Dr. Mohd Ishtyak - Aligarh Muslim University

Moh

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tyakLECTURE NOTES ON

ORDINARY DIFFERENTIAL EQUATIONS

based on the course-contents of MMB-352

(A course of Mathematics (Main/Subsidiary)

prescribed for B. Sc. (Hons.) III Semester)

Complied by

Dr. Mohd IshtyakAssistant Professor,

Department of Mathematics

Aligarh Muslim University, Aligarh-202002, India

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Lecture No. 01

Preliminaries(A topic of Unit I)

In this lecture, we shall revise the basic notions regarding to this course, which we

have already discussed in +2 level Mathematics.

Differential Equations: An equation containing the derivatives or differentials of

one or more dependent variables with respect to one or more independent variables is

called a differential equation.

Ordinary Differential Equations: A differential equation in which each involved

dependent variable is a function of a single independent variable is known as an

ordinary differential equation.

Throughout the course, we shall study of different methods for solving ordinary

differential equations. But for the sake of simplicity, we shall use the term ‘differential

equation’ instead of ‘ordinary differential equation’.

Order of a Differential Equation: The order of a differential equation is the order

of highest order derivative appearing in equation.

Degree of a Differential Equation: The degree of a differential equation is the

power of highest order derivative occurring in the equation, when differential coeffi-

cients are made free from radicals and fractions.

eg, the differential equation d3ydx3− 6(dydx

)2 − 4y = 0 is of order 3 and degree 1.

Classification of Differential Equations: Differential equations are classified into

linear and nonlinear differential equations. An nth order differential equation is called

1

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linear if it can be expressed in the form:

P0dny

dxn+ P1

dn−1y

dxn−1+ P2

dn−2y

dxn−2+ ...Pn−1

dy

dx+ Pny = Q

where P0, P1, ..., Pn and Q are either constants or functions of the variable x and

P0 6= 0.

A differential equation, which is not linear is called a nonlinear differential

equation.

Linear differential equations are further classified into homogeneous and nonho-

mogeneous equations according to Q ≡ 0 and Q 6≡ 0, respectively. In this regards,

we shall study deeply later in Unit II.

Formation of a Differential Equation: Usually, differential equations are derived

to eliminate arbitrary constants (parameters) involved in a family of curves. For

illustration, consider the two-parameter family of straight lines

y = mx+ c (1)

having the slop m and passing through the point (0, c). Differentiating Eq. (2), we

getdy

dx= m.

Differentiating again, we getd2y

dx2= 0 (2)

which is a second order differential equation.

Remark 1: To eliminate n arbitrary constants from n-parameter family of curves,

we obtain a differential equation of order n.

Solution of a Differential Equation: A solution (or integral/primitive) of a dif-

ferential equation is an explicit or implicit relation between the variables involved

that does not contain derivatives and satisfies the differential equation.

Naturally, the solution of a differential equation in one dependent variable y and

one independent variable x are of the form y = f(x) (explicit form) or of the form

φ(x, y) = 0 (implicit form). Geometrically speaking, both forms represents the curves

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in xy-plane. Henceforth, the solutions of such a differential equation are also called

‘integral curves’ of the equation.

For example, Eq. (1) forms an integral curve of differential equation (2).

General and Particular Solutions: A solution of a differential equation which

contains a number of arbitrary constants equal to the order of the differential equation

is called the general solution or the complete solution of the differential equation.

For example, y = a cosx+b sinx is the general solution of the differential equation

y′′ + y = 0. Although, y = a cosx and y = b sinx both satisfy the given equation

yet they are not general solutions as each of them contains only one arbitrary constant.

In lieu of the concept of general solution, it can be highlighted that a single

differential equation can possess an infinite number of solutions corresponding to the

unlimited number of choices for the arbitrary constants. A solution of a differential

equation obtained by giving particular values to arbitrary constants in the general

solution is called a particular solution.

Elementary Methods of Solving of Differential Equations of First Order

and First Degree: In class 12, we have studied the techniques of solving the

following three types of differential equations of order one and degree one:

1. Differential Equations with variable separable

2. Homogeneous Differential Equations

3. Linear Differential Equations

As a continuation, we shall discuss several another methods for solving the differential

equations of order one and degree one in our upcoming lectures.

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Lecture No. 02

Bernoulli Equation(A topic of Unit I)

Recall that the general solution of linear differential equation:

dy

dx+ P (x)y = Q(x)

is

y × I.F. =

∫(Q× I.F.)dx+ c

where c is an arbitrary constant and I.F. = e∫Pdx (called integrating factor).

Sometimes certain nonlinear differential equations can be reduced to linear forms

by making suitable substitutions and hence can be solved easily. Bernoulli equation

is one of such equations. Indeed, an equation of the form

dy

dx+ P (x)y = Q(x)yn (1)

(where n 6= 0 and n 6= 1) is called Bernoulli equation.

Dividing Eq. (1) by yn, we get

y−ndy

dx+ P (x)y−n+1 = Q(x). (2)

Put y−n+1 = u so that (1− n)y−n dydx

= dudx

. Thus, Eq. (2) reduces to

1

1− ndu

dx+ P (x)u = Q(x)

or,du

dx+ (1− n)P (x)u = (1− n)Q(x)

1

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which is a linear differential equation in dependent variable u and independent variable

x. After solving it and then replacing u by y−n+1, we can obtain the general solution

of Eq. (1).

Example 2.24/p.42: Solve (1− x2) dydx

+ xy = xy2.

Solution. Dividing given equation by y2, we get

(1− x2)y−2 dydx

+ xy−1 = x

which can be written as

y−2dy

dx+

x

1− x2y−1 =

x

1− x2.

Put y−1 = u so that −y−2 dydx

= dudx

. Then, above equation reduces to

du

dx− x

1− x2u = − x

1− x2

which is a linear differential equation in dependent variable u and independent variable

x. Its integrating factor is

I.F. = e∫Pdx = e

−∫

x1−x2

dx

= e12

∫ −2x

1−x2dx

= e(1/2) log(1−x2)

= (1− x2)1/2.

Therefore, the solution is

u× I.F. =

∫(Q× I.F.)dx+ c

or,

u(1− x2)1/2 =

∫− x

1− x2(1− x2)1/2dx+ c

=

∫−x(1− x2)−1/2dx+ c

=

∫t−1/2.

1

2dt+ c (put t = 1− x2 so that −2xdx = dt)

=1

2

t1/2

1/2+ c = t1/2 + c

= (1− x2)1/2 + c.

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Putting u = y−1 in above equation, we get

y−1(1− x2)1/2 = (1− x2)1/2 + c

or, √1− x2 = y

√1− x2 + cy

or,

(1− y)√

1− x2 = cy

which is the required solution of given differential equation.

Example 2.25/p.42: Solve x dydx

+ y = y2 log x.

Solution. Dividing given equation by y2, we get

xy−2dy

dx+ y−1 = log x


y−2dy

dx+

1

xy−1 =

log x

x.

Put y−1 = u so that −y−2 dydx

= dudx

. Then, above equation reduces to

−dudx

+1

xu =

log x

x

or,du

dx− 1

xu = − log x

xwhich is a linear differential equation in dependent variable u and independent variable

x. Its integrating factor is

I.F. = e−∫

dxx = e− log x =

1

x.

Therefore, the solution is

u× I.F. =

∫(Q× I.F.)dx+ c

or,

u.1

x= −

∫log x

x

1

xdx+ c

= −∫te−tdt+ c (put t = log x so that 1

xdx = dt and x = et)

= −[−te−t − e−t] + c = (t+ 1)e−t + c

= (log x+ 1).1

x+ c

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u = log x+ 1 + cx.

Putting u = y−1 in above equation, we get

1

y= log x+ 1 + cx

or,

(1 + log x+ cx)y = 1

which is the required solution of given differential equation.

Home Assignments: Solve the following differential equations:

Q.33/p.61: dydx

= 2y tanx+ y2 tan2 x. Ans. − 1y

sec2 x = c+ 12

tan3 x.

Q.34/p.61: cosxdy = y(sinx− y)dx. Ans. y−1 = c cosx+ sinx.

Q.36/p.61: dydx

= x3y3 − xy. Ans. 1y2

= 1 + x2 + cex2.

Q.38/p.61: (x2y3 + xy)dy − dx = 0. Ans. 1x

= 2− y2 + e−y2/2

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Lecture No. 03

Exact Differential Equation(A topic of Unit I)

A differential equation is said to be exact if it can be derived from its primitive

(general solution) directly by differentiation, without any subsequent multiplication,

elimination, etc.

Theorem 0.1. The necessary and sufficient condition for the differential equation

Mdx+Ndy = 0 to be exact is that

∂M

∂y=∂N

∂x

In order to obtain the solution of an exact differential equation, we have to proceed

as follows:

1. Integrate M with respect to x, keeping y as constant.

2. Integrate with respect to y only those terms of N which do not contain x.

3. Add the two expressions obtained in (1) and (2) above and equate the result to

an arbitrary constant.

In other words, the solution of an exact differential equation is∫(y=constant)

Mdx+

∫(terms not having x)

Ndy = c.

Ex.40 / P.61 Solve y sin 2xdx− (1 + y2 + cos2 x)dy = 0

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Solution. The given differential equation is

y sin 2xdx− (1 + y2 + cos2 x)dy = 0

compare the above equation with

Mdx+Ndy = 0,

we get

M = y sin 2x and N = −(1 + y2 + cos2 x)

⇒ ∂M

∂y= sin 2x and

∂N

∂x= 2 cos x sinx = sin 2x

i.e.,∂M

∂y=∂N

∂x

The given equation is exact. Now the solution of the differential equation is∫(y=constant)

Mdx+


Ndy = c

∫y sin 2xdx+

∫(−1− y2)dy = c

−y cos 2x

2+

(−y − y3

3

)= c

So, the solution is

y cos 2x+ 2y +2y3

3= c

Ex.41 / P.61 Solve[y(1 + 1

x

)+ cos y

]dx+ (x+ log x− x sin y)dy = 0.

Solution. The given differential equation is[y

(1 +

1

x

)+ cos y

]dx+ (x+ log x− x sin y)dy = 0


Mdx+Ndy = 0,

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we get

M =

[y

(1 +

1

x

)+ cos y

]and N = (x+ log x− x sin y)

⇒ ∂M

∂y=

(1 +

1

x

)− sin y and

∂N

∂x= 1 +

1

x− sin y

i.e.,∂M

∂y=∂N

∂x


Mdx+


Ndy = c

∫ [y

(1 +

1

x

)+ cos y

]dx+

∫0dy = c

y (x+ log x) + x cos y = c

So, the solution is

xy + y log x+ x cos y = c.

Ex.42 / P.61 Solve (ey + 1) cosxdx+ ey sinxdy = 0.

Solution. The given differential equation is

(ey + 1) cosxdx+ ey sinxdy = 0


Mdx+Ndy = 0,

we get

M = (ey + 1) cosx and N = ey sinx

⇒ ∂M

∂y= ey cosx and

∂N

∂x= ey cosx

i.e.,∂M

∂y=∂N

∂x

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Mdx+


Ndy = c

∫(ey + 1) cosxdx+

∫0dy = c

(ey + 1) sinx = c

So, the solution is

(ey + 1) sinx = c.

Home Assignment

1. (Exp.2.29/P.46) Solve xdx+ ydy + xdy−ydxx2+y2

= 0.

[Sol : x2 − 2 tan−1 xy

+ y2 = 2c]

2. (Exp.2.30/P.46) Solve (1 + exy )dx+ e

xy

(1− x

y

)dy = 0.

[Sol : x + yexy = c]

3. (Exp.2.31/P.47) Solve (sin x cos y + e2x)dx+ (cosx sin y + tan y)dy = 0.

[Sol : 12e2x − cos x cos y + log sec y = c]

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Lecture No. 04

Integrating Factor: Part-I(A topic of Unit I)

The equation

ydx+ (x2y − x)dy = 0 (0.1)

is easily seen to non exact because ∂M∂y

= 1 and ∂N∂x

= 2xy − 1.

However, if we multiply both sides of equation (0.1) by the factor 1x2

, then equation

(0.1) becomesy

x2dx+

(x2y − x)

x2dy = 0, (0.2)

which is exact because∂M

∂y=

1

x2=∂N

∂x.

To what extent can other non exact equations be made exact in this way?. In

other way, if

M(x, y)dx+N(x, y)dy = 0 (0.3)

is not exact, under what conditions can a function µ(x, y) be found with the property

that

µ(Mdx+Ndy) = 0

is exact? Any function µ that act in this way is called an integrating factor (in short,

I.F.) for (0.8). Therefore 1x2

is an integrating factor for (0.1).

A differential equation of the type (0.8) always has an integrating factor but there

is no general method of finding them. Here we shall discuss some of the methods for

finding the integrating factors.

Rule I. If the differential equation (0.8) is homogeneous and Mx + Ny 6= 0, then1

Mx+Nyis the integrating factor.

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Example 2.35/P.50 Solve

x2ydx− (x3 + y3)dy = 0 (0.4)

Solution. From (0.4), we have M = x2y and N = −(x3 + y3), so

Mx+Ny = x3y − x3y − y4 = −y4 6= 0,

and the equation (0.4) is homogeneous. Therefore the integrating factor is − 1y4

.

Multiplying equation (0.4) by this integrating factor, we have

−x2

y3dx+

(x3

y4+

1

y

)dy = 0.

Now

M = −x2

y3N =

x3

y4+

1

y

and hence∂M

∂y=

3x2

y4=∂N

∂x.

Therefore, the differential equation is exact and hence∫y=constant

Mdx = −∫x2

y3dx = − x3

3y3∫terms not having x

Ndy =

∫dy

y= log y.

Thus the required solution is ∫Mdx+

∫Ndy = c

− x3

3y3+ log y = c

i.e.,

x3 = 3y3(log y − c).

Home Assignment:

1. (Q.8/P.60) Solve x2ydx− (x3 + y3)dy = 0.

[Ans : y = cex3/3y3 .]

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2. (Q.39/P.61) Solve (x2 − 2xy − y2)dx− (x+ y)2dy = 0.

[Ans : x3 − y3 + 3xy(x− y) = C]

Rule II. If in the differential equation (0.8), M = yf1(xy) and N = xf2(xy). Then1

(Mx−Ny) is an integrating factor.


(xy + 2x2y2)ydx+ (xy − x2y2)xdy = 0 (0.5)

Solution. From equation (0.5), we have M = (xy + 2x2y2)y = yf1(xy) and N =

(xy − x2y2)x = xf2(xy). Therefore the integrating factor is

1

Mx−Ny=

1

x2y2 + 2x3y3 − x2y2 + x3y3=

1

3x3y3.

Multiplying (0.5) by this integrating factor, we have

y(xy + 2x2y2)

3x3y3dx+

x(xy − x2y2)3x3y3

dy = 0

or,1

3

(1

x2y+

2

x

)dx+

1

3

(1

xy2− 1

y

)dy = 0

Now,∂M

∂y= − 1

x2y2=∂N

∂x

and hence the above differential equation is exact. Therefore∫y=constant

Mdx =1

3

∫ (1

x2y+

2

x

)dx =

1

3

(− 1

xy+ 2 log x

)∫

terms not having xNdy = −1

3

∫1

y= −1

3log y.

Thus the required solution is ∫Mdx+

∫Ndy = c

or,1

3

(− 1

xy+ 2 log x

)+

1

3(− log y) = C

or,

2 log x− log y =1

xy+ C1, where C1 = 3C

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Home Assignment:

1. (Q.48/P.61) Solve (x3y3 + x2y2 + xy + 1)ydx+ (x3y3 − x2y2 − xy + 1)xdy = 0.

[Ans : xy − 1xy− log y2 = C.]

2. (Q.51/P.61) Solve (xy sinxy + cosxy)ydx+ (xy sinxy − cosxy)xdy = 0.

[Ans : Cy cosxy = x]

Rule III. The equation (0.8) has e∫f(x)dx as the integrating factor if 1

N

(∂M∂y− ∂N

∂x

)is a function of x, say f(x).

Remark 1. If 1N

(∂M∂y− ∂N

∂x

)= k(a constant), then the integrating factor is e

∫kdx.


(x2 + y2)dx− 2xydy = 0 (0.6)

Solution. From equation (0.6), we have M = (x2 + y2) and N = −2xy. Therefore∂M∂y

= 2y and ∂N∂x

= −2y, and hence

1

N

(∂M

∂y− ∂N

∂x

)= − 1

2xy(2y − (−2y)) =

−2

x= f(x)(say).

Thus the integrating factor is

e∫f(x)dx = e

∫ −2xdx

= e−2 log x

=1

x2(0.7)

Multiplying the equation (0.6) by this integrating factor, we obtain(x2 + y2

x2

)dx− 2xy

x2dy = 0

or, (1 +

y2

x2

)dx− 2y

xdy = 0

or,

dx+ d

(−y2

x

)= 0

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Now integrating term by term, we get

x− y2

x= C.

which is the required solution.

Home Assignment:

1. (Q.52/P.61) Solve (x3 − 2y2)dx+ 2xydy = 0.

[Ans : x+ y2

x2= C]

2. (Q.53/P.61) Solve (x2 + y2 + 2x)dx+ 2ydy = 0.

[Ans : ex (x2 + y2) = C]

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Lecture No. 05

Integrating Factor: Part-II(A topic of Unit I)

In the previous lecture, we have discussed the integrating factor of the following non

exact differential equation

M(x, y)dx+N(x, y)dy = 0 (0.8)

Now, we also include the following methods to find out the integrating factor of the

differential equation (0.8).

Rule IV. The equation (0.8) has e∫f(y)dy as the integrating factor if 1

M

(∂N∂x− ∂M

∂y

)is a function of y, say f(y).

Remark 2. If 1M

(∂N∂x− ∂M

∂y

)= k(a constant), then the integrating factor is e

∫kdy.

Example 2.41 /P.53 Solve

(xy3 + y)dx+ 2(x2y2 + x+ y4)dy = 0. (0.9)

Solution. From equation (0.9), we have M = (xy3 + y) and N = 2(x2y2 + x+ y4).

Therefore∂M

∂y= 3xy2 + 1 and

∂N

∂x= 2(2xy2 + 1)

and hence

1

M

(∂N

∂x− ∂M

∂y

)=

1

xy3 + y

(4xy2 + 2− 3xy2 − 1

)=

1

xy3 + y

(xy2 + 1

)=

1

y(0.10)

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Thus the integrating factor is e∫f(y)dy = e

∫1ydy = elog y = y. Multiplying (0.9) by y,

we obtain the following exact differential equation

(xy4 + y2)dx+ 2(x2y3 + xy + y5)dy = 0 (0.11)

Now,

M1 = (xy4 + y2) and N1 = 2(x2y3 + xy + y5)

and hence ∫y=constant

M1dx =1

2xy4 + y2x

and ∫terms not having x

N1dy =2

6y6.

Thus the required solution is

1

2xy4 + y2x+

2

6y6 = C

or,

3xy4 + 6y2x+ 2y6 = 6C.

Home Assignment:

1. (Q.54/P.61) Solve (2x2y − 3y2)dx+ (2x3 − 12xy + log y)dy = 0.

[Ans : 6x3y3 − 27xy4 − 3y3 log y − y3 = C.]

2. (Q.55/P.61) Solve (3x2y4 + 2xy)dx+ (2x3y3 − x2)dy = 0.

[Ans : x3y3 + x2 = Cy]

Rule V. If the differential equation (0.8) is of the form

xayb(mydx+ nxdy) + xcyd(pydx+ qxdy) = 0,

where a, b, c, d, m, n, p and q are constants, then xhyk is the integrating factor

of the given differential equation, where h, k are constants and can be obtained by

applying the condition that after multiplication by xhyk the given equation is exact.


(y2 + 2x2y)dx+ (2x3 − xy)dy = 0. (0.12)

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Solution. Equation (0.12) can also be written as

y(y + 2x2)dx+ x(2x2 − y)dy = 0

Let xhyk be the integrating factor. Multiplying (0.12) by this integrating factor, we

have

(xhyk+2 + 2xh+2yk+1)dx+ (2xh+2yk − xh+1yk+1)dy = 0 (0.13)

Now,

M = xhyk+2 + 2xh+2yk+1

and

N = 2xh+2yk − xh+1yk+1

If (0.13) is exact, then ∂M/∂y = ∂N/∂x, that is,

(k + 2)xhyk+1 + 2(k + 1)xh+2yk = 2(h+ 3)xh+2yk − (h+ 1)xhyk+1

Equating the coefficients of xhyk+1 and xh+2yk on both sides and solving, we have

h = −5

2and k = −1

2

Thus the integrating factor is

xhyk = x−52y−

12

Multiplying (0.12) by this integrating factor, we have

(x−52y

32 + 2x−

12y

12 )dx+ (2x

12y−

12 − x−

32y

12 )dy = 0

From this equation, we have

M1 = x−52y

32 + 2x−

12y

12

and

N1 = 2x12y−

12 − x−

32y

12

Since the equation is exact, we have∫y=constant

M1dx = −2

3x−

32y

32 + 4x

12y

12

and ∫terms not having x

N1dy = 0

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Therefore the required solution is∫M1dx+

∫N1dy = C

or,

−2

3x−

32y

32 + 4x

12y

12 = C

Home Assignment:

1. (Q.57/P.61) Solve (3x+ 2y2)ydx+ 2x(2x+ 3y2)dy = 0.

[Ans : x3y4 + x2y6 = C.]

2. (Q.60/P.61) Solve (2x2y − 3y4)dx+ (3x3 + 2xy3)dy = 0.

[Ans : 12x−10/13y15/13 + 5x−36/13y24/13 = C]

Change of Variables: If we are unable to apply the rules/methods, which we have

already discussed to solve a given differential equation, then we shall reduce the given

differential equation by suitable substitution. This procedure of reducing the given

differential equation by substitution is called the change of dependent / independent

variable.


xdx+ ydy =a2(xdy − ydx)

x2 + y2(0.14)

Solution. Let x = r cos θ and y = r sin θ, so r2 = x2 + y2 and tan θ = y/x. Differen-

tiating the above relations, we have

dx = −r sin θdθ, dy = r cos θdθ and rdr = xdx+ ydy.

From (0.14), we obtain

rdr =a2

r2(r2 cos2 θdθ + r2 sin2 θdθ

)=a2

r2r2dθ

= a2dθ (0.15)

On integration, we have

r2 = 2a2θ + C

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or,

x2 + y2 = 2a2 tan−1y

x+ C


Home Assignment:

1. (Q.2.46/P.56) Solve sec2 y(dydx

)+ 2x tan y = x3.

[Ans : tan y = 12(x2 − 1) + Ce−x

2]

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Lecture No. 06

Orthogonal Trajectories(A topic of Unit I)

A family of curves whose members cut every member of another family of curves at

right angles is called an orthogonal trajectories of second family of curves.

Formation of Orthogonal trajectories in Cartesian Coordinates: Let the

given family of curves is represented by the equation

F (x, y, c) = 0 (0.16)

where c is a parameter.

We know that differentiating equation (0.16) w. r. to x and then eliminating c

with the help of (0.16), we get a first order ODE of the form

dy

dx= f(x, y) (0.17)

Equation (0.17) is called the differential equation of the family (0.16).

As dydx

represents the slop of a curve of the family (0.16), therefore the slope of the

curve perpendicular to this curve will be − 1dy/dx

. Thus the equation of the orthogonal

trajectories is

− 1

dy/dx= f(x, y)

ordy

dx= − 1

f(x, y)(0.18)

Solving Differential equation (0.18), we obtain the orthogonal trajectories of the

family (0.16) as desired.

Example 4.48/p.136 Find the orthogonal trajectories of x2 + y2 = cx.

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Solution. The given family of curves is

x2 + y2 = cx (0.19)

Differentiating (0.19) w.r. to x, we get

2x+ 2ydy

dx= c (0.20)

Eliminating c between (0.19) and (0.20), we get

x2 + y2 =

(2x+ 2y

dy

dx

)x

x2 + y2 = 2x2 + 2xydy

dx

y2 − x2 = 2xydy

dx

⇒ dy

dx=

y2 − x2

2xy= f(x, y)

which is the differential equation of family (0.19). Hence Differential equation of

orthogonal trajectories is

dy

dx= − 1

f(x, y)=

2xy

x2 − y2(0.21)

which is a first order differential equation. Putting y = vx so that dydx

= v + x dvdx

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(0.21), we get

v + xdv

dx=

2x(vx)

x2 − (vx)2

v + xdv

dx=

2v

1− v2

xdv

dx=

2v

1− v2− v

xdv

dx=

2v − v(1− v2)1− v2

=2v − v + v3

1− v2

xdv

dx=

v + v3

1− v2

xdv

dx=

v(1 + v2)

1− v21− v2

v(1 + v2)dv =

dx

x

[(1 + v2)− 2v2]

v(1 + v2)dv =

dx

x

dv

v− 2v

1 + v2dv =

dx

x

On integrating, we get

log v − log(1 + v2) = log x+ log av

1 + v2= ax

Putting v = yx

in above equation, we get

y/x

1 + (y/x)2= ax

⇒ yx

x2 + y2= ax

⇒ x2 + y2 =1

ay

or

x2 + y2 = by,

where 1a

= b, which is the desired orthogonal trajectories.

Example 4.49/p.137 Find the orthogonal trajectories of the family y = x+ ce−x

and determine that particular member of each family that passes through (0, 3).

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y = x+ ce−x (0.22)

Differentiating (0.22), we getdy

dx= 1− ce−x (0.23)

From (0.22), ce−x = y − x. Putting this value in (0.23), we get

dy

dx= 1− (y − x) = 1 + x− y

which is Differential equation of family (0.22). Thus Differential equation of Orthogo-

nal trajectories is

dy

dx= − 1

1 + x− y

⇒ dx

dy= −1− x+ y

⇒ dx

dy+ x = y − 1 (0.24)

Equation (0.24) is linear differential equation in which x is dependent variable and y

is independent variable. Hence

I.F. = e∫Pdy = e

∫1dy = ey

The general solution is

x× I.F. =

∫(Q× I.F.)dy + a

xey =

∫((y − 1)× ey)dy + a

=

∫yeydy −

∫eydy + a

= (yey − ey)− ey + a

xey = yey − 2ey + a

(x− y + 2)ey = a, (0.25)

which is orthogonal trajectories of (0.22). This orthogonal trajectories passes through

the point (0, 3). So, putting x = 0 and y = 3 in (0.25), we get

(x− y + 2)ey = a

(0− 3 + 2)e3 = a

⇒ a = −e3

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Hence, (0.25) becomes

(x− y + 2)ey = a

(x− y + 2)ey = −e3

(x− y + 2) = −e3−y

or

x− y + 2 + e3−y = 0

which is the required curve passing through the point (0, 3).

Home Assignment

1. (Q.48/P-157) Find the members of the orthogonal trajectories for x+ y = cey

which passes through (0, 5).

[Ans : y = 2− x + 3e−x]

2. (Que 49/P-157) Find the orthogonal trajectories of the following family of

curves:

• c1x2 + y2 = 1

[Ans : 2 log |y| = x2 + y2 + c2]

• 4y + x2 + 1 + c1e2y = 0

[Ans : y = 14− 1

6x2 + c2x

−4]

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Lecture No. 07

Orthogonal Trajectories in Polar Coordinates(A topic of Unit I)

Consider the equation of the family of curves in polar form:

F (r, θ, c) = 0 (0.26)

where c is a parameter. Differential equation of family (0.26) is of the form:

f

(r, θ,

dr

dθ

)= 0 (0.27)

Replacing drdθ

by −r2 dθdr

in (0.27), we obtain the differential equation of orthogonal

trajectories of the family (0.26) given by

f

(r, θ,−r2dθ

dr

)= 0 (0.28)

On solving (0.28), we get the equation of orthogonal trajectories of (0.26).

Example 4.51/p.138 Find the equation of the orthogonal trajectory of the family

of circles having a polar equation r = f(θ) = 2a cos θ.


r = 2a cos θ (0.29)

Differentiating (0.29) w.r. to θ, we get

dr

dθ= −2a sin θ (0.30)

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From (0.29), we have 2a = rcos θ

. Using this, (0.30) becomes

dr

dθ= −2a sin θ

dr

dθ= −

( r

cos θ

)sin θ

dr

dθ= −r tan θ

which is differential equation of family of (0.29).

Hence, differential equation of orthogonal trajectories is

−r2dθdr

= −r tan θ

rdθ

dr= tan θ

⇒ dθ

tan θ=

dr

r

⇒ cos θ

sin θdθ =

dr

r

On integrating, we get∫cos θ

sin θdθ + c′ =

∫dr

r

log sin θ + log 2a = log r, where c′ = log 2a

⇒ r = 2a sin θ

which is the required equation of orthogonal trajectories.

Example 4.52/p.139 Find the orthogonal trajectories of r = c1(1− sin θ).


r = c1(1− sin θ) (0.31)

Differentiating (0.31) w.r. to θ, we get

dr

dθ= −c1 cos θ (0.32)

Dividing (0.32) by (0.31) (to eliminate c), we have

1

r

dr

dθ= − cos θ

1− sin θ

⇒ dr

dθ= − r cos θ

1− sin θ

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which is differential equation of family of (0.31).

Hence, the differential equation of orthogonal trajectories is

−r2dθdr

= − r cos θ

1− sin θ

⇒ rdθ

dr=

cos θ

1− sin θ

⇒ dr

r=

1− sin θ

cos θdθ

⇒ dr

r= (sec θ − tan θ)dθ


log r = log(sec θ + tan θ) + log cos θ + log a

log r = log[(sec θ + tan θ) cos θa]

⇒ log r = log[a(1 + sin θ)]

r = a(1 + sin θ)

which is the required equation of orthogonal trajectories of family (0.31).

Home Assignment

Find the orthogonal trajectories of the following family of curves:

1. (Q.) r2 = c sin 2θ

[Ans : r2 = a cos 2θ]

2. (Q.) r = c(sec θ + tan θ)

[Ans : r = ae− sin θ]

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Lecture No. 08

Nonlinear ODEs solvable for p(A topic of Unit I)

Differential Equations of first order but not of the first degree: The general form of

differential equation of first order and nth(n > 1) degree, is(dy

dx

)n+ a1

(dy

dx

)n−1+ a2

(dy

dx

)n−2+ · · ·+ an−1

(dy

dx

)+ an = 0 (0.33)

or pn + a1pn−1 + a2p

n−2 + · · ·+ an−1p+ an = 0

where, p = dydx

and aa, a2, cdots, an−1 and an are the functions of x and y.

The equation (0.33) can be written as

f(x, y, p) = 0

Solution of the Differential Equation (0.33): We have the following two cases:

1. Differential Equation (0.33) can be written as product of first degree factors.

2. It can not be resolved into factors of first degree.

Case I: Equations solvable for p: In this case, Differential equation (0.33) is of

the form

[p− f1(x, y)][p− f2(x, y)][p− f3(x, y)] · · · [p− fn(x, y)] = 0. (0.34)

Now, equating each factor to zero,

p− fi(x, y) = 0, i = 1, 2, cdots, n

Let the solutions of these factors be

φ(x, y, ci) = 0, i = 1, 2, · · · , n

Taking, c1 = c2 = c3 = · · · = cn = c, hence required solution is

φ1(x, y, c)φ2(x, y, c) · · ·φn(x, y, c) = 0

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Remark. Since the differential equation (0.33) is of first order, therefore its solution

contains only one arbitrary constant.

Example 3.1/p.64 Solve (p− xy)(p− x2)(p− y2) = 0.

Solution. Here

(p− xy) = 0

p = xydy

dx= xy

dy

y= xdx

dy

y− xdx = 0

on integrating, we have

log y − x2

2− c1 = 0

And,

(p− x2) = 0

dy

dx− x2 = 0

dy − x2 dx = 0

⇒ y − x3

3− c2 = 0,

Also, from the third factor, we have

(p− y2) = 0

dy

dx− y2 = 0

dy

y2− dx = 0

⇒ − 1

y− x− c3 = 0

or1

y+ x+ c3 = 0

Taking, c1 = c2 = c3 = c, hence solution of given differential equation is(log y − x2

2− c)(

y − x3

3− c)(

1

y+ x+ c

)= 0

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Example 3.4/p.65 Solve x2p2 + xyp− 6y2 = 0

Solution. Solving the given equation for p, we have

p =2y

x, − 3y

x

If p = 2yx

, then

log y = 2 log x+ log c1

or y = c1x2

and if p = −3yx

, then

yx3 = c2

Therefore, the required solution is

(y − cx2)(yx3 − c) = 0

Example 3.5/p.65 Solve xy2(p2 + 2) = 2py3 + x3

Solution. The given equation can be written as

(yp− x)[xyp+ (x2 − 2y2)] = 0

If yp− x = 0, then integration yields

y2 − x2 = c1

If xyp+ x2 − y2 = 0, then

xyp+ x2 − y2 = 0

xydy

dx+ x2 − 2y2 = 0

ydy

dx+ x− 2y2

x= 0

2ydy

dx− 4y2

x= −2x

dv

dx− 4

xv = −2x, where v = y2 ⇒ dv

dx= 2y

dy

dx

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This is a linear differential equation in v and its solution is

v

x4= c2 +

1

x2

y2 = c2x4 + x2

Thus, the required solution is

(y2 − x2 − c)(y2 − cx4 − x2) = 0

Home Assignment

1. (Ex. 3.2/p.64) Solve (p+ y + x)(xp+ y + x)(p+ 2x) = 0

[Ans : (1− x− y − ce−x)(2xy + x2 − c)(y + x2 − c) = 0]

2. (Ex. 3.6/p.65) Solve x2p3 + y(1 + x2y)p2 + y3p = 0

[Ans : (y − c)(ye−1x − c)(xy + cy − 1) = 0]

3. (Q. 1/p.73) Solve p2 − 7p+ 12 = 0

[Ans : (y − 4x− c)(y − 3x− c) = 0]

4. (Q. 3/p.73) Solve p3 + 2xp2 − y2p2 − 2xy2p = 0

[Ans : (y − c)(y + x2 − c)(xy + cy + 1) = 0]

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Lecture No. 09

Nonlinear ODEs solvable for y and x(A topic of Unit I)

Case II:- If the differential equation f(x, y, p) = 0 (first order but not first degree)

can not be resolved into linear factors in p.

In this case we will discuss the following two cases:

1. Differential equation solvable for y.

2. Differential equation solvable for x.

(a) Differential equation solvable for y:- In this case the differential equation

f(x, y, p) = 0 (0.35)

can be re-written as

y = g(x, p) (0.36)

Now differentiate with respect to x, we get

dy

dx= p = φ

(x, p,

dp

dx

)(0.37)

And suppose that the solution of Equation (0.37) is

F (x, p, c) = 0 (0.38)

Then, the elimination of p between (0.36) and (0.38) give the required solution. If

this elimination is not possible, then equation (0.36) and (0.38) together give the

required solution.

Example 3.7 / p.67 Solve y + px = x4p2

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Solution. Here

y = x4p2 − px (0.39)

D.w.r. to x, we get

dy

dx= p = 4x3p2 + 2x4p

dp

dx− p− xdp

dx

⇒ (4x3p2 − 2p) +

(2x4p

dp

dx− xdp

dx

)= 0

⇒ (2x3p− 1)

(xdp

dx+ 2p

)= 0

xdp

dx+ 2p = 0, (We take the factor which contains derivative terms)

dp

p+ 2

dx

x= 0

Integrating it, we get

log p+ 2 log x = log c

⇒ p =c

x2

put this value in equation (0.39),

y = x4p2 − px

= x4( cx2

)2− c

x2x

y = c2 − c

x

i.e.,

xy + c = c2x


Example 3.8 / p.67 Solve y = sin p− p cos p.

Solution. Given differential equation is

y = sin p− p cos p (0.40)

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Differentiate w. r. to x, we get

dy

dx= p = cos p

dp

dx−(−p sin p

dp

dx+ cos p

dp

dx

)or p = p sin p

dp

dx

p

(sin p

dp

dx− 1

)= 0

⇒ sin pdp

dx− 1 = 0

sin pdp = dx

On integrating, we have

− cos p+ c = x

cos p = c− x (0.41)

From the given differential equation (0.40)

p =sin p− y

cos p=

√1− cos2 p− y

cos p

=

√1− (c− x)2 − y

c− x, (using(0.41))

=

√1− c2 − x2 + 2cx− y

c− x,

Put this value in equation (0.41), we have

c− x = cos

(√1− c2 − x2 + 2cx− y

c− x

)


Home Assignment

1. (Q.8/p.74) Solve y = 2px+ tan−1(xp2)

[Ans : y = 2c√

x + tan−1 c2]

2. (Ex. 3./p.xx) Solve f

[Ans : ]

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(b) Differential equation solvable for x:- If the differential equation

f(x, y, p) = 0 (0.42)

is solvable for x, then we can write it as

x = h(y, p) (0.43)

Differentiate with respect to y, we get

dx

dy=

1

p= ψ

(y, p,

dp

dy

)(0.44)

If the solution of Equation (0.44) is

G(y, p, c) = 0 (0.45)

Then, the elimination of p from (0.43) and (0.45), we get the required solution. If

this elimination is not possible, then equation (0.43) and (0.45) together give the

required solution.

Example 3.10 / p.68 Solve y2 log y = xyp+ p2

Solution. Given differential equation is

y2 log y = xyp+ p2, (0.46)

or x =1

py log y − p

y(0.47)

Differentiate w. r. to y, we get

dx

dy=

1

p= − 1

p2dp

dy(y log y) +

1

p

(1 · log y + y · 1

y

)− 1

y

dp

dy+

p

y2

1

p= − 1

p2dp

dy· y log y +

1

plog y +

1

p− 1

y

dp

dy+

p

y2(1− y

p

dp

dy

)1

plog y +

p

y2

(1− y

p

dp

dy

)= 0(

1− y

p

dp

dy

)(log y

p+

p

y2

)= 0

1− y

p· dpdy

= 0

⇒ dp

p=

dy

dy

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log p = log y + log c

⇒ p = yc

Put this value in (0.46), we have

y2 log y = xyp+ p2,

y2 log y = xy(yc) + (yc)2

Hence, solution is

log y = xc+ c2

Example 3.11 / p.69 Solve xp3 = a+ bp

Solution. Here,

x =a

p3+

b

p2(0.48)

Differentiate w.r. to y, we get

dx

dy=

1

p= −3a

p4dp

dy+

(−2b

p3

)dp

dy

⇒ 1 = −(

3a

p+ 2b

)1

p2dp

dy

dy +1

p2

(3a

p+ 2b

)dp = 0

On integrating

y +

(− 3a

2p2

)+

(2b

−p

)= c

y =3a

2p2+

2b

p+ c (0.49)

Here, it is not easy to eliminate p from (0.48) and (0.49). Hence, the equations (0.48)

and (0.49) both constitute the solution.

Home Assignment

1. (Ex. 3.12/p.69) Solve x = y + a log p = 0

[Ans : x = c− a log(1− p) + a log p, y = c− a log(1− p)]

2. (Ex. 3./p.xx) Solve f

[Ans : ]

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Lecture No. 10

Clairaut and Lagrange Equation(A topic of Unit I)

The most general form of a differential equation of the first order but not of the first

degree (say nth degree) is(dy

dx

)n+ P1

(dy

dx

)n−1+ P2

(dy

dx

)n−2+ · · ·+ Pn−1

(dy

dx

)+ Pn = 0 (0.50)

or pn + P1pn−1 + P2p

n−2 + · · ·+ Pn−1p+ Pn = 0

where p = dydx

and P1, P2, · · · , Pn are functions of x and y. This equation can also be

written as

F (x, y, p) = 0. (0.51)

The above equation however can not be solved in this general form.

The Clairaut Equation

When the given equation (0.51) is of the first degree in x and y, and of the form

y = px+ f(p) (0.52)

Equation (0.52) is known as Clairaut’s equation. To solve it, we differentiate with

respect to x to obtain

p =dy

dx= [x+ f

′(p)]p

′+ p

or [x+ f′(p)]

dp

dx= 0 (0.53)

If dp/dx = 0, then p = c = constant. Eliminating p between this and equation (0.52),

we get

y = cx+ f(c) (0.54)

which is required solution of Clairaut’s equation.

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Remark 3. Sometimes by a suitable substitution, an equation can be reduced to

Clairaut’s form.

Example (3.13/P-71) Solve (y − px)(p− 1) = p.

Solution. The given equation can be written as

y = xp+p

p− 1,

which is a Clairaut equation. Hence the solution is

y = xc+c

c− 1,

y =xc(c− 1) + c

(c− 1)

y(c− 1) = xc(c− 1) + c

y(c− 1)− xc(c− 1) = c

i.e., which is the required solution.

Example (3.14/P-71) Solve p = log(px− y)

Solution. The given equation is

y = px− ep.

Which is a Clairaut equation. Hence the solution is

y = cx− ec

ec = cx− y,

i.e.,

c = log(cx− y)

Example (3.17/P-73) Solve (px− y)(py + x) = h2p.

Solution. This equation can be written as

p2xy + px2 − py2 − xy = h2p

p2xy − xy + px2 − py2 − h2p = 0

p2xy − xy + p(x2 − y2 − h2) = 0,

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Putting x2 = u⇒ 2xdx = du and y2 = v ⇒ 2ydy = dv, and p =dy

dx=x

y

dv

duthe given

equation takes the form

p2xy − xy + p(x2 − y2 − h2) = 0(x

y

dv

dv

)2

xy − xy +x

y

dv

du(u− v − h2) = 0

x2

y

(dv

dv

)2

− y +1

y

dv

du(u− v − h2) = 0

x2(dv

dv

)2

− y2 +dv

du(u− v − h2) = 0

u

(dv

dv

)2

− v +dv

du(u− v − h2) = 0

or uP 2 + (u− v − h2)P − v = 0, where P =dv

du

or v = uP − h2P

P + 1

which is of Clairaut‘s form and has the solution as

v = uc− h2c

c+ 1

where u = x2 and v = y2. i.e.,

y2 = x2c− h2c

c+ 1.

Home Assignment

1. (Example 3.18/P-73) Solve y = 2px+ y2p3.

[Ans : y2 = cx + 18c3]

2. (Que 21/P-74) sin px cos y = cos px sin y + p,

[Ans : y = cx− sin−1 c]

3. (Que 22/P-74) xy(y − px) = x+ py,

[Ans : y2 = cx2 + (1 + c)]

4. (Que 24/P-74) Solve x2p2 + yp(2x+ y) + y2 = 0 by reducing it to Clairaut’s

form by using the substitution y = u and xy = v.

[Ans : xy = cy + c]

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[RB1] The Lagrange Equation

An equation of the form

y = xf1(p) + f2(p) (0.55)

is known as Lagrange’s Equation. To solve it, we differentiate with respect to x to

obtain

p =dy

dx= f1(p) + xf

′

1(p)dp

dx+ f

′

2(p)dp

dx

p− f1(p) =dp

dx[xf

′

1(p) + f′

2(p)]

dx

dp=

xf′1(p) + f

′2(p)

p− f1(p)

ordx

dp− f

′1(p)

p− f1(p)x =

f′2(p)

p− f1(p)(0.56)

which is a linear equation in x, and hence can be solved in the form

x = φ(p, c) (0.57)

eliminating p from equations (0.55) and (0.57), we get the required solution. If it is

not possible to eliminate p, then the values of x and y in terms of p can be found

from equations (0.55) and (0.57), and these will constitute the required solution.

Note. The Clairaut Equation is the particular case of the Lagrange Equation.

Example 2/ P.46 Solve the equation y = xp2 − 1p.

Solution. This equation is Lagrange equation. Differentiating it w.r.t. x, we get

dy

dx= p = x

(2pdp

dx

)+ p2(1)−

(− 1

p2

)dp

dx

p− p2 =

[2xp+

1

p2

]dp

dx

dx

dp=

2xp+ 1p2

p− p2dx

dp=

2xp

p− p2+

1

p2(p− p2)

⇒ dx

dp+

2x

p− 1= − 1

p3(p− 1)

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This equation is linear in x [An equation is linear if it is of the form dydx

+ Py = Q].

Therefore

I.F. = e∫

2p−1

dp = elog(p−1)2

= (p− 1)2


x.(I.F.) =

∫(I.F.)Qdp

x(p− 1)2 =

∫ {(p− 1)2

(− 1

p3(p− 1)

)}dp

x(p− 1)2 = −∫ (

p− 1

p3

)dp

x(p− 1)2 = −∫ {

1

p2− 1

p3

}dp

x(p− 1)2 = c1 +1

p− 1

2p2

x(p− 1)2 =2c1p

2 + 2p− 1

2p2

x =2c1p

2 + 2p− 1

2p2(p− 1)2

x =cp2 + 2p− 1

2p2(p− 1)2(0.58)

Substituting this value of x in the given equation, we get

y = xp2 − 1

p

y =cp2 + 2p− 1

2(p− 1)2− 1

p(0.59)

Equations (0.58) and (0.59) together give the required solution. Hence,

x =cp2 + 2p− 1

2p2(p− 1)2, y =

cp2 + 2p− 1

2(p− 1)2− 1

p

Example 4 /P.47 Solve the equation y = 2px+ pn.

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Solution. This equation is Lagrange equation. Differentiating it w.r.t. x, we get

dy

dx= p = 2p.(1) + x.

(2dp

dx

)+ npn−1

dp

dx

p = 2p+ (2x+ npn−1)dp

dx

−p = (2x+ npn−1)dp

dxdx

dp=

2x+ npn−1

−p

⇒ dx

dp+

2

px = −npn−2

This equation is linear in x. Therefore

I.F. = e∫(2/p)dp = e2 log p = elog p

2

= p2


x.(I.F.) =

∫(I.F.)Qdp

x.p2 =

∫ {p2.(−npn−2)

}dp

x.p2 = −∫npndp

xp2 = − n

n+ 1pn+1 + c

⇒ x =c

p2− n

n+ 1pn−1 (0.60)

Substituting this value of x in the given equation, we get

y = 2px+ pn

y = 2p

[c

p2− n

n+ 1pn−1

]+ pn

y =2c

p− 2n

n+ 1pn + pn

y =2c

p− (n− 1)

(n+ 1)pn (0.61)

Equation (0.60) and (0.61) together give the required solution. Hence,

x =c

p2− n

n+ 1pn−1, y =

2c

p− (n− 1)

(n+ 1)pn

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Home Assignment

1. (Ex 7/ p-48) Solve y = x(1 + p) + p2

[Ans : x = 2(1− p) + ce−p, y = {2(1− p) + ce−p}(1 + p) + p2]

2. (M Ex.8/ p-48) Solve y = xp2 + p

[Ans : x = (log p− p + c)(p− 1)−2, y = xp2 + p]

3. (M Ex.10/ p-48) Solve y = 32xp+ ep

[Ans : x = cp3 − 2ep

(1p− 2

p2 + 2p3

), y = 3c

2p2 − 2ep(1− 3

p+ 3

p2

)]

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Lecture No. 11

The Singular Solution of a First order ODE1

(A topic of Unit I)

Consider the ordinary differential equation

f(x, y, p) = 0, where p =dy

dx. (0.62)

In Lecture 1, we have already discussed the two types of solution of an ordinary

differential equation namely: general solution and particular solution.

In case of differential equation (0.62), its general solution will contain an arbitrary

constant c, and therefore the general solution of differential equation (0.62) is of the

form

g(x, y, c) = 0. (0.63)

Recall that, any particular solution of equation (0.62) can be obtained from (0.63)

by substituting the suitable values of c.

Besides general solution and particular solution, there is another type of solution

of equation (0.62), which is called Singular solution.

Definition 1. A singular solution of equation (0.62) is a solution, which is free from

arbitrary constants but not a particular solution of the equation.

For example, consider the first order differential equation

y = xp+√

1 + p2.

Here, x2 + y2 = a2 is the singular solution of above equation as it can not be derived

from its general solution y = cx +√

(1 + c2) by assigning values to the arbitrary

constant c.1The Content of this lecture is based on the reference book

”Frank Ayres, Jr: Theory and Problems of Differential Equations (Schaum‘s Outline Series)”

1

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Note. An equation of the first order does not have singular solutions; if it is of the

first degree it can not have singular solutions.

Discriminant Relations:

p-discriminant relation: The p-discriminant relation of the equation (0.62) can be

obtain by eliminating p between equation (0.62) and

∂f

∂p= 0.

c-discriminant relation: The c-discriminant relation of the equation (0.62) can be

obtain by eliminating c between equation (0.63) and

∂g

∂c= 0.

Ex 1/p-70 Find the discriminant relations for each of the following:

1. p3 + px− y = 0,

2. [HA] p3x− 2p2y − 16x2 = 0,

3. y = C(x− C)2

Solution 1. The given equation is

f(x, y, p) = p3 + px− y = 0; (0.64)

and∂f

∂p= 3p2 + x = 0. (0.65)

eliminate p between (0.64) and (0.65) as follows:

Multiply (0.65) with p, we get

3p3 + px = 0 (0.66)

Now, {(0.66)− (0.64)}, we get

(3p3 + px)− (p3 + px− y) = 0

2p3 + y = 0

p =(−y

2

) 13

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substitute the value of p in (0.64), we get

p3 + px− y = 0

−y2

+(−y

2

) 13x− y = 0(

−y2

) 13x− 3

2y = 0(

−y2

) 13x =

3

2y

−y2x3 =

27

8y3

27y3 + 4yx3 = 0

Hence the required p-discriminant relation is

4x3 + 27y2 = 0.

Remark 4. If f(x, y, p) = 0 is of the degree n in p, we eliminate p between nf−p∂f∂p

=

0 and ∂f∂p

= 0.

Solution 2. Do yourself.

Solution 3. The given equation is

g(x, y, c) = cx2 − 2c2x+ c3 − y = 0 (0.67)

and∂g

∂c= x2 − 4cx+ 3c2 = 0 (0.68)

Now we eliminate c from (0.67) and (0.68) as:

Multiply (0.68) with c, we get

cx2 − 4c2x+ 3c3 = 0 (0.69)

Now, {(0.69)− 3× (0.67)}, we get

(cx2 − 4c2x+ 3c3)− 3(cx2 − 2c2x+ c3 − y) = 0

−2cx2 + 2c2x+ 3y = 0 (0.70)

Now, Multiplying {3× (0.70)− 2x× (0.68)} , we have

3(−2cx2 + 2c2x+ 3y)− 2x(x2 − 4cx+ 3c2) = 0

−2cx2 + 2x3 − 9y = 0

⇒ c =2x3 − 9y

2x2

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substitute the value of c in (0.68), we get

x2 − 4cx+ 3c2 = 0

x2 − 4

(2x3 − 9y

2x2

)x+ 3

(2x3 − 9y

2x2

)2

= 0

After simplification, the required c-discriminant relation is

y(4x3 − 27y) = 0.

Extraneous Loci Let h(x, y) = 0 be the singular solution of the differential equation

(0.62). Also suppose that

C(x, y) = 0

and

P (x, y) = 0

are the c-discriminant and p-discriminant relations, respectively of equation (0.62).

Then h(x, y) will be a factor of each of C(x, y) and P (x, y). Thus

C(x, y) = h(x, y)E1(x, y)

and

P (x, y) = h(x, y)E2(x, y).

The relations E1(x, y) = 0 and E2(x, y) = 0 generally do not satisfy the differential

equation (0.62). Such relations are called extraneous. Usually there are three special

types of extraneous loci.

1. Tac Locus

2. Nodal Locus

3. Cusp Locus

Determination of Singular Solutions and extraneous loci: An equation which

posses a singular solution is not considered completely solved until the singular

solution has been found.

Method 1 (Using c-discriminant relation): Whenever we determine the c-

discriminant relation C(x, y) = 0 of equation (0.62) then C(x, y) includes as a factor

(after equating to 0) as follows:

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1. singular solution once.

2. cuspidal locus three times.

3. the nodel locus twice.

Method 2 (Using p-discriminant relation): Whenever we determine the p-

discriminant relation P (x, y) = 0, of the differential equation (0.62); then p-discriminant

P (x, y), includes as a factor (after equating to 0) as follows:

1. singular solution once.

2. the cuspidal locus once.

3. the tac locus twice.

Ex 2/p-70 Solve y = 2xp− yp2 and examine for singular solutions.

Solution. Solving for x

The above equation can be written as

2x =y

p+ yp

Differentiate w.r. to y, we get

2dx

dy=

(1

p· 1− y · 1

p2dp

dy

)+

(p+ y

dp

dy

)2

p=

1

p− y 1

p2dp

dy+ p+ y

dp

dy

2

p− 1

p− p = y

(− 1

p2+ 1

)dp

dy

1− p2

p= y

(1− 1

p2

)dp

dy

(p2 − 1)

[p+ y

dp

dy

]= 0

So, we have

p+ ydp

dy= 0

dy

y+dp

p= 0

⇒ py = c

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substituting the value of p in the given differential equation, we get

y = 2px− p2y

y = 2

(c

y

)x−

(c

y

)2

y

so the solution of the given differential equation is

y2 = 2cx− c2.

Now, for p-discriminant relation, we eliminate p from f(x, y, p) = 0 and ∂f∂p

= 0. i.e.

f(x, y, p) = 2px− p2y − y = 0 (0.71)

and

∂f

∂p= 2x− 2py = 0

⇒ p =x

y

Substitute the value of p in (0.71), we get the p-discriminant relation

2px− p2y − y = 0

2

(x

y

)x−

(x

y

)2

y − y = 0

⇒ x2 − y2 = 0

Now, for c-discriminant relation, we eliminate c from g(x, y, c) = 0 and ∂g∂c

= 0. i.e.

g(x, y, c) = y2 − 2cx+ c2 = 0 (0.72)

and

∂g

∂c= −2x+ 2c = 0

⇒ c = x

Substitute the value of c in (0.72), we get the c-discriminant relation

y2 − 2cx+ c2 = 0

y2 − 2(x)x+ x2 = 0

⇒ x2 − y2 = 0

The p and c-discriminant relations are x2 − y2 = 0. Since both y = x and y = −xsatisfy the given differential equation, they are singular solution.

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Ex 5/p-71 Solve (x2 − 4)p2 − 2xyp− x2 = 0 and examine for singular solutions and

extraneous loci.

Solution. Solvable for y


2y = xp− 4

xp− x

p

Differentiating w.r. to x, we get

2dy

dx=

{p× 1 + x× dp

dx

}−{

4

x

dp

dx+ p

(− 4

x2

)}−{

1

p+ x

(− 1

p2dp

dx

)}2p = p+ x

dp

dx− 4

x

dp

dx+

4p

x2− 1

p+x

p2dp

dx

2p− p+1

p− 4p

x2=

(x− 4

x+x

p2

)dp

dx

⇒ (p2x2 − 4p2 + x2)

[p− xdp

dx

]= 0.

So, we have

p− xdpdx

= 0

1

pdp =

1

xdx

⇒ p = cx

Substituting the value of p in the given differential equation, we get

(x2 − 4)p2 − 2xyp− x2 = 0

(x2 − 4)(cx)2 − 2xy(cx)− x2 = 0

so the solution of the given differential equation is

c2(x2 − 4)− 2cy − 1 = 0.

Now, for p-discriminant relation, we eliminate p from f(x, y, p) = 0 and ∂f∂p

= 0. i.e.

f(x, y, p) = (x2 − 4)p2 − 2xyp− x2 = 0 (0.73)

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and

∂f

∂p= 2p(x2 − 4)− 2xy = 0

⇒ p =xy

(x2 − 4)

Substitute the value of p in (0.73), we get the p-discriminant relation

(x2 − 4)p2 − 2xyp− x2 = 0

(x2 − 4)

(xy

(x2 − 4)

)2

− 2xy

(xy

(x2 − 4)

)− x2 = 0

x2y2

(x2 − 4)− 2x2y2

(x2 − 4)− x2 = 0

⇒ x2(x2 + y2 − 4) = 0

Now, for c-discriminant relation, we eliminate c from g(x, y, c) = 0 and ∂g∂c

= 0. i.e.

g(x, y, c) = c2(x2 − 4)− 2cy − 1 = 0 (0.74)

and

∂g

∂c= 2c(x2 − 4)− 2y = 0

⇒ c =y

(x2 − 4)


c2(x2 − 4)− 2cy − 1 = 0(y

(x2 − 4)

)2

(x2 − 4)− 2

(y

(x2 − 4)

)y − 1 = 0

y2

(x2 − 4)− 2y2

(x2 − 4)− 1 = 0

⇒ x2 + y2 − 4 = 0

Now x2 + y2 = 4 occurs once in the p and c-discriminant relations and satisfies the

given differential equation, it is a singular solution.

Also x = 0 occurs twice in the p-discriminant relation, does not occur in the

c-discriminant relation, and does not satisfy the differential equation, it is tac locus.

Ex 6/p-72 Solve 4xp2 − (3x − 1)2 = 0 and examine for singular solutions and

extraneous loci.

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Solution. Solvable for p


4xp2 = (3x− 1)2

p2 =(3x− 1)2

4x

p =(3x− 1)

2x12

dy

dx= ±

(3

2x

12 − 1

2x−

12

)On integration, we get

y = ±

(3

2× x

32

32

− 1

2× x

12

12

)+ c1

(y + c) = ±x12 (x− 1)

(y + c)2 = x(x− 1)2

So the solution of the given differential equation is

g(x, y, c) = (y + c)2 − x(x− 1)2 = 0.

Now, for p-discriminant relation,

0− 4× 4x× [−(3x− 1)2] = 0

16x(3x− 1)2 = 0

⇒ x(3x− 1)2 = 0

Now, for c-discriminant relation,

g(x, y, c) = (y + c)2 − x(x− 1)2 = 0 (0.75)

and

∂g

∂c= 2(y + c) = 0

⇒ c = −y


g(x, y, c) = (y + c)2 − x(x− 1)2 = 0

(y − y)2 − x(x− 1)2 = 0

⇒ x(x− 1)2 = 0

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Here, x = 0 is common to the two relations and satisfy the differential equation.

Hence, it is a singular solution.

3x− 1 = 0 is a tac locus since it occurs twice in the p-discriminant relation, does

not occur in the c-discriminant relation, and does not satisfy the differential equation.

x− 1 = 0 is a nodel locus since it occurs twice in the c-discriminant relation, does

not occur in the p-discriminant relation, and does not satisfy the differential equation.

Home Assignment

1. (Ex 4/p-71) Examine 6p2y2 + 3px− y = 0 for singular solution.

[Ans: Solution: y3 = 3cx+ 6c2, singular Solution: 3x2 + 8y3 = 0]

2. (Ex 7/p-73) Solve 9yp2 + 4 = 0 and examine for singular solutions and

extraneous loci.

[Ans: Solution: y3 + (x+ c)2 = 0, cusp locus: y = 0]

3. (Ex 8/p-73) Solve x3p2 + x2yp+ 1 = 0 and examine for singular solutions and

extraneous loci.

[Ans: Solution: c2 + cxy + x = 0, singular Solution: xy2 − 4 = 0]

4. (Ex 9/p-74) Examine p3x − 2p2y − 16x2 = 0 for singular solutions and

extraneous loci.

[Ans: Solution: c3x2 − c2y − 2 = 0, singular Solution: 2y3 + 27x4 = 0]

Special Case: Singular Solution of Clairaut’s Equation: Recall the Clairaut

equation

y = px+ f(p),

its p-discriminant relation is

x+ f′(p) = 0

If we eliminate p between equations above, we get a singular solution of Clairaut

equation.

Ex 3/p-71 Examine p3 + px− y = 0 for singular solution.

Solution. The above equation can be written as

y = px+ p3, (0.76)

which is a Clairaut equation. Also, from

x+ f′(p) = 0

i.e., x+ 3p2 = 0 (0.77)

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Now, we eliminate p from (0.76) and (0.77), we get

y = px+ p3

y =(−x

3

) 12x+

(−x

3

) 32

y =2x

3

(−x

3

) 12

3y = 2x(−x

3

) 12

9y2 = 4x2(−x

3

)27y2 + 4x3 = 0

So, the required singular solution is

4x3 + 27y2 = 0.

Home Assignment

1. (Ex.10/p-74) Investigate for singular solutions.

y = px− 2p2

[Ans: General Solution, y = cx− 2c2;, singular solution, x2 = 8y]

2. (Ex.3.15/p.72(Zafar Ahsan)) Find the general and singular solution of the

differential equation

y = px+√

(a2p2 + b2)

[Ans: Singular Solution, y2a2(a2 − x2) = b2(x2 + a2)]

3. (Ex.3.16/p.72(Zafar Ahsan)) Reduce xyp2 − (x2 + y2 + 1)p + xy = 0 to

Clairaut‘s form and find its singular solution.

[Ans: Singular Solution, y2 = (x+ 1)2]

Dr. Mohd Ishtyak - Aligarh Muslim University

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