have the

canonical form is∂2z

∂v2= 0

Integrating,∂z

∂v= φ(u)

Again, integrating

z = φ1(u)

∫dv + φ2(u)

or, z = v φ1(u) + φ2(u)

or, z = (x+ y)φ1(x− y) + φ2(x− y)

which is required general solution, where φ1, φ2 are arbitrary functions.

Example 299.

y2 ∂2z

∂x2− 2xy

∂2z

∂x∂y+ x2 ∂

2z

∂y2=y2

x

∂z

∂x+x2

y

∂z

∂y


y2r − 2xys+ x2t− y2

xp+

x2

yq = 0 (8.67)

Comparing it with the standard form

Rr + Ss+ Tt+ f(x, y, z, p, q) = 0

we obtain

R = y2, S = −2xy, T = x2


Rλ2 + Sλ+ T = 0

becomes

y2λ2 − 2xyλ+ x2 = 0

or, (yλ− x)2 = 0

or, λ =x

y,x

y

which is real and equal. Hence the equation

dy

dx+ λ = 0

becomesdy

dx+x

y= 0


or, ydy + xdx = 0

Integrating,

x2 + y2 = c

where c is arbitrary constant and this relation transform x, y to u,

u = x2 + y2

In this case, we take v an arbitrary function of x and y, which is independent of u.

Let us take v = x2 − y2.

∂u

∂x= 2x,

∂u

∂y= 2y,

∂v

∂x= 2x,

∂v

∂y= −2y

p =∂z

∂x=∂z

∂u

∂u

∂x+∂z

∂v

∂v

∂x= 2x

∂z

∂u+ 2x

∂z

∂v= 2x

(∂z

∂u+∂z

∂v

)

q =∂z

∂y=∂z

∂u

∂u

∂y+∂z

∂v

∂v

∂y= 2y

∂z

∂u− 2y

∂z

∂v= 2y

(∂z

∂u− ∂z

∂v

)

r =∂

∂x

(∂z

∂x

)=

∂

∂x

(2x

(∂z

∂u+∂z

∂v

))= 2

(∂z

∂u+∂z

∂v

)+ 2x

∂

∂x

(∂z

∂u+∂z

∂v

)= 2

(∂z

∂u+∂z

∂v

)+ 2x

[∂

∂u

(∂z

∂u+∂z

∂v

)∂u

∂x+

∂

∂v

(∂z

∂u+∂z

∂v

)∂v

∂x

]= 2

(∂z

∂u+∂z

∂v

)+ 4x2

[∂

∂u

(∂z

∂u+∂z

∂v

)+

∂

∂v

(∂z

∂u+∂z

∂v

)]as

∂u

∂x=∂v

∂x= 2x

= 2

(∂z

∂u+∂z

∂v

)+ 4x2

[∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

]

s =∂

∂x

(∂z

∂y

)=

∂

∂x

(2y

(∂z

∂u− ∂z

∂v

))= 2y

∂

∂x

(∂z

∂u− ∂z

∂v

)= 2y

[∂

∂u

(∂z

∂u− ∂z

∂v

)∂u

∂y+

∂

∂v

(∂z

∂u− ∂z

∂v

)∂v

∂y

]= 4xy

[∂

∂u

(∂z

∂u− ∂z

∂v

)+

∂

∂v

(∂z

∂u− ∂z

∂v

)]as

∂u

∂x=∂v

∂x= 2x

= 4xy

[∂2z

∂u2− ∂2z

∂v2

]


t =∂

∂y

(∂z

∂y

)=

∂

∂y

(2y

(∂z

∂u− ∂z

∂v

))= 2

(∂z

∂u− ∂z

∂v

)+ 2y

∂

∂y

(∂z

∂u− ∂z

∂v

)= 2

(∂z

∂u− ∂z

∂v

)+ 2y

[∂

∂u

(∂z

∂u− ∂z

∂v

)∂u

∂y+

∂

∂v

(∂z

∂u− ∂z

∂v

)∂v

∂y

]= 2

(∂z

∂u− ∂z

∂v

)+ 4y2

[∂

∂u

(∂z

∂u− ∂z

∂v

)− ∂

∂v

(∂z

∂u− ∂z

∂v

)]as

∂u

∂x= 2y,

∂v

∂y= −2y

= 2

(∂z

∂u− ∂z

∂v

)+ 4y2

[∂2z

∂u2− 2

∂2z

∂u∂v+∂2z

∂v2

]Substituting the values of r, s and t in (8.67)

y2

[2

(∂z

∂u+∂z

∂v

)+ 4x2

(∂2z

∂u2+ 2

∂2z

∂u∂v+∂2z

∂v2

)]− 8x2y2

(∂2z

∂u2− ∂2z

∂v2

)+

x2

[2

(∂z

∂u− ∂z

∂v

)+ 4y2

(∂2z

∂u2− 2

∂2z

∂u∂v+∂2z

∂v2

)]=y2

x2x

(∂z

∂u+∂z

∂v

)+

x2

y2y

(∂z

∂u− ∂z

∂v

)

or,∂2z

∂2v= 0

which is the required Canonical form of (8.67). Now to solve this equation, we have

the canonical form is∂2z

∂v2= 0

Integrating,∂z

∂v= φ(u)

Again, integrating

z = φ1(u)

∫dv + φ2(u)

or, z = v φ1(u) + φ2(u)

or, z = (x2 − y2)φ1(x2 + y2) + φ2(x2 + y2)

which is required general solution, where φ1, φ2 are arbitrary functions.


∂2z

∂x2+ x2 ∂

2z

∂y2= 0

to canonical form.



r + x2t = 0 (8.68)

Comparing it with the standard form

Rr + Ss+ Tt+ f(x, y, z, p, q) = 0

we obtain

R = 1, S = 0, T = x2


Rλ2 + Sλ+ T = 0

becomes

λ2 + x2 = 0

or, λ2 = −x2

or, λ = ±ix

which is real and equal. Hence the equations

dy

dx+ λ1 = 0 and

dy

dx+ λ2 = 0

becomedy

dx+ ix = 0 and

dy

dx− ix = 0

or, dy + ixdx = 0 or dy − ixdx = 0

Integrating,

y +i

2x2 = c1, and y − i

2x2 = c2

where c1 and c2 are arbitrary constants and this relations transform x, y to u, v are

u = y +i

2x2, and v = y − i

2x2

Since u and v are complex conjugates of each other, let us take

u = α+ iβ, v = α− iβ

∴ α = y, β =x2

2

∂α

∂x= 0,

∂α

∂y= 1,

∂β

∂x= x,

∂β

∂y= 0

p =∂z

∂x=∂z

∂α

∂α

∂x+∂z

∂β

∂β

∂x= x

∂z

∂β


q =∂z

∂y=∂z

∂α

∂α

∂y+∂z

∂β

∂β

∂y=∂z

∂α

r =∂

∂x

(∂z

∂x

)=

∂

∂x

(x∂z

∂β

)=

∂z

∂β+ x

∂

∂x

(∂z

∂β

)=

∂z

∂β+ x

[∂

∂α

(∂z

∂β

)∂α

∂x+

∂

∂β

(∂z

∂β

)∂β

∂x

]=

∂z

∂β+ x2 ∂

2z

∂β2as

∂β

∂x= 1,

∂α

∂x= 0

t =∂

∂y

(∂z

∂y

)=

∂

∂y

(∂z

∂α

)=

∂

∂α

(∂z

∂α

)∂α

∂y+

∂

∂β

(∂z

∂α

)∂β

∂y

=∂2z

∂α2as

∂α

∂y= 1,

∂β

∂y= 0

Substituting the values of r and t in (8.68)

∂z

∂β+ x2 ∂

2z

∂β2+ x2 ∂

2z

∂α2= 0

or,∂2z

∂β2+∂2z

∂α2= − 1

x2

∂z

∂β

or,∂2z

∂β2+∂2z

∂α2= − 1

2β

∂z

∂β

which is required canonical form of the given equation.

Example 301. Show that, by simple substitution, the equation

Rr + Pp = W

can be reduces to a linear partial differential equation of the first oder, and outline

a procedure for determining the solution of the ordinary equation. Illustrate the

method by finding the solutions of the equations

(a) xr + 2p = 2y

(b) s− q = ex=y

Solution: The given partial dirfferential equation is

Rr + Pp = W

or, R∂

∂x

(∂z

∂x

)+ P

∂z

∂x= W

or, R∂p

∂x+ Pp = W as p =

∂z

∂x

or,∂p

∂x+P

Rp =

W

R(8.69)


which is linear in p, regarding y as a constant. Now

I.F. = e∫PRdx

Multiplying the equation (8.69) by I.F. and integrating with respect to x, taking y

as a constant, we get

p× e∫PRdx =

∫ (W

Re∫PRdx

)dx+ f1(y)

or,∂z

∂x=

1

e∫PR dx

∫ (W

Re∫PRdx

)dx+

f1(y)

e∫PRdx

Again, integrating

z =

∫ (1

e∫PRdx

∫W

Re∫PRdxdx

)dx+ f1(y)

∫1

e∫PRdxdx+ f2(y)

Solution of (a): The given partial differential equation is

xr + 2p = 2y

or,∂p

∂x+

2

xp =

2y

x(8.70)

Integrating factor

I.F. = e∫

2xdx = e2 loge x = eloge x

2= x2


x2 ∂p

∂x+ 2xp = 2xy

or,∂

∂x(x2p) = 2xy

Integrating, with respect to x, taking y as a constant

x2p = 2y

∫xdx+ f1(y)

or, x2p = x2y + f1(y)

or, p = y +1

x2f1(y)

or,∂z

∂x= y +

1

x2f1(y)

Again, integrating,

z = y

∫dx+ f1(y)

∫x−2dx+ f2(y)

or, z = xy − f1(y)

x+ f2(y)

8.5. GENERAL METHOD OF SOLVING RR+ SS + TT = V 461

where f1 and f2 are arbitrary functions, which is required general solution.

Solution of (a): The given partial differential equation is

s− q = ex+y

or,∂q

∂x− q = ex+y (8.71)

Integrating factor

I.F. = e∫

(−1)dx = e−x


e−x∂q

∂x− e−xq = ey

or,∂

∂x(qe−x) = ey

Integrating, with respect to x, taking y as a constant

qe−x = ey∫dx+ f1(y)

or, e−xq = xey + f1(y)

or, q = xex+y + exf1(y)

or,∂z

∂y= xex+y + exf1(y)

Again, integrating taking x as a constant

z = xex∫eydy + ex

∫f1(y)dy + g(x)

or, z = xex+y + exf(y) + g(x)

as

f(y) =

∫f1(y)dy

where f and g are arbitrary functions, which is required general solution.

8.5 General Method of Solving Rr + Ss+ Tt = V

Theorem 32. Solve the second order partial differential equation Rr+Ss+Tt = V ,

where R,S, T and V are functions of x, y, z, p, q by Monge’s method.


Rr + Ss+ Tt = V (8.72)


where R,S, T and V are functions of x, y, z, p, q. We have

dp =∂p

∂xdx+

∂p

∂ydy = rdx+ sdy

∴ r =dp− sdy

dx

dq =∂q

∂xdx+

∂q

∂ydy = sdx+ tdy

∴ t =dq − sdx

dy


Rdp− sdy

dx+ Ss+ T

dq − sdxdy

= V

or,(Rdpdy + Tdqdx− V dxdy)− s(R(dy)2 − Sdxdy + T (dx)2

)= 0 (8.73)

The equations obtained by equating the expression with in brackets to zero, are

called Monge’s subsidiary equations. Hence, the Monge’s subsidiary equations

are

R(dy)2 − Sdxdy + T (dx)2 = 0 (8.74)

Rdpdy + Tdqdx− V dxdy = 0 (8.75)

Obviously, if some relation between x, y, z, p, q satisfying these relations (8.74) and

(8.75) will autometically satisfy (8.73) and hence (8.72).

The equation (8.74) is quadratic and can be resolved into two equations of type

dy −m1dx = 0 (8.76)

dy −m2dx = 0 (8.77)


Case I : When m1 and m2 are distinct. From (8.76) and (8.75), combining if

necessary with dz = pdx+ qdy, we find two integrals

u1 = a and v1 = b

Then

u1 = f1(v1) (8.78)

is an intermidiate integral of (8.72), where f1 is an arbitrary function.

Similarly, taking (8.77) and (8.75) we get another intermidiate integral of (8.72) as

u2 = f2(v2) (8.79)


where f2 is an arbitrary function.

Thus, we have two intermediate integrals (8.78) and (8.79), which on solving them

give values of p and q interm of x and y. Now substituting these in

dz = pdx+ qdy

and integrating, we get the complete integral of (8.72).

Case II When m1 = m2. In this case we get only one intermidiate integral which

is in the form

Pp+Qq = R

Solving this with help of Lagrange’s method, we get the required complete integral

of (8.72).


r = a2t [T.U.2057, 2059, 2061, 2062, 2063, 2067, 2071]

by Monge’s method.


r = a2t (8.80)

Comparing with

Rr + Ss+ Tt = V

we get

R = 1, S = 0, T = −a2, V = 0

The Monge’s subsidiary equations are

Rdpdy + Tdqdx− V dxdy = 0 i.e. dpdy − a2dqdx = 0 (8.81)

and R(dy)2 − Sdxdy + T (dx)2 = 0 i.e. (dy)2 − a2(dx)2 = 0 (8.82)


(dy − adx)(dy + adx) = 0

Hence we consider following two sytem of equations

dpdy − a2dqdx = 0, dy − adx (8.83)

and dpdy − a2dqdx = 0, dy + adx (8.84)

We consider the system (8.83), Integrating

dy − adx = 0

we get

y − ax = c1 (8.85)


Again, putting

dy = adx

in

dpdy − a2dqdx = 0

we get

dp adx− a2dq dx = 0

or,dp− adq = 0

Integrating,

p− aq = b1 (8.86)

From (8.85) and (8.86) we have

p− aq = f1(y − ax) (8.87)

Again, we consider the system (8.84), Integrating

dy + adx = 0

we get

y + ax = c1 (8.88)

Again, putting

dy = −adx

in

dpdy − a2dqdx = 0

we get

dp adx+ a2dq dx = 0

or,dp+ adq = 0

Integrating,

p+ aq = b1 (8.89)


p+ aq = f2(y + ax) (8.90)


p =1

2[f1(y − ax) + f2(y + ax)]


q =1

2a[f1(y − ax)− f2(y + ax)]

Putting the values of p and q in dz = pdx+ qdy, we get

dz =1

2[f1(y − ax) + f2(y + ax)]dx+

1

2a[f1(y − ax)− f2(y + ax)]dy

or,dz =1

2af2(y + ax)(dy + adx)− 1

2af1(y − ax)(dy − adx)

Integrating,

z =1

2aφ2(y + ax)− 1

2aφ2(y − ax)

z = ψ1(y + ax) + ψ2(y − ax)



r = t



r = t (8.91)

Comparing with

Rr + Ss+ Tt = V

we get

R = 1, S = 0, T = −1, V = 0


Rdpdy + Tdqdx− V dxdy = 0 i.e. dpdy − dqdx = 0 (8.92)

and R(dy)2 − Sdxdy + T (dx)2 = 0 i.e. (dy)2 − (dx)2 = 0 (8.93)


(dy − dx)(dy + dx) = 0

Hence we consider following two sytem of equations

dpdy − dqdx = 0, dy − dx (8.94)

and dpdy − dqdx = 0, dy + dx (8.95)

We consider the system (8.94), Integrating

dy − dx = 0


we get

y − x = c1 (8.96)

Again, putting

dy = dx

in

dpdy − dqdx = 0

we get

dp dx− dq dx = 0

or,dp− dq = 0

Integrating,

p− q = b1 (8.97)


p− q = f1(y − x) (8.98)

Again, we consider the system (8.95), Integrating

dy + dx = 0

we get

y + x = c1 (8.99)

Again, putting

dy = −dx

in

dpdy − dqdx = 0

we get

dpdx+ dq dx = 0

or,dp+ dq = 0

Integrating,

p+ q = b1 (8.100)


p+ q = f2(y + x) (8.101)



p =1

2[f1(y − x) + f2(y + x)]

q =1

2[f1(y − x)− f2(y + x)]

Putting the values of p and q in dz = pdx+ qdy, we get

dz =1

2[f1(y − x) + f2(y + x)]dx+

1

2[f1(y − x)− f2(y + x)]

or,dz =1

2f2(y + x)(dy + dx)− 1

2f1(y − x)(dy − dx)

Integrating,

z =1

2φ2(y + x)− 1

2φ2(y − x)

z = ψ1(y + x) + ψ2(y − x)



q2r − 2pqs+ p2t = 0



q2r − 2pqs+ p2t = 0 (8.102)

Comparing with

Rr + Ss+ Tt = V

we get

R = q2, S = −2pq, T = p2, V = 0


Rdpdy + Tdqdx− V dxdy = 0 i.e. q2dpdy + p2dqdx = 0(8.103)

and R(dy)2 − Sdxdy + T (dx)2 = 0 i.e. q2(dy)2 + 2pqdx dy + p2(dx)2 = 0

i.e. (pdx+ qdy)2 = 0(8.104)

From the equation (8.104), we have

pdx+ qdy = 0

or, dz = 0


Integrating,

z = a1 (8.105)

Again, putting qdy = −pdx in (8.103), we get

−pqdpdx+ p2dqdx = 0

or, pdq = qdp

or,dp

p=dq

q

Integrating,

loge p = loge q + loge b1

or, loge p = loge(qb1)

or, p = qb1

or,p

q= b1 (8.106)

From (8.105) and(8.106) we get an intermediate integral

p

q= f(z)

or, p− qf(z) = 0

where f is arbitrary function. This is first order linear partial differential equationof

the form Pp+Qq = R. The Lagrange’s auxiliary equations are

dx

1=

dy

−f(z)=dz

0

From first and last ratios, we get

dz = 0

Integrating,

z = a2 (8.107)

Again from first and second ratios

dx

1= − dy

f(a2)

or, f(a2)dx = −dy

Integrating,

f(a2)x+ y = b2

or, xf(z) + y = b2 (8.108)

From (8.107) and (8.108) we get the general solution

b2 = g(a2)

or, xf(z) + y = g(z)



y2r − 2ys+ t = p+ 6y



y2r − 2ys+ t = p+ 6y (8.109)

Comparing with

Rr + Ss+ Tt = V

we get

R = y2, S = −2y, T = 1, V = p+ 6y


Rdpdy + Tdqdx− V dxdy = 0 i.e. y2dpdy + dqdx− (p+ 6y)dxdy = 0 (8.110)

and R(dy)2 − Sdxdy + T (dx)2 = 0 i.e. y2(dy)2 − 2ydx dy + (dx)2 = 0

i.e. (ydy + dx)2 = 0 (8.111)

From the equation (8.111), we have

ydy + dx = 0

Integrating,

y2

2+ x =

a1

2or, y2 + 2x = a1 (8.112)

Again, putting dx = −ydy in (8.110), we get

y2dpdx− ydqdy + y(p+ 6y)dydy = 0

or, ydp+ pdy − dq + 6ydy = 0

or, d(yp)− dq + 6ydy = 0

Integrating,

yp− q + 3y3 = b1 (8.113)

From (8.112) and(8.113) we get an intermediate integral

yp− q + 3y3 = f(y2 + 2x)

or, yp− q = f(y2 + 2x)− 3y2


where f is arbitrary function. This is first order linear partial differential equationof

the form Pp+Qq = R. The Lagrange’s auxiliary equations are

dx

y=dy

−1=

dz

f(y2 + 2x)− 3y2

From first and second ratios, we get

ydy + dx = 0

Integrating,

y2

2+ x =

a2

2or, y2 + 2x = a2 (8.114)

Again from second and third ratios

or, [f(a2)− 3y2]dy + dz = 0

Integrating,

yf(a3)− y3 + z = b2

or, yf(y2 + 2x)− y3 + z = b2 (8.115)

From (8.114) and (8.115) we get the general solution

b2 = g(a2)

or, yf(y2 + 2x)− y3 + z = g(y2 + 2x)


z(qs− pt) = pq2 [T.U. 2068, 2070, 2074]



z(qs− pt) = pq2 (8.116)

Comparing with

Rr + Ss+ Tt = V

we get

R = 0, S = zq, T = −zp, V = pq2


Rdpdy + Tdqdx− V dxdy = 0 i.e. − zpdpdx− pq2dxdy = 0 (8.117)

and R(dy)2 − Sdxdy + T (dx)2 = 0 i.e. − zqdxdy − zp(dx)2 = 0 (8.118)


From the equation (8.117)

−pdx(zdq + q2dy) = 0

or, zdq + q2dy = 0 (8.119)

Again, from (8.118)

−zdx(qdy + pdx) = 0

or, pdx+ qdy = 0

or, dz = 0 as dz = pdx+ qdy

Integrating,

z = a1 (8.120)

Putting the value of z in (8.119)

a1dq + q2dy = 0

or,dq

q2+ dy = 0

Integrating,

−a1

q+ y = b1

−zq

+ y = b1 (8.121)

From the equations (8.120) and (8.121), we get the intermediate integral

−zq

+ y = F (z)

or,− z ∂y∂z

+ y = F (z)

or,∂y

∂z− y

z= −F (z)

z

or,∂y

∂z− y

z= G(z) (8.122)

where

G(z) = −F (z)

z

The equation (8.122) is linear partial differential equation in y and z. Its integrating

factor is

I.F. = e∫pdx = e−

∫dzz = e− loge z = eloge(z

−1) =1

z



1

z

∂y

∂z− y

z2=G(z)

z

or,∂

∂z

(yz

)=G(z)

z

Integrating,

y

z=

∫G(z)

zdz + f(x, y)

or, y = z[g(z) + f(x, y)]

where

g(z) =

∫G(z)

zdz

and which is the general solution.


pq = x(ps− qr)



pq = x(ps− qr)or, − xqr + xps = pq (8.123)

Comparing with

Rr + Ss+ Tt = V

we get

R = −xq, S = xp, T = 0, V = pq


Rdpdy + Tdqdx− V dxdy = 0 i.e. − xqdpdx− pqdxdy = 0 (8.124)

and R(dy)2 − Sdxdy + T (dx)2 = 0 i.e. − xq(dy)2 − xpdxdy = 0 (8.125)

From the equation (8.124)

qdy(xdp+ pdx) = 0

or,d(xp) = 0

Integrating,

xp = a1 (8.126)


Again, from (8.125)

xdy(qdy + pdx) = 0

or, pdx+ qdy = 0

or, dz = 0 as dz = pdx+ qdy

Integrating,

z = a1 (8.127)

From the equations (8.126) and (8.127), we get the intermediate integral

xp = f(z)

or, x∂z

∂y= F (z)

or,1

f(z)

∂z

∂x=

1

x

Integrating, ∫1

f(z)dz = loge x+ F (y)

or, G(z) = loge x+ F (y)

where

g(z) =

∫1

f(z)dz

and which is the general solution.

8.5.1 Monge’s Method for Rr + Ss+ Tt+ U(rt− s2) = V

Theorem 33. Solve the second order partial differential equation Rr + Ss + Tt +

U(rt = s2) = V , where R,S, T, U and V are functions of x, y, z, p, q by Monge’s

method.


Rr + Ss+ Tt+ U(rt− s2) = V (8.128)

where R,S, T, U and V are functions of x, y, z, p, q. We have

dp =∂p

∂xdx+

∂p

∂ydy = rdx+ sdy

∴ r =dp− sdy

dx

dq =∂q

∂xdx+

∂q

∂ydy = sdx+ tdy

∴ t =dq − sdx

dy



Rdp− sdy

dx+ Ss+ T

dq − sdxdy

+ U

(dp− sdy

dx

dq − sdxdy

− s2

)= V

or,(Rdpdy + Tdqdx+ Udpdq − V dxdy)

−s(R(dy)2 − Sdxdy + T (dx)2 + Udpdx+ Udqdy

)= 0 (8.129)

The equations obtained by equating the expression with in brackets to zero, are

called Monge’s subsidiary equations. Hence, the Monge’s subsidiary equations

are

M = Rdpdy + Tdqdx− V dxdy + Udpdq − V dxdy = 0 (8.130)

N = R(dy)2 − Sdxdy + T (dx)2 + Udpdx+ Udqdy = 0 (8.131)

The equation (8.131) can not be factorized on account of the term Udpdx+ Udqdy

present it, so let us try to factorize N + λM , where λ is some multiplier to be

determine latter. Now

N + λM =(R(dy)2 − Sdxdy + T (dx)2 + Udpdx+ Udqdy

)+λ(Rdpdy + Tdqdx+ Udpdq − V dxdy) = 0

= R(dy)2 + T (dx)2 − (S + λV )dxdy + Udpdx+ Udqdy +

+λRdpdy + λTdqdx+ λUdpdq (8.132)

Also let the factor of N + λM be

αdy + βdx+ γdp and α′dy + β′dx+ γ′dq

so that

N + λM = (αdy + βdx+ γdp)(α′dy + β′dx+ γ′dq) (8.133)

Equating the coefficients of (dy)2, (dy)2, dpdq in (8.132) and (8.133)

αα′ = R, ββ′ = T, γγ′ = λU

Now if we take

α′ = 1, β′ =1

k, γ′ =

λ

m

then

α = R β = kT, γ = mU


Further equating the other terms in (8.132) and (8.133), we get

αβ′ + βα′ = −(S + λV )

or,R

k+ kT = −(S + λV ) (8.134)

γβ′ = U

or,mU

k= U (8.135)

γα′ = λR

or, mU = λR (8.136)

βγ′ = λT

or,kTλ

m= λT (8.137)

From the relation (8.137), we get

m = k

From the relation (8.136), we get

m = k =λR

U

Putting the values of k in From the relation (8.134), we get

λR

UT +R

U

λR= −(S + λV )

or, λ2(RT + UV ) + λUS + U2 = 0 (8.138)

The equation (8.138) gives two values of λ. Let the roots be λ1, λ2

N + λM = (αdy + βdx+ γdp)(α′dy + β′dx+ γ′dq)

Putting values of α, β, γ, α′, β′, γ′

N + λM = (Rdy + kTdx+mUdp)

(dy +

1

kdx+

λ

mdq

)Putting the values of m and k

N + λM =

(Rdy +

λR

UTdx+

λR

Udp

)(dy +

U

λRdx+ λ

U

λRdq

)R

U(Udy + λTdx+ λUdp)

1

λR(λRdy + Udx+ λUdq)

The equation N + λN = 0 corresponding to the roots λ1 and λ2 are

(Udy + λ1Tdx+ λ1Udp) (λ1Rdy + Udx+ λ1Udq) = 0 (8.139)

(Udy + λ2Tdx+ λ2Udp) (λ2Rdy + Udx+ λ2Udq) = 0 (8.140)


Now one of the factor of (8.139) is combined with one factor of (8.140) we get two

intermediate integrals. From these two intermediate integrals the values of p and q

can be determine then integrating

dz = pdx+ qdy

will give the solution of the given PDE.


r + 4s+ t+ (rt− s2) = 2

Solution : Comaparig the given equation with

Rr + Ss+ Tt+ U(rt = s2) = V

we get

R = 1, S = 4, T = 1, U = 1, V = 2

The equation

λ2(RT + UV ) + λUS + U2 = 0

becomes

3λ2 + 4λ+ 1 = 0

Solving this equation

λ1 = −1, λ2 = −1

3

Now the pair of equations becomes

Udy + λ1Rdy + λ2Udq = 0

Udx+ λ2Rdy + λ2Udq = 0

becomes

dy − dx− dp = 0 or dp+ dx− dy = 0

dx− 1

3dy − 1

3dq = 0 or dq + dy − 3dx = 0

Integrating,

P + x− y = c1, q + y − 3x = c2 (8.141)

Thus we get a intermidiate integral

p+ x− y = f(q + y − 3x) (8.142)

Again we consider the pair of equations becomes

Udx+ λ1Rdy + λ1Udq = 0


Udy + λ2Tdy + λ2Udq = 0

becomes

dx− dy − dq = 0 or dq + dx− dx = 0

dy − 1

3dx− 1

3dp = 0 or dq + dx− 3dy = 0

Integrating,

q + y − x = c3, p+ x− 3y = c4 (8.143)

Thus we get a intermidiate integral

p+ x− 3y = g(q + y − x) (8.144)

Let

q + y − 3x = α (8.145)

q + y − x = β (8.146)

we get

p+ x− y = f(α) (8.147)

p+ x− 3y = g(β) (8.148)

Solving (8.145) and (8.146) we get

x =1

2(β − α)

Solving (8.147) and (8.148) we get

y =1

2(f(α)− g(β))

=⇒ dx =1

2(dβ − dα) and dy =

1

2(f ′(α) dα− g′(β) dβ)

Also, from (8.147) and (8.146)

p = y − x+ f(α), q = x− y + β

dz = pdx+ qdy

= [y − x+ f(α)]dx+ [x− y + β]dy

= (ydx+ xdy)− xdx− ydy + f(α)dx+ βdy

= d(xy)− xdx− ydy + f(α)1

2(dβ − dα) + β

1

2(f ′(α) dα− g′(β) dβ)

= d(xy)− xdx− ydy +1

2[f(α)dβ + βf ′(α)dα]− 1

2f(α)dα− 1

2βg′(β)dβ

2dz = 2d(xy)− 2xdx− 2ydy + [f(α)dβ + βf ′(α)dα]− f(α)dα− βg′(β)dβ

2dz = 2d(xy)− 2xdx− 2ydy + d(βf(α))− f(α)dα− βg′(β)dβ


Integrating,

2z = 2xy − x2 − y2 + βf(α)−∫f(α)dα−

∫βg′(β)dβ

Using the integration by part in the last term

2z = 2xy − x2 − y2 + βf(α)−∫f(α)dα−

(β

∫g′(β)dβ −

∫ [dβ

dβ

∫g′(β)dβ

]dβ

)

or, 2z = 2xy − x2 − y2 + βf(α)− βg(β)−∫f(α)dα+

∫g(β)dβ

or, 2z = 2xy − x2 − y2 + β[f(α)− g(β)]−∫f(α)dα+

∫g(β)dβ

Let ∫f(α)dα = φ(α) and

∫g(β)dβ = ψ(β)

Then

f(α) = φ′(α) and g(β) = ψ′(β)

∴ 2z = 2xy − x2 − y2β[φ′(α)− ψ′(β)]− φ(α) + ψ(β)

where

2x = β − α, 2y = φ′(α)− ψ′(β)

This is required solution in parametric form with α, β as parameters and φ, ψ being

arbitrary functions.

Chapter 9

Partial Differential Equations

and Fourier Series

9.1 Boundary Value Problem

9.1.1 Two Points Boundary Balue Problem

A boundary value problem is a differential equation together with a set of additional

constraints, at boundary. A differential equation and suitable boundary conditions

which specifies at two boundaries is forms a two-points boundary value problem.

Example 309.

A typical example is the differential equation

y′′ + p(x)y′ + q(x)y = g(x) (9.1)

with boundary conditions

y(α) = y0, y(β) = y1 (9.2)

A solution to a boundary value problem is a solution to the differential equation

which also satisfies the boundary conditions.

9.1.2 Homogeneous and Non-homogeneous Boundary Value Pro-

belms

Consider a two point boundary value problem

y′′ + p(x)y′ + q(x)y = g(x) (9.3)

with boundary conditions

y(α) = y0, y(β) = y1 (9.4)

479

480CHAPTER 9. PARTIAL DIFFERENTIAL EQUATIONS AND FOURIER SERIES

If g(x) = 0 for each x, and if y0 = 0 and y1 = 0 then the problem (9.3) together

with (9.4) is called homogeneous. Otherwise it is called non-homogenous.

Boundary value problems under similar conditions may have a unique solution, they

may have no solution, or in some cases they may have infinitely many solutions.

The following are examples of boundary value problem with unique solution.

Example 310.

Solve the boundary value problem

y′′ + 2y = 0, y(0) = 1, y(π) = 0


y′′ + 2y = 0, y(0) = 1, y(π) = 0 (9.5)


m2 + 2 = 0

m = 0±√

2i

The general solution of the given differntial equation is

y = e0x(c1 cos√

2x+ c2 sin√

2x)

y = c1 cos√

2x+ c2 sin√

2x (9.6)

Using the first boundary condition y(0) = 1 in (9.6), we have

c1 cos√

2 0 + c2 sin√

2 0 = 1

or, c1 = 1

Again, using the second boundary condition y(π) = 0 in (9.6)

c1 cos√

2π + c2 sin√

2π = 0

or, cos√

2π + c2 sin√

2π = 0 as c1 = 1

c2 =cos√

2π

sin√

2π= − cot

√2π = −0.2762 approximate.

Putting the values of c1 and c2 in (9.6), we get unique solution

y = cos√

2x− cot√

2π sin√

2x

Example 311.


y′′ + y = 0, y(0) = 0, y′(π) = 2

9.1. BOUNDARY VALUE PROBLEM 481


y′′ + y = 0, y(0) = 0, y′(π) = 2 (9.7)


m2 + 1 = 0

m = 0± i


y = e0x(c1 cosx+ c2 sinx)

y = c1 cosx+ c2 sinx (9.8)


c1 cos 0 + c2 sin 0 = 0

or, c1 = 0

Differentiating the equation (9.7) with respect to x

y′ = −c1 sinx+ c2 cosx

Again, using the second boundary condition y′(π) = 2 in (9.8)

−c1 cosπ + c2 sinπ = 2

or, c1 = −2


y = −2 cosx

Example 312.


y′′ + 2y = 0, y′(0) = 1, y′(π) = 0


y′′ + 2y = 0, y′(0) = 1, y′(π) = 0 (9.9)


m2 + 2 = 0

m = 0±√

2i



y = e0x(c1 cos√

2x+ c2 sin√

2x)

y = c1 cos√

2x+ c2 sin√

2x (9.10)


y′ = −√

2c1 sin√

2x+√

2c2 cos√

2x

Using the first boundary condition y′(0) = 1 in (9.8), we have

−√

2c1 sin 0 +√

2c2 cos 0 = 1

or, c2 =1√2

Again, using the second boundary condition y′(π) = 0 in (9.10)

−c1

√2 sin

√2π +

√2c2 cos

√2π = 0

or, − c1

√2 sin

√2π +

√2

1√2

cos√

2π = 0

or, − c1

√2 sin

√2π + cos

√2π = 0

or, c1 =cot√

2π√2


y =cot√

2π√2

cos√

2x+1√2

sin√

2x

Example 313.


y′′ + y = 0, y′(0) = 2, y(L) = 0


y′′ + y = 0, y′(0) = 2, y(L) = 0 (9.11)


m2 + 1 = 0

m = 0± i






y′ = −c1 sinx+ c2 cosx

Using the first boundary condition y′(0) = 2 in (9.12), we have

−c1 sin 0 + c2 cos 0 = 2

or, c2 = 2

Again, using the second boundary condition y(L) = 0 in (9.10)

c1 cosL+ c2 sinL = 0

or, c1 cosL+ 2 sinL = 0

or, c1 = −2 sinL

cosL= −2 tanL


y = −2 tanL cosx+ 2 sinx where cosL 6= 0

becaues if cosL = 0, then c1 is not defined and the equation has no solution.

Example 314.


y′′ + 2y = x, y(0) = 0, y(π) = 0


y′′ + 2y = x, y(0) = 0, y(π) = 0

or, (D2 + 2)y = x y(0) = 0, y(π) = 0


m2 + 2 = 0

m = 0±√

2i


C.F. = e0x(c1 cos√

2x+ c2 sin√

2x)

y = c1 cos√

2x+ c2 sin√

2x

For particular integral:

P.I. =1

D2 + 2x

=1

2(1 + D2

2 )x

=1

2

(1 +

D2

2

)−1

x

=1

2

(1− D2

2+D4

4− · · ·

)x

=1

2x


Rquired solution is

y = C.F.+ P.I.

y = c1 cos√

2x+ c2 sin√

2x+1

2x (9.13)


c1 cos 0 + c2 sin 0 +1

2· 0 = 0

or, c1 = 0


c1 cos√

2π + c2 sin√

2π +1

2π = 0

or, c2 sin√

2π +1

2π = 0

or, c2 = − π

2 sin√

2π


y = −π sin√

2x

2 sin√

2π+x

2

or, y =−π sin

√2x+ x sin

√2π

2 sin√

2π

Example 315.


x2y′′ − 2xy′ + 2y = 0, y(1) = −1, y(2) = 1


x2y′′ − 2xy′ + 2y = 0, y(1) = −1, y(2) = 1

or, x2 d2y

dx2− 2x

dy

dx+ 2y = 0 y(1) = −1, y(2) = 1 (9.14)

Let us put x = ez i.e. z = loge x. Then

dz

dx=

1

x

Nowdy

dx=dy

dz

dz

dx=

1

x

dy

dz

xdy

dx=dy

dz


Again

d2y

dx2=

d

dx

(dy

dx

)or,d2y

dx2=

d

dx

(1

x

dy

dz

)or,d2y

dx2=

d

dx

(x−1

) dydz

+1

x

d

dx

(dy

dz

)or,d2y

dx2= − 1

x2

dy

dz+

1

x

d

dz

(dy

dz

)dz

dx

or,d2y

dx2= − 1

x2

dy

dz+

1

x2

d

dz

(dy

dz

)or, x2 d

2y

dx2= −dy

dz+d2y

dz2

Putting values of x dydx , and x2 d2ydx2

in (9.14), we get

d2y

dz2− dy

dz− 2

dy

dz+ 2y = 0

d2y

dz2− 3

dy

dz+ 2y = 0


m2 − 3m+ 2 = 0

m = 1, m = 2


y. = c1ez + c2e

2z

or, y = c1elog x + c2e

2 loge x

y = c1eloge x + c2e

loge x2

y = c1x+ c2x2 (9.15)

Using the first boundary condition y(1) = −1 in (9.15), we have

c1 + c2 = −1 (9.16)

Again, using the second boundary condition y(2) = 1 in (9.13)

2c1 + 4c2 = 1 (9.17)


c1 = −5

2, c2 =

3

2



y = −5

2x+

3

2x2

The following example illustrate that a nonhomogeneous boundary problem may

have no solution.

Example 316.


y′′ + y = 0, y(0) = 1, y(π) = 3


y′′ + y = 0, y(0) = 1, y(pi) = 3 (9.18)


m2 + 1 = 0

m = 0± i





c1 cos 0 + c2 sin 0 = 1

or, c1 = 1


c1 cosπ + c2 sinπ = 3

or, c1 = −3

Thus, we get different value of c1 by the given two boundary conditions. Hence, the

given differentilal equation has no solution.

The following example shows that a boundary value problem may have trivial solu-

tion.

Example 317.



y′′ + 2y = 0, y(0) = 0, y(π) = 0


y′′ + 2y = 0, y(0) = 0, y(π) = 0 (9.20)


m2 + 2 = 0

m = 0±√

2i


y = e0x(c1 cos√

2x+ c2 sin√

2x)

y = c1 cos√

2x+ c2 sin√

2x (9.21)


c1 cos 0 + c2 sin 0 = 0

or, c1 = 0


c1 cos√

2π + c2 sin√

2π = 0

or, c2 sin√

2π = 0

Since sin√

2π 6= 0, so c2 = 0.


y = 0

This solution y = 0 is trivial solution of the given boundary value problem.

The next exmples are of boundary value problem with infinitely many solutions.

Example 318.


y′′ + y = 0, y(0) = 0, y(π) = 0


y′′ + y = 0, y(0) = 0, y(π) = 0 (9.22)


m2 + 1 = 0


m = 0± i





c1 cos 0 + c2 sin 0 = 0

or, c1 = 0


c1 cosπ + c2 sinπ = 0

or, c2 sinπ = 0 as c1 = 0

Since sinπ = 0, the second boundary condition is also satisfied when c1 = 0, for all

values of c2. Thus, the solition of the system is

y = c2 sinx where c2 is arbitrary.

Hence, the given boundary value problem has infinitely many solutions.

Example 319.


y′′ + 4y = cosx, y′(0) = 0, y′(π) = 0


y′′ + 4y = cosx, y′(0) = 0, y′(π) = 0

(D2 + 4)y = cosx, y′(0) = 0, y′(π) = 0 (9.24)


m2 + 4 = 0

m = 0± 2i

The completementrary function of

C.F. = e0x(c1 cos 2x+ c2 sin 2x)

C.F. = c1 cos 2x+ c2 sin 2x (9.25)

Particular integral is

P.I. =1

D2 + 4cosx

=1

−12 + 4cosx =

cosx

3



y = C.F.+ P.I.

or, y = c1 cos 2x+ c2 sin 2x+cosx

3(9.26)

Differentiating with respect to x

y′(x) = −2c1 sin 2x+ 2c2 cos 2x− sinx

3

Using the first boundary condition y′(0) = 0 we have

−2c1 sin 0 + 2c2 cos 0− 2

3sin 0 = 0

or, c2 = 0

Again, using the second boundary condition y′(π) = 0, we get

−2c1 sin 2π + c2 cos 2π − 2

3sin 2π = 0

or, c2 = 0

Both boundary conditions are also satisfied when for all values of c1. Thus, the

solution of the system is

y = c1 cos 2x+cosx

3where c2 is arbitrary.

Hence, the given boundary value problem has infinitely many solutions.

9.1.3 Eigenvalue Problems

Let A be a square matrix. Let us consider a matrix equation

Ax = λx (9.27)

The equation (9.27) has the solution x = 0 for every value of λ. But for certain values

of λ, there are non zero solutions, in this case the value of λ is called eigenvalues

and the nonzero solutions x are called eigen vectors. There is similar conditions for

the boundary value problems.

Consider the problem of differential equation

y′′ + λy = 0, y(0) = 0, y(π) = 0 (9.28)

The problem has only trivail solution y = 0, when λ = 1. We extend the terminolgy

associated with the equation (9.27), to the differential equation (9.28).

The values of λ for which nontrivial solutions of (9.28) occurs are called eigenvalues,

and the nontrivial solutions themselves are called eigenfunctions of the equation.


Example 320. Find the eigenvalue and eigenfunction for the differential equation

y” + λy = 0, y(0) = 0, y(π) = 0


y” + λy = 0 y(0) = 0, y(π) = 0 (9.29)

The auxiliary equation of the given differentail equation is (9.29) is

m2 + λ = 0

or, m2 = −λ

Case I If λ = 0, then m2 = 0 which implies m = 0, 0. The general solution of

(9.29)

y = e0x(c1 + c2x)

y = c1 + c2x (9.30)

Using the boundary condition

y(0) = 0 =⇒ c1 = 0

and

y(π) = 0 =⇒ c1 + c2π = 0

c2π = 0

c2 = 0

Hence, the solution (9.30) becomes

y = 0

Which is trivial solution.

Case II If λ < 0 then −λ > 0.

m2 = −λm = ±

√−λ

The general solution of (9.29) is

y = c1e√−λx + c2e

−√−λx (9.31)


y(0) = 0 =⇒ c1 + c2 = 0 (9.32)


and

y(π) = 0 =⇒ c1e√−λ π + c2e

−√−λ π = 0 (9.33)


c1 = 0, c2 = 0

Putting the values of c1 and c2 in (9.31)

y = 0

which is trivial solution.

Case III If λ > 0, then −λ < 0

∴ m = 0± i√λ


y = e0x(c1 cos√λ x+ c2 sin

√λ x)

or, y = c1 cos√λ x+ c2 sin

√λ x (9.34)

Using the buondary conditions

y(0) = 0 =⇒ c1 cos 0 + c2 sin 0 = 0

c1 = 0

and

y(π) = 0 =⇒ c1 cos√λπ + c2 sin

√λπ = 0

or, c2 sin√λπ = 0 as c1 = 0

For c2 6= 0, we must have

sin√λx = 0

∴√λπ = nπ

or,√λ = n

or, λ = n2

Thus, λ1 = 12, λ2 = 22, λ3 = 32, · · · are eigenvalues of the given differential equation

and the corresponding non-trivial solutions are

y1(x) = sinx, y2(x) = sin 2x, y3(x) = sin 3x, y4(x) = sin 4x, · · · , yn = sinnx, · · ·

are eigenfunctions of the given differential equation.


Example 321. Find the eigenvalues and eigenfunctions for the boundary value

problem

y” + λy = 0, y(0) = 0, y(L) = 0


y” + λy = 0 y(0) = 0, y(L) = 0 (9.35)


m2 + λ = 0

or, m2 = −λ

Case I If λ = 0, then m2 = 0 which implies m = 0, 0. The general solution of (9.35)

y = e0x(c1 + c2x)

y = c1 + c2x (9.36)


y(0) = 0 =⇒ c1 = 0

and

y(π) = 0 =⇒ c1 + c2L = 0

c2L = 0

c2 = 0


y = 0



m2 = −λm = ±

√−λ



−√−λx (9.37)


y(0) = 0 =⇒ c1 + c2 = 0 (9.38)


and

y(L) = 0 =⇒ c1e√−λ L + c2e

−√−λ L = 0 (9.39)


c1 = 0, c2 = 0


y = 0



∴ m = 0± i√λ

Then the general solution of (9.29) is


√λ x)


√λ x (9.40)

Using the buondary condition

y(0) = 0 =⇒ c1 cos 0 + c2 sin 0 = 0

c1 = 0

and

y(L) = 0 =⇒ c1 cos√λL+ c2 sin

√λL = 0

or, c2 sin√λL = 0 as c1 = 0


sin√λL = 0

∴√λL = nπ

or,√λ =

nπ

L

or, λ =n2π2

L2

Thus, λn = n2π2

L2 n = 1, 2, 3 · · · are eigenvalues of the given differential equation


yn(x) = sinnπ

Lx n = 1, 2, 3 · · ·



Example 322. Find the eigenvalues and eigenfunctions for the differential equation

y” + λy = 0, y(0) = 0, y′(π) = 0

Solution: The given differentialequation is

y” + λy = 0 y(0) = 0, y′(π) = 0 (9.41)


m2 + λ = 0

or, m2 = −λ


y = e0x(c1 + c2x)

y = c1 + c2x (9.42)

and y′ = c2


y(0) = 0 =⇒ c1 = 0

and

y′(π) = 0 =⇒ c2 = 0

(9.43)


y = 0



m2 = −λm = ±

√−λ



−√−λx (9.44)

y′ =√−λ(c1e

√−λx − c2e

−√−λx) (9.45)


y(0) = 0 =⇒ c1 + c2 = 0 (9.46)


and

y′(π) = 0 =⇒ c1e√−λ π − c2e

−√−λ π = 0 (9.47)


c1 = 0, c2 = 0


y = 0



∴ m = 0± i√λ



√λ x)


√λ x (9.48)

or, y′ =√λ(−c1 sin

√λ x+ c2 cos

√λ x) (9.49)

Using the boundary conditions

y(0) = 0 =⇒ −c1 cos 0 + c2 sin 0 = 0

c1 = 0

and

y′(π) = 0 =⇒√λ(−c1 sin

√λπ + c2 cos

√λπ) = 0

or, c2 cos√λπ = 0 as c1 = 0


cos√λπ = 0

∴√λπ =

(2n− 1)π

2

or,√λ = n− 1

2

or, λ =

(n− 1

2

)2

Thus, λn =(n− 1

2

)2n = 1, 2, 3 · · · are eigenvalues of the given boundary value

problem and the corresponding non-trivial solutions are

yn(x) = sin

(n− 1

2

)x n = 1, 2, 3 · · ·

are eigenfunctions of the given boundary value problem.



y” + λy = 0, y′(0) = 0, y(π) = 0.


y” + λy = 0 y′(0) = 0, y(π) = 0 (9.50)


m2 + λ = 0

or, m2 = −λ


y = e0x(c1 + c2x)

y = c1 + c2x (9.51)

and y′ = c2


y′(0) = 0 =⇒ c2 = 0

and

y(π) = 0 =⇒ c1 + c2π = 0

=⇒ c1 = 0


y = 0



m2 = −λm = ±

√−λ



−√−λx (9.52)

y′ =√−λ(c1e

√−λx − c2e

−√−λx)


y′(0) = 0 =⇒ c1 − c2 = 0 (9.53)


and

y(π) = 0 =⇒ c1e√−λ π + c2e

−√−λ π = 0 (9.54)


c1 = 0, c2 = 0


y = 0



∴ m = 0± i√λ



√λ x)


√λ x (9.55)


√λ x+ c2 cos

√λ x) (9.56)


y′(0) = 0 =⇒ −c1 sin 0 + c2 cos 0 = 0

c2 = 0

and

y(π) = 0 =⇒ (c1 cos√λπ + c2 sin

√λπ) = 0

or, c1 cos√λπ = 0 as c2 = 0


cos√λπ = 0

∴√λπ =

(2n− 1)π

2

or,√λ = n− 1

2

or, λ =

(n− 1

2

)2

Thus, λn =(n− 1

2

)2n = 1, 2, 3 · · · are eigenvalues of the given boundary value

problem and the corresponding non-trivial solutions are

yn(x) = cos

(n− 1

2

)x n = 1, 2, 3 · · ·

are eigenfunctions of the given boundary value problem.



y” + λy = 0, y′(0) = 0, y′(L) = 0


y” + λy = 0 y′(0) = 0, y′(π) = 0 (9.57)


m2 + λ = 0

or, m2 = −λ

Case I If λ = 0, then m2 = 0 which implies m = 0, 0 The general solution of (9.57)

y = e0x(c1 + c2x)

y = c1 + c2x (9.58)

and y′ = c2


y′(0) = 0 =⇒ c2 = 0

and

y′(π) = 0 =⇒ c2 = 0


y = c1

Hence λ0 = 0 is eigenvalue with corresponding eigenfunction y0 = 1.


m2 = −λm = ±

√−λ



−√−λx (9.59)

y′ =√−λ(c1e

√−λx − c2e

−√−λx)


y′(0) = 0 =⇒ c1 − c2 = 0 (9.60)

and

y′(L) = 0 =⇒ c1e√−λ L − c2e

−√−λ L = 0 (9.61)



c1 = 0, c2 = 0


y = 0



∴ m = 0± i√λ



√λ x)


√λ x (9.62)


√λ x+ c2 cos

√λ x) (9.63)

Using the buondary condition

y′(0) = 0 =⇒ −c1 sin 0 + c2 cos 0 = 0

c2 = 0

and

y′(L) = 0 =⇒ (−c1 sin√λL+ c2 cos

√λL) = 0

or, c1 sin√λL = 0 as c2 = 0


sin√λL = 0

∴√λL = nπ

or,√λ =

nπ

L

or, λ =n2π2

L2

Thus, λn = n2π2

L2 n = 1, 2, 3 · · · are eigenvalues of the given differential equation


yn(x) = cosn2π2

L2x n = 1, 2, 3 · · ·



9.2 Fourier series

9.2.1 Periodic Functions

A function f(x) is said to be periodic with period T > 0 if the domain of f contains

x+ T whenever it contains x, if

f(x+ T ) = f(x) (9.64)

for every value of x.

The graph of such periodic function is obatained by periodic repeatition of its graph

in any interval of length T .

From (9.64) it follow that, if n is any integer, then

f(x+ nT ) = f(x) for all x

Hence 2T, 3T, 4T, · · · are periods of f(x). The smallest value of T for which (9.64)

holds is called the fundamental period of f .

A constant function is a periodic function with an arbitrary period but no funda-

mental period.

Theorem 34. If f and g are any two functions with common period T , then

(i)fg

(ii) for any c1 and c2, c1f + c2g are periodic functions with period T .

Proof: Here f and g are periodic function with period T . Therefore,

f(x+ T ) = f(x), g(x+ T ) = g(x) for all x

(fg)(x+ T ) = f(x+ T )g(x+ T ) = f(x)g(x)

Hence fg is periodic function with period T .

(c1f + c2g)(x+ T ) = c1f(x+ T ) + c2g(x+ T ) = c1f(x) + c2g(x)

Thus, c1 f + c2 g is periodic function with period T .

Examples:

1. Each of the function sinx, cosx, secx, cscx is a periodic function with period

2π.

2. tanx, cotx are periodic function with period π.

3. sinnx, cosnx both are periodic functions with period 2πn .

Proof:

sinn

(x+

2π

n

)= sin (nx+ 2π) = sinnx for all x,

cosn

(x+

2π

n

)= cos (nx+ 2π) = cosnx for all x.

9.2. FOURIER SERIES 501

4. sin πxl , cos πxl are periodic functions with period 2l.

Proof:

sinπ

l(x+ 2l) = sin

(πxl

+ 2π)

= sinπx

l

cosπ

l(x+ 2l) = cos

(πxl

+ 2π)

= cosπx

l

5. The sum of a number of periodic functions is a periodic functions.

∞∑n=1

an cosnx,∞∑n=1

bn sinnx,∞∑n=1

(an cosnx+ bn sinnx)

all are periodic functions with period 2π.

Example 325. Determine whether the given function f(x) = sin πxL is periodic or

not. If periodic find fundamental period.

Solution: Here f(x) = sin πxL . Let T be the period of f(x). Then

f(x+ T ) = f(x)

or, sinπ(x+ T )

L= sin

πx

L

or, sin

(πx

L+πT

L

)= sin 7x

or, sinπx

Lcos

πT

L+ cos

πx

Lsin

πT

L= sin

πx

L+ 0 · cos

πx

L

Comparing coefficients

cosπT

L= 1, sin

πT

L= 0

or, cosπT

L= cos 2π, sin

πT

L= sin 2π least positive

∴πT

L= 2π

or, T =2L

T

Example 326. Determine whether the given function f(x) = sin 7x is periodic or

not. If periodic find fundamental period.

Solution: Here f(x) = sin 7x. Let T be the period of f(x). Then

f(x+ T ) = f(x)

or, sin 7(x+ T ) = sin 7x

or, sin(7x+ 7T ) = sin 7x

or, sin 7x cos 7T + cos 7x sin 7T = sin 7x+ 0 cos 7x


Comparing coefficients

cos 7T = 1, sin 7T = 0

or, cos 7T = cos 2π, sin 7T = sin 2π least positive

∴ 7T = 2π

or, T =2π

7

9.2.2 Orthogonality of sine and cosine Functions

The inner product of two real valued functions u, v is denoted by (u, v) and is defined

on the interval a ≤ x ≤ b, is given by

(u, v) =

∫ b

au(x)v(x)dx

The function u and v are said to be mutually orthogonal on a ≤ x ≤ b if their inner

product is zero that is if

(u, v) = 0

or,

∫ b

au(x)v(x)dx = 0 (9.65)

Example 327. The function u(x) = x, v(x) = x4 are orthogonal on the interval

−1 ≤ x ≤ 1.

Solution: Here

(u, v) =

∫ 1

−1u(x)v(x)dx

=

∫ 1

−1xx4dx

=

∫ 1

−1x5dx

=

[x6

6

]1

−1

=1

6− 1

6= 0

A set of functions is said to be mutually orthogonal if each pair of functions in the

set are orthogonal. Let u1, u2, · · · be a set of real valued functions defined on

a ≤ x ≤ b are said to be orthogonal if

For m 6= n, (un, um) = 0

i.e.

∫ b

aunumdx = 0 (9.66)


Example 328. The functions

sinmπx

L, cos

mπx

L, m = 1, 2, 3 · · ·

form a mutually orthogonal set of functions on the interval −L ≤ x ≤ L. In fact,

they satisfy the following relations

∫ L

−Lsin

mπx

Lsin

nπx

Ldx =

0 m 6= n

L m = n(9.67)

∫ L

−Lcos

mπx

Lcos

nπx

Ldx =

0 m 6= n

L m = n(9.68)∫ L

−Lsin

mπx

Lcos

nπx

Ldx = 0 for all m,n (9.69)

Proof: To derive the Eq.(9.67), for m 6= n,

∫ L

−Lsin

mπx

Lsin

nπx

Ldx =

1

2

∫ L

−L2 sin

mπx

Lsin

nπx

Ldx

=1

2

∫ L

−L

[cos

(m− n)πx

L− cos

(m+ n)πx

L

]dx

=1

2

[sin (m−n)πx

L(m−n)π

L

−sin (m+n)πx

L(m+n)π

L

]L−L

=L

2π

[sin (m−n)πL

L

(m− n)−

sin (m+n)πLL

(m+ n)

]

− L

2π

[sin (m−n)π(−L)

L

m− n−

sin (m+n)π(−L)L

m+ n

]

=L

2π

[sin(m− n)π

(m− n)− sin(m+ n)π

(m+ n)

]− L

2π

[−sin(m− n)π

(m− n)+

sin(m+ n)π

(m+ n)

]= 0 as sin(m− n)π = 0, sin(m+ n)π = 0


For m = n∫ L

−Lsin

mπx

Lsin

nπx

Ldx =

∫ L

−Lsin2 mπx

Ldx

=1

2

∫ L

−L2 sin2 mπx

Ldx

=1

2

∫ L

−L

[1− cos

2mπx

L

]dx

=1

2

[x−

sin 2mπxL

2mπL

]L−L

=1

2

[L−

sin 2mπLL

2mπL

−

(−L−

sin 2mπ(−L)L

2mπL

)]

=1

2

[2L− sin 2mπ

2mπL

+sin 2mπ

2mπL

]=

1

2[2L− 0− 0] = L

Hence

∫ L

−Lsin

mπx

Lsin

nπx

Ldx =

0 m 6= n

L m = n

To derive the Eq.(9.68), for m 6= n,

∫ L

−Lcos

mπx

Lcos

nπx

Ldx =

1

2

∫ L

−L2 cos

mπx

Lcos

nπx

Ldx

=1

2

∫ L

−L

[cos

(m− n)πx

L+ cos

(m+ n)πx

L

]dx

=1

2

[sin (m−n)πx

L(m−n)π

L

+sin (m+n)πx

L(m+n)π

L

]L−L

=L

2π

[sin (m−n)πL

L

(m− n)+

sin (m+n)πLL

(m+ n)

]

− L

2π

[sin (m−n)π(−L)

L

(m− n)+

sin (m+n)π(−L)L

(m+ n)

]

=L

2π

[sin(m− n)π

(m− n)+

sin(m+ n)π

(m+ n)

]− L

2π

[−sin(m− n)π

(m− n)− sin(m+ n)π

(m+ n)

]= 0 as sin(m− n)π = 0, sin(m+ n)π = 0


For m = n∫ L

−Lcos

mπx

Lcos

nπx

Ldx =

∫ L

−Lcos2 mπx

Ldx

=1

2

∫ L

−L2 cos2 mπx

Ldx

=1

2

∫ L

−L

[1 + cos

2mπx

L

]dx

=1

2

[x+

sin 2mπxL

2mπL

]L−L

=1

2

[L+

sin 2mπLL

2mπL

−

(−L+

sin 2mπ(−L)L

2mπL

)]

=1

2

[2L− sin 2mπ

2mπL

+sin 2mπ

2mπL

]=

1

2[2L− 0− 0] = L

Hence ∫ L

−Lcos

mπx

Lcos

nπx

Ldx =

0 m 6= n

L m = n

To derive the Eq.(9.69), let

f(x) = sinmπx

Lcos

nπx

L

Then

f(−x) = sinmπ(−x)

Lcos

nπ(−x)

L= − sin

mπx

Lcos

nπx

L= −f(x)

Thus, f(x) is odd functions∫ L

−Lf(x)dx = 0

or,

∫ L

−Lsin

mπx

Lcos

nπx

Ldx = 0 for all m,n

Note We have ∫ L

−Lsin

mπx

Lsin

nπx

Ldx =

0 m 6= n

L m = n(9.70)

∫ L

−Lcos

mπx

Lcos

nπx

Ldx =

0 m 6= n

L m = n(9.71)


Since

f(x) = sinnπx

Lsin

mπx

Land f(x) = cos

nπx

Lcos

mπx

L

are even functions so∫ L

−Lsin

mπx

Lsin

nπx

Ldx = 2

∫ L

0sin

mπx

Lsin

nπx

Ldx

∫ L

−Lcos

mπx

Lcos

nπx

Ldx = 2

∫ L

0cos

mπx

Lcos

nπx

Ldx

Hence, the equations (9.70) and (9.71) becomes

∫ L

0sin

mπx

Lsin

nπx

Ldx =

0 m 6= nL2 m = n∫ L

0cos

mπx

Lcos

nπx

Ldx =

0 m 6= nL2 m = n

9.2.3 Trigonometric Series

A series of the form

a0

2+

∞∑n=1

(an cosnx+ bn sinnx)

is said to be a trigonometric series, where a0, an, bn are constants.

Note

1. When an = 0, the series will be pure sine series.

2. When bn = 0, the series will be pure cosine series.

3. Sum of sines of n angles in A.P.

sinα+ sin(α+ β) + sin(α+ 2β) + · · ·+ sin(α+ (n− 1)β)

=sinn.diff.

2

sin diff.2

sinfirst angle+last angle

2

=sin nβ

2

sin β2

sinα+ α+ (n− 1)β

2

=sin nβ

2

sin β2

sin

(α+

(n− 1)β

2

)

4. Sum of cosines of n angles in A.P.


cosα+ cos(α+ β) + cos(α+ 2β) + · · ·+ cos(α+ (n− 1)β)

=sinn.diff.

2

sin diff.2

cosfirst angle+last angle

2

=sin nβ

2

sin β2

cosα+ α+ (n− 1)β

2

=sin nβ

2

sin β2

cos

(α+

(n− 1)β

2

)

9.2.4 Fourier Series

If the trigonometric series

a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)

converge to a function f(x), at each value of x then the trigonometric series is called

Fourier series of f . The constant terms a0, am, an are called Fourier coefficients.

9.2.5 Determination of Fourier Coefficients

Let us suppose the series of the form

a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)

converges and let its sum be f(x) in −L ≤ x ≤ L. Then

f(x) =a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)(9.72)


The coefficients a0, am, bm can be related to f(x). Integrating, the Eq.(9.72) with

respect to x from −L to L.∫ L

−Lf(x)dx =

a0

2

∫ L

−Ldx+

∞∑m=1

(am

∫ L

−Lcos

mπx

Ldx+ bm

∫ L

−Lsin

mπx

Ldx

)

or,

∫ L

−Lf(x)dx =

a0

2[x]L−L +

∞∑m=1

(amL

mπ

[sin

mπx

L

]L−L

+bmL

mπ

[− cos

mπx

L

]L−L

)

or,

∫ L

−Lf(x)dx =

a0

22L+

∞∑m=1

(amL

mπ[sinmπ − sin(−mπ)] +

bmL

mπ[− cosmπ + cos(−mπ)]

)

or,

∫ L

−Lf(x)dx = La0 +

∞∑n=1

(amL

nπ[0 + 0] +

bmL

mπ[− cosmπ + cosmπ]

)or,

∫ L

−Lf(x)dx = La0

or, a0 =1

L

∫ L

−Lf(x)dx (9.73)

Multiplying, the Eq.(9.72)by cos nπxL , where n > 0 is fixed integer and integrating

with respect to x from −L to L.∫ L

−Lf(x) cos

nπx

Ldx =

a0

2

∫ L

−Lcos

nπx

Ldx+

∞∑m=1

(am

∫ L

−Lcos

mπx

Lcos

nπx

Ldx+

bm

∫ L

−Lsin

mπx

Lcos

nπx

Ldx

)(9.74)

From orthoganality relations we have

∫ L

−Lcos

mπx

Lcos

nπx

Ldx =

0 m 6= n

L m = n

∫ L

−Lsin

mπx

Lcos

nπx

Ldx = 0 for all m,n

and ∫ L

−Lcos

nπx

Ldx = 0

Using these results in Eq.(9.74) we get∫ L

−Lf(x) cos

mπx

Ldx = amL

or, am =1

L

∫ L

−Lf(x) cos

mπx

Ldx (9.75)


Again, multiplying, the Eq.(9.72)by sin nπxL , where n > 0 is fixed integer and inte-

grating with respect to x from −L to L.∫ L

−Lf(x) sin

nπx

Ldx =

a0

2

∫ L

−Lsin

nπx

Ldx+

∞∑m=1

(am

∫ L

−Lcos

mπx

Lsin

nπx

Ldx+

bm

∫ L

−Lsin

mπx

Lsin

nπx

Ldx

)(9.76)

From orthoganality relations we have∫ L

−Lsin

mπx

Lsin

nπx

Ldx =

0 m 6= n

L m = n∫ L

−Lcos

mπx

Lsin

nπx

Ldx = 0 for all m,n

and ∫ L

−Lsin

nπx

Ldx = 0

Using these results in Eq.(9.76) we get∫ L

−Lf(x) sin

mπx

Ldx = bmL

or, bm =1

L

∫ L

−Lf(x) sin

mπx

Ldx (9.77)

The equations (9.75) and (9.77) are know as the Euler-Fourier formulas for the

coefficients in a Fourier series.

Example 329. Suppose that there is a Fourier series converging to the function f

defined by

f(x) =

−x −2 ≤ x < 0

x 0 ≤ x < 2

f(x+ 4) = f(x)

Then, determine the coefficients in this Fourier series and hence, find the Fourier

series of the function.

Solution: Given function is

f(x) =

−x −2 ≤ x < 0

x 0 ≤ x < 2

f(x+ 4) = f(x)

This function represents a triangular wave and it has period 4.

∴ 2L = 4


Figure 9.1: Triangular wave

or, L = 2

Let the Fourier series of the given function be

f(x) =a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)or, f(x) =

a0

2+∞∑m=1

(am cos

mπx

2+ bm sin

mπx

2

)(9.78)

Then

a0 =1

L

∫ L

−Lf(x)dx

=1

2

∫ 2

−2f(x)dx

=1

2

∫ 0

−2f(x)dx+

1

2

∫ 2

0f(x)dx

=1

2

∫ 0

−2(−x)dx+

1

2

∫ 2

0xdx

= −1

2

[x2

2

]0

−2

+1

2

[x2

2

]2

0

= −1

2

(02

2− 22

2

)+

1

2

[22

2− 02

2

]= 2


am =1

L

∫ L

−Lf(x) cos

mπx

2dx

=1

2

∫ 2

−2f(x) cos

mπx

2dx

=1

2

∫ 0

−2f(x) cos

mπx

2dx+

1

2

∫ 2

0f(x) cos

mπx

2dx

=1

2

∫ 0

−2(−x) cos

mπx

2dx+

1

2

∫ 2

0x cos

mπx

2dx

=1

2

[−x∫

cosmπ

2dx

]0

−2

−∫ 0

−2

(d(−x)

dx

∫cos

mπx

2dx

)dx

+1

2

[x

∫cos

mπx

2dx

]2

0

−∫ 2

0

(dx

dx

∫cos

mπx

2dx

)dx

=1

2

[−x 2

πmsin

mπx

2

]0

−2

−∫ 0

−2(−1)

2

πmsin

mπx

2dx

+1

2

[x

2

πmsin

mπx

2

]2

0

−∫ 2

0

2

πmsin

mπx

2dx

=1

2(0− 0) +

1

πm

∫ 0

−2sin

mπx

2dx+

1

2(0− 0)− 1

πm

∫ 2

0sin

mπx

2dx

=2

m2π2

[− cos

mπx

2

]0

−2− 2

m2π2

[− cos

mπx

2

]2

0

=2

m2π2[−1 + cosmπ]− 2

m2π2[− cosmπ + 1]

=4

m2π2(cosmπ − 1), m = 1, 2, · · ·

=

− 8m2π2 m odd

0 even(9.79)

bm =1

L

∫ L

−Lf(x) sin

mπx

2dx

=1

2

∫ 2

−2f(x) sin

mπx

2dx

=1

2

∫ 0

−2f(x) sin

mπx

2dx+

1

2

∫ 2

0f(x) sin

mπx

2dx

=1

2

∫ 0

−2(−x) sin

mπx

2dx+

1

2

∫ 2

0x sin

mπx

2dx


=1

2

[−x∫

sinmπ

2dx

]0

−2

−∫ 0

−2

(d(−x)

dx

∫sin

mπx

2dx

)dx

+1

2

[x

∫sin

mπx

2dx

]2

0

−∫ 2

0

(dx

dx

∫sin

mπx

2dx

)dx

=1

2

[x

2

πmcos

mπx

2

]0

−2

−∫ 0

−2(−1)

(−2)

πmcos

mπx

2dx

+1

2

[−x 2

πmcos

mπx

2

]2

0

−∫ 2

0

(−2)

πmcos

mπx

2dx

=1

2

(0 +

4

mπcosmπ

)− 1

πm

∫ 0

−2cos

mπx

2dx+

1

2

(− 4

mπcosmπ + 0

)+

1

πm

∫ 2

0cos

mπx

2dx

= − 1

πm

∫ 0

−2cos

mπx

2dx+

1

πm

∫ 2

0cos

mπx

2dx

= − 2

m2π2

[sin

mπx

2

]0

−2+

2

m2π2

[sin

mπx

2

]2

0

= − 2

m2π2[sin 0− sinmπ] +

2

m2π2[sinmπ − sin 0] = 0

Substituting the values of a0, am and bm in (10.17), we get

f(x) =2

2+∞∑m=1

(am cos

mπx

2+ 0 cos

mπx

2

)= 1 + a1 cos

πx

2+ a2 cos

2πx

2+ a3 cos

3πx

2+ a4 cos

4πx

2+ · · ·

= 1− 8

π2cos

πx

2− 8

32π2cos

3πx

2− 8

52π2cos

5πx

2− · · ·

= 1− 8

π2

(cos

πx

2+

1

32cos

3πx

2+

1

52cos

5πx

2+ · · ·

)= 1− 8

π2

∞∑m=1,3,5

1

m2cos

mπx

2

= 1− 8

π2

∞∑m=1

1

(2m− 1)2cos

(2m− 1)πx

2replacing m by (2m− 1)

which is required Fourier series.

Example 330. Let

f(x) =

0 −3 < x < −1

1 −1 < x < 1

0 1 < x < 3.

f(x+ 6) = f(x)

Sketch the graph, find the coefficients in the Fourier series and the series.



f(x) =

0 −3 < x < −1

1 −1 < x < 1

0 1 < x < 3.

f(x+ 6) = f(x)

∴ 2L = 6

or, L = 3

The graph of the given function is

Figure 9.2: Graph of Example 330


f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)or, f(x) =

a0

2+

∞∑m=1

(am cos

mπx

3+ bm sin

mπx

3

)(9.80)

Then

a0 =1

L

∫ L

−Lf(x)dx

=1

3

∫ 3

−3f(x)dx

=1

3

∫ −1

−3f(x)dx+

1

3

∫ 1

−1f(x)dx+

1

3

∫ 3

1f(x)dx

= 0 +1

3

∫ 1

−1dx+ 0

=1

3[x]1−1 =

1

3[1 + 1] =

2

3


Aslo

am =1

L

∫ L

−Lf(x) cos

mπx

3dx

=1

3

∫ 3

−3f(x) cos

mπx

3dx

=1

3

∫ −1

−3f(x) cos

mπx

3dx+

1

3

∫ 1

−1f(x) cos

mπx

3dx+

1

3

∫ 3

1f(x) cos

mπx

3

= 0 +1

3

∫ 1

−1cos

mπx

3dx+ 0

=1

3

3

mπ

[sin

mπx

3

]1

−1

=1

mπ

[sin

mπ

3− sin

(−mπ)

3

]=

2

mπsin

mπ

3for m = 1, 2, 3, · · ·

Again,

bm =1

L

∫ L

−Lf(x) sin

mπx

3dx

=1

2

∫ 3

−3f(x) sin

mπx

3dx

=1

3

∫ −1

−3f(x) sin

mπx

3dx+

1

3

∫ 1

−1f(x) sin

mπx

3dx+

1

3

∫ 3

1f(x) sin

mπx

3

= 0 +1

3

∫ 1

−1sin

mπx

3dx+ 0

=1

3

∫ 1

−1sin

mπx

3dx = 0

As sin mπx3 is odd an function.

Putting the values of a0, an, and bn in (9.80)

f(x) =1

3+

∞∑m=1

2

mπsin

mπ

3cos

mπx

3

=1

3+

2

πsin

π

3cos

πx

3+

2

2πsin

2π

3cos

2πx

3+

2

3πsin

3π

3cos

3πx

3+ · · ·

=1

3+

√3

π

(cos

πx

3+

1

2cos

2πx

3− 1

4cos

4πx

3− 1

5cos

5πx

3+ · · ·

)As

sinπ

3= sin

2π

3=

√3

2, sinπ = 0, sin

4π

3= sin

5π

3= −√

3

2

Example 331. Find the Fourier series of

f(x) =

1 −L ≤ x < 0

0 0 ≤ x < L


f(x+ 2L) = f(x)


f(x) =

1 −L ≤ x < 0

0 0 ≤ x < L

f(x+ 2L) = f(x)


f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)(9.81)

Then

a0 =1

L

∫ L

−Lf(x)dx

=1

L

∫ 0

−Lf(x)dx+

1

L

∫ L

0f(x)dx

=1

L

∫ 0

−Ldx+ 0

=1

L[x]0−L =

1

L[0 + L] = 1

Aslo

am =1

L

∫ L

−Lf(x) cos

mπx

Ldx

=1

L

∫ 0

−Lf(x) cos

mπx

Ldx+

1

L

∫ L

0f(x) cos

mπx

Ldx

=1

L

∫ 0

−Lcos

mπx

Ldx+ 0

=1

L

L

mπ

[sin

mπx

L

]0

−L

=1

mπ

[sin 0− sin

(−mπL)

L

]=

1

mπ[sin 0 + sinmπ] = 0


Again,

bm =1

L

∫ L

−Lf(x) sin

mπx

Ldx

=1

L

∫ 0

−Lf(x) sin

mπx

Ldx+

1

L

∫ L

0f(x) sin

mπx

Ldx

=1

L

∫ 0

−Lsin

mπx

Ldx+ 0

=1

L

L

mπ

[− cos

mπx

L

]0

−L

=1

mπ[−1 + cosmπ]

=

0 if n = even

− 2mπ if n = odd

Putting the values of a0, an and bn in (9.81)

f(x) =1

2+

∞∑m=odd

(− 2

mπ

)sin

mπx

L

=1

2− 2

π

∞∑n=1

sin (2n−1)πxL

(2n− 1)

as m is odd, so put m = (2n− 1).


f(x) =

x+ L −L ≤ x < 0

L 0 ≤ x < L

f(x+ 2L) = f(x)


f(x) =

x+ L −L ≤ x < 0

L 0 ≤ x < L

f(x+ 2L) = f(x)


f(x) =a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)(9.82)


Then

a0 =1

L

∫ L

−Lf(x)dx

=1

L

∫ 0

−Lf(x)dx+

1

L

∫ L

0f(x)dx

=1

L

∫ 0

−L(x+ L)dx+

1

L

∫ L

0Ldx

=1

L

[x2

2+ Lx

]0

−L+L

L[x]L0

=1

L

[0− L2

2+ L2

]+ L

=L

2+ L =

3L

2

Aslo

am =1

L

∫ L

−Lf(x) cos

mπx

Ldx

=1

L

∫ 0

−Lf(x) cos

mπx

Ldx+

1

L

∫ L

0f(x) cos

mπx

Ldx

=1

L

∫ 0

−L(x+ L) cos

mπx

Ldx+

1

L

∫ L

0L cos

mπx

Ldx

=1

L

[(x+ L)

∫cos

mπx

L

]0

−L−∫ 0

−L

[d

dx(x+ L)

∫cos

mπx

Ldx

]dx

+L

mπ

[sin

mπx

L

]L0

=1

L

[(x+ L)

L

mπsin

mπx

L

]0

−L− L

mπ

∫ 0

−Lsin

mπx

L

+

L

mπ(sinmπ − sin 0)

=1

L

[0− 0]− L2

m2π2

[− cos

mπx

Ldx]0

−L

+ 0

=1

L

− L2

m2π2[−1 + cosmπ]

=

L

m2π2(1− cosmπ)

=

0 if m is even2Lm2π2 if m is odd


Again,

bm =1

L

∫ L

−Lf(x) sin

mπx

Ldx

=1

L

∫ 0

−Lf(x) sin

mπx

Ldx+

1

L

∫ L

0f(x) sin

mπx

Ldx

=1

L

∫ 0

−L(x+ L) sin

mπx

Ldx+

1

L

∫ L

0L sin

mπx

Ldx

=1

L

[(x+ L)

∫sin

mπx

L

]0

−L−∫ 0

−L

[d

dx(x+ L)

∫sin

mπx

Ldx

]dx

+L

mπ

[− cos

mπx

L

]L0

=1

L

[−(x+ L)

L

mπcos

mπx

L

]0

−L+

L

mπ

∫ 0

−Lcos

mπx

L

+

L

mπ(− cosmπ + cos 0)

=1

L

[− L2

mπ+

(−L+ L)

mπcosmπ

]− L2

m2π2

[sin

mπx

L

]0

−L

+

L

mπ(1− cosmπ)

= − L

mπ− L

m2π2(sin 0− sin(−mπ)) +

L

mπ− L

mπcosmπ

= −L cosmπ

mπ=

(−1)m+1L

mπ


f(x) =3L

4+

∞∑m=1

(L

m2π2(1− cosmπ)cos

mπx

L− L cosmπ

mπsin

mπx

L

)

=3L

4+

∞∑m=odd

2L

m2π2cos

mπx

L+∞∑m=1

(−1)m+1L

mπsin

mπx

L

=3L

4+∞∑n=1

2L

(2n− 1)2π2cos

(2n− 1)πx

L+∞∑n=1

(−1)n+1L

nπsin

nπx

L

=3L

4+

∞∑n=1

(2L

(2n− 1)2π2cos

(2n− 1)πx

L+

(−1)n+1L

nπsin

nπx

L

)(9.83)



f(x) =

x+ 1 −1 ≤ x < 0

1− x 0 ≤ x < 1

f(x+ 2) = f(x)



f(x) =

x+ 1 −1 ≤ x < 0

1− x 0 ≤ x < 1

f(x+ 2) = f(x)

∴ 2L = 2 =⇒ L = 1


Figure 9.3: Graph of the function

f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)or f(x) =

a0

2+∞∑m=1

(am cosmπx+ bm sinmπx) (9.84)

Then

a0 =1

L

∫ 1

−1f(x)dx

=

∫ 0

−1f(x)dx+

∫ 1

0f(x)dx

=

∫ 0

−1(x+ 1)dx+

∫ 1

0(1− x)dx

=

[x2

2+ x

]0

−1

+

[x− x2

2

]1

0

= 0− 1

2+ 1 + 1− 1

2− 0 = 1


Aslo

am =1

L

∫ L

−Lf(x) cos

mπx

Ldx

=

∫ 1

−1f(x) cosmπxdx

=

∫ 0

−1f(x) cosmπxdx+

∫ 1

0f(x) cosmπxdx

=

∫ 0

−1(x+ 1) cosmπxdx+

∫ 1

0(1− x) cosmπxdx

=

[(x+ 1)

∫cosmxdx

]0

−1

−∫ 0

−1

d

dx(x+ 1)

∫cosmxdx

dx+

[(1− x)

∫cosmxdx

]0

−1

−∫ 1

0

d

dx(1− x)

∫cosmπxdx

dx

=

[(x+ 1)

sinmπx

mπ

]0

−1

− 1

mπ

∫ 0

−1sinmπxdx+

[(1− x)

sinmπx

mπ

]1

0

+1

mπ

∫ 1

0sinmπxdx

= 0− 0 +1

m2π2[cosmπx]0−1 + 0− 0− 1

m2π2[cosmπx]10

=1

m2π2(1− cosmπ)− 1

m2π2(cosmπ − 1)

=2

m2π2(1− cosmπ) =

0 if m is even4

m2π2 if m is odd


Again,

bm =1

L

∫ L

−Lf(x) sin

mπx

Ldx

=

∫ 1

−1f(x) sinmπxdx

=

∫ 0

−1f(x) sinmπxdx+

∫ 1

0f(x) sinmπxdx

=

∫ 0

−1(x+ 1) sinmπxdx+

∫ 1

0(1− x) sinmπxdx

=

[(x+ 1)

∫sinmxdx

]0

−1

−∫ 0

−1

d

dx(x+ 1)

∫sinmxdx

dx+

[(1− x)

∫sinmxdx

]0

−1

−∫ 1

0

d

dx(1− x)

∫sinmπxdx

dx

=[−(x+ 1)

cosmπx

mπ

]0

−1+

1

mπ

∫ 0

−1cosmπxdx+

[−(1− x)

cosmπx

mπ

]1

0

− 1

mπ

∫ 1

0cosmπxdx

= − 1

mπ+ 0 +

1

m2π2[sinmπx]0−1 + 0 +

1

mπ− 1

m2π2[sinmπx]10

= 0

Putting the values of a0, am and bm in (9.84) we get

f(x) =1

2+

∞∑m=odd

4

m2π2cosmπx

=1

2+

4

π2

∞∑n=1

cos(2n− 1)πx

(2n− 1)2


f(x) =

0 −2 ≤ x < −1

2x −1 ≤ x < 1

0 1 ≤ x < 2

f(x+ 4) = f(x)


f(x) =

0 −2 ≤ x < −1

2x −1 ≤ x < 1

0 1 ≤ x < 2

f(x+ 4) = f(x)


∴ 2L = 4 =⇒ L = 2


f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)or, f(x) =

a0

2+

∞∑m=1

(am cos

mπx

2+ bm sin

mπx

2

)(9.85)

Then

a0 =1

L

∫ L

−Lf(x)dx

=1

2

∫ 2

−2f(x)dx

=1

2

∫ −1

−2f(x)dx+

1

2

∫ 1

−1f(x)dx+

1

2

∫ 2

1f(x)dx

= 0 +1

2

∫ 1

−12xdx+ 0

= [x2]1−1 =1

2[1− 1] = 0

Aslo

am =1

L

∫ L

−Lf(x) cos

mπx

2dx

=1

2

∫ 2

−2f(x) cos

mπx

2dx

=1

2

∫ −1

−2f(x) cos

mπx

2dx+

1

2

∫ 1

−1f(x) cos

mπx

2dx+

1

2

∫ 2

1f(x) cos

mπx

2

= 0 +2

2

∫ 1

−1x cos

mπx

2dx+ 0 = 0

As

x cosmπx

2

9.3. THE FOURIER CONVERGENCE THEOREM 523

is an odd function. Again,

bm =1

L

∫ L

−Lf(x) sin

mπx

2dx

=1

2

∫ 2

−2f(x) sin

mπx

2dx

=1

2

∫ −1

−2f(x) sin

mπx

2dx+

1

2

∫ 1

−1f(x) sin

mπx

2dx+

1

2

∫ 2

1f(x) sin

mπx

2dx

= 0 +1

2

∫ 1

−12x sin

mπx

2dx+ 0

=2

2

∫ 1

−1x sin

mπx

2dx

=

[x

∫sin

mπx

2dx

]1

−1

−∫ 1

−1

(dx

dx

∫sin

mπx

2dx

)dx

=2

mπ

[−x cos

mπx

2

]1

−1+

2

mπ

∫ 1

−1cos

mπx

2dx

=2

mπ

[− cos

mπ

2− (−(−1)) cos

mπ(−1)

2

]+

4

m2π2

[sin

mπx

2

]1

−1

= − 4

mπcos

mπ

2+

4

m2π2

[sin

mπ

2− sin

mπ(−1)

2

]= − 4

mπcos

mπ

2+

8

m2π2sin

mπ

2

= − 4

mπcos

mπ

2+ 2

(2

mπ

)2

sinmπ

2


f(x) =∞∑m=1

[− 4

mπcos

mπ

2+ 2

(2

mπ

)2

sinmπ

2

]sin

mπx

2

9.3 The Fourier Convergence Theorem

In previous section, we showed that if the Fourier series

a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)(9.86)

converges and thereby defines a function f , then f is periodic with period 2L, and

the coefficient am and bm are related to f(x) by Euler-Fourier formulas

am =1

L

∫ L

−Lf(x) cos

mπx

Ldx m = 0, 1, 2, 3 · · · (9.87)

bm =1

L

∫ L

−Lf(x) sin

mπx

Ldx m = 0, 1, 2, 3 · · · (9.88)


If the given function f is periodic with periodic with period 2L and integrable on

the interval [−L,L], then we compute a set of coefficients am and bm from (9.87) and

(9.88), and formally construct a series of the form (9.86). The quaestion is whether

this series converges for each value of x and, if so, whether its sum is f(x). The

Fourier seies corresponding to a function f may not converge to f(x).

To guarantee convergence of a Fourier series to the function from which its coeffi-

cients were computed, it is essential to place additional conditions on the function.

Before stating convergece theorem for Fourier series, we define following terms.

Definition 1. Piecewise Continuous:

A function f is said to be piecewise continuous on an interval a ≤ x ≤ b if the

interval can be partitioned by a finite number of points a = x0 < x1 < x2 < · · · <xn = b so that

1. f is continuous on each open sub-interval xi−1 < x < xi for i = 1, 2, 3 · · · .2. f approaches a finite limit at the endpoints of each subinterval are approached

from within the sub interval.

Notation: The notation f(c+) is usud to denote the limit of f(x) as x→ c from

right i.e.

f(c+) = limx→c+

f(x)

and f(c+) is usud to denote the limit of f(x) as x→ c from left i.e.

f(c+) = limx→c−

f(x)

Definition 2. Piecewise Smooth:

A function f is said to be piecewise smooth on an interval a ≤ x ≤ b if f(x) and

f ′(x) both are piecewise continuous on [a, b].

Theorem 35. Suppose that f and f ′ are piecewise continuous on the interval −L ≤x ≤ L. Further, suppose that f is defined outside the interval −L ≤ x ≤ L so that

it is periodic with period 2L. Then f has a Fourier series

f(x) =a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)whose coefficients are given by

am =1

L

∫ L

−Lf(x) cos

mπx

Ldx m = 0, 1, 2, 3 · · ·

bm =1

L

∫ L

−Lf(x) sin

mπx

Ldx m = 0, 1, 2, 3 · · ·


The Fourier converges to f(x) at all points where f is continuous, and to

f(x+) + f(x−)

2

at all points where f is discontinuous.

Example 335. Parseval’s equation, let

f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)Then show that

1

L

∫ L

−L[f(x)]2dx =

a20

2+

∞∑n=1

(a2n + b2n)

Proof: We have

f(x) =a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)Multiplying both sides by f(x), we get

[f(x)]2 =a0

2f(x) +

∞∑m=1

(amf(x) cos

mπx

L+ bmf(x) sin

mπx

L

)Integrating, and dividing both sides by L

1

L

∫ l

−L[f(x)]2dx =

a0

2

1

L

∫ L

−Lf(x) +

∞∑m=1

(am

1

L

∫ L

−Lf(x) cos

mπx

Ldx+ bm

1

L

∫ L

−Lf(x) sin

mπx

Ldx

)

or,1

L

∫ L

−L[f(x)]2dx =

a0

2a0 +

∞∑n=1

(anan + bnbn)

or,1

L

∫ L

−L[f(x)]2dx =

a20

2+∞∑n=1

(a2n + b2n)

Example 336. Assume that the given function is periodically extended outside the

original interval. Find the Fourier series for the extended function. Sketch the graph

of the function to which the series converges for the period.

f(x) =

−1 −1 ≤ x < 0

1 0 ≤ x < 1


f(x) =

−1 −1 ≤ x < 0

1 0 ≤ x < 1


Figure 9.4: The graph of given function

The given function ia assume to be periodic with 2L = 2. It is discontinous at x = 0

as f(0−) = −1 and f(0+) = 1.


f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)or, f(x) =

a0

2+

∞∑m=1

(am cosmπx+ bm sinmπx) (9.89)

a0 =1

L

∫ L

−Lf(x)dx

=

∫ 1

−1f(x)dx

=

∫ 0

−1f(x)dx+

∫ 1

0f(x)dx

=

∫ 0

−1(−1)dx+

∫ 1

01dx

= −[x]0−1 + [x]10

= −(0− (−1)) + 1− 0 = −1 + 1 = 0


am =1

L

∫ L

−Lf(x) cos

mπx

Ldx

=

∫ 1

−1f(x) cosmπxdx

=

∫ 0

−1f(x) cosmπxdx+

∫ 1

0f(x) cosmπxdx

=

∫ 0

−1(−1) cosmπxdx+

∫ 1

0cosmπxdx

= −[

sinmπx

mπ

]0

−1

+

[sinmπx

mπ

]1

0

= − 1

mπ[sin 0− sinmπ] +

1

mπ[sinmπ − sin 0]

= 0

and

am =1

L

∫ L

−Lf(x) sin

mπx

Ldx

=

∫ 1

−1f(x) sinmπxdx

=

∫ 0

−1f(x) sinmπxdx+

∫ 1

0f(x) sinmπxdx

=

∫ 0

−1(−1) sinmπxdx+

∫ 1

0sinmπxdx

= −[−cosmπx

mπ

]0

−1+[−cosmπx

mπ

]1

0

=1

mπ[cos 0− cosmπ]− 1

mπ[cosmπ − cos 0]

=2(1− cosmπ)

mπ

=

0 if m is even4mπ if m is odd

Putting the values of a0, am and bm in (9.89), the Fourier series is

f(x) =4

π

∞∑m=odd

1

msinmπx

=4

π

∞∑n=1

sin(2n− 1)πx

2n− 1

The function is piecewise continous on each finite interval. The point of discontinu-

ties are at integer values of x. At the points, the series converges to

f(x+) + f(x−)

2= 0



f(x) =

L+ x −L ≤ x < 0

L− x 0 ≤ x < L


f(x) =

L+ x −L ≤ x < 0

L− x 0 ≤ x < L


f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)(9.90)

Then

a0 =1

L

∫ L

−Lf(x)dx

=1

L

∫ 0

−Lf(x)dx+

1

L

∫ L

0f(x)dx

=1

L

∫ 0

−L(L+ x)dx+

1

L

∫ L

0(L− x)dx

=1

L

[Lx+

x2

2

]0

−L+

1

L

[Lx− x2

2

]L0

=1

L

[L2 − L2

2

]+

1

L

[L2 − L2

2

]=

L

2+L

2= L


am =1

L

∫ L

−Lf(x) cos

mπx

Ldx

=1

L

∫ 0

−Lf(x) cos

mπx

Ldx+

1

L

∫ L

0f(x) cos

mπx

Ldx

=1

L

∫ 0

−L(x+ L) cos

mπx

Ldx+

1

L

∫ L

0(L− x) cos

mπx

Ldx

=1

L

[(x+ L)

∫cos

mπx

Ldx

]0

−L−∫ 0

−L

[d

dx(x+ L)

∫cos

mπx

Ldx

]dx

+1

L

[(L− x)

∫cos

mπx

Ldx

]L0

−∫ L

0

[d

dx(L− x)

∫cos

mπx

Ldx

]dx

=1

L

[(x+ L)

L

mπsin

mπx

L

]0

−L− L

mπ

∫ 0

−Lsin

mπx

Ldx

+

+1

L

[(L− x)

L

mπsin

mπx

L

]L0

+L

mπ

∫ L

0sin

mπx

Ldx

=1

L

[0− 0]− L2

m2π2

[− cos

mπx

L

]0

−L

+

1

L

[0− 0] +

L2

m2π2

[− cos

mπx

L

]L0

=

1

L

− L2

m2π2[−1 + cosmπ]

+

1

L

L2

m2π2[− cosmπ + 1]

=

L

m2π2(1− cosmπ) +

L

m2π2(1− cosmπ)

=2L

m2π2(1− cosmπ)

=

0 if m is even4Lm2π2 if m is odd


Again,

bm =1

L

∫ L

−Lf(x) sin

mπx

Ldx

=1

L

∫ 0

−Lf(x) sin

mπx

Ldx+

1

L

∫ L

0f(x) sin

mπx

Ldx

=1

L

∫ 0

−L(L+ x) sin

mπx

Ldx+

1

L

∫ L

0(L− x) sin

mπx

Ldx

=1

L

[(L+ x)

∫sin

mπx

Ldx

]0

−L−∫ 0

−L

[d

dx(L+ x)

∫sin

mπx

Ldx

]dx

+1

L

[(L− x)

∫sin

mπx

Ldx

]L0

−∫ L

0

[d

dx(L− x)

∫sin

mπx

Ldx

]dx

=1

L

[−(x+ L)

L

mπcos

mπx

L

]0

−L+

L

mπ

∫ 0

−Lcos

mπx

Ldx

+

1

L

[−(L− x)

L

mπcos

mπx

L

]L0

− L

mπ

∫ L

0cos

mπx

Ldx

=1

L

[− L2

mπ+

(−L+ L)

mπcosmπ

]− L2

m2π2

[sin

mπx

L

]0

−L

+

1

L

[0 +

L2

mπ

]+

L2

m2π2

[sin

mπx

L

]L0

= − L

mπ− L

m2π2(sin 0− sin(−mπ)) +

L

mπ+

L

mπ[sinmπ − sin 0]

= 0


f(x) =L

2+

∞∑m=odd

4L

m2π2cos

mπx

L

=L

2+

4L

π2

∞∑n=1

1

(2n− 1)2cos

(2n− 1)πx

L


The function f(x) is continuous. By the Fourier convergence theorem, the series

converges to the continuous function f(x).

Example 338. Let

f(x) =

0 −π ≤ x < −π

2

1 −π2 ≤ x <

π2

0 π2 ≤ x <

π2 .

Find the coefficients in the Fourier series and the series.



f(x) =

0 −π ≤ x < −π

2

1 −π2 ≤ x <

π2

0 π2 ≤ x <

π2 .

The given function is assumed to be periodic with 2L = 2π. Let its Fourier series

be

f(x) =a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)or, f(x) =

a0

2+∞∑m=1

(am cos

mπx

π+ bm sin

mπx

π

)or,f(x) =

a0

2+

∞∑m=1

(am cosmx+ bm sinmx) (9.91)

Then

a0 =1

L

∫ L

−Lf(x)dx

=1

π

∫ π

−πf(x)dx

=1

π

∫ −π2

−πf(x)dx+

1

π

∫ π2

−π2

f(x)dx+1

π

∫ π

π2

f(x)dx

= 0 +1

π

∫ π2

−π2

dx+ 0

=1

π[x]

π2

−π2

=1

π

[π2

+π

2

]=

1

ππ = 1


Also

am =1

L

∫ L

−Lf(x) cos

mπx

Ldx

=1

π

∫ π

−πf(x) cosmxdx

=1

π

∫ −π2

−πf(x) cosmxdx+

1

π

∫ π2

−π2

f(x) cosmxdx+1

π

∫ π

π2

f(x) cosmxdx

= 0 +1

π

∫ π2

−π2

cosmxdx+ 0

=1

π

1

m[sinmx]

π2

−π2

=1

mπ

[sin

mπ

2− sin

(−mπ

2

)]=

2

mπsin

mπ

2for m = 1, 2, 3, · · ·

=2

mπ

0 if m = 2n

(−1)n−1 if m = 2n− 1

=

0 if m = 2n i.e. m is even

2(−1)n−1

(2n−1)π if m = 2n− 1 i.e. m is odd

Again,

an =1

L

∫ L

−Lf(x) sin

mπx

Ldx

=1

π

∫ π

−πf(x) sinmxdx

=1

π

∫ −π2

−πf(x) sinmxdx+

1

π

∫ π2

−π2

f(x) sinmxdx+1

π

∫ π

π2

f(x) sinmxdx

= 0 +1

π

∫ π2

−π2

sinmxdx+ 0

=1

mπ[− cosmπ]

π2

−π2

=1

mπ

[− cosm

π

2+ cos

(−mπ

2

)]=

1

mπ

[− cos

mπ

2+ cos

mπ

2

]= 0


f(x) =1

2+

2

π

∞∑n=1

(−1)n−1

(2n− 1)cos(2n− 1)x

The graph of the function to which the series converges is

The given function is piecewise discontinuous with discontinuities at odd multiple


of π/2 i.e. at

xd =(2k − 1)π

2k = 1, 2, 3 · · ·

and at this point the series converges to

f(xd+) + f(xd−)

2=

1

2

Example 339. Find the general solution of the initial value problem

y′′ + ω2y = sinnt, y(0) = 0, y′(0) = 0

when n is a positive integer and ω2 6= n2. What happens if ω2 = n2?


y′′ + ω2y = sinnt (9.92)

The auxiliaray equation is

m2 + ω2 = 0

or,m = 0± ωi

Hence complemntary function is

C.F. = e0x(c1 cos(ωt) + c2 sin(ωt)) = c1 cos(ωt) + c2 sin(ωt)

Let

ω2 6= n2

Then particular integral

P.I. =1

D2 + ω2sinnt

=1

−n2 + ω2sinnt =

sinnt

ω2 − n2

Therefore the solution of (9.92) is

y = C.F.+ P.I

y(t) = c1 cos(ωt) + c2 sin(ωt) +sinnt

ω2 − n2(9.93)

Diffentiating with respect to t

y′(t) = c1ω cos(ωt)− c2ω sin(ωt) + ncosnt

ω2 − n2(9.94)

Using the initial condition y(0) = 0, we get from (9.93)

0 = c1


and y′(0) = 0, we get from (9.94)

or, 0 = c1ω +n

ω2 − n2

c1 = − n

ω(ω2 − n2)

Putting values of c1 and c2 in (9.93), the solution is

y(t) = − n cosωt

ω(ω2 − n2)+

sinnt

ω2 − n2

or, y(t) =ω sinnt− n cosωt

ω(ω2 − n2)

Let ω2 = n2, the particulat integral is

P.I. =1

D2 + ω2sinnt

= t1

2Dsinnt

=t

2

∫sinntdt

= − t cosnt

2n

Therefore the solution of (9.92) is

y = C.F.+ P.I

y(t) = c1 cos(ωt) + c2 sin(ωt)− t cosnt

2n(9.95)

Diffentiating with respect to t

y′(t) = c1ω cos(ωt)− c2ω sin(ωt)− cosnt

2n+

sinnt

2(9.96)

Using the initial condition y(0) = 0, we get from (9.93)

0 = c1

and y′(0) = 0, we get from (9.96)

or, 0 = c2ω −1

2n

c2 =1

2nωBut ω = n, so

c2 =1

2n2

Putting values of c1 and c2 in (9.93), the solution is

y(t) =1

2n2sin(ωt)− t cosnt

2n=

sinωt− nt cosωt

2n2

Putting ω = n required solution is

y(t) =sinnt− nt cosnt

2n2

9.4. ODD AND EVEN FUNCTIONS 535

9.4 Odd and Even Functions

A function f(x) is said to be even function on the domain D if x ∈ D =⇒ −x ∈ Dand f(−x) = f(x) for all x ∈ D. The even function is symmetrical about y−axis.

Figure 9.5: Even function

Example 340. f(x) = cosx is an even function.

Solution: Here f(x) = cosx. Then

f(−x) = cos(−x) = cosx = f(x)

Therefore, f(x) is even function.

Example 341. f(x) = |x|5 is an even function.

Solution: Here f(x) = |x|5. Then

f(−x) = | − x|5 = |x|5 = f(x)

Hence f(x) is even function.

Following functions are even functions

(i) cosx (ii) secx (iii) x2


A function f(x) is said to be odd on the domain D if x ∈ D =⇒ −x ∈ D and

f(−x) = −f(x) for all x ∈ D. The odd function is symmetrical about origin.

Figure 9.6: Odd function

Example 342. f(x) = sinx is an odd function.

Solution: Here f(x) = sinx. Then

f(−x) = sin(−x) = − sinx = −f(x)

Therefore, f(x) is an odd function.

Example 343. f(x) = |x|x is an odd function.

Solution: Here f(x) = |x|x. Then

f(−x) = | − x|(−x) = −|x|x = −f(x)

Hence f(x) is an odd function.

Following functions are odd functions

(i)x3 (ii) tanx (iii) x+ sinx

Properties of Odd and Even Functions

1. The sum, difference, product and quotient of two even functions are even.

Proof: Let f1(x) and f2(x) be two even functions. Then

f1(−x) = f1(x), f2(−x) = f2(x)

(f1 + f2)(−x) = f1(−x) + f2(−x) = f1(x) + f2(x) = (f1 + f2)(x)


This shows that f1 + f2 is an even function.

(f1 − f2)(−x) = f1(−x)− f2(−x) = f1(x)− f2(x) = (f1 − f2)(x)

This shows that f1 − f2 is an even function.

(f1f2)(−x) = f1(−x)f2(−x) = f1(x)f2(x) = (f1f2)(x)

This shows that f1f2 is an even function.(f1

f2

)(−x) =

f1(−x)

f2(−x)=f1(x)

f2(x)=

(f1

f2

)(x)

This shows that f1f2

is an even function.

2. The sum and difference of the two odd functions are odd, and the product and

quotient of two odd function is even.

Proof: Let f1(x) and f2(x) be two odd functions. Then

f1(−x) = −f1(x), f2(−x) = −f2(x)

(f1 + f2)(−x) = f1(−x) + f2(−x) = −f1(x)− f2(x) = −(f1 + f2)(x)

This shows that f1 + f2 is an odd function.

(f1 − f2)(−x) = f1(−x)− f2(−x) = −f1(x) + f2(x) = −(f1 − f2)(x)

This shows that f1 − f2 is an odd function.

(f1f2)(−x) = f1(−x)f2(−x) = (−f1(x))(−f2(x)) = f1(x)f2(x) = (f1f2)(x)

Thus, f1f2 is an even function.(f1

f2

)(−x) =

f1(−x)

f2(−x)=f1(x)

f2(x)=

(f1

f2

)(x)

Hence f1f2


3. The product and quotient of odd and even functions are odd.

Proof: Let f1(x) be an odd and f2(x) be an even function. Then

f1(−x) = −f1(x), f2(−x) = f2(x)

(f1f2)(−x) = f1(−x)f2(−x) = (−f1(x))(f2(x)) = −f1(x)f2(x) = −(f1f2)(x)

Thus, f1f2 is an odd function.(f1

f2

)(−x) =

f1(−x)

f2(−x)=−f1(x)

f2(x)= −

(f1

f2

)(x)

Hence, −f1f2



4. If f is an even function, then∫ L

−Lf(x)dx = 2

∫ L

0f(x)dx

Proof: Here f(x) is an even function, so

f(−x) = f(x) for all x ∈ [−L,L] (9.97)

From properties of definite ingeral∫ b

af(x)dx =

∫ b

af(s)ds (9.98)∫ b

af(x)dx = −

∫ a

bf(x)dx (9.99)

and

∫ L

−Lf(x)dx =

∫ 0

−Lf(x)dx+

∫ L

0f(x)dx (9.100)

Letting x = −s and dx = −ds in the first term on the right side (9.100). When

x = 0 then s = 0 and when x = −L then s = L∫ L

−Lf(x)dx = −

∫ 0

Lf(−s)ds+

∫ L

0f(x)dx

or,

∫ L

−Lf(x)dx = −

∫ 0

Lf(s)ds+

∫ L

0f(x)dx using (9.97)

or,

∫ L

−Lf(x)dx = −

∫ 0

Lf(x)dx+

∫ L


or,

∫ L

−Lf(x)dx =

∫ L

0f(x)dx+

∫ L


or,

∫ L

−Lf(x)dx = 2

∫ L

0f(x)dx

5. If f is odd, then ∫ L

−Lf(x)dx = 0

Proof: Here f(x) is an odd function, so

f(−x) = −f(x) for all x ∈ [−L,L] (9.101)

From properties of definite ingeral∫ b

af(x)dx =

∫ b

af(s)ds (9.102)∫ b

af(x)dx = −

∫ a

bf(x)dx (9.103)


Let

I =

∫ L

−Lf(x)dx

Putting x = −s in I. Then dx = −ds, when x = −L, we get s = L and when

x = L, then s = −L.

I =

∫ L

−Lf(x)dx = −

∫ −LL

f(−s)ds

=

∫ −LL

f(s)ds using (9.101)

=

∫ −LL

f(x)dx using (9.102)

= −∫ L

−Lf(x)dx = −I using (9.103)

∴ 2I = 0

or, I = 0

or,

∫ L

−Lf(x)dx = 0

6. Prove that the derivatives of an even function is odd and the derivative of an

odd function is even.

Solution: Let f(x) be an even function Then

f(−x) = f(x)

Differentiating both sides with respect to x

−f ′(−x) = f ′(x)

or, f ′(−x) = −f ′(x)

This shows that f ′(x) is an odd function.

Let f(x) be an odd function. Then

f(−x) = −f(x)

Differentiating both sides with respect to x

−f ′(−x) = −f ′(x)

or, f ′(−x) = f ′(x)

Hence, f ′(x) is an even function.


9.4.1 Fourier sine series or Fourier series of odd functions

Let f(x) be a function such that

1. f(x) is odd function.

2. f(x) and f ′(x) are piecewise continuous on −L ≤ x ≤ L.

3. periodic function with period 2L.

Let the Fourier series of f(x) be

f(x) =a0

2+∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)(9.104)

Since f(x) is odd function so

a0 =1

L

∫ L

−Lf(x)dx = 0

We also know that cos mπxL is an even function. The the product of even and odd is

odd so f(x) cos mπxL is odd.

∴ bm =1

L

∫ L

−Lf(x) cos

mπx

Ldx = 0

Also, the product of two odds functions is even, so f(x) sin mπxL is even,. Hence

bm =1

L

∫ L

−Lf(x) cos

mπx

L=

2

L

∫ L

0f(x) cos

mπx

Ldx

Putting the values a0, am and bm in (9.104)

f(x) =∞∑m=1

bm sinmπx

L

Hence, the Fourier series for any odd function consists only of the odd trigonometric

function sin mπxL . This series is called Fourier sine series.

9.4.2 Fourier cosine series or Fourier series of even functions

Let f(x) be a function such that

1. f(x) is even.

2. f(x) and f ′(x) are piecewise continuous on −L ≤ x ≤ L.

3. periodic function with period 2L.


Let the Fourier series of f(x) be

f(x) =a0

2+

∞∑m=1

(am cos

mπx

L+ bm sin

mπx

L

)(9.105)

Since f(x) is an even function so

a0 =1

L

∫ L

−Lf(x)dx =

2

L

∫ L

0f(x)dx

We also know that cos mπxL is an even function.The the product of two even is even

so f(x) cos mπxL is even.

∴ bm =1

L

∫ L

−Lf(x) cos

mπx

Ldx =

2

L

∫ L

0f(x) cos

mπx

Ldx

Also, the product of even and odds functions is odd, so f(x) sin mπxL is odd,. Hence

bm =1

L

∫ L

−Lf(x) cos

mπx

L= 0

Putting the values a0, am and bm in (9.105)

f(x) =a0

2+∞∑m=1

am cosmπx

L

Hence, the Fourier series for any any function consists only of the trigonometric

function cos mπxL and constant term . This series is called Fourier cosine series.

Example 344. Let

f(x) = x, −L < x < L

and let f(−L) = f(L) = 0. Let f be defined elsewhere so that it is periodic of period

2L. The function defined in this manner is know as a sawtooth wave. Find the

Fourier series for this function.

Solution: Here

f(x) = x

is odd, so

am = 0, m = 0, 1, 2, · · ·


and

bm =2

L

∫ L

0f(x) sin

mπx

Ldx

=2

L

∫ L

0x sin

mπx

Ldx

=2

L

[x

∫sin

mπx

Ldx

]L0

−∫ L

0

(dx

dx

∫sin

mπx

Ldx

)dx

=2

L

[− Lxmπ

cosmπx

L

]L0

+L

mπ

∫ L

0cos

mπx

Ldx

=2

L

− L2

mπcosmπ − 0 +

L2

m2π2

[sin

mπx

L

]L0

=

2

L

− L2

mπcosmπ − 0 +

L2

m2π2[sinmπ − sin 0]

=

−2L

mπcosmπ

=−2L

mπ(−1)m, m = 1, 2, 3 · · ·

=2L

mπ(−1)m+1

Therefore, the Fourier sine series of the given sawtooth wave function f is

f(x) =∞∑m=1

bm sinmπx

L

=

∞∑m=1

2L

mπ(−1)m+1 sin

mπx

L

=2L

m

∞∑m=1

(−1)m+1

msin

mπx

L

9.4.3 Even and Odd Extension

We can defined a function

fe(x) =

f(x) 0 ≤ x ≤ Lf(−x) −L < x < 0

and

fe(x+ 2L) = fe(x)

Then the function fe is thus even periodic extension of f . It Fourier series, which is

a cosine series represents f on [0, L]. i.e.

fe(x) =a0

2+∞∑n−1

an cosnπx

L


where

an =

∫ L

0f(x) cos

nπx

Ldx

Again, we can defined a function

f0(x) =

f(x) 0 < x < L

0 x = 0, L

−f(−x) −L < x < 0

and

fo(x+ 2L) = fo(x)

Then the function fo is thus odd periodic extension of f . It Fourier series, which is

a cosine series represents f on [0, L]. i.e.

fo(x) =∞∑n=1

bn sinnπx

L

where

bn =

∫ L

0f(x) sin

nπx

Ldx

Example 345. Find the even and odd periodic extension of

f(x) =

0 0 ≤ x < 1

1 1 ≤ x < 2


f(x) =

0 0 ≤ x < 1

1 1 ≤ x < 2

For the periodic extension, we define

fe(x) =

f(x) 0 ≤ x ≤ 2

f(−x) −2 < x < 0

so fe(x) has even periodic extension of the period 4 whose graph is

f0(x) =

f(x) 0 < x < 2

0 x = 0, 2

−f(−x) −2 < x < 0


Figure 9.7: Odd Periodic Extension

Figure 9.8: Even Periodic Extension

Then the function fo is thus odd periodic extension of f of the period 4 whose

graph is

Example 346. Find Fourirer seiries for the given function

f(x) = 1, 0 ≤ x ≤ π, cosine series, period 2π

Solution: Here L = π. For an even extension of the function, the sine coeffi-

cients are zero. The cosine coefficient are given by

a0 =2

L

∫ L

0f(x)dx

=2

π

∫ π

0dx =

2

π[x]π0 =

2

ππ = 2


and

an =2

L

∫ L

0f(x) cos

nπx

Ldx

=2

π

∫ π

0cos

nπx

πdx

=2

π

∫ π

0cosnxdx

=2

π

[sinnx

n

]π0

=2

π

[sinnπ

n− sinn0

n

]= 0

The even extension of the constant function is constant. The Fourier cosine series is

f(x) =a0

2= 1

Example 347. Find Fourirer seiries for the given function and sketch the graph of

the function to which the given function converges over three periods

f(x) =

x 0 ≤ x < 1

1 1 ≤ x < 2sine series, period 4

Solution: Here L = 2. For an odd extension of the function, the cosine

coefficients are zero. The sine coefficients are given by

bn =2

L

∫ L

0f(x) sin

nπx

Ldx

=2

2

∫ 2

0f(x) sin

nπx

2dx

=

∫ 1

0f(x) sin

nπx

2dx+

∫ 2

1f(x) sin

nπx

2dx

=

∫ 1

0x sin

nπx

2dx+

∫ 2

1sin

nπx

2dx

=

[x

∫sin

nπx

2dx

]1

0

−∫ 1

0

(dx

dx

∫ 1

0sin

nπx

2dx

)dx− 2

nπ

[− cos

nπx

2

]2

1

=

[− 2x

nπcos

nπx

2

]1

0

+2

nπ

∫ 2

1cos

nπx

2dx+

2

nπ

[− cosnπ + cos

nπ

2

]= − 2

nπcos

nπ

2+ 0 +

4

n2π2

[sin

nπx

2

]1

0− 2

nπcosnπ

=4

n2π2

[sin

nπ

2− sin 0

]− 2

nπcosnπ

=4

n2π2sin

nπ

2− 2

nπcosnπ

=2

nπ

(− cosnπ +

2

nπsin

nπ

2

)


Therefore, the Fourier sine series of the given function is

f(x) =∞∑n=1

bn sinnπx

2

=

∞∑n=1

[2

nπ

(− cosnπ +

2

nπsin

nπ

2

)]sin

nπx

2



f(x) =

x 0 ≤ x < π

0 π ≤ x < 2πcosine series, period 4π

Solution: Here L = 2π. For an even extension of the function, the sine

coefficients are zero.

bn = 0

The constant is

a0 =1

L

∫ L

0f(x)dx

=1

π

∫ π

0xdx

=1

π

[x2

2

]π0

=1

π

[π2

2− 02

2

]=π

2


The cosine coefficients are given by

an =2

L

∫ L

0f(x) cos

nπx

Ldx

=2

2π

∫ 2π

0f(x) cos

nπx

2πdx

=1

π

∫ π

0f(x) cos

nx

2dx+

∫ 2π

πf(x) cos

nx

2dx

=

1

π

∫ π

0x cos

nx

2dx+

∫ 2π

π0 cos

nx

2dx

=

1

π

[x

∫cos

nx

2dx

]π0

−∫ π

0

(dx

dx

∫cos

nx

2dx

)dx

=

1

π

[2x

nsin

nx

2

]π0

− 2

n

∫ 2

1sin

nx

2dx

=

1

π

2π

nsin

nπ

2+ 0 +

4

n2

[cos

nx

2

]π0

=

1

π

2π

nsin

nπ

2+

4

n2

[cos

nπ

2− cos 0

]=

1

π

2π

nsin

nπ

2+

4

n2

[cos

nπ

2− 1]


f(x) =a0

2+

∞∑n=1

an cosnπx

L

=π

4+

1

π

∞∑n=1

2π

nsin

nπ

2+

4

n2

(cos

nπ

2− 1)

cosnx

2



f(x) =

1 0 < x < 1

0 1 < x < 2cosine series, period 4

Solution: Here L = 2. For an even extension of the function, the sine coeffi-

cients are zero.

bn = 0


The constant is

a0 =1

L

∫ L

0f(x)dx

=1

2

∫ 2

0F (x)dx

=1

2

∫ 1

0f(x)dx+

∫ 2

1f(x)dx

=

1

2

∫ 1

0dx+

∫ 2

10dx

=

1

2[x]10 =

1

2

The cosine coefficients are given by

an =2

L

∫ L

0f(x) cos

nπx

Ldx

=2

2

∫ 2

0f(x) cos

nπx

2dx

=

∫ 1

0f(x) cos

nπx

2dx+

∫ 2

1f(x) cos

nπx

2dx

=

∫ 1

0cos

nπx

2dx+

∫ 2

10 cos

nπx

2dx

=2

nπ

[sin

nπx

2

]1

0

=2

nπ

[sin

nπ

2− sin 0

]=

2

nπ

0 if n = 2, 4, 6 · · ·1 if n = 1, 5, 9 · · ·−1 if n = 3, 7, 11, · · ·

=2

nπ

0 if n = 2, 4, 6 · · ·

(−1)n−1 if n = 1, 5, 9 · · ·(−1)n−1 if n = 3, 7, 11, · · ·

=2

nπ

0 if is n even

(−1)n−1 if is n odd

=2(−1)n−1

(2n− 1)π


f(x) =a0

2+

∞∑n=odd

an cosnπx

L

=1

2+

2

π

∞∑n=1

(−1)n−1

(2n− 1)cos

(2n− 1)πx

2

Chapter 10

Separation of Variables

10.1 One Dimensional Heat Equation

10.1.1 Fourier ’s Law of Heat Conduction

Let us cosider a uniform rod insulated on the lateral surfaces so that heat flow on the

lateral surface only in the axial direction. Let us consider two parallel cross sections

of the same area A and different temperatures T1 and T2 respectively, are separated

by a small distance d, an amount of heat per unit time will pass from the warmer

section to the coolar one. The amount of heat flow per unit time is proportinal to

the area A, and to the temperature difference |T2 − T1| and inversely proportional

to the separation distance d.

Amount of heat flow per unit time =κA|T2 − T1|

d(10.1)

where the positive proportionality factor κ is called the termal conductivity and

depends on the materal of the rod. The relation (10.1) is called Fourier’s law of heat

conduction. Let the temperature depends on distance x and time t i.e.

T = u(x, t)

The the equation (10.1) become

Amount of heat flow per unit time =κA|u(x1, t)− u(x2, t)|

d

10.1.2 Derivation of One Diamensional heat Equation

1 Let us consider the element of rod lying between the cross section x and x + δx,

where x is arbitrary and δx is small. The instantaneous heat transfer H(x, t) from

1Student may leave the deduction

549

550 CHAPTER 10. SEPARATION OF VARIABLES

left to right across the section x is given by

H(x, t) = − limd→0

κAu(x+ d/2, t)− u(x− d/2, t)

d

= −κA ∂

∂xu(x, t) = −κAux(x, t) (10.2)

The minus sign appears in this equation because there will be a positive flow of the

heat from left to right only if the temperature is greater to the left of x than to the

right; in this case ∂u∂x is negative.

Similarly, the rate at which heat passes from left to right through the cross section

x+ δx is given by

H(x+ δx, t) = −κAux(x+ δx, t) (10.3)

The net rate at which the heat flows into the line segment of the bar between x and

x+ δx is thus given by

Q = H(x, t)−H(x+ δx) = κA [ux(x+ δx, t)− ux(x, t)]

The amount of heat entering this bar element in time δt

Qδt = κA [ux(x+ δx, t)− ux(x, t)] δt (10.4)

The average change in temperature δu in the interval δt, is proportional to the

amount of heat Qδt introduce and inversely proportinal to the mass δm of the

element. Thus

δu =Qδt

sδm=

Qδt

sρAδx(10.5)

where s is proportionality constant, is known as the specific heat of the metarial of

the bar, and ρ is its density.

The average temperature change δu in the bar element under consideration is the

actual temperature change at some intermediate point x+ θδx where 0, θ < 1.

δu = u(x+ θδx, t+ δt)− u(x+ θδx, t)

Now from (10.5)

u(x+ θδx, t+ δt)− u(x+ θδx, t) =Qδt

sρAδx

or, Qδt = [u(x+ θδx, t+ δt)− u(x+ θδx, t)] sρAδx (10.6)

10.1. ONE DIMENSIONAL HEAT EQUATION 551

From (10.4) and (10.6), we have

κA [ux(x+ δx, t)− ux(x, t)] δt = [u(x+ θδx, t+ δt)− u(x+ θδx, t)] sρAδx

Dividing both sides by δxδt

κ [ux(x+ δx, t)− ux(x, t)]

δx=

[u(x+ θδx, t+ δt)− u(x+ θδx, t)] sρ

δt

Letting δx→ 0 and δt→ 0

limδx→0

κ [ux(x+ δx, t)− ux(x, t)]

δx= lim

δt→0

[u(x+ θδx, t+ δt)− u(x+ θδx, t)] sρ

δt

or, κuxx = sρut

or,κ

sρuxx = ut

or, α2uxx = ut (10.7)

where α2 = κρs is called the termal diffusivity and it depends on the metarial.

10.1.3 Initial and boundary conditions

Let us assume that the initial temperature distribution in the rod is given by

u(x, 0) = f(x), 0 ≤ x ≤ L

and we consider the rod has fixed temperature 0 at the two ends of the rod at x = 0,

and x = L.

u(0, t) = 0, u(L, t) = 0 t > 0

Hence, a one diamensional heat equation with initial and boundary condition is

Partial differential equation: ut = α2uxx

Initial condition: u(x, 0) = f(x); 0 ≤ x ≤ LBoundary condition: u(0, t) = u(L, t) = 0, t > 0

10.1.4 Separation of Variables

Example 350. Find the ordinary differential equations to replace the partilal dif-

ferential equation

tuxx + xut = 0

by method of separation of variables. [B.Sc.2075]


Solution: The given partial differentialequation is

tuxx + xut = 0 (10.8)

Let u(x, t) = X(x)T (t) be the separation solution. Then

ut =∂u

∂t= XT ′ and uxx =

∂2u

∂x2= X ′′T

where primes refer to ordinary differentiation with respect to independent variables

x or t. Therefore, the equation (10.8) becomes

tX ′′ + xXT ′ = 0

or, tX ′′T = −xXT ′

or,X ′′

xX=−T ′

tT= λ say

which gives

X ′′ − λxX = 0, T ′ + λtT = 0


ferential equation

xuxx + 2ut = 0

by method of separation of variables.


xuxx + 2ut = 0 (10.9)

Let u(x, t) = X(x)T (t) be the separation solution. Then

ut =∂u

∂t= XT ′ and uxx =

∂2u

∂t2= X ′′T




xX ′′ + 2XT ′ = 0

or, xX ′′T = −2XT ′

or,xX ′′

X=−2T ′

T= λ say

which gives

xX ′′ − λX = 0, 2T ′ + λT = 0


ferential equation

uxx + uyy + xu = 0



uxx + uyy + xu = 0 (10.10)

Let u(x, t) = X(x)Y (y) be the separation solution. Then

uxx =∂2u

∂x2= X ′′Y and uyy =

∂2u

∂y2= XY ′′



X ′′Y +XY ′′ + xXY = 0

or, − Y (X ′′ + xX) = XY ′′

or, − X ′′ + xX

X=Y ′′

Y= λ say

which gives

X ′′ + xX + λX = 0, Y ′′ − λY = 0

or, X ′′ + (x+ λ)X = 0, Y ′′ − λY = 0


ferential equation

[p(x)ux]x − r(x)utt = 0



[p(x)ux]x − r(x)utt = 0 (10.11)


Let u(x, t) = X(x)Y (y) be the separation solution. Then

ux = X ′T

or, p(x)ux = p(x)X ′T

or, [p(x)ux]x = Td

dx(p(x)X) = Tp′(x)X ′ + Tp(x)X ′′

and utt = XT ′′



Tp′(x)X ′ + Tp(x)X ′′ − r(x)XT ′′

or, Tp′(x)X ′ + Tp(x)X ′′ = r(x)XT ′′

or, T [p′(x)X ′ + p(x)X ′′] = r(x)XT ′′

or,p′(x)X ′ + p(x)X ′′

r(x)X=T ′′

T= −λ say

which gives

p′(x)X ′ + p(x)X ′′ + λr(x)X = 0, T ′′ + λT = 0

or, (p(x)X ′)′ + λr(x)X = 0, T ′′ + λT = 0

10.1.5 Solution of One Diamensional Heat Equation by Separation

of Variables

The one diamesional heat equation is

ut = α2uxx

or,∂u

∂t= α2∂

2u

∂x2(10.12)

where

α2 =κ

ρs

along with boundary conditions

u(0, t) = 0 and u(L, t) = 0, t > 0

and initial condition

u(x, 0) = f(x), 0 ≤ x ≤ L

Let u(x, t) = X(x)T (t) be a variable separation solution of the equation (10.12) with

the given boundary and initial conditions. Then

∂u

∂t= XT ′ and

∂2u

∂x2= X ′′T (10.13)



x or t. Subsituting the values in (10.12) we get

XT ′ = α2X ′′T

or,X”

X=

1

α2

T ′

T= −λ (say) where λ > 0 (10.14)

Hence we obtain the following two ordinary differential equations

X ′′ + λX = 0 (10.15)

and T ′ + α2λT = 0 (10.16)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ

Hence, the general solution of (10.15) is

X(x) = c1 cos√λx+ c2 sin

√λx (10.17)


u(0, t) = 0, and u(L, t) = 0

or, X(0)T (t) = 0 and X(L)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(L) = 0

Again, using X(0) = 0 in (10.17), we get

0 = c1 cos 0 + c2 sin 0

or, c1 = 0

and using X(L) = 0 in (10.17), we get

0 = c1 cos√λL+ c2 sin

√λL

or, c2 sin√λL = 0


sin√λL = 0

=⇒√λL = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

L, n = 1, 2, 3, · · ·

or, λ =n2π2

L2, n = 1, 2, 3, · · ·


Hence the equation (10.17) becomes

X = c2 sinnπx

L, n = 1, 2, 3, · · ·

Letting c2 = 1, the eigenfunction (non-zero solutions ) are

Xn(x) = sinnπx

L, n = 1, 2, 3, · · ·

associated with the eigenvalues

λn =n2π2

L2, n = 1, 2, 3 · · ·

Now turning to the equation (10.16), putting value of λ

T ′ + α2n2π2

L2T = 0

or,dT

dt= −α2n

2π2

L2T

or,dT

T= −α

2n2π2

L2dt

loge T = −α2n2π2

L2t integrating without arbitrary constant

or, T = e−α2n2π2

L2 t, n = 1, 2, 3, · · ·

or, Tn(t) = e−α2n2π2

L2 t, n = 1, 2, 3, · · ·

The value of T depends on n, so we suppose it as Tn. Therefore, the functions

u(x, t) = Xn(x)Tn(x)

= sinnπx

Le−

α2n2π2

L2 t, n = 1, 2, 3, · · ·

are solutions of the one diamensional heat equation under the given conditions. The

functions un are fundamental solutions of the given heat equation. It remains to

satisfy the initial condition

u(x, 0) = f(x), 0 ≤ x ≤ L

Now, we consider linear combination of all these solutions

u(x, t) =∞∑n=1

cn sinnπx

Le−

α2n2π2

L2 t (10.18)

Using the initial condition u(x, 0) = f(x), assuming the series convergent, we get

∞∑n=1

cn sinnπx

L= f(x)


which is a Fourier sine series of f(x) with period 2L, so that the series on the left

converge to the initial temperature f(x). Therefore,

cn =2

L

∫ L

0f(x) sin

nπx

Ldx

Hence, solution of the one dimensional heat equation is

u(x, t) =∞∑n=1

cn sinnπx

Le−

α2n2π2

L2 t

where

cn =2

L

∫ L

0f(x) sin

nπx

Ldx, n = 1, 2, 3, · · ·

Example 354. Find the solution of the heat conduction problem

100uxx = ut, 0 < x < 1, t > 0

u(0, t) = 0, u(1, t) = 0, t > 0

u(x, 0) = 2(sin 2πx− sin 5πx), 0 ≤ x ≤ 1

Solution: Let u(x, t) = X(x)T (t) be a variable separation solution of the given

equation with the given boundary and initial condition. Then

∂u

∂t= XT ′ and

∂2u

∂x2= X ′′T (10.19)


x or t. Subsituting the values in the given heat equation, we get

100X ′′T = XT ′

or,X”

X=

1

100

T ′

T= −λ (say) where λ > 0 (10.20)


X ′′ + λX = 0 (10.21)

and T ′ + 100λT = 0 (10.22)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.23)



u(0, t) = 0, and u(1, t) = 0

or, X(0)T (t) = 0 and X(1)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(1) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0

and using X(1) = 0 in (10.23), we get

0 = c1 cos√λ+ c2 sin

√λ

or, c2 sin√λ = 0


sin√λ = 0

=⇒√λ = nπ, n = 1, 2, 3, · · ·

or,√λ = nπ, n = 1, 2, 3, · · ·

or, λ = n2π2, n = 1, 2, 3, · · ·


X = c2 sinnπx, n = 1, 2, 3, · · ·


Xn(x) = sinnπx, n = 1, 2, 3, · · ·


λn = n2π2, n = 1, 2, 3 · · ·


T ′ + 100n2π2T = 0

or,dT

T= −100n2π2dt

or, loge T = −100n2π2t integrating without arbitrary constant

or, T = e−100n2π2t, n = 1, 2, 3, · · ·or, Tn(t) = e−100n2π2t, n = 1, 2, 3, · · ·

Therefore, the functions

u(x, t) = Xn(x)Tn(x)

= e−100n2π2t sinnπx, n = 1, 2, 3, · · ·



functions un are fundamental solutions of the given equation. It remains to satisfy

the initial condition

u(x, 0) = 2(sin 2πx− sin 5πx), 0 ≤ x ≤ 1

Now we consider linear combinations of all these solutions

u(x, t) =

∞∑n=1

cn sinnπx e−100n2π2t (10.24)

Using the initial conditions u(x, 0) = 2(sin 2πx− sin 5πx), assuming the series con-

vergent, we get

∞∑n=1

cn sinnπx = 2(sin 2πx− sin 5πx)

which is a Fourier sine series of with period 2L = 2 × 1 = 2, so that the series on

the left converge to the initial temperature 2(sin 2πx− sin 5πx). Therefore,

cn =2

L

∫ L

0f(x) sin

nπx

Ldx

=2

1

∫ 1

02(sin 2πx− sin 5πx) sinnπxdx

= 4

∫ 1

0sin 2πx sinnπxdx− 4

∫ 1

0sin 5πx sinnπxdx (10.25)

Using the orthogonality properties∫ 1

0sin 2πx sinnπxdx =

0 if n 6= 212 if n = 2

and ∫ 1

0sin 5πx sinnπxdx =

0 if n 6= 512 if n = 5

Hence from (10.25)

c2 = 4 · 1

2= 2, c5 = 0− 4 · 1

2= −2, cn = 0 for n 6= 2, 5.

Hence the solution (10.24) becomes

u(x, t) = 2e−400π2t sin 2πx− 2e−2500π2t sin 5πx

Example 355. Find the temperature u(x, t) at any time in a metrial rod 50cm long,

insulated on the sides, which initially has a uniform temperature of 200 throughout

and whose ends are maintained at 00C


Solution: The heat conduction problem is formulated as

α2uxx = ut, 0 < x < 50, t > 0 (10.26)

u(0, t) = 0, u(50, t) = 0, t > 0

u(x, 0) = 200C, 0 ≤ x ≤ 50


the given boundary and initial condition. Then

∂u

∂t= XT ′ and

∂2u

∂x2= X ′′T and (10.27)

where primes refer to ordinary differentiation with respect to idependent variables


α2X ′′T = XT ′

or,X”

X=

1

α2

T ′

T= −λ (say) where λ > 0 (10.28)


X ′′ + λX = 0 (10.29)

and T ′ + α2λT = 0 (10.30)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.31)


u(0, t) = 0, and u(50, t) = 0

or, X(0)T (t) = 0 and X(50)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(50) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0

and using using X(50) = 0 in (10.31), we get

0 = c1 cos(50√λ) + c2 sin(50

√λ)

or, c2 sin(50√λ) = 0



sin 50√λ = 0

=⇒ 50√λ = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

50, n = 1, 2, 3, · · ·

or, λ =n2π2

2500, n = 1, 2, 3, · · ·


X = c2 sinnπx

50, n = 1, 2, 3, · · ·


Xn(x) = sinnπx

50, n = 1, 2, 3, · · ·


λn =n2π2

2500, n = 1, 2, 3 · · ·


T ′ +α2n2π2

2500T = 0

or,dT

T= −α

2n2π2

2500dt

or, loge T = −α2n2π2

2500t integrating without arbitrary constant

or, T = e−α2n2π2

2500t, n = 1, 2, 3, · · ·

or, Tn(t) = e−−α2n2π2

2500t, n = 1, 2, 3, · · ·


un(x, t) = Xn(x)Tn(x)

= e−α2n2π2

2500t sin

nπx

50, n = 1, 2, 3, · · ·


functions un are fundamental solutions of the given equations. It remains to satisfy


u(x, 0) = 20 0 ≤ x ≤ 1

Now we consider linear combination of all these solutions

u(x, t) =

∞∑n=1

cn sinnπx

50e−−

α2n2π2

2500t (10.32)


Using the initial condition u(x, 0) = 20, assuming the series convergent, we get

∞∑n=1

cn sinnπx = 20

which is a Fourier sine series of with period 2L = 2 × 50 = 100, so that the series

on the left converge to the initial temperature 20. Therefore,

cn =2

L

∫ L

0f(x) sin

nπx

Ldx

=2

50

∫ 50

020 sin

nπx

50dx

=4

5

∫ 5

0sin

nπx

50dx

=4

5

50

nπ

[− cos

nπx

50

]50

0

=40

nπ(1− cosnπ)

=

80nπ , n odd

0, n even(10.33)


u(x, t) =80

π

∑n=odd

1

nsin

nπx

50e−

α2n2π2

2500t

Example 356. Consider the conduction of hea in a rod 40cm in lenght whose ends

are maintained at 0oC for all t > 0 . Find an expression for the temperature u(x, t) in

the following cases if the initial temperature distribution in the rod is given function

. Suppose that α2 = 1.

1. u(x, 0) = 50, 0 < x < 40.

2. u(x, 0) = x, 0 < x < 40.

3.

u(x, 0) =

0, 0 ≤ x < 10

50 10 ≤ x ≤ 30

0 30 ≤ x ≤ 40

Solution: The heat conduction problem is formulated as, taking α2 = 1

uxx = ut, 0 < x < 40, t > 0 (10.34)

with initial conditions u(0, t) = 0, u(40, t) = 0, t > 0




∂u

∂t= XT ′ and

∂2u

∂x2= X ′′T and (10.35)



X ′′T = XT ′

or,X”

X=T ′

T= −λ (say) where λ > 0 (10.36)


X ′′ + λX = 0 (10.37)

and T ′ + λT = 0 (10.38)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.39)


u(0, t) = 0, and u(40, t) = 0

or, X(0)T (t) = 0 and X(40)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(40) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0


0 = c1 cos 40√λ+ c2 sin 40

√λ

or, c2 sin 40√λ = 0


sin 40√λ = 0

=⇒ 40√λ = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

40, n = 1, 2, 3, · · ·

or, λ =n2π2

1600, n = 1, 2, 3, · · ·



X = c2 sinnπx

40, n = 1, 2, 3, · · ·


Xn(x) = sinnπx

40, n = 1, 2, 3, · · ·


λn =n2π2

1600, n = 1, 2, 3 · · ·


T ′ +n2π2

1600T = 0

or,dT

T= −n

2π2

1600dt

or, loge T = −n2π2


or, T = e−n2π2

1600t, n = 1, 2, 3, · · ·

or, Tn(t) = e−−n2π2

1600t, n = 1, 2, 3, · · ·



= e−n2π2

1600t sin

nπx

40, n = 1, 2, 3, · · ·


functions un are fundamental solutions of the given equation.

Now we consider linear combination of all these solutions

u(x, t) =

∞∑n=1

cn sinnπx

40e−

n2π2

1600t (10.40)

(1) Using the initial condition u(x, 0) = 50, assuming the series convergent, we get

∞∑n=1

cn sinnπx

40= 50


which is a Fourier sine series of with period 2L = 2× 40 = 80, so that the series on

the left converges to the initial temperature 50. Therefore

cn =2

L

∫ L

0f(x) sin

nπx

Ldx

=2

40

∫ 40

050 sin

nπx

40dx

=5

2

∫ 40

0sin

nπx

40dx

=5

2

40

nπ

[− cos

nπx

40

]40

0

=100

nπ(1− cosnπ)


u(x, t) =100

π

∑n=1

1− cosnπ

nsin

nπx

40e−

n2π2

1600t

(2) Using the initial conditions u(x, 0) = x, 0 < x < 40, in (10.40) assuming the

series convergent, we get

∞∑n=1

cn sinnπx

40= x


the left converge to the initial temperature u(x, 0) = x. Therefore,

cn =2

L

∫ L

0f(x) sin

nπx

Ldx

=2

40

∫ 40

0x sin

nπx

40dx

=1

20

[x

∫sin

nπx

40

]40

0

−∫ 40

0

(dx

dx

∫sin

nπx

40dx

)dx

=1

20

[−40x

nπcos

nπx

40

]40

0

+40

nπ

∫ 40

0cos

nπx

40dx

=1

20

[−1600

nπcos

40nπ

40+ 0

]+

1600

n2π2

[sin

nπx

40

]40

0

=

1

20

[−1600

nπcosnπ + 0

]+

1600

n2π2[sinnπ − 0]

= − 80

nπcosnπ = − 80

nπ(−1)n =

80

nπ(−1)n+1


u(x, t) =80

π

∑n=1

(−1)n+1

nsin

nπx

40e−

n2π2

1600t


(3) Using the initial conditions ,

u(x, 0) =

0, 0 ≤ x < 10

50 10 ≤ x ≤ 30

0 30 ≤ x ≤ 40

assuming the series convergent, we get

∞∑n=1

cn sinnπx

40= u(x, 0)


the left converge to the initial temperature f(x) = u(x, 0). Therefore,

cn =2

L

∫ L

0f(x) sin

nπx

Ldx

=2

40

∫ 40

0u(x, 0) sin

nπx

40dx

=1

20

∫ 10

0u(x, 0) sin

nπx

40dx+

1

20

∫ 20

10u(x, 0) sin

nπx

40dx+

1

20

∫ 40

20u(x, 0) sin

nπx

40dx

=1

20

∫ 10

00 sin

nπx

40dx+

1

20

∫ 30

1050 sin

nπx

40dx+

1

20

∫ 40

300 sin

nπx

40dx

= 0 +50

20

40

nπ

[− cos

nπx

40

]30

10+ 0

=100

nπ

(cos

nπ

4− cos

3nπ

4

)Hence the solution (10.40) becomes

u(x, t) =100

π

∑n=1

1

n

(cos

nπ

4− cos

3nπ

4

)e−

n2π2

1600t sin

nπx

40

10.2 Other Heat Conduction Problem

We considered the the problem consisting of the heat conduction equation

ut = α2uxx (10.41)

the boundary condition

u(0, t) = 0, u(L, t) = 0, t > 0 (10.42)

and the initial condition

u(x, 0) = f(x); 0 ≤ x ≤ L (10.43)

10.2. OTHER HEAT CONDUCTION PROBLEM 567

We found the solution

u(x, t) =∞∑n=1

cne−n2π2α2t

L2 sinnπx

L(10.44)

where the coefficients cn are the same as in the series

f(x) =

∞∑n=1

cn sinnπx

L(10.45)

The series is Fourier sine series for the function f , and its coefficients are given by

cn =2

L

∫ L

0f(x) sin

nπx

Ldx (10.46)

Therefore, the solution of the heat conduction problem (10.41), (10.42), (10.43) is

given by the equation (10.44) with the coefficients computed by the equation (10.46).

10.2.1 Non-homogeneous Boundary conditions, and its Solution

Suppose that one end of the bar is held at a constant temperature T1 i.e. u(0, t) = T1

and the other end at T2, i.e. u(L, t) = T2, t > 0. Then the heat problem becomes

ut = α2uxx (10.47)

the boundary condition

u(0, t) = T1, u(L, t) = T2, t > 0 (10.48)


u(x, 0) = f(x); 0 ≤ x ≤ L (10.49)

This problem is slightly more difficult, because of the nono-homogeneous boundary

condition. We can solve it by reducing it to a problem having homogenous boundary

conditions, which can be solved.

Suppose that after long time as t → ∞ a steady temperature distribution v(x) is

reached, which is independent of the time t and initial conditions. Since v(x) satisfies

(10.47), we have

0 = α2v′′(x), 0 < x < L, as ut = 0 for steady state

v′′(x) = 0

Integrating,

v′(x) = c1

v(x) = c1x+ c2 (10.50)


Hence the steady state temperature distribution v(x) is a linear function of x. Fur-

ther v(x) must satisfy the boundary conditions

v(0) = T1, v(L) = T2


c2 = T1, T2 = c1L+ c2

=⇒ c1 =T2 − T1

L


v(x) =T2 − T1

Lx+ T1 (10.51)

Now returning to the original problem, we will try to express u(x, t) as the sum of

the steady state distribution and another transient (time dependent) temperature

distribution w(x, t). Then

u(x, t) = v(x) + w(x, t) (10.52)

Now from (10.47)

(v + w)t = α2(v + w)xx

or, vt + wt = α2(vxx + wxx)

wt = α2wxx as vt = 0, vxx = 0 (10.53)

Aslo ,

w(0, t) = u(0, t)− v(0) = T1 − T1 = 0

w(L, t) = u(L, t)− v(L) = T2 − T2 = 0

w(x, 0) = u(x, 0)− v(x) = f(x)− v(x)

Hence the solution of the equation (10.53) is

w(x, t) =∞∑n=1

bne−n2π2α2

L2 t sinnπx

L(10.54)

where

bn =2

L

∫ L

0[f(x)− v(x)] sin

nπx

Ldx n = 1, 2, 3, · · · (10.55)

Hence from the (10.51),(10.54) and (10.55), the general solution of the heat conuc-

tion (10.47), with non-homogeneous boundary conditions (10.48), and the initial

condition (10.49) is

u(x, t) = v(x) + w(x, t)

= (T2 − T1)x

L+ T1 +

∞∑n=1

bne−n2π2α2t

L2 sinnπx

L


where

bn =2

L

∫ L

0

[f(x)− T2 − T1

Lx− T1

]sin

nπx

Ldx n = 1, 2, 3,

Example 357. Consider the heat conduction problem

ut = uxx, 0 < x < 30, t > 0

u(0, t) = 20, u(30, t) = 50, t > 0

u(x, 0) = 60− 2x; 0 < x < 30 (10.56)

Find the steady state temperature distribution and the boundary value problem that

determines the transient distribution.

Solution: The steady state temperature v(x) satisfies uxx = ut, so

v′′(x) = 0 (10.57)

and the boundary conditions

v(0) = 20 and v(30) = 50

Integrating, the equation (10.57)

v′(x) = c1, v(x) = c1x+ c2

Since,

v(0) = 20 =⇒ c2 = 20

Again, v(30) = 50, so

50 = 30c1 + 20

or, c1 = 1

Putting the values of c1 and c2 in v(x) we get

v(x) = x+ 20

which is steady state temperature distribution.

Consider u(x, t) as the sum of the steady state temperature distribution v(x) and

another transient temperature distribution w(x, t). Then

u(x, t) = v(x) + w(x, t)

The transient distribution w(x, t) satisfies the heat conduction equation

wxx = wt

with the homogeneous boundary conditions

w(0, t) = u(0, t)− v(0) = 20− 20 = 0


w(30, t) = u(30, t)− v(30) = 50− 50 = 0

and the modified initial condition is

w(x, 0) = u(x, 0)− v(x)

= 60− 2x− (20 + x)

= 40− 3x

Hence, the required temperature distribution is

wt = wxx, 0 < x < 30, t > 0

w(0, t) = 0, w(30, t) = 0, t > 0

w(x, 0) = 40− 3x; 0 < x < 30 (10.58)

Example 358. Find the steady-state solution of the heat conduction equation ut =

α2uxx that satisfies the given set of the boundary conditions

1. u(0, t) = 10, u(50, t) = 60

2. u(0, t) = 30, u(40, t) = −10

3. u(0, t) = T, ux(L, t) = 0

4. u(0, t) = 0, ux(L, t) = 0

5. u(0, t) = 0, ux(L, t) + u(L, t) = 0

Solution: 1. The steady state temperature v(x) satisfies α2uxx = ut, so

v′′(x) = 0 (10.59)


v(0) = 10 and v(50) = 60


v′(x) = c1, v(x) = c1x+ c2

Since,

v(0) = 10 =⇒ c2 = 10

Again, v(50) = 60, so

60 = 50c1 + 10

or, c1 = 1


v(x) = x+ 10


which is steady state temperature distribution.


v′′(x) = 0 (10.60)


v(0) = 30 and v(40) = −10


v′(x) = c1, v(x) = c1x+ c2

Since,

v(0) = 30 =⇒ c2 = 30

Again, v(40) = −10, so

−10 = 40c1 + 30

or, c1 = −1


v(x) = −x+ 30


v′′(x) = 0 (10.61)


v(0) = T and v′(L) = 0


v′(x) = c1, v(x) = c1x+ c2

Since,

v(0) = T =⇒ c2 = T

Again, v′(L) = 0, so

c1 = 0


v(x) = T


v′′(x) = 0 (10.62)



v(0) = 0 and v′(L) = 0


v′(x) = c1, v(x) = c1x+ c2

Since,

v(0) = 0 =⇒ c2 = 0

Again, v′(L) = 0, so

c1 = 0


v(x) = 0


v′′(x) = 0 (10.63)


v(0) = T and v(L) + v′(L) = 0


v′(x) = c1, v(x) = c1x+ c2

Since,

v(0) = T =⇒ c2 = T

Again, v(L) + v′(L) = 0, so

c1L+ c2 + c1 = 0

or, c1(1 + L) + T = 0

or, c1 = − T

1 + L


v(x) = − T

1 + Lx+ T =

T (1 + L− x)

1 + L


10.2.2 Bar with Insulated Ends

We consider a bar, with ends are insulated, so there is no passage of heat through

it. We know that, the rate of flow of heat across a cross section is proportional to

the rate of change of temperature in the x direction. Thus, in this case there is no

flow of heat. Hence the boundary conditions are

ux(0, t) = 0, ux(L, t) = 0, t > 0

The heat conduction problem becomes

ut = α2uxx (10.64)

the boundary conditions

ux(0, t) = 0, ux(L, t) = 0, t > 0 (10.65)


u(x, 0) = f(x); 0 ≤ x ≤ L (10.66)



∂u

∂t= XT ′ and

∂2u

∂x2= X ′′T (10.67)



XT ′ = α2X ′′T

or,X”

X=

1

α2

T ′

T= −λ (say) where λ > 0 (10.68)


X ′′ + λX = 0 (10.69)

and T ′ + α2λT = 0 (10.70)


ux(0, t) = 0, and ux(L, t) = 0

or, X ′(0)T (t) = 0 and X ′(L)T (t) = 0

Since T (t) 6= 0, =⇒ X ′(0) = 0, X ′(L) = 0


m2 + λ = 0 (10.71)


Case I If λ < 0, then we may suppose λ = −µ2 where µ is real and positive. Then

by (10.71)

m2 = µ2

or, m = ±µ


X(x) = c1eµx + c2e

−µx

X ′(x) = µc1eµx − µc2e

−µx

Now X ′(0) = 0 =⇒ 0 = µc1 − µc2 (10.72)

or, c1 = c2

and X ′(L) = 0 =⇒ 0 = µc1eµL − µc2e

−µL (10.73)


c1 = c2 = 0

Hence X(x) = 0, which is not acceptable and hence λ can not be negative.

Case II If λ > 0, then suppose that λ = µ2, where µ is real and positive. The

auxiliary equation of (10.71) is

m2 + µ2 = 0

or, m2 = −µ2

or, m = ±iµ


X(x) = c1 cosµx+ c2 sinµx (10.74)

X ′(x) = µ(−c1 sinµx+ c2 cosµx) (10.75)

Using X ′(0) = 0 in (10.75), we get

0 = −µ(c1 sin 0 + c2 cos 0)

or, c2 = 0

and using X ′(L) = 0 in (10.75), we get

0 = µ(−c1 sinµL+ c2 cosµL)

or, c1 sinµL = 0


sinµL = 0

=⇒ µL = nπ, n = 1, 2, 3, · · ·or, µ =

nπ

L, n = 1, 2, 3, · · ·

or, λ = µ2 =n2π2

L2, n = 1, 2, 3, · · ·



X = c1 cosnπx

L, n = 1, 2, 3, · · ·


Xn(x) = cosnπx

L, n = 1, 2, 3, · · ·


λn =n2π2

L2, n = 1, 2, 3 · · ·


T ′ + α2n2π2

L2T = 0

or,dT

T= −α

2n2π2

L2dt


L2t integrating without arbitrary constant

T = e−α2n2π2

L2 t, n = 1, 2, 3, · · ·

Tn(t) = e−α2n2π2

L2 t, n = 1, 2, 3, · · ·


u0(x, t) = 1


= cosnπx

Le−

α2n2π2

L2 t, n = 1, 2, 3, · · ·


functions un are fundamental solutions of the given equation. It remains to satisfy


u(x, 0) = f(x), 0 ≤ x ≤ L


u(x, t) =c0

2+∞∑n=1

cn cosnπx

Le−

α2n2π2

L2 t (10.76)

Using the initial conditions u(x, 0) = f(x), assuming the series convergent, we get

c0

2+

∞∑n=1

cn cosnπx

L= f(x)


which is a Fourier cosine series of f(x) with period 2L, so that the series on the left

converge to the initial temperature f(x). Therefore,

cn =2

L

∫ L

0f(x) cos

nπx

Ldx, n = 0, 1, 2, 3, · · ·

Hence, solution of the one dimensional heat equation is

u(x, t) =c0

2+

∞∑n=1

cn cosnπx

Le−

α2n2π2

L2 t

where

cn =2

L

∫ L

0f(x) cos

nπx

Ldx, n = 0, 1, 2, 3, · · ·

Example 359. Find the temperature u(x, t) of the heat equation α2uxx = ut in a

metal rod of length 25cm that is insulated on the ends as well as on the sides and

whose initial temperature distribution is u(x, 0) = x for 0 < x < 25.

Solution: Here the both ends of the rod of length 25cm are insulated, there is

no flow of heat through them.

ux(0, t) = 0, ux(25, t) = 0, t > 0

The heat conduction problem becomes

ut = α2uxx (10.77)


ux(0, t) = 0, ux(25, t) = 0, t > 0 (10.78)

and the initial conditions

u(x, 0) = x; 0 ≤ x ≤ 25 (10.79)



∂u

∂t= XT ′ and

∂2u

∂x2= X ′′T (10.80)



XT ′ = α2X ′′T

or,X”

X=

1

α2

T ′

T= −λ (say) where λ > 0 (10.81)



X ′′ + λX = 0 (10.82)

and T ′ + α2λT = 0 (10.83)


ux(0, t) = 0, and ux(25, t) = 0

or, X ′(0)T (t) = 0 and X ′(25)T (t) = 0

Since T (t) 6= 0, =⇒ X ′(0) = 0, X ′(25) = 0


m2 + λ = 0 (10.84)

Case I If λ < 0, then we may suppose λ = −µ2 where µ is real and positive. Then

by (10.84)

m2 = µ2

or, m = ±µ


X(x) = c1eµx + c2e

−µx

X ′(x) = µc1eµx − µc2e

−µx

Now X ′(0) = 0 =⇒ 0 = µc1 − µc2 (10.85)

or, c1 = c2

and X ′(25) = 0 =⇒ 0 = µc1eµ25 − µc2e

−µ25 (10.86)


c1 = c2 = 0

Hence X(x) = 0, which is not acceptable and hence λ can not be negative.

Case II If λ > 0, then suppose that λ = µ2, where µ is real and positive. The

auxiliary equation of (10.71) is

m2 + µ2 = 0

or, m2 = −µ2

or, m = ±iµ


X(x) = c1 cosµx+ c2 sinµx (10.87)

X ′(x) = µ(−c1 sinµx+ c2 cosµx) (10.88)


Using X ′(0) = 0 in (10.88), we get

0 = −µ(c1 sin 0 + c2 cos 0)

or, c2 = 0

and using X ′(25) = 0 in (10.75), we get

0 = µ(−c1 sin 25µ+ c2 cos 25µ)

or, c1 sin 25µ = 0


sin 25µ = 0

=⇒ 25µ = nπ, n = 1, 2, 3, · · ·or, µ =

nπ

25, n = 1, 2, 3, · · ·

or, λ = µ2 =n2π2

625, n = 1, 2, 3, · · ·


X(x) = c1 cosnπx

25, n = 1, 2, 3, · · ·


Xn(x) = cosnπx

25, n = 1, 2, 3, · · ·


λn =n2π2

625, n = 1, 2, 3 · · ·


T ′ + α2n2π2

625T = 0

or,dT

T= −α

2n2π2

625dt



T = e−α2n2π2

625t, n = 1, 2, 3, · · ·

Tn(t) = e−α2n2π2

625t, n = 1, 2, 3, · · ·


u0(x, t) = 1


= cosnπx

25e−

α2n2π2

625t, n = 1, 2, 3, · · ·



functions un are fundamental solutions of the given equations. It remains to satisfy


u(x, 0) = f(x), 0 ≤ x ≤ 25


u(x, t) =c0

2+

∞∑n=1

cn cosnπx

25e−

α2n2π2

625t (10.89)

Using the initial conditions u(x, 0) = x, assuming the series convergent, we get

c0

2+∞∑n=1

cn cosnπx

25= x

which is a Fourier series of x with period 2L = 50, so that the series on the left

converge to the initial temperature x. Therefore,

cn =2

25

∫ 25

0x cos

nπx

Ldx, n = 0, 1, 2, 3, · · ·

For n = 0, c0 =2

25

∫ 25

0x · 1dx

=2

25

[x2

2

]25

0

=2

25

[625

2− 02

2

]= 25

For n 6= 0, cn =2

L

∫ L

0f(x) cos

nπx

Ldx

=2

25

∫ 25

0x cos

nπx

25dx

=2

25

[x

∫cos

nπx

25dx

]25

0

−∫ 25

0

(dx

dx

∫cos

nπx

25dx

)dx

=2

25

[25

nπx sin

nπx

25

]25

0

− 25

nπ

∫ 25

0sin

nπx

25dx

=2

25

−0 +

252

n2π2

[cos

nπx

25

]25

0

=

2

25

252

n2π2(cosnπ − 1)

=50

n2π2(cosnπ − 1)

=

− 100n2π2 for n is odd

0 for n is even


Hence, solution heat equation is

u(x, t) =25

2− 100

π2

∞∑n=1,3,5,···

1

n2cos

nπx

25e−

α2n2π2

625t

10.3 The Wave Equation: Vibrations of an Elastic String

2 Let us consider a string with

1. perfectly flexible, homogenoeus i.e. its mass per unit lenght ρ is a constant ,

2. the vertical displacement u(x, t) is measured from the x−axis at any time t is

small in comparision to the length,

3. tension acts tangent to the sring, its magnitude is same at all points and it is

large compared with the force of gravity.

4. no other external forces act on the string ie. air resistance and friction.

Again let the string stretched to its length L and aligned along the x−axis so that

its two end points are at x = 0 and x = L. Let us consider two points A and B

at distances x and x + δx from the end x = 0. Let T1 and T2 be tensions at the

ends points A and B respectively. Let α and β be the angle of inclination of T1 and

T2 with the horizontal direction. Since there is no displacement in the horizontal

direction, so the component of the tensions T1 and T2 along horizontal direction

must be balance i.e

T1 cosα = T2 cosβ = T say (10.90)

2Deduction of wave equation may escape

10.3. THE WAVE EQUATION: VIBRATIONS OF AN ELASTIC STRING 581

In vertical direction, we have, from Newton’s law of motion

T2 sinβ − T1 sinβ = ρ δx∂2u

∂t2

or,T

cosβsinβ − T

cosαsinα = ρδx

∂2u

∂t2from (10.90)

or,ρδx

T

∂2u

∂t2= tanβ − tanα

or,ρδx

T

∂2u

∂t2=∂u

∂x

∣∣∣∣x+δx

− ∂u

∂x

∣∣∣∣x

or,ρ

T

∂2u

∂t2=

∂u(x+δx)∂x − ∂u

∂x

δx

Letting δ → 0, we get

ρ

T

∂2u

∂t2=∂2u

∂t2

or, utt = a2uxx (10.91)

where

a2 =T

ρ

and a is a constant and its unit is length/time, i.e. of the velocity. It is known as

velocity propagation of the waves along the string. Equation (10.91) is known as

one diamensional wave equation.

For the complete description of the motion of the string, we specify the suitable

initial and boundary conditions for the displacement u(x, t).

The end point are assume to be fixed, and therefore the boundary conditions are

u(0, t) = 0, u(L, t) = 0 t ≥ 0 (10.92)

Sine the differential equation is of second order with respect to t, so we need two

initial conditions . These are (a) initial position of the string

u(x, 0) = f(x) 0 ≤ x ≤ L


(b) initial velocity of the string

ut(x, 0) = g(x) 0 ≤ x ≤ L

where f and g are given functions. In order to consistent results we have

f(0) = f(L) = 0 and g(0) = g(L) = 0 (10.93)

10.3.1 General Solution of One Dimensional Wave Equation

The vertical displacement u(x, t) must satisfy the wave equation

utt = a2uxx, 0 < x < L, t > 0 (10.94)



∂2u

∂t2= XT ′′ and

∂2u

∂x2= X ′′T (10.95)



XT ′ = a2X ′′T

or,X”

X=

1

a2

T ′

T= −λ (10.96)


X ′′ + λX = 0 (10.97)

and T ′′ + a2λT = 0 (10.98)

Now we have the flowing cases

Case I: When λ = 0, the equations (10.97) and (10.98)

X ′′ = 0, T ′′ = 0

Integrating,

X ′(x) = a1, T ′(t) = a2

and X(x) = a1x+ a3, T (t) = a2t+ a4

Case II : When λ < 0, let λ = −µ2. The roots of the charateristic equation of

(10.97) and (10.98) are

±µ and ± aµ

The solutions of (10.97) and (10.98) are

X(x) = b1eµx + c2e

−µx, and T (t) = b3eaµt + b4e

−aµt


Case III: When λ > 0, let λ = −µ2. The roots of the charateristic equation of

(10.97) and (10.98) are

±iµ and ± iaµ

The solutions of (10.97) and (10.98) are

X(x) = c1 cosµx+ c2 sinµx and T (t) = c3 cos aµt+ c4 sin aµt

Thus the various possible solutions are

u(x, t) = (a1x+ a3)(a2t+ a4)

u(x, t) =(b1e

µx + c2e−µx) (b3eaµt + b4e

−aµt)u(x, t) = (c1 cosµx+ c2 sinµx) (c3 cos aµt+ c4 sin aµt)

We have to choose that solution which consistent with the the physical nature of

the problem. The vibration u(x, t) must be periodic function of x and t. Thus, the

true solution is, which contains trigonometric functions i. e.

u(x, t) = (c1 cosµx+ c2 sinµx) (c3 cos aµt+ c4 sin aµt)

For this we chose λ > 0.

10.3.2 Vibration of Elastic String with Non-zero Initial Displace-

ment

Let us suppose that the string is distrubed from its equalibrium position and then

released at time t = 0 with initial velocity to vibrate freely. Then the vibration must

satisfy the wave equation

or, utt = a2uxx 0 < x < L, t > 0 (10.99)


u(0, t) = 0, u(L, t) = 0, t ≥ 0 (10.100)

and initial conditions

u(x, 0) = f(x), ut(x, 0) = 0, 0 ≤ x ≤ L (10.101)

where f is a given function describing the confugration of the string at t = 0.


the given boundary and initial conditions. Then

∂2u


∂2u

∂x2= X ′′T (10.102)




α2X ′′T = XT ′′

or,X”

X=

1

a2

T ′′

T= −λ (say) where λ > 0 (10.103)


X ′′ + λX = 0 (10.104)

and T ′′ + a2λT = 0 (10.105)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.106)


u(0, t) = 0, and u(L, t) = 0

or, X(0)T (t) = 0 and X(L)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(L) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0

and using using X(L) = 0 in (10.106), we get


√λL



sin√λL = 0

=⇒√λL = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

L, n = 1, 2, 3, · · ·

or, λ =n2π2

L2, n = 1, 2, 3, · · ·



X = c2 sinnπx

L, n = 1, 2, 3, · · ·


Xn(x) = sinnπx

L, n = 1, 2, 3, · · ·


λn =n2π2

L2, n = 1, 2, 3 · · ·


m2 + a2λ = 0

m = ±a√λi = ±anπ

Li = ±nπa

Li

The general solution (10.105) is

T (t) = c1 cosnπat

L+ c2 sin

nπat

L(10.107)

T ′(t) = −nπaL

k1 sinnπat

L+nπa

Lk2 cos

nπat

L(10.108)

Also, we have

ut(x, 0) = 0 =⇒ X(x)T ′(0) = 0

=⇒ T ′(0) = 0


T ′(0) = −nπaL

k1 sinnπa0

L+nπa

Lk2 cos

nπa0

L= 0

=⇒ k2 = 0

Thus, from (10.107), taking k1 = 1

T = cosnπat

L

Thus, the function

un(x, t) = sinnπx

Lcos

nπat

L(10.109)

consitute the fundamental solutions of (10.99) is along with the boundary conditions

and the initial condition ut(x, 0) = 0. It remains to satisfy the non-homogeous initial

condition

u(x, 0) = f(x), 0 ≤ x ≤ L


We assume u(x, t) has the form

u(x, t) =

∞∑n=1

bnun(x, t) =

∞∑n=1

bn sinnπx

Lcos

nπat

L, n = 1, 2, 3 · · · (10.110)

where

u(x, 0) =∞∑n=1

bn sinnπx

L(10.111)

This is the Fourier sine series of f(x) of period 2L and bn are given by

bn =2

L

∫ L

0f(x) sin

nπx

L, n = 1, 2, 3 · · · (10.112)

Example 360. Consider a vibrating string of length L = 50 that satisfy the wave

equation

4uxx = utt

Assume that the ends of the string are fixed and that the string is set in motion with

no initial velocity from the initial postion.

u(x, 0) = f(x) =

x10 , 0 ≤ x ≤ 10

30−x20 , 10 ≤ x ≤ 30

Find the displacement u(x, t) of the string and describe its motion through one period.

Solution: Here L = 30. The given wave equation is

or, utt = 4uxx 0 < x < 30, t > 0 (10.113)


u(0, t) = 0, u(30, t) = 0, t ≥ 0 (10.114)


u(x, 0) = f(x) =

x10 , 0 ≤ x ≤ 10

30−x20 , 10 ≤ x ≤ 30

ut(x, 0) = 0, 0 ≤ x ≤ 30 (10.115)

Let u(x, t) = X(x)T (t) be a variable separation solution of the equation (10.113)

with the given boundary and initial conditions. Then

∂2u


∂2u

∂x2= X ′′T (10.116)




4X ′′T = XT ′′

or,X”

X=

1

4

T ′′

T= −λ (say) where λ > 0 (10.117)


X ′′ + λX = 0 (10.118)

and T ′′ + 4λT = 0 (10.119)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.120)


u(0, t) = 0, and u(30, t) = 0

or, X(0)T (t) = 0 and X(30)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(30) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0


0 = c1 cos(30√λ) + c2 sin(30

√λ)

or, c2 sin(30√λ) = 0


sin(30√λ) = 0

=⇒ 30√λ = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

30, n = 1, 2, 3, · · ·

or, λ =n2π2

900, n = 1, 2, 3, · · ·



X = c2 sinnπx

30, n = 1, 2, 3, · · ·


Xn(x) = sinnπx

30, n = 1, 2, 3, · · ·


λn =n2π2

900, n = 1, 2, 3 · · ·


m2 + 4λ = 0

m = ±2√λi = ±2

nπ

30i = ±2nπ

30i


T (t) = k1 cos2nπt

30+ k2 sin

2nπt

30(10.121)

T ′(t) = −2nπ

30k1 sin

2nπt

30+

2nπ

30k2 cos

2nπt

30(10.122)

Also, we have

ut(x, 0) = 0 =⇒ X(x)T ′(0) = 0

=⇒ T ′(0) = 0


T ′(0) = −2nπ

30k1 sin

2nπ0

30+

2nπ

30k2 cos

2nπ0

30= 0

=⇒ k2 = 0


T = cos2nπt

30

Thus, the function

un(x, t) = sinnπx

30cos

2nπt

30(10.123)

consititute the fundamental solutions of (10.113) is along with the boundary condi-

tions and the initial condition ut(x, 0) = 0. It remains to satisfy the non-homogeous

initial condition

u(x, 0) = f(x) =

x10 , 0 ≤ x ≤ 10

30−x20 , 10 ≤ x ≤ 30



u(x, t) =

∞∑n=1

bnun(x, t) =

∞∑n=1

bn sinnπx

30cos

2nπt

30, n = 1, 2, 3 · · · (10.124)

where

u(x, 0) =

∞∑n=1

bn sinnπx

30= f(x)


bn =2

30

∫ L

0f(x) sin

nπx

30dx, n = 1, 2, 3 · · · (10.125)

=2

30

∫ 10

0f(x) sin

nπx

30dx+

2

30

∫ 30

10f(x) sin

nπx

30dx

=2

30

∫ 10

0

x

10sin

nπx

30dx+

2

30

∫ 30

10

30− x20

sinnπx

30dx

=2

300

[x

∫sin

nπx

30dx

]10

0

−∫ 10

0

[dx

dx

∫sin

nπx

30dx

]dx

+2

600

[(30− x)

∫sin

nπx

30dx

]30

10

−∫ 30

10

[d(30− x)

dx

∫sin

nπx

30dx

]dx

=2

300

[−30x

nπcos

nπx

30

]10

0

+30

nπ

∫ 10

0cos

nπx

30dx

+2

600

[−30(30− x)

nπcos

nπx

30

]30

10

− 30

nπ

∫ 30

10cos

nπx

30dx

=2

300

[−30× 10

nπcos

10nπ

30− 0

]+

900

n2π2

[sin

nπx

30

]10

0

+

2

600

[−0 +

600

nπcos

10nπ

30

]− 900

n2π2

[sin

nπx

30

]30

10

=

2

300

−300

nπcos

nπ

3+

900

n2π2

[sin

10nπ

30− 0

]+

2

600

600

nπcos

nπ

3− 900

n2π2

[sin

30nπ

30− sin

10nπ

30

]=

2

300

−300

nπcos

nπ

3+

900

n2π2

[sin

nπ

3− 0]

+2

600

600

nπcos

nπ

3− 900

n2π2

[sinnπ − sin

nπ

3

]= − 2

nπcos

nπ

3+

6

n2π2sin

nπ

3+

2

nπcos

nπ

3+

3

n2π2sin

nπ

3

=9

n2π2sin

nπ

3


Hence the rquired solution is

u(x, t) =

∞∑n=1

bnun(x, t) =

∞∑n=1

bn sinnπx

30cos

2nπt

30, n = 1, 2, 3 · · ·

where

bn =9

n2π2sin

nπ

3

u(x, t) gives the displacement of the string at any point x at any time t. The motion

is periodic in time with period 30, so it is suffices to analyze for 0 ≤ t ≤ 30.

Example 361. Consider an elastic string of length L whose ends are held fixed. The

string is set in motion with no initial velocity from an initial position u(x, 0) = f(x).

Let Find the displacement u(x, t) for the given initial position f(x).

(1) f(x) =

4xL 0 ≤ x ≤ L

4

1 L4 ≤ x <

3L4

4(L−x)L

3L4 ≤ x ≤ L

(2) f(x) =

1 L

2 − 1 < x < L2 + 1 (L > 2)

0 otherwise

Solution: The wave equation is

utt = a2uxx 0 < x < L, t > 0 (10.126)


u(0, t) = 0, u(L, t) = 0, t ≥ 0 (10.127)


u(x, 0) = f(x) from 1 or 2

ut(x, 0) = 0, 0 ≤ x ≤ L (10.128)


with the given boundary and initial condition. Then

∂2u


∂2u

∂x2= X ′′T (10.129)



a2X ′′T = XT ′′

or,X”

X=

T ′′

a2T= −λ (say) where λ > 0 (10.130)



X ′′ + λX = 0 (10.131)

and T ′′ + a2λT = 0 (10.132)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.133)


u(0, t) = 0, and u(L, t) = 0

or, X(0)T (t) = 0 and X(L)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(L) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0


0 = c1 cos(L√λ) + c2 sin(L

√λ)

or, c2 sin(L√λ) = 0


sin(L√λ) = 0

=⇒ L√λ = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

L, n = 1, 2, 3, · · ·

or, λ =n2π2

L2, n = 1, 2, 3, · · ·


X = c2 sinnπx

L, n = 1, 2, 3, · · ·


Xn(x) = sinnπx

L, n = 1, 2, 3, · · ·



λn =n2π2

L2, n = 1, 2, 3 · · ·


m2 + a2λ = 0

m = ±a√λi = ±nπa

Li = ±nπa

Li


T (t) = k1 cosnπat

L+ k2 sin

nπat

L(10.134)

T ′(t) = −nπaL

k1 sinnπat

L+nπa

Lk2 cos

nπat

L(10.135)

Also, we have

ut(x, 0) = 0 =⇒ X(x)T ′(0) = 0

=⇒ T ′(0) = 0


T ′(0) = −nπaL

k1 sinnπ0

L+nπa

Lk2 cos

nπ0

L= 0

=⇒ k2 = 0


T = cosnπat

L

Thus, the function

un(x, t) = sinnπx

Lcos

nπat

L(10.136)

consitute the fundamental solutions of (10.126) is along with the boundary condi-

tions and the initial condition ut(x, 0) = 0. It remains to satisfy the non-homogeous

initial condition. We assume u(x, t) has the form

u(x, t) =∞∑n=1

bnun(x, t) =∞∑n=1

bn sinnπx

Lcos

nπat

L, n = 1, 2, 3 · · · (10.137)

where

u(x, 0) =∞∑n=1

bn sinnπx

L= f(x)


bn =2

L

∫ L

0f(x) sin

nπx

Ldx, n = 1, 2, 3 · · · (10.138)


1. Let

f(x) =

4xL 0 ≤ x ≤ L

4

1 L4 < x < 3L

4

4(L−x)L

3L4 ≤ x ≤ L

From (10.138)

bn =2

L

∫ L/4

0f(x) sin

nπx

Ldx+

2

L

∫ 3L/4

L/4f(x) sin

nπx

Ldx+

2

L

∫ L

3L/4f(x) sin

nπx

Ldx

=2

L

∫ L/4

0

4x

Lsin

nπx

Ldx+

2

L

∫ 3L/4

L/4sin

nπx

Ldx+

2

L

∫ L

3L/4

4(L− x)

Lsin

nπx

Ldx

=8

L2

[x

∫sin

nπx

Ldx

]L/40

−∫ L/4

0

(dx

dx

∫sin

nπx

L

)dx

− 2

L

L

nπ

[cos

nπx

L

]3L/4

L/4

+8

L2

[(L− x)

∫sin

nπx

Ldx

]L3L/4

−∫ L

3L/4

[d(L− x)

dx

∫sin

nπx

Ldx

]dx

=8

L2

L

nπ

[−x cos

nπx

L

]L/40

+L

nπ

∫ L/4

0cos

nπx

Ldx

− 2

nπ

[cos

3nπ

4− cos

nπ

4

]

+8

L2

L

nπ

[−(L− x) cos

nπx

L

]L3L/4− L

nπ

∫ L

3L/4cos

nπx

Ldx

=8

L2

L

nπ

(−L

4cos

nπ

4+ 0

)+

L2

n2π2

[sin

nπx

L

]L/40

− 2

nπ

[cos

3nπ

4− cos

nπ

4

]+

8

L2

L

nπ

[0 +

L

4cos

3nπ

4

]− L2

n2π2

[sin

nπx

L

]L3L/4

= − 2

nπcos

nπ

4+

8

n2π2

[sin

nπ

4− 0]− 2

nπ

[cos

3nπ

4− cos

nπ

4

]+

2

nπcos

3nπ

4− 8

n2π2

[sinnπ − sin

3nπ

4

]=

8

n2π2

(sin

nπ

4+ sin

3nπ

4

)(10.139)

Hence from (10.137)

u(x, t) =8

π2

∞∑n=1

1

n2

(sin

nπ

4+ sin

3nπ

4

)sin

nπx

Lcos

nπat

L

which is required expression for the displacement.

2. Let

f(x) =

1 L

2 − 1 < x < L2 + 1 (L > 2)

0 otherwise


From (10.138)

bn =2

L

∫ L

0f(x) sin

nπx

Ldx

=2

L

∫ L/2−1

0f(x) sin

nπx

Ldx+

2

L

∫ L/2+1

L/2−1f(x) sin

nπx

Ldx+

2

L

∫ L

L/2+1f(x) sin

nπx

Ldx

=2

L

∫ L/2+1

L/2−1sin

nπx

Ldx

=2

L

L

nπ

[− cos

nπx

L

]L/2+1

L/2−1

=2

nπ

[− cos

(nπ2

+nπ

L

)+ cos

(nπ2− nπ

L

)]=

2

nπ

[cos(nπ

2− nπ

L

)− cos

(nπ2

+nπ

L

)]=

4

nπsin

nπ

2sin

nπ

L

Hence from (10.137)

u(x, t) =4

π

∞∑n=1

1

nsin

nπ

2sin

nπ

Lsin

nπx

Lcos

nπat

L

which is required expression for the displacement.

10.3.3 Vibration of Elastic String with Non-zero Initial Velocity

Let us suppose that the string is distrubed from its equalibrium position and then

released at time t = 0 with initial velocity g(x). Then the vibration must satisfy the

wave equation

or, utt = a2uxx 0 < x < L, t > 0 (10.140)


u(0, t) = 0, u(L, t) = 0, t ≥ 0 (10.141)


u(x, 0) = 0, ut(x, 0) = g(x), 0 ≤ x ≤ L (10.142)

where g is initial velocity at the point x of the string.



∂2u


∂2u

∂x2= X ′′T (10.143)




α2X ′′T = XT ′′

or,X”

X=

1

a2

T ′′

T= −λ (say) where λ > 0 (10.144)


X ′′ + λX = 0 (10.145)

and T ′′ + a2λT = 0 (10.146)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.147)


u(0, t) = 0, and u(L, t) = 0

or, X(0)T (t) = 0 and X(L)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(L) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0



√λL



sin√λL = 0

=⇒√λL = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

L, n = 1, 2, 3, · · ·

or, λ =n2π2

L2, n = 1, 2, 3, · · ·



X = c2 sinnπx

L, n = 1, 2, 3, · · ·


Xn(x) = sinnπx

L, n = 1, 2, 3, · · ·


λn =n2π2

L2, n = 1, 2, 3 · · ·


m2 + a2λ = 0


Li = ±nπa

Li


T (t) = k1 cosnπat

L+ k2 sin

nπat

L(10.148)

As u(x, 0) = 0 =⇒ X(x)T (0) = 0 =⇒ T (0) = 0

(10.149)


T (0) = k1 cosnπa0

L+ k2 sin

nπa0

L

or, 0 = k1

=⇒ k1 = 0


T = sinnπat

L

Thus, the function

un(x, t) = X(x)T (t) = sinnπx

Lcos

nπat

L(10.150)

consititute the fundamental solutions of (10.140) is along with the boundary condi-

tions and the initial condition u(x, 0) = 0. It remains to satisfy the non-homogeous

initial condition

ut(x, 0) = g(x), 0 ≤ x ≤ L


u(x, t) =

∞∑n=1

bnun(x, t) =

∞∑n=1

bn sinnπx

Lsin

nπat

L(10.151)


Differentiating partially with respect to t, we get

ut(x, t) =∞∑n=1

bn∂

∂xun(x, t) =

L

nπa

∞∑n=1

bn sinnπx

Lcos

nπat

L, (10.152)

∴ ut(x, 0) = g(x)

or,∞∑n=1

L

nπabn sin

nπx

L= g(x)

Thus, the quantities nπaL bn are the coefficients in the Fourier sine series of the period

2L for g. Therefore,

nπa

Lbn =

2

L

∫ L

0g(x) sin

nπx

Ldx, n = 1, 2, 3 · · ·

or, bn =2

nπa

∫ L

0g(x) sin

nπx

Ldx, n = 1, 2, 3, · · ·

Hence, the required solution is

u(x, t) =∞∑n=1

bn sinnπx

Lsin

nπat

L

where

bn =2

nπa

∫ L

0g(x) sin

nπx

Ldx

Example 362. Consider an elastic string of length L, where both ends are held fixed.

The string is set in motion from its equilibrium position with an initial velocity

ut(x, 0) = g(x) =

2xL for 0 ≤ x ≤ L

2

2(L−x)L for L

2 < x ≤ L

Find the initial displacement u(x, t).

Solution: Here, the string is distrubed from its equalibrium position and then

released at time t = 0 with initial velocity g(x). Then the vibration must satisfy the

wave equation

or, utt = a2uxx 0 < x < L, t > 0 (10.153)


u(0, t) = 0, u(L, t) = 0, t ≥ 0 (10.154)


u(x, 0) = 0, ut(x, 0) = g(x) =

2xL for 0 ≤ x ≤ L

2

2(L−x)L for L

2 < x ≤ L(10.155)


where g is initial velocity at the point x of the string.



∂2u


∂2u

∂x2= X ′′T (10.156)



α2X ′′T = XT ′′

or,X”

X=

1

a2

T ′′

T= −λ (say) where λ > 0 (10.157)


X ′′ + λX = 0 (10.158)

and T ′′ + a2λT = 0 (10.159)


m2 + λ2 = 0

or, m2 = −λ2

or, m = ±i√λ



√λx (10.160)


u(0, t) = 0, and u(L, t) = 0

or, X(0)T (t) = 0 and X(L)T (t) = 0

Since T (t) 6= 0, =⇒ X(0) = 0, X(L) = 0


0 = c1 cos 0 + c2 sin 0

or, c1 = 0

and using X(L) = 0 in (10.147), we get


√λL




sin√λL = 0

=⇒√λL = nπ, n = 1, 2, 3, · · ·

or,√λ =

nπ

L, n = 1, 2, 3, · · ·

or, λ =n2π2

L2, n = 1, 2, 3, · · ·


X = c2 sinnπx

L, n = 1, 2, 3, · · ·


Xn(x) = sinnπx

L, n = 1, 2, 3, · · ·


λn =n2π2

L2, n = 1, 2, 3 · · ·


m2 + a2λ = 0


Li = ±nπa

Li


T (t) = k1 cosnπat

L+ k2 sin

nπat

L(10.161)

As u(x, 0) = 0 =⇒ X(x)T (0) = 0 =⇒ T (0) = 0

(10.162)


T (0) = k1 cosnπa0

L+ k2 sin

nπa0

L

or, 0 = k1

=⇒ k1 = 0


T = sinnπat

L

Thus, the function

un(x, t) = X(x)T (t) = sinnπx

Lcos

nπat

L(10.163)


consitute the fundamental solutions of (10.153) is along with the boundary condi-

tions and the initial condition u(x, 0) = 0. It remains to satisfy the non-homogeous

initial condition

ut(x, 0) = g(x), 0 ≤ x ≤ L


u(x, t) =∞∑n=1

bnun(x, t) =∞∑n=1

bn sinnπx

Lsin

nπat

L(10.164)

Differentiating partially with respect to t, we get

ut(x, t) =∞∑n=1

bn∂

∂xun(x, t) =

L

nπa

∞∑n=1

bn sinnπx

Lcos

nπat

L, (10.165)

∴ ut(x, 0) = g(x)

or,∞∑n=1

L

nπabn sin

nπx

L= g(x)

Thus, the quantities nπaL bn are the coefficients in the Fourier sine series of the period

2L for g. Therefore,

nπa

Lbn =

2

L

∫ L

0g(x) sin

nπx

Ldx, n = 1, 2, 3 · · ·

or, bn =2

nπa

∫ L

0g(x) sin

nπx

Ldx, n = 1, 2, 3, · · ·

Hence, the required solution is

u(x, t) =

∞∑n=1

bn sinnπx

Lsin

nπat

L(10.166)


where

bn =2

nπa

∫ L

0g(x) sin

nπx

Ldx

=2

nπa

∫ L/2

0g(x) sin

nπx

Ldx+

2

nπa

∫ L

L/2g(x) sin

nπx

Ldx

=2

nπa

∫ L/2

0

2x

Lsin

nπx

Ldx+

2

nπa

∫ L

L/2

2(L− x)

Lsin

nπx

Ldx

=4

nπaL

∫ L/2

0x sin

nπx

Ldx+

4

nπaL

∫ L

L/2(L− x) sin

nπx

Ldx

=4

nπaL

[x

∫sin

nπx

Ldx

]L/20

−∫ L/2

0

(dx

dx

∫sin

nπx

Ldx

)dx

+4

nπaL

[(L− x)

∫sin

nπx

Ldx

]LL/2

−∫ L

L/2

(d

dx(L− x)

∫sin

nπx

Ldx

)dx

=4

nπaL

[− L

nπx cos

nπx

L

]L/20

+

∫ L/2

0

L

nπcos

nπx

Ldx

+4

nπaL

[− L

nπ(L− x) cos

nπx

L

]LL/2

−∫ L

L/2

L

nπcos

nπx

Ldx

=4

nπaL

[− L2

2nπcos

nπ

2+ 0

]+

L2

n2π2

[sin

nπx

L

]L/20

+

4

nπaL

[0 +

L2

2nπcos

nπ

2

]− L2

n2π2

[sin

nπx

L

]LL/2

=

4L

n3π3a

[sin

nπ

2− 0]− 4L

n3π3a

[sinnπ − sin

nπ

2

]=

8L

n3π3asin

nπ

2


u(x, t) =8L

π3a

∞∑n=1

1

n3sin

nπ

2sin

nπx

Lsin

nπat

L

Example 363. Dimensionless variables s = xL introduced to the wave equation

utt = a2uxx

Show that the wave equation becomes

a2uss = L2utt

Solution: The wave equation is

utt = a2uxx (10.167)


We have

s =x

L=⇒ x = sL

∴∂x

∂s=

1

L

uxx =∂

∂x

(∂u

∂x

)=

∂

∂x

(∂u

∂s

∂s

∂x

)=

∂

∂x

(∂u

∂s

1

L

)=

1

L

∂

∂x

(∂u

∂s

)=

1

L

∂

∂s

(∂u

∂s

)∂s

∂x

=1

L2

∂

∂s

(∂u

∂s

)=

1

L2

∂2u

∂s2=

1

L2uss


utt =a2

L2uss

or, L2utt = a2uss

10.4 Laplace Equation

One of the important of all partial differential equations occuring in applied math-

ematics is Laplace’s equation. In two dimensions the Laplace equation is

uxx + yyy = 0 (10.168)

and in three dimensions

uxx + uyy + uzz = 0 (10.169)

10.4.1 Dirichlet Problem for a Rectangule

Let us consider two dimensional Laplace equation in Cartesian form

uxx + uyy = 0 (10.170)

in a rectangle 0 < x < a, 0 < y < b, and also satisfying the boundary conditions

u(x, 0) = 0, u(x, b) = 0, 0 < x < a (10.171)

u(0, y) = 0, u(a, y) = f(y) 0 ≤ y ≤ b, (10.172)

10.4. LAPLACE EQUATION 603

Figure 10.1: Dirichlet Problem for a rectangle

where f is a given function on 0 ≤ y ≤ b. Let us suppose that

u(x, y) = X(x)Y (y) (10.173)

be solution of (10.170). Then

uxx = X ′′Y, uyy = XY ′′

and the equation (10.170) becomes

X ′′Y +XY ′′ = 0X ′′

X= −Y

′′

Y= λ

where λ is the separation constants. Thus we obtained two ordinary differential

equations

X ′′ − λX = 0 (10.174)

Y ′′ + λY = 0 (10.175)

Now using the boundary conditions

u(0, y) = 0 =⇒ X(0)Y (y) = 0 for all y

=⇒ X(0) = 0 (10.176)

u(x, 0) = 0 =⇒ X(x)Y (0) = 0 for all x

=⇒ Y (0) = 0 (10.177)

u(x, b) = 0 =⇒ X(x)Y (b) = 0 for all x

=⇒ Y (b) = 0 (10.178)


Now we find the solution of (10.175), the auxiliary equation is

r2 + λ = 0

or, r = i√λ

∴ Y (y) = c1 cos(√λy) + c2 sin(

√λy) (10.179)


Y (0) = 0 we get c1 = 0

and

Y (b) = 0 we get c2 sin√λb = 0

=⇒√λb = nπ

=⇒ λ =n2π2

b2

Y (y) is proportional to eighen function sin nπyb .

Putting the values of Y in (10.174), we obtained

X ′′ − nπ

nX = 0 (10.180)

The auxiliary equation of(10.175) is

r2 − n2π2

b2= 0

or, r =nπ

b

∴ X(x) = k1 coshnπx

b+ k2 sinh

nπx

b

Using the boundary condition X(0) = 0, we get

k1 cosh 0 + k2 sinh 0 = 0

or, k1 = 0 as cosh 0 = 1, sinh 0 = 0

Therefore,

X(x) = k2 sinhnπx

b

Thus, the solution X(x) is proportional to sinh nπxb . The Fundamental solution of

the given problem is

un(x, y) = sinhnπx

bsin

nπy

b

These functions satisfy the differenrial equation (10.170) and all the homogeneous

boundary conditions for each value of n.

To satisfy the remaining non-homogeous boundary conditions at x = a, we assuse

the solution u(x, y) in the form

u(x, y) =

∞∑n=1

cnun(x, y) =

∞∑n=1

cn sinhnπx

bsin

nπy

b(10.181)


The coefficient cn are determined by the boundary condition

u(a, y) = f(y)

or,

∞∑n=1

cn sinhnπa

bsin

nπy

b= f(y)

Therefore the quantities cn sinh nπab must be the coefficients in the Fourier sine series

of the period 2b for f and are given by

cn sinhnπa

b=

2

b

∫ b

0f(y) sinh

nπy

bdy

or, cn =2

b sinh nπab

∫ b

0f(y) sinh

nπy

bdy

Hence the fundamental solution is

u(x, y) =

∞∑n=1

cn sinhnπx

bsin

nπy

b

where

cn =2

b sinh nπab

∫ b

0f(y) sinh

nπy

bdy.

Example 364.

Find the solution of the Laplace equation

uxx + uyy = 0 (10.182)

in a rectangle 0 < x < a, 0 < y < b, and satisfying the boundary conditions

u(x, 0) = h(x), u(x, b) = 0, 0 ≤ x ≤ a (10.183)

u(0, y) = 0, u(a, y) = 0 0 < y < b, (10.184)

where h is a given function on 0 ≤ y ≤ b.Solution: Let us suppose that

u(x, y) = X(x)Y (y) (10.185)




X ′′Y +XY ′′ = 0X ′′

X= −Y

′′

Y= −λ


Figure 10.2: Dirichlet Problem for a rectangle

where λ is the separation constants. Thus we obtained two ordinary differential

equations

X ′′ + λX = 0 (10.186)

Y ′′ − λY = 0 (10.187)


u(0, y) = 0 =⇒ X(0)Y (y) = 0 for all y

=⇒ X(0) = 0 (10.188)

u(x, b) = 0 =⇒ X(x)Y (b) = 0 for all x

=⇒ Y (b) = 0 (10.189)

u(a, y) = 0 =⇒ X(a)Y (y) = 0 for all y

=⇒ X(a) = 0 (10.190)


r2 + λ = 0

or, r = i√λ

∴ X(y) = c1 cos(√λx) + c2 sin(

√λx) (10.191)


X(0) = 0 we get c1 = 0

and

X(a) = 0 we get c2 sin√λa = 0


=⇒√λa = nπ

=⇒ λ =n2π2

a2

X(x) is proportional to eighen function sin nπxa .

Putting the values of Y in (10.187), we obtained

Y ′′ − nπ

aY = 0 (10.192)


r2 − n2π2

b2= 0

or, r = ±nπa

∴ Y (y) = k1 coshnπy

a+ k2 sinh

nπy

a(10.193)

Using the boundary condition Y (b) = 0, we get

k1 coshnπb

a+ k2 sinh

nπa

b= 0

or,k2

k1= −

cosh nπba

sinh nπab

Thus from (10.193)

Y (y) = k1

(cosh

nπy

a+k2

k1sin

nπy

a

)

or, Y (y) = k1

(cosh

nπy

a−

cosh nπba

sinh nπba

sinnπy

a

)

or, Y (y) =k1

sinh nπba

(sinh

nπb

acosh

nπy

a− cosh

nπb

asinh

nπy

a

)or, Y (y) = dn sinh

nπ(b− y)

a

where

dn =k1

sinh nπba

Thus, the solution Y (x) is proportional to sinh nπ(b−y)a . The Fundamental solution

of the given problem is

un(x, y) = sinnπx

asinh

nπ(b− y)

a

These functions satisfy the differenrial equation (10.182) and all the homogeneous

boundary conditions for each value of n.


To satisfy the remaining non-homogeous boundary conditions at u(x, 0) = h(x), we

assuse the solution u(x, y) in the form

u(x, y) =∞∑n=1

cnun(x, y) =∞∑n=1

cn sinnπx

asinh

nπ(b− y)

a(10.194)

The coefficient cn are determined by the boundary condition

u(x, 0) = h(x)

or,

∞∑n=1

cn sinhnπx

asinh

nπb

a= h(x)

Therefore the quantities cn sinh nπba must be the coefficients in the Fourier sine series

of the period 2a for h and are given by

cn sinhnπb

a=

2

a

∫ a

0h(x) sin

nπx

adx

or, cn =2

a sinh nπba

∫ b

0h(x) sin

nπx

adx

Hence the fundamental solution is

u(x, y) =∞∑n=1

cn sinnπx

asinh

nπ(b− y)

a

where

cn =2

a sinh nπba

∫ b

0h(x) sin

nπx

adx.

10.4.2 Laplace’s Equation in Polar form

Example 365. From the Laplace equation of two dimension uxx+uyy = 0, obtained

the Laplace equation for polar form

urr +1

rur +

1

r2uθθ = 0

Solution: The two dimensional Laplace’s equation in Cartesian form is

uxx + uyy = 0

or,∂2u

∂x2+∂2u

∂y2= 0 (10.195)

Let

x = r cos θ, y = r sin θ

Then

∂x

∂r= cos θ,

∂x

∂θ= −r sin θ

∂y

∂r= sin θ,

∂x

∂θ= r cos θ


So,

∂u

∂r=

∂u

∂x

∂x

∂r+∂u

∂y

∂y

∂r

=∂u

∂xcos θ +

∂u

∂ysin θ

Also,

∂2u

∂r2=

∂

∂r

(∂u

∂xcos θ +

∂u

∂ysin θ

)= cos θ

∂

∂r

(∂u

∂x

)+ sin θ

∂

∂r

(∂u

∂y

)= cos θ

[∂

∂x

(∂u

∂x

)∂x

∂r+

∂

∂y

(∂u

∂x

)∂y

∂r

]+ sin θ

[∂

∂x

(∂u

∂y

)∂x

∂r+

∂

∂y

(∂u

∂y

)∂y

∂r

]= cos θ

[cos θ

∂2u

∂x2+ sin θ

∂2u

∂x∂y

]+ sin θ

[cos θ

∂2u

∂x∂y+ sin θ

∂2u

∂y2

]= cos2 θ

∂2u

∂x2+ 2 cos θ sin θ

∂2u

∂x∂y+ sin2 θ

∂2u

∂y2(10.196)

Again,

∂u

∂θ=

∂u

∂x

∂x

∂θ+∂u

∂y

∂y

∂θ

= −r sin θ∂u

∂x+ r cos θ

∂u

∂y


Now,

∂2u

∂θ2=

∂

∂θ

(∂u

∂θ

)=

∂

∂θ

(−r sin θ

∂u

∂x+ r cos θ

∂u

∂y

)= −r ∂

∂θ

(sin θ

∂u

∂x

)+ r

∂

∂θ

(cos θ

∂u

∂y

)= −r

[sin θ

∂

∂θ

(∂u

∂x

)+∂u

∂x

∂

∂θ(sin θ)

]+ r

[cos θ

∂

∂θ

(∂u

∂y

)+∂u

∂y

∂

∂θ(cos θ)

]= −r

[sin θ

∂

∂θ

(∂u

∂x

)+ cos θ

∂u

∂x

]+ r

[cos θ

∂

∂θ

(∂u

∂y

)− sin θ

∂u

∂y

]= −r

[sin θ

∂

∂x

(∂u

∂x

)∂x

∂θ+

∂

∂y

(∂u

∂x

)∂y

∂θ

+ cos θ

∂u

∂x

]+r

[cos θ

∂

∂x

(∂u

∂y

)∂x

∂θ+

∂

∂y

(∂u

∂y

)∂y

∂θ

− sin θ

∂u

∂y

]= −r

[sin θ

−r sin θ

∂2u

∂x2+ r cos θ

∂2u

∂x∂y

+ cos θ

∂u

∂x

]+r

[cos θ

−r sin θ

∂2u

∂x∂y+ r cos θ

∂2u

∂y2

− sin θ

∂u

∂y

]= −r

(cos θ

∂u

∂x+ sin θ

∂u

∂y

)+ r2

(sin2 θ

∂2u

∂x2− 2 sin θ cos θ

∂2u

∂x∂y+ cos2 θ

∂2u

∂y2

)= −r∂u

∂r+ r2

(sin2 θ

∂2u


∂2u

∂x∂y+ cos2 θ

∂2u

∂y2

)∴

1

r2

∂2u

∂θ2= −1

r

∂u

∂r+

(sin2 θ

∂2u


∂2u

∂x∂y+ cos2 θ

∂2u

∂y2

)(10.197)

Adding (10.196) and (10.197)

∂2u

∂r2+

1

r2

∂2u

∂θ2= −1

r

∂u

∂r+ (cos2 θ + sin2 θ)

∂2u

∂x2+ (sin2 θ + cos2 θ)

∂2u

∂y2

or,∂2u

∂r2+

1

r2

∂2u

∂θ2+

1

r

∂u

∂r=∂2u

∂x2+∂2u

∂y2

or,∂2u

∂r2+

1

r2

∂2u

∂θ2+

1

r

∂u

∂r= 0 from (10.195)

or, urr +1

rur +

1

r2uθθ = 0

10.4.3 Dirichlet Problem for a Circle

Consider the problem of solving Laplace’s equation in a circular region r < a subject

to the boundary conditions

u(a, θ) = f(θ) (10.198)


where f is a given function on 0 ≤ θ < 2π. In polar coordinates Laplace’s equation

has the form

urr +1

rur +

1

r2uθθ = 0

or, r2urr + rur + uθθ = 0 (10.199)

where x = r cos θ, y = r sin θ and r =√x2 + y2. We assume u(r, θ) is a single valued

function, periodic in θ with period 2π and bounded for r ≤ a.

Assume the following separation

u(r, θ) = R(r)Θ(θ) (10.200)

Then

ur = R′Θ, urr = R′′Θ, uθθ = RΘ′′

Using (10.200) in (10.199) we get

r2R′′Θ + rR′Θ +RΘ′′ = 0

or, r2R′′

R+ r

R′

R= −Θ′′

Θ= λ

where λ is a separation constant. This gives the following two ordinary differential

equations

r2R′′ + rR′ − λR = 0 (10.201)

and Θ′′ + λΘ = 0 (10.202)


Case I If λ < 0, then let λ = −µ2 where µ > 0 . The equation (10.202) becomes

Θ′′ − µ2Θ = 0 (10.203)


m2 − µ2 = 0

or, m = ±µ

Thus solution of (10.203) is

Θ(θ) = c1eµθ + c1e

−µθ (10.204)

which is periodic if and only if c1 = c2 = 0 and we conclude that λ can to be

negative.

Case II If λ = 0 then the equation (10.202) becomes

Θ′′ = 0


Integrating,

Θ′ = c3

Θ′′ = c3θ + c4

For periodic we must have c4 = 0, so that Θ(θ) is constant and in particular, we can

take

Θ(θ) = 1, we assume c4 = 1

Again, for λ = 0, equation (10.201) becomes

r2R′′ + rR′ = 0

or, r2d2R

dr2+ r

dR

dr= 0

Let us put y = dRdr . Then the above equation becomes

r2dy

dr+ ry = 0

or, rdy

dr+ y = 0

or,d

dr(ry) = 0

Integrating,

ry = k1

or, rdR

dr= k1

or, dR =k1

rdr

Integrating

R = k1 loge r + k2

when r → 0, loge → −∞, so bounded value of u(r, θ), we must have c1 = 0. Hence R

must be constant i.e. R = k2 In particular we take k2 = 1 and u(r, θ) = R(r)Θ(θ) = 1

Case III: Finally let λ > 0. Then λ = µ2for some values of λ. Then the equations

(10.201) and (10.202) becomes

r2R′′ + rR′ − µ2R = 0 (10.205)

Θ′′ + µ2Θ = 0 (10.206)

The equation (10.205) is Euerian, has solution

R(r) = d1rµ + d2r

−µ (10.207)

At the same time the auxiliary equation of (10.206) is

m2 + µ2 = 0


or, m = ±iµ


Θ(θ) = e1 sinµθ + e2 cosµθ

For Θ to be periodic function with period 2π, we must have µ = n a positive integer.

But this again make r−µ = r−n unbounded as r → 0. So we must have d2 = 0.

Hence the appropriate solution (10.199) is

un(r, θ) = rn cosnθ, vn(r, θ) = rn sinnθ, n = 1, 2, 3, · · · (10.208)

These functions together with u0(r, θ) = 1 form a set of fundamental solution. We

now assume that u can be expressed as a linear combination of the fundamental

solutions; that is

u(r, θ) =c0

2+∞∑n=1

rn(cn cosnθ + kn sinnθ) (10.209)

But this solution must satisfy the boundary condition

u(a, θ) = f(θ)

or, u(a, θ) =c0

2+∞∑n=1

an(cn cosnθ + kn sinnθ) (10.210)

for 0 ≤ θ < 2π. The function f may be extended outside this interval so that it is

periodic with period 2π and therefore has a Fourier series of the form (10.210) Then

the coefficients are given by

ancn =1

π

∫ 2π

0f(θ) cosnθ n = 1, 2, 3 · · ·

ankn =1

π

∫ 2π

0f(θ) sinnθ n = 1, 2, 3 · · ·

Example 366. Find the solution u(r, θ) of Laplace equation in the semicircular

region r < a, 0 < θ < π, that satisfies the boundary conditions

u(r, 0) = f(θ), u(r, π) = 0, 0 ≤ r < a

u(a, θ) = f(θ), 0 ≤ θ ≤ π

Assume that u is single-valued and bounded in the given region.

Find the solution if f(θ) = θ(π − θ).

Solution: In polar coordinates Laplace’s equation has the form

urr +1

rur +

1

r2uθθ = 0

or, r2urr + rur + uθθ = 0 (10.211)


where x = r cos θ, y = r sin θ and r =√x2 + y2. We assume u(r, θ) is a single valued

function, periodic in θ with period 2π and bounded for r ≤ a.

Assume the following separation

u(r, θ) = R(r)Θ(θ) (10.212)

Then

ur = R′Θ, urr = R′′Θ, uθθ = RΘ′′

Using (10.212) in (10.211) we get

r2R′′Θ + rR′Θ +RΘ′′ = 0

or, r2R′′

R+ r

R′

R= −Θ′′

Θ= λ

where λ is a separation constant. This gives the following two ordinary differential

equations

r2R′′ + rR′ − λR = 0 (10.213)

and Θ′′ + λΘ = 0 (10.214)

The boundary conditions

u(r, 0) = 0, u(r, π) = 0, 0 ≤ r < a

=⇒ R(r)Θ(0) = 0, R(r)Θ(π) = 0 0 ≤ r < a

=⇒ Θ(0) = 0, Θ(π) = 0 0 ≤ r < a (10.215)

Let λ > 0. The equation (10.213) is Euerian, has solution

R(r) = d1r√λ + d2r

−√λ (10.216)

At the same time the auxiliary equation of (10.214) is

m2 + λ = 0

or, m = ±i√λ


Θ(θ) = c1 sin√λθ + c2 cos

√λθ


Θ(0) = 0 =⇒ c2 = 0

and Θ(π) = 0 =⇒ c1 sin√λπ + c2 cos

√λπ = 0

or, c1 sin√λπ = 0

For c1 6= 0, sin√λπ = 0

or,√λπ = nπ

or, λ = n2


Hence the eigenvalues are λn = n2 and corresponding eigen functions are Θn(θ) =

sinnθ.

Again the solution (10.216) becomes

R(r) = d1rn + d2r

−n

(10.217)

For boudedness at r = 0, we must have

limr→0

R(r) = limr→0

(d1r

n +d2

rn

)<∞

For this d2 = 0. Hence

R(r) = d1rn

Hence the fundamental solutions are

un(r, θ) = rn sinnθ

The most general solution is

u(r, θ) =∞∑n=1

cnun(r, θ) =

∞∑n=1

cnrn sinnθ

Using the condition

u(a, θ) = f(θ)

we get

∞∑n=1

cnrn sinnθ = f(θ)

which is a Fourier sine series of f(θ) with period 2π. So the coefficients are given by

cnan =

2

π

∫ π

0f(θ) sinnθdθ

cn =2

anπ

∫ π

0f(θ) sinnθdθ


Let f(θ) = θ(π − θ) = πθ − θ2. Then cn are given by

cn =2

anπ

∫ π

0f(θ) sinnθdθ

=2

anπ

∫ π

0(πθ − θ2) sinnθdθ

=2

anπ

[(πθ − θ2)

∫sinnθdθ

]π0

−∫ π

0

[d

dθ(πθ − θ2)

∫sinnθdθ

]dθ

=

2

anπ

[−(πθ − θ2)

cosnθ

n

]π0

+1

n

∫ π

0(π − 2θ) cosnθdθ

=

2

nanπ

0 +

[(π − 2θ)

∫cosnθdθ

]π0

−∫ π

0

[d

dθ(π − 2θ)

∫cosnθdθ

]dθ

=

2

nanπ

[(π − 2θ)

sinnθ

n

]π0

+2

n

∫ π

0sinnθdθ

=

2

n2anπ

−π sinnπ − π sin 0− 2

n[cosnθ]π0

= − 4

n3anπ(cosnπ − 1) =

4(1− cosnπ)

n3anπ

Thus the solution is

u(r, θ) =4

π

∞∑n=1

rn(1− cosnπ) sinnθ

ann3

10.4.4 Laplace Equation with Numann Boundary Conditions

Consider the problem of finding a solution u(x, y) of Laplace equation

uxx + uyy = 0 (10.218)

in a rectangle 0 < x < a, 0 < y < b, and satisfying the boundary conditions

ux(0, y) = 0, ux(a, y) = f(y), 0 < y < b (10.219)

uy(x, 0) = 0, uy(x, b) = 0 0 ≤ x ≤ a, (10.220)

This is example of a Neumann problem.

Show that Laplace equation and the homogeneous boundary condition determine

the fundamental set of solutions

u0(x, y) = c0, un(x, y) = cn coshnπx

bcos

nπy

b, n = 1, 2, 3, · · ·

Solution: Laplace equation is

uxx + uyy = 0 0 < x < a, 0 < y < b, (10.221)


and satisfying the boundary conditions

ux(0, y) = 0, ux(a, y) = f(y), 0 < y < b (10.222)

uy(x, 0) = 0, uy(x, b) = 0 0 ≤ x ≤ a, (10.223)

Let us suppose that

u(x, y) = X(x)Y (y) (10.224)




X ′′Y +XY ′′ = 0X ′′

X= −Y

′′

Y= λ

where λ > 0 is the separation constants. we neglect the λ ≤ 0 because the general

solutions obtained in this cases are not consitent with the physical nature of the

problem. Thus we obtained two ordinary differential equations

X ′′ − λX = 0 (10.225)

Y ′′ + λY = 0 (10.226)


ux(0, y) = 0 =⇒ X ′(0)Y (y) = 0 for all y

=⇒ X ′(0) = 0 (10.227)

uy(x, 0) = 0 =⇒ X(x)Y ′(0) = 0 for all x

=⇒ Y ′(0) = 0 (10.228)

uy(x, b) = 0 =⇒ X(x)Y ′(b) = 0 for all y

=⇒ Y ′(b) = 0 (10.229)


r2 + λ = 0

or, r = ±i√λ

(10.230)

The genral solution of (10.226) is

Y (y) = c1 cos(√

λy)

+ c2 sin(√

λy)

(10.231)


Differentiating with respect to y

Y ′(y) =√λ(−c1 sin

√λy + c2 cos

√λy)

(10.232)


Y ′(0) = 0 we get c2 = 0

and

Y ′(b) = 0 we get c1 sin√λb = 0

For c1 6= 0

sin√λb = 0

=⇒√λb = nπ

=⇒ λ =n2π2

b2

Y (y) is proportional to eighen function cos nπyb . Let it be YnPutting the values of λ in (10.225), we obtained

X ′′ − n2π2

b2X = 0 (10.233)


r2 − n2π2

b2= 0

or, r = ±nπb

(10.234)

Thus the genral solution is

X(x) = k1 coshnπx

b+ k2 sinh

nπx

b(10.235)

Differentiating both sides with respect to x, we get

X ′(x) =bk1

nπsinh

nπx

b+bk2

nπcosh

nπx

b(10.236)

Using the boundary condition X ′(0) = 0, we get

bk1

nπsinh 0 +

bk2

nπcosh 0 = 0

or, k2 = 0 as sinh 0 = 0, cosh0 = 0 (10.237)

Thus the solution (10.235) becomes

X(x) = k1 coshnπx

b


Letting k1 = 1, we have a solution depending on n so let it be Xn

Xn(x) = coshnπx

b, n = 0, 1, 2, · · ·

The fundamental solution of the Neumann problem is

un(x, y) = XnYn = coshnπx

bcos

nπy

b(10.238)

u0(x, y) = cosh 0 cos 0 = 1


u(x, y) =∞∑n=0

cnun(x, y)

or, u(x, y) =

∞∑n=0

cn coshnπx

bcos

nπy

b

or, u(x, y) = c0 cosh 0 cos 0 +

∞∑n=1

cn coshnπx

bcos

nπy

b

or, u(x, y) = c0 +∞∑n=1

cn coshnπx

bcos

nπy

b

Differentiating both sides partially with respect to x

ux(x, y) =∞∑n=1

nπcnb

sinhnπx

bcos

nπy

b

It remains to show that the solution satisfy ux(a, y) = f(y)

∞∑n=1

nπcnb

sinhnπa

bcos

nπy

b= f(y)

This is Fourier cosine series of f(y) with period 2b, and the constant are given by

nπcnb

sinhnπa

b=

∫ b

0f(y) cos

nπy

bdy

or, cn =2

nπ sinh nπab

∫ b

0f(y) cos

nπy

bdy

Thus the required solution is

u(x, y) = c0 +

∞∑n=1

cn coshnπx

bcos

nπy

b

where

cn =2

nπ sinh nπab

∫ b

0f(y) cos

nπy

bdy

Bibliography

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