Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Laplace Transforms Solutions for Various Integral Equations, Study notes of Applied Mathematics

Solutions to various integral equations using laplace transforms. It covers topics such as convolution, inverse laplace transforms, and the second shifting theorem. The solutions are presented in a step-by-step manner, making it useful for students studying maths at sathyabama university.

Typology: Study notes

2022/2023

Uploaded on 03/27/2024

shreyas-adki
shreyas-adki 🇮🇳

1 / 173

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
SCHOOL OF SCIENCE AND HUMANITIES
DEPARTMENT OF MATHEMATICS
UNIT I LAPLACE TRANSFORM SMT1401
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f
pf20
pf21
pf22
pf23
pf24
pf25
pf26
pf27
pf28
pf29
pf2a
pf2b
pf2c
pf2d
pf2e
pf2f
pf30
pf31
pf32
pf33
pf34
pf35
pf36
pf37
pf38
pf39
pf3a
pf3b
pf3c
pf3d
pf3e
pf3f
pf40
pf41
pf42
pf43
pf44
pf45
pf46
pf47
pf48
pf49
pf4a
pf4b
pf4c
pf4d
pf4e
pf4f
pf50
pf51
pf52
pf53
pf54
pf55
pf56
pf57
pf58
pf59
pf5a
pf5b
pf5c
pf5d
pf5e
pf5f
pf60
pf61
pf62
pf63
pf64

Partial preview of the text

Download Laplace Transforms Solutions for Various Integral Equations and more Study notes Applied Mathematics in PDF only on Docsity!

SCHOOL OF SCIENCE AND HUMANITIES
DEPARTMENT OF MATHEMATICS

UNIT – I – LAPLACE TRANSFORM – SMT

Sathyabama University

CHAPTER - IV
LAPLACE TRANSFORMS

1. Introduction :

A transformation is mathematical operations, which transforms a mathematical expressions into another

equivalent simple form. For example, the transformation logarithms converts multiplication division, powers

into simple addition, subtraction and multiplication respectively.

The Laplace transform is one which enables us to solve differential equation by use of algebraic methods.

Laplace transform is a mathematical tool which can be used to solve many problems in Science and Engineeing.

This transform was first introduced by Laplace, a French mathematician, in the year 1790, in his work on

probability theory. This technique became very popular when heaveside funcitons was applied ot the solution

of ordinary differential equation in electrical Engeneering problems.

Many kinds of transformation exist, but Laplace transform and fourier transform are the most well known.

The Laplace transform is related to fourier transform, but whereas the fourier transform expresses a function

or signal as a series of mode of vibrations, the Laplace transform resolves a function into its moments.

Like the fourier transfrom, the Laplace transform is used for solving differential and integral equations. In

Physics and Engineering it is used for analysis of linear time invariant systems such as electrical circuits,

harmonic oscillators, optical devices and mechanical systems. In such analysis, the Laplace transform is

often interpreted as a transformation form the time domain in which inputs and outputs are functions of

time, to the frequency domain, where the same inputs and outputs are functions of complex angular frequency

in radius per unit time. Given a simple mathematical or functional discription of an input or output to a

system, the Laplace transform provides an alternative functional discription that often simplifies the process

of analyzing the behaviour of the system or in synthesizing a new system based on a set of specification. The

Laplace transform belongs to the family of integral transforms. The solutions of mechanical or electrical

problems involving discontinuous force function are obtained easily by Laplace transforms.

1.1 DEFINITION OF LAPLACE TRANSFORMS

Let f t ( ) be a functions of the variable t which is defined for all positive values of t. Let s be the real constant.

If the integral

0

st

e f t dt

exist and is equal to F(s), then F(s) is called the Laplace transform of f t ( )

and is

denoted by the symbol L f t [ ( )]

.

i.e

0

[ ( )] ( )

st

L f t e f t dt F s

The Laplace Transform of f t ( )

is said to exist if the integral converges for some values of s, otherwise it does

not exist.

Here the operator L is called the Laplace transform operator which transforms the functions f t ( ) into F(s).

Remark : im ( ) 0

S

L F s



1.2. Piecewise continuous function :

A function f t ( ) is said to be piecewise continuous in any interval [ , a b ] if it is defined on that interval, and

the interval can be divided into a finite number of sub intervals in each of which f t ( ) is continuous.

UNIT I

Sathyabama University

Apply L - Hospital Rule

im

st

t

t

L

e s



which is indeterminate form

Again apply L - Hospital Rule.

2 2

im im 0

st

st

t t

L L e

s e s

 

(finite)

2

im 0

st

t

L e t



(finite numbers)

Hence

2

f t ( )  t

is exponential order..

2

t

b f te

Solution :

By the definition of exponential order.

2 2

im ( ) 0

im im

st

t

st st t

t t

t

L e f t

L e e L e e



   

 

2

t

f te

is not of exponential order..

2. Laplace Transform of Standard functions :

(1) Prove that

[ ]

at

L e

s a

where sa  0

or s   a

Proof :

By definition

0

[ ( )] ( )

st

L f t e f t dt

0

( )

0

( )

0

0

[ ]

at st at

t s a

t s a

L e e e dt

e dt

e

e e

s a s a

s a

  

 

 



Hence

[ ]

at

L e

s a

Sathyabama University

  1. Prove that
[ ]

at

L e

s a

where

s  a

Proof :

By the defn of

0

[ ( )] ( )

st

L f t e f t dt

0

( )

0

( )

0

0

[ ]

at st at

s a t

s a t

L e e e dt

e dt

e

s a

e e

s a

s a

 

 

 



Hence

[ ]

at

L e

s a

0

2 2

0

2 2

2 2

(cos ) cos

( cos sin )

st

st

L at e at dt

e

s at a at

s a

s

s a

s

s a

2 2

2 2

sin sin cos

cos cos sin

ax

ax

ax

ax

e

e bxdx a bx b bx

a b

e

e bxdx a bx b bx

a b

Hence

2 2

(cos )

s

L at

s a

0

2 2

0

2 2

2 2

(sin ) sin

( sin cos )

(sin )

st

st

L at e at dt

e

s at a at

s a

a

s a

a

L at

s a

Sathyabama University

8..

0

1

0 0

1

0

1

1

1 2

3 2

2

n St n

St St

n n

St n

n

n n

n n

n

L t e t dt

e e

t nt dt

s s

n

e t dt

s

n

L t

s

n

L t L t

s

n

L t L t

s

L t L t

s

L t L t

s

n n n

L t

s s s s

  

 

 

1 1

n n

n

n n

L

s s

n n

L

s s s

n n

L t or

s s

 

In particular n 1, 2,3.....

we get

2

2

3

3

4

L t

s

L t

s

L t

s

2.1. Linear property of Laplace Transform

  1. L f t ( ( )  g t ( ))  L f t ( ( )  L g t ( ( ))
  2. L Kf t ( ( ))  KL f t ( ( ))

Proof (1) : By the defn of L.T

Sathyabama University

0

0

0 0

st

st

st st

L f t e f t dt

L f t g t e f t g t dt

e f t dt e g t dt

L f t L g t

 

 

Hence

L f t ( )  g t ( )  L f t ( )  L g t ( )

(2)

L Kf t ( )  KL f t ( )

By the defn of L.T

0

0

st

st

L Kf t e Kf t dt

K e f t dt

KL f t

Hence

L Kf t ( )  KL f t ( )

2.2. Recall

  1. 2sin A cos B  sin( AB )  sin( AB )

  2. 2 cos A sin B  sin( AB )  sin( AB )

  3. 2 cos A cos B  cos( AB )  cos( AB )

2sin A sin B  cos( AB )  cos( AB )

2

1 cos 2

sin

A
A

2

1 cos 2

cos

A
A

3

sin 3 A  3sin A 4 sin A

3

cos3 A  4 cos A 3cos A

  1. sin( AB )  sin A cos B cos A sin B

  2. sin( AB )  sin A cos B cos A sin B

  3. cos( AB )  cos A cos B sin A sin B

cos( AB )  cos A cos B sin A sin B

Sathyabama University

  1. Find

3 4

t t

L e e

Solution :

3 4 3 4

t t t t

L e e L L e L e

s s s

 

  1. Find

6

(3 sin 2 5cos 3 )

t

Lett

Solution :

6 6

2 2

(3 sin 2 5cos 3 ) 3 (1) ( ) (sin 2 ) 5 (cos 3 )

t t

L e t t L L e L t L t

s

s s s s

  1. Find L (sin(2 t 3))

Solution :

2 2

(sin(2 3)) (sin 2 cos 3 sin 3cos 2 )

cos 3 (sin 2 ) sin 3 (cos 2 )

cos 3 sin 3

L t L t t

L t L t

s

s s

  1. Find

5

(sin 4 3sin 2 4cos 5 )

t

L t h t h t e

Solution :

5

5

2 2 2

2 2 2

(sin 4 3sin 2 4 cos 5 )

(sin 4 ) 3 (sin 2 ) 4 (cos 5 ) ( )

t

t

L t h t h t e

L t L h t L h t L e

s

s s s s

s

s s s s

  1. Find

2

L ((1  t ) )

Solution :

2 2

2

2 3

L t L t t

L L t L t

s s s

Sathyabama University

  1. Find the Laplace Transform of

sin 0

t t

f t

t

Solution :

By definition,

0

0

0

0

2 2 2 2

0

sin (0)

sin

( sin cos sin (

st

st st

st st

st

st ax

ax

L f t e f t dt

e f t dt e f t dt

e t dt e dt

e t dt

e e

s t t e bx dx

s a b

     

 

 

a sin bxb cos bx )

0

2 2

2 2

2

( sin co s ) (0 1)

s

s

s

e e

s

s s

e

s s

e

s

  1. Find the Laplace Transform of

t

e t

f t

t

Solution :

By definition,

 

0

1

0 1

1

0 1

1

( 1)

0

1

(1 )

0

1

st

st st

st t st

s t

s t

s

L f t e f t dt

e f t dt e f t dt

e e dt e dt

e dt

e

s

e

s

 

 

 

Sathyabama University

4.3. Problems :

  1. Find

2

t

L te

Solution :

2

2

2 2

2

t

s s

s s

L te L t

s s

 

 

  1. Find

5

t

L t e

Solution :

5 5

1

6

1

6

t

s s

s s

L t e L t

s

s

 

 

  1. Find

2

( sin 3 )

t

L e t

Solution :

2

2

2

2

2

( sin 3 ) (sin 3 )

t

s s

s s

L e t L t

s

s

 

 

  1. Find ( cos 4 )

t

L e h t

Solution :

1

2

1

2

( cos 4 ) (cos 4 )

t

s s

s s

L e h t L h t

s

s

s

s

 

 

  1. Find

3 2

( sin 4 )

t

L e t

Solution :

3 2 2

3 3 3 2 3 2

( sin 4 ) (sin 4 )

1 cos 8

(1) (cos8 )

t

s s

s s

s s

S S

L e t L t

t

L

L L t

s

s s

s

s s

 

 

 

 

Sathyabama University

  1. Find

2

( sin 4 cos 6 )

t

L e t t

Solution :

2

2

2

2

( sin 4 cos 6 ) (sin 4 cos 6 )

(2sin 4 cos 6 )

(sin(4 6 ) (sin 4 6 )

(sin10 sin

t

s s

s s

s s

L e t t L t t

L t t

L t t t t

L t

 

 

 

2

2 2

2

2 2

s s

s s

t

s s

s s

 

 

  1. Find  

4 3 3

(sin 3 cosh 3 )

t

L e tt

Solution:

   

4 3 3 3 3

4

4

3 3

(sin 3 cosh 3 ) sin 3 cosh 3

3sin 3 sin 9 3cosh 3 cosh 9

3sin sin 3 3cosh cosh 3

sin , cosh

(sin 3 ) (sin 9 ) (cosh 3 ) (cosh 9 )

t

s s

s s

L e t t L t t

t t t t

L

L t L t L t t

 

 

4

2 2 2 2

4

2 2 2 2

s s

s s

s s

s s s s

s s

s s s s

 

 

  1. Find L (cosh cos 2 ) t t

Solution :

1 1

2 2

1 1

(cos cos 2 ) cos 2

( cos 2 cos 2 )

(cos 2 ) (cos 2 )

t t

t t

s s s s

s s s s

e e

L ht t L t

L e t e t

L t L t

s s

s s

   

   

2 2

s s

s s

Sathyabama University

3 3

2

2

t t

d

L te L e

ds

d

ds s

s

s

s

  1. Find L t ( sin 3 ) t

Solution :

2

2

2 2

2 2

( ( )) (sin 3 )

d

L tf t L f t

ds

d

L tf t L t

ds

d

ds s

s s

s

s

s

  1. Find

2

L t ( cos 3 ) t

Solution :

2 2

2

2

2 2 2

( cos 3 ) (cos 3 )

1 cos 6

(1) (cos 6 )

d

L t t L t

ds

d t

L

ds

d

L L t

ds

d s

ds s s

s s s

s s

2

2 2 2

2

2 2 2

s

s s

s

s s

Sathyabama University

  1. Find

2

( sin 3 )

t

L te t

Solution :

2 2 2 2 2 2

2 2

2

( ( sin 3 )) ( sin 3 )

( (sin 3 )

t

s s

s s

s s

s s

L e t t L t t

d

L t

ds

d

ds s

s s

s

 

 

 

 

2 2

s

s

  1. Find

2

( sin 2 sin 3 )

t

L te t t

Solution :

2

2

2

2

( sin 2 sin 3 )

( sin 2 sin 3 )

( 2sin 2 sin 3 )

(cos(2 3 ) cos(2 3 ))

t

s s

s s

s s

L te t t

L t t t

L t t t

L t t t t t

L

 

 

 

2

2

2 2

2

2

2 2

cos cos 5

(cos ) (cos 5 )

s s

s s

s s

t t t t

d d

L t L t

ds ds

d s d s

ds s ds s

s s s

s

 

 

 

2

2 2

2

2 2

2 2 2 2

2

2 2

2 2 2 2

2

s s

s s

s s

d s s s

ds s

s s

s s

s s

s s

 

 

 

   

2 2

2 2

2 2

s s

s s

Sathyabama University

0

st

s

e f t ds dt

 

(Changing the order of integration since ‘s’ and

‘t’ are independent variable)

0

0

0

0

st

s

st

s

st

st

s

f t e ds dt

e

f t dt

t

f t dt e

t

f t

e dt

t

f t

L

t

f t

L L f t ds

t

 

      

Similarly we can prove that

2

s s

f t

L L f t ds ds

t

 

In general

n

s s s n

n

f t

L L f t ds ds ds

t

  

times

times

Recall :

  1. log( AB )  log A log B

log log log

A
A B
B

log log

B

A  B A

log1  0

  1. log 0  

  2. log   

dx log x

x

1

2 2

tan

dx x

a x a a

1

tan ( )

1 1

cot tan

s s

a a

 

Sathyabama University

Problems :

  1. Find

2

t

e

L

t

Solution :

2

0

t

t

e

Lim

t

(Indeterminate form)

Apply L - Hospital Rule

2

0

t

t

e

Lim

the given function exists in the limit t  0

2

2

2

log log( 2)

t t s t s s s

e

L L e ds

t

L L e ds

ds

s s

s s

1

log

log log

0 log

log

log

s

s

S

s

s

s

s

s s s s s s s s

  1. Find

1 cos at

L

t

Solution :

0

1 cos 0

t

at

Lim

t

(Indeterminate form)

Apply L - Hospital Rule.

0

sin

t

a at

Lim

(finite)