Laplace Transformations I

Engineering Mathematics - I Semester – 1 By Dr N VNagendram

UNIT – IV Laplace Transformations Class 1

Section I

Introduction

The knowledge of Laplace Transformations has in recent yearsbecame an essential part of Mathematical background requiredof engineers and scientists.

This is because the transform methods provide an easy and effective means for the solution of many problems arising in engineering.

This subject originated from the operational methods appliedby the English engineer Oliver Heaviside (1850 – 1925) toproblems in electrical engineering.

Unfortunately, Heaviside’s treatment was unsystematic andlacked rigour, which was placed on sound mathematical footingby Bromwich and carson during 1916 – 1917.

It was found that Heaviside’s operational calculus is bestintroduced by means of a particular type of definite integralscalled “Laplace Transforms”.

The method of Laplace Transforms has the advantage of directlygiving the solution of differential equations with givenboundary values without the necessary of first finding thegeneral solution and then evaluating from it the arbitraryconstants.

Moreover, the ready tables of Laplace Transforms reduce theproblems of solving differential equations to mere algebraicmanipulation.

Definition: Integral transform

Let K(s, t) be a function of two variables s and t where s isa parameter [s R or C] independent of t. Then the functionf(s) defined by an Integral which is convergent. i.e.,

is called the Integral Transform of thefunction F(t) and is denoted by [T{F(t)], K(s, t) is kernel ofthe transformation.

If kernel K(s, t) is defined as

then

is called “Laplace

Transform” of the function F(t) and is also denoted by L{ F(t)} or .

L { F(t) } = =

Definition: Laplace Transformation

P f(P)

f is real valued /complex valued function

Domain Range

Let F(t) be a real or complex valued function defined on [0,

). Then the function f(s) defined by is

called Laplace Transformation of F(t) if the integral existsand f(s) = L{ F(t) }.

Note: If L { F(t) } = F(t) = L-1 { }. Then F(t) iscalled the inverse Laplace Transform of which transformsF(t) into is called “The Laplace Transforms Operator”.

Linearity Property of Laplace Transformation:

A transformation T is said to be linear if a1, a2 constantsand F1(t), F2(t) be any functions of F. Then L{ a1 F1(t) + a2

F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) } or F1(t), F2(t), F3(t) thereexists a1, a2, a3 constants such thatL {a1 F1(t) + a2 F2(t) – a3 F3(t)} = a1 L{ F1(t) + a2 L{ F2(t) } –a3 L{ F3(t)}.

Theorem: Laplace transformation is a linear transformationi.e., a1, a2 constants L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) +a2 L{ F2(t) }.

Proof: By definition,

L.H.S. = L{ a1 F1(t) + a2 F2(t) }

=

= +

= a1 L{ F1(t) + a2 L{ F2(t) }= R.H.S.

L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) }Hence Laplace Transformation is a linear Transformation.This completes the proof of the theorem.Definition: Piece-wise or sectionally continuous

A function F(t) is said to be piece-wise or sectionallycontinuous on a closed interval a t b, if it is defined onthat interval can be divided into finite number of sub-intervals in each of which F(t) is continuous and has finiteleft limit and right hand limits.i.e., Lim F(t) = Lim F(t) = Lim F(t) = finite F say

a t b. t - 0 t + 0 t 0

therefore, F is continuous.

Geometrically: F(t)

0 a t1 t2 t3 bt I1 I2 I3 I4

Figure. piece-wise or sectionally continuous.


UNIT – IV Laplace Transformations Class 2

Definition: Functions of an Exponential order

A function F(t) is said to be an exponential order as ttends to if there exist a positive real number M a number

and a finite number t0 such that | F(t) | < M e t or | e-t

F(t) | < M, t t0.

Note: If a function F(t) is of an exponential order . It isalso of such that > .

Definition: A function of class A

A function which is piece-wise continuous or sectionallycontinuous on every finite interval in the range t 0 and isof an exponential order , as t is known as “A function ofclass A”.

Theorem: Existence of Laplace Transformation

If F(t) is a function which is piecewise or sectionallycontinuous on every interval (finite) in the range t 0 andsatisfies | F(t) | M. eat, t 0 where a and M areconstants. Then the Laplace Transform exists for every s > a.

Proof:

- 0 t t0 t1

|---------------0-------|------I1--------|---------- I2---------------|

By definition,

= +

....................... (1)

exists and piece-wise continuous in interval 0

t t0 so that

(since, |F(t) M eat)

= (since, s > a implies s – a

> 0 0)

=

for any s > a

If t0 is too large then R.H.S. is too small or infinite small therefore L{ F(t) } exists for s > a.

This completes the proof of the theorem.


UNIT–IV Table of General Properties of Laplace Transform Class 3

F(s) =

S No. Name Laplace Transform Inverse Laplace Transform

01. Definition L{f(t)} = f(s) L-1{f(s)} = f(t)02. Linearity af1(t) +b f2(t) a f1(s)+bf2(s)

03. Change of scale f(at)

04. First shifting Th. eat f(t) f(s-a)05. second shifting u(t-a) = { e-as f(s)06. Derivative (multiply by s) f (t)

sf(s)-f(0)07. Second derivative (multiply by s2) f (t) s2f(s) – s f(0) - f (0)08. n th derivative (multiply by sn) f (n)(t) snf(s) – sn-1 f(0) - sn-2

f (0).....– f n-1(0)

09.Integral(division by s)

10. Multiple integral

(division by sn)11. Multiply by t - t f(t) f (s)12. Multiply by t2 t2 f(t) f (s)

13. Multiply by tn (-1)n tn f(t) f n(s)

14. Division by t

15. Convolution f(t) * g(t) =

=

f(s)*g(s)=L(f * g)= L-1(f(s) g(s)}

16. f-periodic with period p f(t) = f(t + p)

Table of some Important Laplace Transforms By Dr N V Nagendram

________________________________________________________________S No. Laplace Transform Inverse Laplace Transform01. L{f(t)} = f(s) L-

1{f(s)} = f(t)

02. 1

03. t

04. t2

05. tn n = 0,1,2,3,...

06. eat

07. e-at

08. n =

1,2,3,.....

09. k =

1,2,3,.....

10. sin at

11. cos at

12. ebt sin at

13. ebt cos at

14. sinh at

15. cosh at

16. ebt sinh at

17. ebt cosh at

18. u(t - a)

19. f(t – a). u(t – a) e-as f(s)

20. ,a b

Table of some Important Laplace Transforms By Dr N V Nagendram________________________________________________________________S No. Laplace Transform Inverse Laplace Transform

21.

22.

23.

24.

25.

26. t cos at

27.

28.

29. (sinh at + at cosh at)/ 2a

30. (cosh at + 1/2at sinh at)

31. t cosh at

32.

33.

34.

35.

36.

37.

38. (1) – log t


UNIT–IV Laplace Transformations and its applicationsClass 4

Definition: Laplace Transform of Periodic function

A function f(t) is said to be a periodic function of period T > 0 if F(t) = f(T + t) = f(2T + t) = f(3T + t ) = ..... = f(nT + t).

Sin t , cos t are periodic functions of period 2.

The Laplace transform of a piecewise periodic function f(t) with period p is

L{ f(t) } = ; s > 0

Note: (1) = , ( ) = ,( ) =( +1) = ( ) = ,

,( ) =( +1) = ( ) = and (- ) =(- ) =

-2.


UNIT–IV Problems Vs Solutions on Laplace transformationsClass 5

Problem Solution

01. L{ 3t - 5}

02. L{ 2t3 – 6t + 8 }

03. L { 6 sin 2t – 5 cos 2t }

04. L { 3 cosh 5t – 4 sinh 5t }

05. L{ (t2 + 1)2 }

06. L[ cos2 t }

07. L{ 3t4-2t3 + 4 e-3t – 2 sin 5t + 3 cos 2t }

08. L{ sin2 at }

09. L{ 2 e3t – e-3t }

10. L{ sin 5t + cos 3t }

11. L{ e-2t – e-3t }

12. L{ F(t) } if F(t) =

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

14. L{ F(t) } if F(t) =

15. L{ F(t) } if F(t) =

16. L{ F(t) } if F(t) =

17. L{ F(t) } if F(t) =

18. L{

19. L{ e2t + 4 t3 – 2 sin 3t + 3 cos 3t }

20. L{ 1 + 2 +3 }

21. L{ cosh at – cos at }

22. L{ cos (at+b) }

23. L{ (sint – cos t)2 }

24. L{ sin 2t cos 3t }

25. L{ sin at sin bt }


UNIT–IV Problems Vs Solutions on Laplace transformationsClass 6

Section II

. First shifting / Translation Lemma

. Second shifting / Translation Lemma

. Change of scale property

. Multiplication by tn

. Problems Vs Solutions


UNIT – IV Laplace TransformationsClass 6

Section II

Lemma: First shifting / Translation Lemma.

If L ( F(t) ) where s > then L ( eat F(t) )

, s > a or If is a Laplace transformation ofF(t) then is a Laplace transformation of e

at 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) }.


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 everyt (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) }




Section II

Problems Vs Solutions

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 Inverse Laplace Function Solution

.

NO

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


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 partialfractions

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)(s

22a

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







Section III




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 = =

L-1 = is required

solution.


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


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 thesolution of linear non-homogeneous ordinary differentialequations with constant coefficients, when all the boundaryconditions are specified for the unknown function and itsderivatives at a single 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 ofpartial fractions.

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 Sin2 L{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 differentialequation by 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 assumey(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,and solving 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. Assume y(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.y=e-2t + 2e-t -2te-t

– Cos t +2 Sin t]14. [

]

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 familyof m simultaneous equations in m dependent variables which arefunctions of the independent variable t. consider a family oftwo simultaneous 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 givendifferential equation 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 givendifferential equation 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 +Sinh t

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 I

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