Laplace Transformations II

Engineering Mathematics - I Semester – 1 By Dr N V Nagendram

UNIT – IV Laplace Transformations and ItsApplications

. First shifting / Translation Lemma

. Second shifting / Translation Lemma

. Change of scale property

. Multiplication by tn

. Problems Vs Solutions


UNIT – IV Laplace Transformations Class 1

Lemma: First shifting / Translation Lemma.

If L ( F(t) ) where s > then L ( eat F(t) ) , s > a or If is a Laplace transformation of

F(t) then is a Laplace transformation of eat F(t).

Proof: Y

O L ( eat F(t) ) L

( F(t) ) X

Figure

Given is a Laplace transformation. Then =

= L { F(t) }

So, = = = eat

= eat

= eat since, = = L

{ F(t) }

is a Laplace transformation of eat L{ F(t) }.

This completes the proof of the theorem.

Lemma: Second Translation / shifting Lemma.

If L { F(t) } and G(t) = then, L{ G(t) } e-a s .

Proof:

a

|------------------------ t < a ------------|0------|--- t > a -----|

Given = = L { F(t) } and given G(t) =

So that, = =

Put t – a = x implies dt = dx , since da/dx = 0 implies a = c.

= = = e-a s = e – a s

This completes the proof of lemma.

Note: for every x (a, ) such that x (0, ) and for every t (0, ), a = 0

=

For any t, e-as

L { G(t) } = e-as L { F(t) }

e-as = L { G(t) } since, =

L { G(t) } = e-as

L { G(t) } = e-as L { F(t) }


Lemma: Change of scale property

If L { F(t) } then L { F(at) } = .

Proof: Given = = L { F(t) }

Let us consider L { F(at) } =

Put at = x implies t = x/a so dt = (1/a) dx

L { F(at) } = =

=

= [since,

]

= [since, =

]

L { F(at) } = or L { F(at) } =

This completes the proof of the lemma.

Lemma: Multiplication by tn

If L { F(t) } then L { tn F(t) } (-1)n for

n = 1,2,3,………………We have =

On differentiation w.r.t. s, = { }

By Leibnitz principle for differentiation under integral sign,

= { } = { }

= { }

for n = m, = ( 1)m { }

This completes the proof of lemma on multiplication by tn.

Lemma: Division by t

If L { F(t) } then L { F(t) } provided

integral exists.

Proof: since =

On integrating both sides w.r.t. s from s we get,

=

= = here, since t is

independent of s

= = = L { F(t) } by

definition of Laplace transformation.

L { F(t) }




Problem #1: Find L { t sin at }

Problem #2: Find L { t2 sin at }

Problem #3: Find L { t3 e

-3t }

Problem #4: Find L { t e-t sin 3t

}

Problem #5: Find L

Problem #6: Find L

Problem #7: Find and evaluate at s = 2

*Problem #8: Find

Problem #9: Find L

Problem #1: Find L { t sin at }

Solution: Since L { sin at } =

L { t sin at } = { } =

L { t sin at } = is required solution.

Problem #2: Find L { t2 sin at }

Solution: Since L { sin at } =

L { t2 sin at } = ( 1)2 { }

= =

= =

= = =

= =

L { t2 sin at } = = is required solution.

Problem #3: Find L { t3 e

-3t }

Solution: since, L { e-3t } =

L { t3 e

-at } = ( 1)3

= -

= -

= -

= -

=

L { t3 e

-at } = is required solution.

Problem #4: Find L { t e-t

Sin 3t }

Solution: since, L { Sin 3t } =

And L { t Sin 3t } = - =

So, L { t e-t

Sin 3t } =

L { t e-t

Sin 3t } = is required solution.

Problem #5: Find L

Solution: since L { 1- et } = =

Now L = = =

=

=

=

L = is required solution.

Problem #6: Find L

Solution : since L { Cos at – Cos bt } =

L = =

=

= 0 =

L = is required solution.

Problem #7: Find and evaluate at s = 2

Solution : Let

= = - 1. = at s = 2

=

= and s = 2 value is Is required

solution.

Problem #8: Find

Solution: L { Sin mt } =

Now =

= as s 0

= if m > 0

= if m < 0

= is required solution.

Problem #9: Find L

Solution: since L { } =

L { } =L { et Cot

-1s} = Cot

-1 (s – 1) [

by shifting lemma ]

L = .Cot-1 (s 1). Hence the solution.



Section II

Problems Vs SolutionsS

NO

F(t) L { F(t) } Solution

01 Sin t Cos t L{Sin t Cos t}

= ½ L{ 2 sin t cos t}

= ½ L { Sin 2t }

, for s>|2|

02 Cosh2 2t L{Cosh2 2t }

= ½. L{1+Cosh 4t} ,s > 0

03 Sinh at – Sin at L{Sinh at – Sin at}, for s>|a|

04 Sin (at + b) L{ Sin (at+b) }

=L{Sin at cos b +

Cos at Sin b}

,

for s>|a|

05 Cos( t + ) L { Cos( t + ) }

for s>||

06 Cos3 3t = 1/4 cos3t + 3/4

cos t

L{ Cos3 3t }

= L{ 1/4 cos3t + 3/4 cos

t}

07 Cos3 t L{ Cos3 t }

08 Sin3 2t

(Sin 3t = 3 sin t – 4 sin3

t)

L { Sin3 2t }

09 Sin 2t Sin 3t L { Sin 2t Sin 3t }

=½ L{ Cos t - Cos 5t }

10 Cos 5t Cos 2t L { Cos 5t Cos 2t }

=½ L{ Cos 7t + Cos 3t }

11F(t) =

L { F(t) }

12F(t) =

L { F(t) }

13 F(t) = L { F(t) }

14 e-t[3 cos 5t – 4 sin 5t] L{e-t[3 cos 5t – 4 sin

5t] }

15 e2t(3 sinh 2t – 5 cosh 2t) L{ e2t(3 sinh 2t – 5 cosh

2t)}

16L { }

17 e-t cos2 t L (e-t cos2 t } =

L { e-t }

18 e-at Sin bt L { e-at Sin bt }

19 Cosh at – Cos at L { Cosh at – Cos at }

20 ( 1 + t e-t )3 L { ( 1 + t e-t )3 }

21 t e-4t Sin 3t L [t e-4t Sin 3t }

22 (t-1)3[u(t – 1) L { (t-1)3 [u(t – 1) }6

23 eat [ u( t – a ) } L [eat [ u( t – a ) }

24 e-2t {1- u(t – 1) } L { e-2t } – L { e-2t (1-

u(t – 1) )}

25 If L { F(t) }

then

L { F(t/a) } a. Change of scale

property.

26 If L [ F(t) } L[3e2t sin

t – 4 e2t cos 4t }

L[ F(3t) }

27If L { } Tan-1 L { } Tan-1

28 F(t) L { F(t) }

29 F(t) L { F(t) }

30 F(t) t2 at b L { F(t) }

31 F(t) t3 5 Cos t L { F(t) }

32. Evaluate [Ans. L{

}

33. Evaluate [Ans. L{

}

34. Evaluate [Ans. L{ }





UNIT – IV Inverse Laplace Transformations Class 9

Section III

Having found Laplace Transformation of a new functions let us now

determine the inverse Laplace Transformations of given functions

of S.

We have seen L { F(t) } is an algebraic function which is rational.

Hence to find inverse laplace transforms, we have to express the given

function of S into partial fractions which will, then to recognize as

one of the following standard forms:

Sl

.

NO

Inverse Laplace Function Solution

1 L-1 { } 1

2 L-1 { } e

at

3 L-1 { } n1,2,3

..

4 L-1 { } e

atn1,2

,3..

5 L-1 { }

6 L-1 { }

7 L-1 { }

8 L-1 { }

9 L-1 { }

Cosh at

10 L-1 { } e

at Cos bt

11 L-1 { }

12 L-1 { }

Note: Reader is strongly advised to commit these results to

memory.



Section III

Problems Vs Solutions

Problem #1: Evaluate L-1












Solution: L-1

L-1

L-1

L-1

L-1

L-1

L-1

L-1

1 – 3 t 4

L-1

1 – 3 t 4 is required solution.


Solution: L-1

L-1

L-1

L-1

L-1

L-1

is required solution.


Solution: L-1

By using synthetic division method, we can get factors as

S3 s

2 s c

S - 1 1 - 6 11 - 6

0 1 -5 6

1 -5 6 0

S 2 0 2 - 6

1 - 3 0

(S – 1) (S – 2) (S – 3) factor so by partial fractions

on solving We get A ½, B -

1, C 5/2

L-1 L

-1- L

-1 L

-1

½ et - e

-2t 5/2 e

3t

L-1 ½ e

t - e

-2t 5/2 e

3t is required

solution.


Solution: To find L-1 by using partial fractions

4s5 A( s – 1 ) ( s 2 ) B ( s 2 )Put s 1 9 3B B 3 and co efficient of s

2, 0 A – 1/3

A 1/3

L-1 L

-1

L-1

L-1 L-1

1/3 et 3 t e

t – 1/3 e

-2t

L-1 1/3 e

t 3 t e

t – 1/3 e

-2t is required

solution.


Solution: L-1

5s 3 A (s 1)( )

On solving we get, s 1 8 8A A 1

s 0 3 5A C C 2

L-1 L

-1

L-1 L

-1 L-1

et - e

- t Cos 2t 3/2 e

-t Sin 2t

L-1 e

t - e

- t Cos 2t 3/2 e

-t Sin 2t

Is required solution.


Solution: To find L-1

For that by known formula, s4 4a

4 (s

22a

2)2-(2as)

2(s

22a

22as)

(s22a

2-2as)

By partial fractions

s (As b) (Cs d) Co efficient of s3 A – C 0Co efficient of s2 B D 0

Co efficient of s A C 0 and on solving, B ; D –

B so D

L-1 L

-1 L

-1

L-1 e-a t sin at ea t sin at is required

solution.


Solution: L-1 L

-16 L

-1 3 L

-1

6

L-1 6 Is required solution.


Solution: L-1

L-1 L

-1

L-1



Solution: L-1

L-1

L-1

L-1

[Since, L-1

e-3t Cos 2t - 3/2 e

-3t Sin 2t

L-1

e-3t Cos 2t - 3/2 e

-3t Sin 2t is required

solution.


Solution: L-1

L-1

L-1

L-1

3 e-t Cosh 3t 3. .Sinh 3t.e

-t

3 e-t

e-t

e-t

L-1



Solution: L-1

L-1

L-1

L-1

10 L-1

3 et 6 Sinh 2t

5 L

-1

3 et 5 Sinh 2t e

t

3 5

L-1




Section III

Problems Vs Solutions



By partial fractions, L-1

L-1

L-1

L-1

We get A = ; B = ; C =

L-1

L-1

L-1

L-1

L-1

L-1

L-1

e-3t

e2t

L-1

e-3t

e2t is required solution.


Solution: L-1

L-1

By partial fractions =

We get, A = ; B = 2 ; C =

L-1

L-1

+ 2 L-1

L-1

L-1

L-1

+ 2 L-1

L-1

e7t + 2 e

3t e

2t

L-1

e7t + 2 e

3t e

2t is required

solution.


Solution: L-1 L

-1

L-1 L

-1

L-1 L

-1

= L-1 L

-1

= t et - t e-t

L-1 = t et - t e-t is required solution.


Solution: L-1 = L

-1

= L-1 L

-1

= L-1 L

-1 = L

-1 L

-1

= L-1 -L

-1 L

-1 -L

-1

= L-1 - L

-1 L

-1 - L

-1

= L-1 - L

-1 L

-1 - L

-1

= t et - t e-2t

L-1 = = t et - t e-2t is required

solution.



By partial fractions =

Put s = 3 A =

Put s = 0 C =

Put s = 1 B =

L-1 L

-1 L

-1

L-1 L

-1 L

-1

L-1 L

-1 L

-1

= L-1 L

-1 L

-1

= e3t Cos 2t Sin 2t

L-1 = e

3t Cos 2t Sin 2t is required

solution.


Solution: L-1 = L

-1

= L-1

= L-1 L

-1

= L-1 L

-1

= L-1 L

-1

= Sin t - t e-t

L-1 = Sin t - t e-t is required solution.


Solution: L-1 = L

-1

= L-1

= L-1 + L

-1

= L-1 + L

-1

= Cosh at + Cos at

L-1 = Cosh at + Cos at is required solution.


Solution: L-1 = L

-1 = L

-1

L-1 = L

-1 by partial

fractions

Here 1= A[(s-a)2+3as] + (Bs + c) (s-a) ; Put s = a A = ;

Put s = 0 C = ; Put s = 1 B =

L-1 = L

-1

= L-1

= L-1 + L

-1 - L

-1

= + L-1 L

-1

= + L-1 L

-1 L

-1

+ L-1

= e-at t e-at e-at t e-at Cos at

L-1 = e

-at t e-at e-at

t e-at Cos at



Solution: L-1

= L-1

= L-1

+ L-1

+ L-1

= L-1

+ L-1

+ L-1

= L-1

+ L-1

+ L-1

= L-1

+ L-1

+ L-1

L-1

= L-1

+ L-1

+ L-1

L-1

= Sin 2t + Sin t + Sin t - Sin 2t

= Sin 2t + Sin t + Sin t - Sin 2t

= Sin t - Sin 2t

= [5 Sin t - Sin 2t]

L-1

= [5 Sin t - Sin 2t] is required

solution.


Solution: L-1

= L-1

= L-1

= L-1

L-1

= L-1

L-1

= 2 e-2t Cos 3t 7 e

-2t sin 3t

L-1

= 2 e-2t Cos 3t 7 e

-2t sin 3t is required

solution.


Solution: L-1

= L-1

= e-2t Cos t

L-1

= e-2t Cos t is required solution.


Solution: L-1

= L-1

= L-1

= L-1

+ L-1

- L-1

= L-1

+ L-1

- 3 L-1

= Sin t + e-t Sin t - 3 L

-1

L-1

= Sin t + e-t Sin t - 3 L

-1



Solution: L-1

= L-1

= L-1

=

= L-1

L-1

= e-1/2t

Cos e-1/2t

Sin

L-1

= e-1/2t

Cos e-1/2t

Sin



Solution: L-1

= L-1

= L-1

L-1

= L-1

L-1

=a Cos 2a2t a Sin 2a

2t

L-1

=a Cos 2a2t a Sin 2a

2t is required solution.


Solution: L-1 = L

-1

= L-1

= L-1 + L

-1 + L

-1

= L-1 + L

-1 + L

-1

= e2t + 4 t e

2t + 4 t

2 e

2t

L-1 = e

2t + 4 t e

2t + 4 t

2 e

2t is required solution.


Solution: L-1 = L

-1

= L-1 = L

-1 +L

-1

= e2t Cos 3t + 5/3 e

2t Sin 3t

L-1 = e

2t Cos 3t + 5/3 e

2t Sin 3t is required

solution.


Solution: Since, L-1 =

L-1 = = =

L-1 = is required solution.


Solution: L-1 = L

-1 = e

-at L

-1

Here we have, L-1

= = ;

L-1

= = and

L-1

= =

L-1 = =



Solution: Let F (t) = L-1

Hence, L = = = since,

after applying limits. So, = =



Solution: L-1

= L-1

=

=

=

=

=



Solution: Let F (t) = L-1

So, t. F(t) = - L-1

= - L-1

= - L-1

= - L-1 + L

-1

= - L-1 + L

-1

= - e-t

+ et

t. F(t) = - e-t

+ et

F(t) = L-1 =2 is required solution.


Solution: t. F(t) L-1

L-1

L-1 L

-1 L

-1

L-1 L

-1 L

-1

t. F(t) 2 Cos t 1 t e-t

F(t) e-t

F(t) is required solution.


Solution: t F(t) L-1 L

-1

L-1 L

-1

L-1 L

-1

L-1

L-1

L-1 L

-1

L-1 L

-1 L

-1 L

-1

et Sin t e

- t Sin t 2 Sin t 2 sinh t

sin t

t F(t) 2 Sinh t sin t



Solution: Let t F(t) L-1

L-1

L-1

t F(t) Sin 2t




Section III Convolution theorem:

If L-1{ } F(t) and L

-1{ } G(t) then L

-1{ }

F * G is called the convolution or falting of

F and G.

u

u t

tu t

O u 0

t

Let, (t)

L{ (t) }

here put t-u v

then dv dt

F * G.

L-1{ } F * G

This completes the proof of the theorem.Exercises: Try yourself….. By Dr N V Nagendram

Problem #01 Evaluate L-1 [Ans. t Sin at

]

Problem #02 Evaluate L-1 [Ans. t Sin at

]

Problem #03 Evaluate L-1

[Ans.

]


[Ans.

]

Problem #05 Evaluate L-1 [Ans. Cos t

]

Problem #06 Evaluate L-1 [Ans. ( t- Sin at)

]

Problem #07 Evaluate L-1 [Ans.

]

Problem #08 Evaluate L-1 [Ans. e

-at(1 – at)

]


]

Problem #10 Evaluate L-1 [Ans. e-at

]

Problem #11 Evaluate L-1 [Ans. (1 - e-at)

]

Problem #12 Evaluate L-1 [Ans. (e-b t - e-at)

]

Problem #13 Evaluate L-1 [Ans. e

- t e-2t e-

3t ]

Problem #14 Evaluate L-1 [Ans. (Cos at – Cos

bt) ]

Problem #15 Evaluate L-1 [Ans. (1 – Cosh

at) ]

Problem #16 Evaluate L-1 [Ans. (et – Cos t)

]


]


]


]


Solution: L-1 = L

-1

[Since, L-1 and L

-1 ]

By convolution theorem,

L-1 = =

=

L-1 = = t Sin at

L-1 = t Sin at is required solution.



Since, L-1 and L

-1

by convolution theorem,

L-1 = L

-1

= =

=

=

=

L-1 = is required

solution.


UNIT – IV Application of Laplace Transform Class 13

Section III

Application of Laplace Transform to differential equations with

constant coefficients:

Laplace transform is especially suitable to obtain the solutionof linear non-homogeneous ordinary differential equations withconstant coefficients, when all the boundary conditions arespecified for the unknown function and its derivatives at asingle point.

Consider the initial value problem …………………………..

(1)

y(t = 0) = k0, y1(t=0) = k1 ………………………………………………………………… (2)

Where a,b,k0,k1 are all constants and r(t) is a function of t.

Method of Solution to differential equation (D.E) by Laplace Transform(L.T.):

1. Apply Laplace transform on both sides of the given differential equation (1), resulting in a subsidiary equation as

[s2Y – s y(0) - y(0)] +a[sY-y(0)]+bY = R(s) …………………………………………. (3) where Y = L{y(t)} and R(s) = L{r(t)}.

Replace y(0), y(0) using given initial conditions (2).

2. Solve (3) algebraically for Y(s), usually to a sum of partialfractions.

3. Apply inverse Laplace transform to Y(s) obtained in 2. Thisyields the solution of O.D.E. (1) satisfying the initialconditions (2) as y(t) = L-1[Y(s)].

Problem #1 Solve the differential equation by using Laplacetransformation y-2y-8y = 0, y(0) = 3, y(0) = 6.

Solution: On applying Laplace transform (s2Y - 3s - 6) – 2( sY –3 ) - 8Y = 0

On solving, Y(s) = =

By using partial fractions, Y(s) = +

Applying 1 L.T. y(t) = L-1(Y(s)) = 2 L-1 +L-1 = 2 e4t +

e-2t.

y(t) = 2 e4t + e-2t is required solution.

Problem #2 Solve the differential equation by using Laplacetransformation y+2y+5y = e-t Sin t, y(0) = 0, y(0) = 1.Solution: Using L.T. On applying Laplace transform we get thegiven differential equation as, (s2Y - 0 - 1) + 2( sY – 0 ) =

L{ e-t Sin t } =

On solving Y =

By partial fractions, = +

= (As + B) ( )+ (Cs + D ) ( )= s3(A + C) + s2 (2A + 2C + B + D ) + s (2A + 5C + 2B +

2D )+ 2B + 5DEquating coefficients of s on either side,A + C = 0 ; 2A + 2C + B + D = 1; 2A + 5C + 2B + 2D = 2; 2B + 5D= 3

A = 0, B = , C = 0, D =

Y(s)= + = +

Applying I.T. y(t) = L-1{Y}= L-1 + L-1

using first shifting theorem, y(t)= e-t L-1 + e-t L-1

y(t)= is required solution.

Problem #3 Solve the differential equation by using Laplacetransformation y+n2y = a Sin (nt+2), y(0) = 0, y(0) = 0.

Solution: y+n2y = a Sin (nt+2)

= a [Sin nt Cos 2 + Cos nt Sin 2]

Applying L.T. L{ y} + n2 L { y } = a cos 2. L{Sin nt} + a Sin 2L{Cos nt}

Using L.T. On applying Laplace transform we get the givendifferential equation as,

[s2Y – sy(0) - y(0)] + n2Y = .a.Cos 2 + .a. Sin 2

Solving Y, Y(s) = .a. Cos 2 + .a. Sin 2

Applying I.T. we get

y(t) = n. a. Cos 2. L-1 + a. Sin 2. L-1 …… (1)

From I.L.T. tables, we know that 2nd term in R.H.S.

L-1 = …………………………………………………………………… (2) to find

first term in R.H.S.

L-1 = L-1 = =

….(3)

Thus substituting (2) and (3) in (1), we get

y(t) = a.n. Cos 2. + a Sin 2

= [ + Cos 2. Sin nt + nt. Sin 2 .

Sin nt]

= [ ]

= [ ]

y(t) = [ ] is required

solution.Problem #4 Find the general solution of the differential equationby using Laplace transformation y - 3y +3y - y =t2 et,y(0) = 1, y(0) = 0,y(0) = - 2.

Solution: Since the initial conditions are arbitrary assume y(0)= a, y(0) =b, y(0) =c.

Then Using L.T. On applying Laplace transform we get the givendifferential equation as, (s2Y – as2 – bs - c) - 3( s2Y – as -b )

+ 3(sY – a) - Y =

Y =

By partial fractions Y = where c1,

c2, c3 are constants depending on a, b, c.

Applying I.T. and using first shifting theorem

y(t) = c1 et + c2 t et + c3 et + is required

solution.

Problem #5 Solve the differential equation by using Laplacetransformation y - 3y +3y - y =t2 et, y(0) = 1, y(0) =0,y(0) = - 2.

Solution: Applying L.T. to D.E. L{ y - 3y +3y - y} =L{ t2 et }

L{ y} - 3L{ y} +3 L { y} - L{ y} =L{ t2 et

}

[s3Y – s2y(0) - sy(0) - y(0) ] – 3 [s2Y – sy(0) - y(0)] + 3[sY

– y(0)] – Y =

Using the initial condition y(0) = 1, y(0) = 0,y(0) = - 2, andsolving for Y

(s3 – 3a2 + 3s – 1)Y – s2 + 3s – 1=

Y = = =

Y =

On applying I L.T. we get

y(t) = L-1 { Y } = L-1 L-1 L-1 +2 L-1

y(t) = L-1 { Y } = et t et + is required solution.



Section III Exercise Try yourself……..

Use L.T. to solve each of the following I.V.P. consisting of aD.E. with I.C.

01. , general solution [Ans. Assumey(0)=0=A=constant; y=Aet]02. [Ans. y=(3et +e3t)/2]03. [Ans.G.S. y=C + De-t, C=A +B,D= B ]

04. [Ans.G.S. y=et+ Cos t + Sin t]05. [Ans.G.S. y=(3t + 2)e3t

]

06. [Ans.G.S. y=

]

07. [Ans.G.S. y=e-2t-2e-3t+e-5t

]

08. [Ans.G.S.

y=4e2t+3te2t+3e3t-2e5t ]09. [Ans.G.S. y=

]

10. [Ans. y=

]

11. [Ans.G.S. y=

]

12. [Ans.y=

]

13. [Ans.G.S. y=e-2t + 2e-t -

2te-t – Cos t + 2 Sin t]14. [Ans.

]

15. [Ans.G.S. y= 2(Sin 2t –

Sin t ]



Section IV

Application of Laplace Transform to system of simultaneous

differential equations:

Laplace transform can also be used to solve a system or family ofm simultaneous equations in m dependent variables which arefunctions of the independent variable t. consider a family of twosimultaneous differential equations in the two dependentvariables x, y which are functions of t.

……………………………………….. (1)

……………………………………….. (2)

Initial conditions x(0) = c1; y(0) = c2; x(0) = c3; y(0) = c4

………………………… (3)

Here a1, a2, a3, a4, a5, a6,b1, b2, b3, b4, b5, b6 and c1, c2, c3, c4

are all constants and R1(t), R2(t) are functions of t.

Method of solution to system of differential equation (D.E.):

I. Apply Laplace transform on both sides of each of the twodifferential equation (1) and (2) above. This reduces (1) and (2)to two algebraic equations in X(s) and Y(s) where X(s) =L{ x(t) } and Y(s) = L{ y(t) }. a1[s

2X-sx(0) - x(0)] + a2[ s2Y – sy(0) - y(0)] + a3[sX – x(0) ] + a4[sY- y(0) ]+ a5 X + a6 Y = Q1(s)…..(4)

b1[s2X-sx(0) - x(0)] + b2[ s2Y – sy(0) - y(0)] + b3[sX – x(0) ] + b4[sY- y(0) ]

+ b5 X + b6 Y = Q2(s)… (5)

use the initial conditions (3) and substitute for x(0) , y(0) ,x(0) , y(0). II. Solve (4) and (5) for X(s) and Y(s).

III. The required solution is obtained by taking the inverseLaplace Transform of X(s) and Y(s) as

x(t) = L-1{ X(s) } and y(t) = L-1 { Y(s) }.

Problem #1 Solve ;

x(0) = 0; y(0) = 6.5; x(0) = 0;

Solution: Taking Laplace Transformation of the given differentialequation we have

[s2X-sx(0) - x(0)] -3 [sX – x(0) ] - [sY- y(0) ]+ 2Y = 14 +3

[sX – x(0) ] – 3X + [sY- y(0) ] =

Use given initial conditions (I.C.) x(0) = 0; y(0) = 6.5=13/2;x(0) = 0;

S(s - 3)X +(2 – s)Y = and (s – 3 )X + sY =

On solving we get Y(s) = and X(s) =

Taking inverse Laplace Transform y(t) = L-1{Y(s)} = L-1

= L-1

= L-1

y(t) = 7t + 5 – et + e-2t

Similarly, x(t) = L-1{X(s)} = L-1 by partial

fractions

= L-1

= 2 et e3t e-2t

x(t) =2 et e3t e-2t Is required solution.

Problem #2 Solve ;

x(0) = 2; y(0) = 1;

Solution: Taking Laplace Transformation of the given differentialequation we have

2[sX(s) - x(0)] + sY(s)- y(0) –X(s)- Y(s) =

sX(s) – x(0) + sY(s) - y(0)+ 2 X(s) + Y(s) =

Use given initial conditions (I.C.) x(0) = 2; y(0) = 1.

(2s - 1)X(s) +(s – 1)Y(s) = and (s + 2 )X(s) + (s+1)Y(s)

=

On solving we get Y(s) = =

and X(s) =

Taking inverse Laplace Transform y(t) = L-1{Y(s)} = L-1

= Cos t – 13 Sin t + Sinht

y(t) = Cos t – 13 Sin t + Sinh t

Similarly, x(t) = L-1{X(s)} = L-1 by partial fractions

= 2 Cos t + 8 Sin t

x(t) =2 Cos t + 8 Sin t Is required solution.



Exercise Solve the following system ofequations Try yourself ……..

01. ; , x(0) = 8, y(0) = 3

[Ans. x(t) = 5 e-t + 3 e4t; y(t) = 5 e-t – 2 e4t]

02. ; ,

x(0) = 35,y(0) = 27; x(0) = -48,y(0) = -55.

[Ans. x(t) = 30 Cos t – 15 Sin 3t + 3 e-t + 2 Cos 2t; y(t) = 30 Cos t – 60 Sin t - 3 e-t + Sin 2t]

03. ; , x(0) = y(0) = 1; x(0) =

y(0) = 0

[Ans. x(t) = ;

y(t) = ]

04. ; , x(0) = 1, y(0) = 0

[Ans. x(t) = ; y(t) =

]

05. ; , x(0) = 3, y(0) = 0

[Ans.x(t)= ; y(t) = ]

06. ; , x(0) = y(0)= y(0) = 0

[Ans. x(t)=1+ e-t – e-at-e-bt; y(t)=1+e-t– be-at a e-bt ]

Note: a = ; b =

07. ; , x(0) = 0, x(0) = 0

[Ans. x(t) = 22e-t (1+t); y(t) = 2t-2e-t (1+t)]

08. ; , x(0) =1, y(0) =0

[Ans. x(t) = e-2t - t et; y(t) = et +tet ]

09. ; ,x(0) =-1, y(0) = 0

[Ans. x(t) = -2 et + e4t; y(t)= et + e4t]

10. ; , x(0) = 2, y(0) = 0

[Ans. x(t)=e-t + et = 2Cosh t; y(t)=Sin t – 2 Sinh t]

Laplace Transformations II

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