SCHOOL OF SCIENCE AND HUMANITIES
DEPARTMENT OF MATHEMATICS
UNIT – I – LAPLACE TRANSFORM – SMT1401
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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.
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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 t e
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 t e
is not of exponential order..
2. Laplace Transform of Standard functions :
(1) Prove that
[ ]
at
L e
s a
where s a 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
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- 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
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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
- L f t ( ( ) g t ( )) L f t ( ( ) L g t ( ( ))
- L Kf t ( ( )) KL f t ( ( ))
Proof (1) : By the defn of L.T
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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
2sin A cos B sin( A B ) sin( A B )
2 cos A sin B sin( A B ) sin( A B )
2 cos A cos B cos( A B ) cos( A B )
2sin A sin B cos( A B ) cos( A B )
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
sin( A B ) sin A cos B cos A sin B
sin( A B ) sin A cos B cos A sin B
cos( A B ) cos A cos B sin A sin B
cos( A B ) cos A cos B sin A sin B
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- 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
- Find
6
(3 sin 2 5cos 3 )
t
L e t t
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
- 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
- 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
- Find
2
L ((1 t ) )
Solution :
2 2
2
2 3
L t L t t
L L t L t
s s s
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- 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 bx b 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
- 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
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4.3. Problems :
- Find
2
t
L te
Solution :
2
2
2 2
2
t
s s
s s
L te L t
s s
- 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
- 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
- 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
- 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
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- 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
- Find
4 3 3
(sin 3 cosh 3 )
t
L e t t
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
- 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
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3 3
2
2
t t
d
L te L e
ds
d
ds s
s
s
s
- 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
- 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
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- 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
- 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
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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 :
log( AB ) log A log B
log log log
A
A B
B
log log
B
A B A
log1 0
log 0
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
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Problems :
- 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
- 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)