Introduction:

A transformation is a mathematical device, which converts one function into

another.
For example, when a differential operator
|

\
|

dx
d
D operates on ( ) x sin x f = , it gives a
new function ( ) ( ) [ ] x cos x f D x g = = .
Laplace transform or Laplace transformation is widely used by scientists and
engineers. It is particularly effective in solving linear differential equations- ordinary as
well as partial. It reduces an ordinary differential equation into an algebraic equation.
Laplace transform directly gives the solution of differential equations with given
initial conditions without the necessity of first finding the general solution and then
evaluating the arbitrary constants.
LAPLACE TRANSFORM:
Definition: Let f(t) be a function of t defined for all positive values of t.
Then, the Laplace transforms of f(t), denoted by L{f(t)} is defined by
1 11 1
st st st st
Topic Topic Topic Topic
Laplace Transforms Laplace Transforms Laplace Transforms Laplace Transforms
Definitions, Properties, Transforms of
elementary functions
Linearity property
Shifting property
Change of scale property
Prepared by:
Dr. Sunil
NIT Hamirpur (HP)
(Last updated on 25-10-2007)
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
2
( ) { } ( )dt t f e t f L
st
0

= .
The parameter s is a real or complex number. In general, the parameter s is taken to be
real positive number.
Now, since L{f(t)} is a function of s , then it can be briefly written as ) s ( f .
i.e. ( ) { } ( ) s f t f L = .
Sometimes, we use symbol p for the parameter s.
Thus, ( ) { } ( ) ( ) p f dt t f e t f L
pt
0
= =

.
Existence of Laplace transforms:
The Laplace transforms is said to be exist, if the integral ( ) { } ( )dt t f e t f L
st
0

= is
convergent for some values of s.
Otherwise, we may use the following theorem:
Sufficient conditions for the existence of Laplace Transform of f(t):
If f(t) is continuous and ( ) { } t f e Lt
at
t

is finite, or in other words,
a. If f(t) is piecewise (or sectionally) continuous i.e. f(t) is continuous in every
subinterval and has finite limit at end points of each of these sub-intervals and
b. If f(t) is of exponential order of i.e., there exists M, such that ( )
t
Me t f

< . In
other words, functions of exponential order do not grow faster than
t
e

,
then the Laplace transform of f(t), i.e. ( )dt t f e
st
0

exists for a s > .

Remarks: It should be noted that, the above conditions are sufficient rather than
necessary. For example, |

\
|
t
1
L exists, though
t
1
is infinite at t = 0.
Similarly, a function f(t) for which ( ) { } t f e Lt
at
t

is finite and having a finite discontinuity
will have a Laplace transform for a s > .
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
3
INVERSE LAPLACE TRANSFORM:
( ) { } ( ) s f t f L = can also be written as ( ) ( ) { } s f L t f
1
= .
Then f(t) is called the inverse Laplace transform of ) s ( f .

Laplace transformation operator:
The symbol L, which transforms f(t) into ) s ( f , is called the Laplace
transformation operator. The operation of multiplying ( ) t f by
st
e

and integrating
from 0 to is called Laplace transformation.

Notation: In various textbooks, authors follow two types of notations:
(i). Functions are denoted by lower case letter ( ) ( ) ( ),......... t h , t g , t f .
and their Laplace transforms are denoted by
( ) ( ) ( ),......... s h , s g , s f respectively or by ( ) ( ) ( ),......... p h , p g , p f .
(ii). Functions are denoted by capital letters ( ) ( ) ( ),......... t H , t G , t F
and their Laplace transforms are denoted by corresponding lower case letters
( ) ( ) ( ),......... s h , s g , s f respectively or by ( ) ( ) ( ),......... p h , p g , p f .

Applications of Laplace Transform:
Laplace transform is very useful in obtaining solution of linear differential
equations, both ordinary and partial, solution of system of simultaneous differential
equations, solution of integral equations, solution of linear difference equations and in the
evaluation of definite integrals.
Advantages of Laplace Transform:
1. With the application of Laplace transform, particular solution of differential
equation is obtained directly without the necessity of first determining general
solution and then obtaining the particular solution (by substitution of initial
conditions).
2. Laplace transform solves non-homogeneous differential equation without the
necessity of first solving the corresponding homogeneous differential equation.
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
4
3. Laplace transform is applicable not only to continuous function but also the piecewise
continuous functions, complicated periodic functions, step function and impulse
functions.
4. Laplace transform of various functions are readily available (in tabulated form).

General properties of Laplace transform:
Although theoretically ) s ( f , the Laplace transform of f(t) is obtained from the
definition, but in practice, most of the time Laplace transforms are obtained by the
judicial application of some of the following important properties. In a nutshell, they are:
1. Linearity property states that Laplace transform of a linear combination (sum) is the
linear combination (sum) of Laplace transforms.
2. First shift theorem proves that multiplication of f(t) by
at
e amounts to replacement s
by a s in ) s ( f .
3. In change of scale, where the argument t of f is multiplied by constant a, s is replaced
by
a
s
in ) s ( f and then multiplied by
a
1
.
4. Laplace transform of a derivative f amounts to multiplication of ) s ( f by s
(approximately but for the constant ) 0 ( f .
5. Laplace transform of an integral f amounts to division of ) s ( f by s.
6. Multiplication of f(t) by t
n
amounts to differentiation of ) s ( f n times w.r.t s (with
( )
n
1 as sign).
7. Division of f(t) by t amounts to integration of ) s ( f between the limits s to .
8. Second shift theorem proves that Laplace transform of shifted function
( ) ( ) a t u a t f is obtained by multiplying ) s ( f by
at
e

.

LAPLACE TRANSFORMS OF SOME ELEMENTARY FUNCTIONS:
The direct application of the definition gives the following formulae:
(1) ( )
s
1
1 L = [ ] 0 s >
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
5
(2) ( )
1 n
n
s
! n
t L
+
= , when n =0, 1, 2, 3, .........
( )
(

+
+1 n
s
1 n
Otherwise
(3) ( )
a s
1
e L
at

= [ ] a s >
(4) ( )
2 2
a s
a
at sin L
+
= [ ] 0 s >
(5) ( )
2 2
a s
s
at cos L
+
= [ ] 0 s >
(6) ( )
2 2
a s
a
at sinh L

= [ ] a s >
(7) ( )
2 2
a s
s
at cosh L

= [ ] a s >
Proofs:
(1) To show: ( )
s
1
1 L = , [ ] 0 s >
.

Proof: ( )
s
1
s
e
dt 1 . e 1 L
0
st
st
0
= = =

, if 0 s > .
(2) To show: ( )
1 n
n
s
! n
t L
+
= , when n =0, 1, 2, 3, ......... .
Proof: ( )
s
dp
s
p
. e dt t . e t L
n
p
0
n st
0
n
|

\
|
= =

, (on putting st = p, dt = dp/s)

( )
1 n
n p
0
1 n
s
1 n
dp p . e
s
1
+

+
+
= =

, if 0 s and 1 n > > .

In particular, ( )
s
s
2
1
t L
2 / 1
2 / 1

=
|

\
|

; ( )
2 / 3 2 / 3
2 / 1
s . 2 s
2
3
t L

=
|

\
|

= .
If n be a positive integer, ( ) ! n 1 n = + .
Therefore, ( )
1 n
n
s
! n
t L
+
= .
(3) To show: ( )
a s
1
e L
at

= , [ ] a s > .
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
6
Proof: ( )
( )
dt e dt e . e e L
t a s
0
at st
0
at

= =
( )
( ) a s
1
a s
e
0
t a s

=

=

, if 0 s > .
(4) To show: ( )
2 2
a s
a
at sin L
+
= , [ ] 0 s >
.

Proof: ( ) atdt sin e at sin L
st
0

= ( )
2 2
0
2 2
st
a s
a
at cos a at sin s
a s
e
+
=
+
=

.
(5) To show: ( )
2 2
a s
s
at cos L
+
= , [ ] 0 s >
.

Proof: ( ) atdt cos e at cos L
st
0

= ( )
2 2
0
2 2
st
a s
s
at sin a at cos s
a s
e
+
=
+
=

.
(6) To show: ( )
2 2
a s
a
at sinh L

= , [ ] a s > .
Proof: ( ) dt . at sinh e at sinh L
st
0

= dt
2
e e
e
at at
st
0
|
|

\
|

=

( ) ( )
(
(

=
+

dt e dt e
2
1
t a s
0
t a s
0


2 2
a s
a
a s
1
a s
1
2
1

=
(

= , for a s > .
(7) To show: ( )
2 2
a s
s
at cosh L

= , [ ] a s > .
Proof: ( ) dt . at cosh e at cosh L
st
0

= dt
2
e e
e
at at
st
0
|
|

\
|
+
=

( ) ( )
(
(

+ =
+

dt e dt e
2
1
t a s
0
t a s
0


2 2
a s
s
a s
1
a s
1
2
1

=
(

+
+

= , for a s > .


Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
7
PROPERTIES OF LAPLACE TRANSFORMATIONS:
(1). Linearity property: If a, b, c be any constants and f, g, h any functions of t, then
( ) ( ) ( ) [ ] ( ) { } ( ) { } ( ) { } t h L c t g L b t f L a t h c t g b t f a L + = + .
or
Linearity property states that Laplace transform of a linear combination (sum) is the
linear combination (sum) of Laplace transforms.
Proof: L.H.S. = ( ) ( ) ( ) [ ] t h c t g b t f a L +
( ) ( ) ( ) { }dt t ch t bg t af e
st
0
+ =

(by definitions)
( ) ( ) ( )dt t h e c dt t g e b dt t f e a
st
0
st
0
st
0

+ =
( ) { } ( ) { } ( ) { } t h cL t g bL t f aL + = .
Thus, linear transform is a linear operator, additive, subtractive and homogeneous.
This result can easily be generalized to more than three functions.
Because of the above property of L, it is called a linear operator.
(2). First shifting or first translation property or s-shift theorem:
(Replacement of s by s-a in transform)
If ( ) { } ( ) s f t f L = , then ( ) { } ( ) a s f t f e L
at
= .
or
First shift theorem proves that multiplication of f(t) by
at
e amounts to replacement s
by a s in ) s ( f .
Proof: Given ( ) { } ( ) s f t f L = .
To show: ( ) { } ( ) a s f t f e L
at
= .
Since ( ) { } ( ) ( )
( )
( )dt t f e dt t f e e t f e L
t a s
0
at st
0
at

= = (by definition)
( )dt t f e
t r
0

= , where a s r = .
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
8
( ) ( ) a s f r f = = .
Thus, if we know the transform ) s ( f of f(t), then we can write the transform of ( ) t f e
at

simply replacing s by a s to get ( ) a s f .
Application of this property leads to the following useful results:
(1) ( )
a s
1
e L
at

= ( )
(

=
s
1
1 L Q
(2) ( )
( )
1 n
n at
a s
! n
t e L
+

= ( )
(

=
+1 n
n
s
! n
t L Q
(3) ( )
( )
2 2
at
b a s
b
bt sin e L
+
= ( )
(

+
=
2 2
b s
b
bt sin L Q
(4) ( )
( )
2 2
at
b a s
a s
bt cos e L
+

= ( )
(

+
=
2 2
b s
s
bt cos L Q
(5) ( )
( )
2 2
at
b a s
b
bt sinh e L

= ( )
(

=
2 2
b s
b
bt sinh L Q
(6) ( )
( )
2 2
at
b a s
a s
bt cosh e L

= ( )
(

=
2 2
b s
s
bt cosh L Q
where in each case a s > .
3. Change of scale property: If ( ) { } ( ) s f t f L = , then ( ) { } |

\
|
=
a
s
f
a
1
at f L .
or
In change of scale, where the argument t of f is multiplied by constant a, s is replaced
by
a
s
in ) s ( f and then multiplied by
a
1
.
Proof: ( ) { } ( ) ( )
a
du
. u f e dt at f e at f L
a / su
0
st
o

= =
(

= =
a
du
dt u at Put
( ) |

\
|
= =

a
s
f
a
1
du u f e
a
1
a / su
0
.

Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
9
Now let us find Laplace transform of simple functions where only linearity
property is using:

Q.No.1.: Find the Laplace transforms of
(i) t 3 sin t 2 sin . (ii) t 2 cos
2
. (iii) t 2 sin
3
.
Sol. (i). Since ( ) t 5 cos t cos
2
1
t 3 sin t 2 sin = .
( ) ( ) ( ) [ ] t 5 cos L t cos L
2
1
t 3 sin t 2 sin L = (by linearity property)

(

+
=
2 2 2 2
5 s
s
1 s
s
2
1

( )( ) 25 s 1 s
s 12
2 2
+ +
= . Ans.
(ii) Since ( ) t 4 cos 1
2
1
t 2 cos
2
+ = .
( ) ( ) ( ) { } t 4 cos L 1 L
2
1
t 2 cos L
2
+ = (by linearity property)

|
|

\
|
+
+ =
16 s
s
s
1
2
1
2
. Ans.
(iii) Since t 6 sin
4
1
t 2 sin
4
3
t 2 sin t 2 sin 4 t 2 sin 3 t 6 sin
3 3
= = .
( ) ( ) ( ) t 6 sin L
4
1
t 2 sin L
4
3
t 6 sin
4
1
t 2 sin
4
3
L t 2 sin L
3
= |

\
|
= (by linearity property)

( )( ) 36 s 4 s
48
6 s
6
.
4
1
2 s
2
.
4
3
2 2 2 2 2 2
+ +
=
+

+
= . Ans.
Q.No.2.: Find the Laplace transform of t 3 cos 3 t 3 sin 2 t 4 e
3 t 2
+ + .
Sol.: ( ) t 3 cos 3 t 3 sin 2 t 4 e L
3 t 2
+ + ( ) ( ) ( ) ( ) t 3 cos L 3 t 3 sin L 2 t L 4 e L
3 t 2
+ + =
(by linearity property)

9 s
s 3
9 s
3 . 2
s
! 3 . 4
2 s
1
2 2 4
+
+
+
+

=

9 s
s 3
9 s
6
s
24
2 s
1
2 2 4
+
+
+
+

=
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
10

( )
9 s
2 s 3
s
24
2 s
1
2 4
+

+ +

= . Ans.
Q.No.3.: Find the Laplace transform of
t
3
t 2 1 + + .
Sol.: |

\
|
+ +
t
3
t 2 1 L ( ) ( ) |

\
|
+ + =
t
1
L 3 t L 2 1 L (by linearity property)

2 / 1 2 / 3
s
1
2
1
3
s
1
2
1
2
s
1
|

\
|
+
+
|

\
|
+
+ =
2 / 1 2 / 3
s
3
s
2
1
. 2
s
1
+

+ =
|

\
|
+

+ =
s
3
s
s
1
2 / 3
. Ans.
Q.No.4.: Find the Laplace transform of at cos at cosh .
Sol.: ( ) at cos at cosh L ( ) ( ) at cos L at cosh L = (by linearity property)

4 4
2
2 2 2 2
a s
s a 2
a s
s
a s
s

=
+

= . Ans.
Q.No.5.: Find the Laplace transform of ( ) b at cos + .
Sol.: ( ) [ ] b at cos L + ( ) b sin at sin b cos at cos L = (by linearity property)
( ) ( ) at sin bL sin at cos bL cos =
2 2 2 2
a s
a
b sin
a s
s
b cos
+

+
=

2 2
a s
b sin a b cos s
+

= . Ans.
Q.No.6.: Find the Laplace transform of ( )
2
t cos t sin .
Sol.: Since ( )
2
t cos t sin t cos t sin 2 t cos t sin
2 2
+ = t 2 sin 1 t cos t sin 2 1 = = .
( ) ( ) ( ) ( ) t 2 sin L 1 L t 2 sin 1 L = (by linearity property)

( ) 4 s s
s 2 4 s
4 s
2
s
1
2
2
2
+
+
=
+
=
( ) 4 s s
4 s 2 s
2
2
+
+
= . Ans.
Q.No.7.: Find the Laplace transform of t 3 cos t 2 sin .
Sol.: Since t 3 cos t 2 sin ( ) t sin t 5 sin
2
1
= .
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
11
( ) ( ) [ ] t sin L t 5 sin L
2
1
2
t sin t 5 sin
L = |

\
|
(by linearity property)

(

+
=
1 s
1
25 s
5
2
1
2 2


( )( )
( )
( )( ) 25 s 1 s
5 s 2
25 s 1 s
20 s 4
.
2
1
2 2
2
2 2
2
+ +

=
+ +

= . Ans.
Q.No.8.: Find the Laplace transform of bt sin at sin .
Sol.: ( ) ( ) ( ) [ ] t b a cos t b a cos L
2
1
bt sin at sin L + =
( ) ( ) ( ) ( ) [ ] t b a cos L t b a cos L
2
1
+ = (by linearity property)

( ) ( ) (
(

+ +

+
=
2 2 2 2
b a s
s
b a s
s
2
1
( ) [ ] ( ) [ ]
2 2 2 2
b a s b a s
abs 2
+ + +
= . Ans.
Q.No.9.: Find the Laplace transform of t 3 sin
2
.
Sol.: Since ( ) t 6 cos 1
2
1
t 3 sin
2
= .
( ) ( ) ( ) ( ) ( ) t 6 cos L 1 L
2
1
t 6 cos 1 L
2
1
t 3 sin L
2
= = (by linearity property)

( ) 36 s s
18
36 s
s
s
1
2
1
2 2
+
=
(

+
= . Ans.
Q.No.10.: Find the Laplace transform of t 2 cos
3
.
Sol.: Since t 2 cos
3
[ ] t 2 cos 3 t 6 cos
4
1
+ = .
( ) t 2 cos L
3
[ ] t 2 cos 3 t 6 cos L
4
1
+ = ( ) ( ) [ ] t 2 cos 3 L t 6 cos L
4
1
+ = (by linearity property)

(

+
+
+
=
4 s
s 3
36 s
s
4
1
2 2
( )( )
(
(

+ +
+ + +
=
4 s 36 s
s 108 s 3 s 4 s
4
1
2 2
3 3


( )( )
(
(

+ +
+
=
4 s 36 s
s 112 s 4
4
1
2 2
3
( )
( )( ) 4 s 36 s
28 s s
2 2
2
+ +
+
= . Ans.
Q.No.11.: Find the Laplace transform of t 2 sinh t .
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
12
Sol.: ( ) ( ) ( ) t 2 sinh L t L t 2 sinh t L = (by linearity property)

( )
2 2
2
2 2
s 4 s
s 4
4 s
2
s
1

+
=

= . Ans.
Q.No.12.: Find the Laplace transforms of t 2 cosh
3
.
Sol.: Since t 2 cosh 3 t 2 cosh 4 t 6 cosh
3
=
t 6 cosh
4
1
t 2 cosh
4
3
t 2 cosh
3
+ = .
( ) ( ) ( ) t 6 cosh L
4
1
t 2 cosh L
4
3
t 6 cosh
4
1
t 2 cosh
4
3
L t 2 cosh L
3
+ = |

\
|
+ =
(by linearity property)

( )
( )( ) 36 s 4 s
28 s s
6 s
s
.
4
1
2 s
s
.
4
3
2 2
2
2 2 2 2

= .
Q.No.13.: Find the Laplace transforms of
bt at
e e .
Sol.: { } { } ( )
bt at bt at
e L e L e e L = (by linearity property)

( )( ) b s a s
b a
b s
1
a s
1

= . Ans.
Q.No.14.: Find the Laplace transforms of kt cos
2
.
Sol.: { }
)
`

+
=
2
kt 2 cos 1
L kt cos L
2
{ } { } kt 2 cos L
2
1
1 L
2
1
+ = (by linearity property)

2 2
k 4 s
s
2
1
s
1
2
1
+
+ = . Ans.
Q.No.15.: Find the Laplace transforms of { }
2
t 2
3 e 5 .
Sol.: ( ) { } 9 30 e 25 L 3 e 5 L
t 2 t 4
2
t 2
+ =
)
`

{ } { } { } 1 L 9 e L 30 e L 25
t 2 t 4
+ = (by linearity property)

s
1
. 9
2 s
1
30
4 s
1
. 25 +

= . Ans.
Q.No.16.: Find the Laplace transforms of t 2 cos 3 t 5 sin 2 e 4 t 2 t 3
t 3 3 4
+ +

.
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
13
Sol.: { } t 2 cos 3 t 5 sin 2 e 4 t 2 t 3 L
t 3 3 4
+ +


{ } { } { } { } { } t 2 cos L 3 t 5 sin L 2 e L 4 t L 2 t L 3
t 3 3 4
+ + =

(by linearity property)

2 2 2 2 4 5
2 s
s
. 3
5 s
5
. 2
3 s
1
4
t
! 3
. 2
t
! 4
. 3
+
+
+

+
+ = .
Q.No.17.: Find the Laplace transforms of t cos .
Sol.: Expanding in series
( )
( )! n 2
t 1
t cos
n 2
2
1
n
0 n
|
|
|

\
|

=
( )
( )
n
n
t
! n 2
1
=

.
{ }
( )
( )
(
(

=

n
n
t
! n 2
1
L t cos L
( )
( )
(
(

=

n
n
t
! n 2
1
L (by linearity property)

( )
( )
( )
n
n
0 n
t L
! n 2
1
=

=
( )
( )
1 n
n
0 n
s
! n
! n 2
1
+

=

.
Q.No.18.: Find the Laplace transform of
3
t
1
t |

\
|
.
Sol.: Since ( ) ( ) ( ) ( )
2 / 3 2 / 1 2 / 1 2 / 3
3
t t 3 t 3 t
t
1
t

+ =
|
|

\
|

( ) ( ) ( ) ( )
2 / 3 2 / 1 2 / 1 2 / 3
3
t L t L 3 t L 3 t L
t
1
t L

+ =
|
|

\
|
(by linearity property)

1 2 / 3 1 2 / 1 1 2 / 1 1 2 / 3
s
1
2
3
s
1
2
1
3
s
1
2
1
3
s
1
2
3
+ + + +
|

\
|
+

\
|
+
+
|

\
|
+

\
|
+
=

2 / 1 2 / 1 2 / 3 2 / 5
s
2
1
s
2
1
3
s
2
1
2
1
3
s
2
3
2
3

\
|

\
|

+
|

\
|

\
|

=
s 2
s
3
s
2
3
s
4
3
2 / 1 2 / 3 2 / 5
+

=
(

= |

\
|
= |

\
|
2
2
1
,
2
1
Q
|

\
|
+ +

=
2 / 1 2 / 1 2 / 3 2 / 5
s
8
s
12
s
6
s
3
4
. Ans.
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
14
Now let us find Laplace transform of some functions where shifting
property is also using:
Q.No.19.: Show that (i) ( )
( )
2
2 2
a s
as 2
at sin t L
+
= , (ii) ( )
( )
2
2 2
2 2
a s
a s
at cos t L
+

= .
Sol.: Since ( )
2
s
1
t L = .
( )
( )
( )
( )( ) [ ]
2
2
2
t a i
ia s ia s
ia s
ia s
1
t e L
+
+
=

= . (by shifting property)

( ) [ ]
( ) ( )
( )
2
2 2
2 2
a s
as 2 i a s
at sin i at cos t L
+
+
= + .
Equating the real and imaginary parts from both sides, we get
( )
( )
2
2 2
a s
as 2
at sin t L
+
= and ( )
( )
2
2 2
2 2
a s
a s
at cos t L
+

= .
Q.No.20.: Find the Laplace transform of bt sinh e
t a
.
Sol.: ( )
( )
2 2
t a
b a s
b
bt sinh e L
+
=

. Ans. (by shifting property)

Q.No.21.: Find the Laplace transform of
t 3 3
e . t

.
Sol.: ( )
( )
( ) ( ) ( )
4 4 4
t 3 3
3 s
6
3 s
! 3
3 s
4
e . t L
+
=
+
=
+

. Ans. (by shifting property)

Q.No.22.: Find the Laplace transform of t 4 sin e
t 2
.
Sol.: ( )
( ) 20 s 4 s
4
16 2 s
4
t 4 sin e L
2 2
t 2
+ +
=
+ +
=

. Ans. (by shifting property)

Q.No.23.: Find the Laplace transforms of at sin t
2
.
Sol.: Since ( )
3 3
2
s
2
s
! 2
t L = = .
{ }
( )
( )
( )( ) [ ]
3
3
3
iat 2
ia s ia s
ia s 2
ia s
2
e t L
+
+
=

= (by shifting property)

Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
15
( ) { }
( ) ( ) [ ]
( )
3
2 2
3 2 2 3
2
a s
a as 3 i s a 3 s 2
at sin i at cos t L
+
+
= +
Equating the imaginary parts on both sides, we get
{ }
( )
( )
3
2 2
2 2
2
a s
a s 3 a 2
at sin t L
+

= . Ans.
Q.No.24.: Find the Laplace transforms of t 3 sin te
t 4
.
Sol.: Since { }
2
t
1
t L = .
{ }
( )
( )
( )( ) [ ]
2
2
2
it 3
i 3 s i 3 s
i 3 s
i 3 s
1
te L
+
+
=

= (by shifting property)

( ) { }
( )
( )
2
2
2
9 s
is 6 9 s
t 3 sin i t 3 cos t L
+
+
= + .
Equating the imaginary parts on both sides, we get
[ ]
( )
2
2
9 s
s 6
t 3 sin t L
+
= .
Again applying the first shifting theorem, we have
{ }
( )
( ) [ ]
( )
( )
2
2
2
2
t 4
25 s 8 s
4 s 6
9 4 s
4 s 6
t 3 sin t . e L
+ +
+
=
+ +
+
=

. Ans.
Q.No.25.: Find the Laplace transforms of ( )
t 3
2
7
e t t f = .
Sol.:
2
9
2
9
1
2
7
2
7
s 16
105
s
2
1
2
1
.
2
3
.
2
5
.
2
7
s
1
2
7
t L

=
|

\
|

=
|

\
|
+
=

+
.
3 - s s at
2
9
2
7
t 3
s 16
105
t . e L
=

(by shifting property)

( )
2
9
3 s 16
105

= .
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
16
Q.No.26.: Find the Laplace transforms of
(i) ( ) t 5 sin 3 t 5 cos 2 e
t 3

. (ii) t sin e
2 t 3
. (iii) t cos t 2 sin e
t 4
.
Sol.: (i) ( ) { } ( ) ( ) t 5 sin e L 3 t 5 cos e L 2 t 5 sin 3 t 5 cos 2 e L
t 3 t 3 t 3
= (by linearity property)

( ) ( )
2 2 2 2
5 3 s
5
. 3
5 3 s
3 s
. 2
+ +

+ +
+
= (by shifting property)

34 s 6 s
9 s 2
2
+ +

= . Ans.
(ii) Since ( ) ( ) ( ) ( ) ( ) t 2 cos L 1 L
2
1
t 2 cos 1 L
2
1
t sin L
2
= = (by linearity property)

)
`

+
=
4 s
s
s
1
2
1
2
.
( )
( )
)

=
4 3 s
3 s
3 s
1
2
1
t sin e L
2
2 t 3
. Ans. (by shifting property)
(iii) Since ( ) ( ) ( ) ( ) ( ) t sin L t 3 sin L
2
1
t sin t 3 sin L
2
1
t cos t 2 sin L + = + = (by linearity property)

)
`

+
+
+
=
2 2 2 2
1 s
1
3 s
3
2
1
.
( )
( ) ( )
)

+
+
+
=
1 4 s
1
9 4 s
3
2
1
t cos t 2 sin e L
2 2
t 4
. Ans. (by shifting property)
Q. No.27.: If ( ) { } ( ) s f t f L = , show that
( ) ( ) [ ] ( ) ( ) [ ] a s f a s f
2
1
t f at sinh L + =
( ) ( ) [ ] ( ) ( ) [ ] a s f a s f
2
1
t f at cosh L + + =
Hence evaluate (i) ( ) t 3 sin t 2 sinh L (ii) ( ) t 2 cos t 3 cosh L .
Sol. We have ( ) ( ) { } ( ) ( )
(

=

t f e e
2
1
L t f at sinh L
at at

( ) { } ( ) { } [ ] t f e L t f e L
2
1
at at
= (by linearity property)
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
17
( ) ( ) [ ] a s f a s f
2
1
+ = . (by shifting property)
Similarly, ( ) ( ) { } ( ) ( )
(

+ =

t f e e
2
1
L t f at cosh L
at at

( ) { } ( ) { } [ ] t f e L t f e L
2
1
at at
+ = (by linearity property)
( ) ( ) [ ] a s f a s f
2
1
+ + = . (by shifting property)
(i) Since ( )
2 2
3 s
3
t 3 sin L
+
= , the first result gives
( )
( ) ( ) 169 s 10 s
s 12
3 2 s
3
3 2 s
3
2
1
t 3 sin t 2 sinh L
2 4 2 2 2 2
+ +
=

+ +

+
= . Ans.
(ii) Since ( )
2 2
2 s
s
t 2 cos L
+
= , the second result gives
( )
( ) ( )
( )
169 s 10 s
5 s s 2
2 3 s
3 s
2 3 s
3 s
2
1
t 2 cos t 3 cosh L
2 4
2
2 2 2 2
+

+ +
+
+
+

= . Ans.
Q.No.28.: Find the Laplace transform of ( )
t 2
e . 2 t + .
Sol.: ( ) [ ] ( ) ( ) ( ) 4 e L t 4 e L t e L e 4 t 4 t L
t t 2 t t 2
+ + = + + (by linearity property)

( ) ( )
( ) 1 s
4
1 s
4
1 s
! 2
2 3

= (by shifting property)

( ) ( )
( ) 1 s
4
1 s
4
1 s
2
2 3

=

( ) ( )
( )
3
2
1 s
1 s 4 1 s 4 2

+ +
=
( )
( )
3
2
1 s
1 s 2 s 2 2

+
= . Ans.
Q.No.29.: Find the Laplace transform of t sin e
2 t
.
Sol.: ( ) t sin e L
2 t
( ) { } t 2 cos 1 e L
2
1
t
=


( ) ( ) ( ) t 2 cos e L 1 e L
2
1
t t
= (by linearity property)
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
18

( )
( ) (
(

+ +
+

+
=
4 1 s
1 s
1 s
1
2
1
2
(by shifting property)

( )( ) 5 s 2 s 1 s
2
2
+ + +
= . Ans.
Q.No.30.: Find the Laplace transform of at sin at cosh .
Sol.: ( ) ( ) [ ] at sin e e L
2
1
at sin at cosh L
at at
+ =
( ) ( ) ( ) at sin e L at sin e L
2
1
at at
+ = (by linearity property)

( ) ( ) (
(

+ +
+
+
=
2 2 2 2
a a s
a
a a s
a
2
1
(by shifting property)

( )
4 4
2 2
a 4 s
a 2 s a
+
+
= . Ans.
Q.No.31.: Find the Laplace transform of t cos t 3 sinh
2
.
Sol.: Since t cos t 3 sinh
2
(

+
(
(

2
t 2 cos 1
2
e e
t 3 t 3
.
( ) t cos t 3 sinh L
2
[ ] t 2 cos e e t 2 cos e e L
4
1
t 3 t 3 t 3 t 3
+ =
( ) ( ) ( ) ( ) [ ] t 2 cos e L e L t 2 cos e L e L
4
1
t 3 t 3 t 3 t 3
+ =
(by linearity property)

( ) ( ) (
(

+ +
+

+
+

=
4 3 s
3 s
4 3 s
3 s
3 s
1
3 s
1
4
1
2 2
(by shifting property)

( )
(
(

=
169 s 10 s
1 s 6
9 s
6
4
1
2 4
2
2
(
(

=
169 s 10 s
1 s
9 s
1
2
3
2 4
2
2
. Ans.
Q.No.32.: Find the Laplace transforms of ( ) t 4 sin 3 t 4 cos e
t 3
+

.
Sol.: ( ) { } { } { } t 4 sin e L 3 t 4 cos e L t 4 sin 3 t 4 cos e L
t 3 t 3 t 3
+ = + (by linearity property)

( ) ( )
2 2 2 2
4 3 s
4
. 3
4 3 s
3 s
+ +
+
+ +
+
= (by shifting property)
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
19

25 s 6 s
15 s
2
+ +
+
= .
Q.No.33.: Find the Laplace transforms of { }
t 2 t 5 4 5
e t 4 cos 4 t 6 sin 3 e 4 t 2 t 3 + +

.
Sol.: { } t 4 cos 4 t 6 sin 3 e 4 t 2 t 3 L
t 5 4 5
+ +


{ } ( ) { } ( ) { } t 4 cos L 4 t 6 sin L 3 e L 4 t L 2 t L 3
t 5 4 5
+ + =

(by linearity property)

16 s
s
4
36 s
6
3
5 s
1
4
s
! 4
2
s
! 5
3
2 2 5 6
+
+
+

+
+ = .
{ }
t 2 t 5 4 5
e t 4 cos 4 t 6 sin 3 e 4 t 2 t 3 L + +



2 - s by replaced s with
2 2 5 6
16 s
s 4
36 s
18
5 s
4
s
48
s
360
+
+
+

+
+ = . (by shifting property)

( ) ( ) ( )
( )
( ) 16 2 s
2 s 4
36 2 s
18
3 s
4
2 s
48
2 s
360
2 2 5 6
+

+
+

+
+

= .
Q.No.34.: Find the Laplace transforms of ( ) bt cos . at cosh t f = .
Sol.: ( ) { } { } ( )
)
`

+ = =

bt cos e e
2
1
L bt cos . at cosh L t f L
at at

{ } { } bt cos e L
2
1
bt cos e L
2
1
at at
+ = (by linearity property)

a s s
2 2
a s s
2 2
b s
s
2
1
b s
s
2
1
+ = =
+
+
+
= (by shifting property)

( ) ( )
2 2 2 2
b a s
a s
2
1
b a s
a s
2
1
+ +
+
+
+

= .
Now let us find Laplace transform of some functions where change of scale
property is using:
Q.No.35.: Find
)
`

t
at sin
L , given that |

\
|
=
)
`

s
1
tan
t
t sin
L
1
.
Sol.: Since given |

\
|
=
)
`

s
1
tan
t
t sin
L
1
.
)
`

=
)
`

a / s
1
tan
a
1
at
at sin
L
1
(by change of scale property)
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
20
|

\
|
=

s
a
tan
a
1
1

Thus,
|

\
|
=
)
`

s
a
tan
t
at sin
L
1
. Ans.
Q.No.36.: If ( ) { }
s
e
t f L
s
1

= , find ( ) { } t 3 f e L
t
.
Sol.: Given ( ) { }
s
e
t f L
s
1

= .
( ) { }
s
e
3
s
e
3
1
t 3 f L
s
3
s
3

= = . (by change of scale property) (a = 3, replace s by
3
s
).
( ) { }
( )
1 s
e
t 3 f e L
1 s
3
t
+
=
+

. (by shifting property)

Q.No.37.: Find the Laplace transforms of ( ) t f defined as
( ) ,
t
t f

= when < < t 0

, 1 = when > t
Sol.: ( ) { } dt 1 . e dt
t
. e t f L
st st
0

+
(
(
(

s
e
dt
s
e
. 1
s
e
. t
1
st st
0 0
st


s
e 0
s
e
s
0 e 1
s
0
2
s s

+
(
(
(

2
s s
2
s s
s
e 1
s
e
s
1 e
s
e

= +

=

. Ans.
Q.No.38.: Find the Laplace transform of ( )

>
< <
=
1 t 0,
1 t 0 , e
t f
t
.
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
21
Sol.: ( ) [ ] dt . 0 dt e e t f L
1
t st
1
0

+ = dt e e
t st
1
0

=

( )
[ ] 1 e
s 1
1
s 1
1
s 1
e
s 1
s 1

=
(
(

. Ans.
Q.No.39.: Find the Laplace transform of ( )

>
< <
=
n t 0,
1 t 0 , t sin
t f .
Sol.: ( ) [ ] dt . 0 tdt sin e t f L
1
st
0

+ = tdt sin e
st
0

= .
Let tdt sin e I
st

= dt
s
e
t cos
s
e
t sin
st st

dt te cos
s
1
s
e
t sin
st
st

+ =

(
(

dt
s
e
t sin
s
e
t cos
s
1
s
e
t sin
st st st

tdt sin e
s
1
s
te cos
s
te sin
st
2 2
st st

=

2 2
st st
s
I
s
te cos
s
te sin
=


(
(

=
(

+

2
st st
2
s
te cos
s
te sin
s
1
1 I [ ]
st st
2
te cos te sin . s
1 s
1


+
= .
( ) [ ] [ ]

+
=
0
st st
2
te cos te sin . s
1 s
1
t f L
[ ]
0 s
2
e e
1 s
1
+
+
=

1 s
1 e
2
s
+
+
=

. Ans.
Q.No.40.: Find the Laplace transform of ( )

>
< <
< <
=
3 t 7,
3 t 2 1, t
2 t 0 , t
t f
2
.
Sol.: ( ) [ ] ( ) dt e 7 dt e 1 t dt e t t f L
st
3
st
3
2
st 2
2
0

+ + =
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
22
( )


(
(

+
(
(

+
(
(

=
3
st
3
2
2
st st
2
0
3
st
2
st st 2
s
e 7
s
e
s
e 1 t
s
e 2
s
te 2
s
e t

(
(

|
|

\
|

|
|

\
|

+
(
(

|
|

\
|
+
|
|

\
|
+

=

2
s 2 s 2
2
s 3 s 3
3 3
s 2
2
s 2 s 2
s
e
s
e
s
e
s
e 2
s
2
s
e 2
s
e 4
s
e 4
(
(

+ +

s
e
0 7
s 3

s
e 7
s
e
s
e
s
e
s
e 2
s
2
s
e 4
s
e 2
s
e 4
s 3
2
s 2 s 2
2
s 3 s 3
3 3
s 2
2
s 2 s 2
+
(
(

+ + +
(
(

+ =
[ ] [ ] s 7 1 s 2
s
e
s
2
s s 2 s 4 s 4
s
e
2
s 3
3
2 2
3
s 2
+ + + =


( ) ( ) 1 s 5
s
e
2 s 3 s 3
s
e
s
2
2
s 3
2
3
s 2
3
+ + =

. Ans.
Q.No.41.: Find the Laplace transform of ( ) 1 t 1 t t f + + = 0 t .
Sol.: Given ( ) 1 t 1 t t f + + = .
( )
( ) ( )

>
=
< + +
=
1 t 2t,
1 t 2,
1 t , 1 t 1 t
t f
( ) [ ] tdt 2 . e dt 2 . e t f L
st
1
st
1
0

+ =

(
(

+
(
(

=
1
2
st st
1
0
st
s
e
s
te
2
s
e
2
[ ]
(
(

+ =

2
s s
s
s
e
s
e
2 1 e
s
2
(
(

+
+
(
(

=

2
s s
s
s
s
e se
2
e
e 1
s
2


( ) ( )
2
s
s
s
s
1 s e 2
se
1 e 2 +
+

|
|

\
|
+ = + =

s
e
1
s
2
s
e 2
s
2
s
2
s
. Ans.
Q.No.42.: Find the Laplace transform of
( )

<

> |

\
|

=
3
2
t 0,
3
2
t ,
3
2
t cos
t f
.
Sol.: ( ) [ ] dt
3
2
t cos e t f L
st
|

\
|
=

.
Let dt
3
2
t cos e I
st
|

\
|
=

dt e
3
2
t sin
s
1
s
e
3
2
t cos
st
st

\
|
|

\
|
=

Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
23

(
(

\
|

\
|

|

\
|

=

dt
s
e
3
2
t cos
s
e
3
2
t sin
s
1
s
e
3
2
t cos
st st
st


2
st
2
st
s
I
e
3
2
t sin
s
1
s
e
3
2
t cos
|

\
|
+
|

\
|

=

.
(
(
(
(

\
|

+
|

\
|

=
(

+

2
st st
2
s
e
3
2
t sin
s
e
3
2
t cos
s
1
1 I
( ) [ ]

(
(
(
(

\
|

+
|

\
|

+
=
3 / 2
2
st st
2
2
s
e
3
2
t sin
s
e
3
2
t cos
1 s
s
t f L

( )
(
(

+ +
+
=

0
s
e
0
1 s
s
s 3 / 2
2
2 ( )
1 s
s
. e
s
e
.
1 s
s
2
3 / s 2
s 3 / 2
2
2
+
=
+
=


. Ans.
Q.No.43.: Find the Laplace transform of at cos at at sin .
Sol.: [ ] at cos at at sin L ( ) ( ) at cos t aL at sin L =
( ) ( ) at cos L
ds
d
1 a
a s
a
1
2 2
2

+
=
( )
( )
2
2 2
2 2
2 2 2 2 2 2
a s
s 2 . s 1 . a s
a
a s
a
a s
s
ds
d
a
a s
a
+
+
+
+
= |

\
|
+
+
+
=
( )
( )
( ) ( )
( ) ( )
2
2 2
3
2
2 2
2 2 2 2
2 2 2
2 2
2 2
2 2
a s
a 2
a s
s a a s a a
a s
a
a s
s a a
a s
a
+
=
+
+ +
+
+
=
+

+
+
= . Ans.
Q.No.44.: Find the Laplace transform of ( ) t 2 cos t 2 t 2 sin e
t

.
Sol.: ( ) ( ) ( ) t 2 cos t 2 L t 2 sin L t 2 cos t 2 t 2 sin L =
( ) t 2 cos t L 2
4 s
2
2

+
= ( ) ( ) t 2 cos L
ds
d
1 2
4 s
2
1
2

+
=
( )
( )
2
2
2
2 2 2
4 s
s 2 . s 4 s
. 2
4 s
2
4 s
s
ds
d
2
4 s
2
+
+
+
+
=
+
+
+
=
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
24
( ) ( )
( )
2
2
2 2
2
2
2
4 s
s 2 8 4 s 2
4 s
s 4 2
4 s
2
+
+ +
=
+

+
+
=
( )
) s ( f
4 s
16
2
2
=
+
= , (say).
Therefore, ( ) [ ] t 2 cos t 2 t 2 sin e L
t

( ) [ ] ( )
2
2
2
2
5 s 2 s
16
4 1 s
16
+ +
=
+ +
= . Ans.
Q.No.45.: Find the Laplace transform of t
2
3
sin
2
t
sinh .
Sol.:
|
|

\
|
t
2
3
sin
2
t
sinh L
(
(

|
|

\
|

=

t
2
3
sin
2
e e
L
2 / 1 2 / 1

(
(

|
|

\
|

(
(

|
|

\
|
=

t
2
3
sin e L
2
1
t
2
3
sin e L
2
1
2 / 1 2 / 1

4
3
2
1
s
2
3
.
2
1
4
3
2
1
s
2
3
2
1
2 2
+ |

\
|
+

(
(
(
(
(

+ |

\
|

=

(
(
(
(

+
=
|
|

\
|
4
3
s
2
3
t
2
3
sin L
2

( ) ( )
( )( ) ys 1 s s 1 s
1 s s 1 s s
.
4
3
2 2
2 2
+ + +
+ + +
=
( )
( )
2 4 2 2 4
2
2
2
s 1 s
s
2
3
s s 2 1 s 2
s . 3
s 1 s
s 2
.
4
3
+ +
=
+ +
=
+
= .Ans.
Q.No.46.: Find the Laplace transform of at sin te
at
.
Sol.: Consider
2 2
a s
a
) at (sin L
+
=
Now ( ) ( ) ( ) ( ) at sin L
ds
d
1 at sin t L
1
=
( ) ( )
) s ( f
a s
as 2
a s
s 2 . 1
) a (
a s
a
ds
d
2
2 2
2
2 2
2 2
=
+
=
+

=
+
=
Therefore ( ) at sin t e L
at
( )
( ) ( )
( )
( )
( )
( )
2
2 2
2
2 2 2
2
2 2
a 2 as 2 s
a s a 2
a as 2 a s
a s a 2
a a s
a s a 2
+

=
+ +

=
+

= .Ans.
Q.No.47.: If ) s ( f )) t ( f ( L = , show that ( ) as f a
a
t
f L =
|
|

\
|
|

\
|
.
Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
25
Sol.: dt )) a ( t ( f . e
a
t
f L
st
0

=
|
|

\
|
|

\
|

(
(

= = =
0 z 0, t , z , t when
adz dt , az t , z
a
t
Put

dz ) z ( f e a adz ) z ( f e
z ) as (
0
saz
0

= =
( ) as f a = . Ans.
(
(

= =

) s ( f dt ) t ( f e )) t ( f ( L
st
0

Q.No.48.: Show that ( )
4 2
2
k 4 s
s k 2
kt sinh kt sin L
+
= .
Sol.:
( ) ( ) ( ) kt sin e L
2
1
kt sin e L
2
1
kt sin
2
e e
L kt sinh kt sin L
kt kt
kt kt

=
(
(

=
(
(

=

2
e e
sinh


( ) ( ) (
(

+ +

+
=
2 2 2 2
k k s
k
k k s
k
2
1


( ) ( )
( )( )
|
|

\
|
+ + +
+ +
=
ks 2 k 2 s ks 2 k 2 s
k k s k k s
2
k
2 2 2 2
2 2 2 2


( )
( )
4 4
2
4 4
2 2
2
2 2
2 2 2 2
k 4 s
s k 2
k 4 s
ks 4
2
k
s k 4 k 2 s
ks 2 k 2 s ks 2 k 2 s
2
k
+
= |

\
|
+
=
+
+ + +
= .Ans.
*** *** *** *** ***
*** *** ***
***
Home Assignments
Find the Laplace Transforms of the following functions:
Q.No. Function Answer
1. at cos at cosh
( )
4 4
2 2
a 4 s
a 2 s a
+
+

2.
( )

>
< <
=
4 t 5,
4 t 0 , t
t f
|
|

\
|
+

2
s 4
2
s
1
s
1
e
s
1

Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
26
3.
( )

>
< <
=
2 t 0,
2 t 0 , t cos
t f
1 s
e 1
2
s
+
+


4. t 2 2
e t


( )
3
2 s
2
+

5.
( )
3
t
te 1

+
( ) ( ) ( )
4 3 2
3 s
6
2 s
6
1 s
3
s
1
+
+
+
+
+
+
6.
kt sin
2

( )
2 2
2
k 4 s s
k 2
+

7.
t 2 cos 2 t 4 sin 3 t 6 e 4
3 t 5
+ +
4 s
s 2
16 s
12
s
36
5 s
4
2 2 4
+
+
+
+

8.
at cos
3
( )
( )( )
2 2 2 2
2 2
a 9 s a s
a 7 s s
+ +
+

9. t cos t 2 cos . t 3 cos
|
|

\
|
+
+
+
+
+
+
s
1
4 s
s
16 s
s
36 s
s
4
1
2 2 2

10.
t sin
Hint: Use
( )

= |

\
|
+
n
2
1 n 2 .... 5 . 3 . 1
2
1
n
for n
positive integer.
( )
|

\
|

= |

\
|

s 4
1
1 n
1 n
e
s s 2
s 4
1
! 1 n
1
s s 2
. Ans
11.
If ( ) { }
20 s 4 s
s 4 20
t f L
2
+

= .
Find ( ) { } t 3 f L
( )
( ) 180 s 12 s
s 15 4
2
+

12.
If { }
1 s
1
t sin L
2
+
= .
Find { } t 3 sin L
9 s
3
2
+

13.
( )
t 2
e 2 t + ( )
( )
3
2
1 s
2 s 4 s 4

+

14.
t 2 cosh e
t 4

( )
( ) 12 s 8 s
4 s
2
+ +
+

Laplace Transforms: Definitions, Properties, Transforms of elementary functions
Prepared by: Dr. Sunil, NIT Hamirpur (HP)
27
15. at sin at sinh
( )
4 2
2
a 4 s
s a 2
+

16.
( ) t 4 cos 4 t 4 sin 3 e
t 2

( )
( ) 20 s 4 s
s 4 20
2
+

17.
t
1 n
e 1
t

( )
( )
n
0 m
m s
n
+

=

18.
t sin e
3 t 2

13 s 4 s
1
4
3
5 s 4 s
1
4
3
2 2
+ +

+ +

19. t 2 4
e . t sin ( )
( ) ( ) (
(

+
+

16 2 s
2 s
4 2 s
2 s 4
2 s
3
8
1
2 2


*** *** *** *** ***
*** *** ***
***

2 22 2
nd nd nd nd
Topic Topic Topic Topic
Laplace Tra Laplace Tra Laplace Tra Laplace Transforms nsforms nsforms nsforms
Laplace Transforms of Derivatives, Integrals, Multiplication and
Division by t

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