This document provides an overview of the Laplace transform and its properties and applications. Specifically, it defines the Laplace transform and inverse Laplace transform, lists several Laplace transform pairs, and provides examples of using the Laplace transform to solve initial value problems involving differential equations with constant coefficients. It also includes a problem set with exercises calculating Laplace transforms and using them to solve initial value problems.

1. BMM 104: ENGINEERING MATHEMATICS I Page 1 of 8
CHAPTER 9: THE LAPLACE TRANSFORM
Laplace Transformation and its Inverse
Definition:
Let f (t ) be defined for 0 ≤ t ≤ ∞ and let s denote an arbitrary real variable. The
Laplace transform of f (t ) , designated by either L[ f ( t ) ] or F ( s ) , is
∞
F ( s ) = L[ f (t )] = ∫e −st f (t )dt
0
for all values of s for which the improper integral converges. Convergence occurs when
the limit
∞ a
∫e
−st
f ( t ) dt = lim ∫ e −st f ( t ) dt
a→∞
0 0
exists. If the limit does not exist, the improper integral does not exist, the improper
integral diverges and f (t ) has no Laplace transform. When evaluating the integral, the
variable s is treated as a constant because the integration is with respect to t.
On the other hand, we may write f ( t ) = L−1 [ F ( s ) ] with L−1 is called inverse Laplace
transformation operator and f (t ) is called the inverse Laplace transformation for
F (s) .
Example:
Find the Laplace transform by definition.
(a) L[1] (b) [ ]
L e at
Properties ( Linearity of Laplace Operator L and its inverse L−1 )
Suppose c1 and c 2 are arbitrary constants, then
(i) L{ c1 f 1 ( t ) + c 2 f 2 ( t ) } = c1 L{ f 1 ( t ) } + c 2 L{ f 2 ( t ) } = c1 F1 ( s ) + c 2 F2 ( s )
(ii) L−1 { c1 F1 ( s ) + c 2 F2 ( s )} = c1 L−1 { F1 ( s )} + c 2 L−1 { F2 ( s )} = c1 f 1 ( t ) + c 2 f 2 ( t )
Example:

2. BMM 104: ENGINEERING MATHEMATICS I Page 2 of 8
Find the Laplace transform for the following functions.
(a) {
L 2 + 3e −2 t }
 1 
(b) L − 2t − sin 3t + 4 cos t 
 2 
(c) { −2 t
L 2 cosh 5t − 7 e cos 4t }
Example:
−1  3 5 6 
(a) Find L  + + 2 .
s − 3 s s + 4
−1  2 4 
(b) Find L  + 2 2 .
s − 3 s − 3 
−1  2s + 3 
(c) Find L  2 .
 s − 4 s + 20 
−1  2s − 3 
(d) Find L  2 .
 s − 4s + 6 
−1  10 
(e) Find L  .
(
( s − 2 ) s + 1 
2
)
Theorem:
Suppose f (t ) is a continuous function for t ≥ 0 that has Laplace transform F ( s ) . If
f ' ( t ) and f '' ( t ) have Laplace transformation, then
* L{ f ' ( t )} = sF ( s ) − f ( 0 )
* L{ f '' ( t )} = s 2 F ( s ) − sf ( 0 ) − f ' ( 0 )
NOTE:
The above Theorem is applied in solving initial value problem.
We apply,
 dy  d 2 y 
L   = sY ( s ) − y ( 0 ) and L  2  = s 2 Y ( s ) − sy ( 0 ) − y ' ( 0 )
 dt   dt 
where Y ( s ) = L{ y ( t )} .
Example:

3. BMM 104: ENGINEERING MATHEMATICS I Page 3 of 8
By taking Laplace transformation on both of the following differential equations, find
Y ( s ).
(a) y ' = cos 5t ; y ( 0 ) = 1
(b) y '' + 4 y ' + 13 y = 0 ; y ( 0 ) = 2; y ' ( 0 ) = −7
(c) y '' + y = 16 cos t ; y ( 0 ) = 0 ; y ' ( 0 ) = 0
First Shift Theorem
If L{ f ( t )} = F ( s ) then L{e at f ( t )} = F ( s − a ) where a is a real constant.
Example:
Determine the following.
(a) {
L e at cos bt } (b) {
L e at sin bt }
Multiplication by t Theorem
d
If L{ f ( t )} = F ( s ) then L{tf ( t )} = − F ( s) .
ds
Example:
Obtain the following.
(a) L{te t } (b) L{t 2 e t } (c) L{t sin t } (d) L{t 2 sin t }
Solving Linear Initial-Value Problems with Constant Coefficients

4. BMM 104: ENGINEERING MATHEMATICS I Page 4 of 8
Laplace transform for derivatives of a function contain terms that need the values of the
function and its derivative at t = 0. By having these (initial) conditions, the approach
using Laplace transformation become very suitable to solve initial value problem that
involving constant coefficients.
Example:
(a) Solve y '' + y = 16 cos t ; y ( 0 ) = y ' ( 0 ) = 0 .
(b) Solve y '' + 4 y ' + 13 y = 0 ; y ( 0 ) = 2; y ' ( 0 ) = −7 .
(c) Solve y '' − 3 y + 2 y ' = 12e 4 t ; y ( 0 ) = 1; y ' ( 0 ) = 0 .
TABLE OF LAPLACE TRANSFORMS

5. BMM 104: ENGINEERING MATHEMATICS I Page 5 of 8
f (t ) F ( s ) = L{ f ( t )}
1 1 1
s >0
s
2 t 1
s >0
s2
3 tn , n = 1,2 ,... n!
s >0
s n +1
4 e at 1
s>a
s −a
5 sin at a
s >0
s + a2
2
6 cos at s
s >0
s + a2
2
7 sinh at a
s >a
s − a2
2
8 cosh at s
s >a
s − a2
2
9 e at sin bt b
s>a
( s − a) 2 + b2
10 e at cos bt s−a
s>a
( s − a) 2 + b2
11 t n e at n!
s>a
( s − a ) n +1
12 t sin at 2as
s >0
(s 2
+ a2 ) 2
13 t cos at s2 − a2
s >0
(s 2
+ a2 ) 2
14 t sinh at 2as
s> a
(s 2
− a2 ) 2
15 t cosh at s2 + a2
s >a
(s 2
− a2 ) 2
16 y ' (t ) sY ( s ) − y (0 ) with Y ( s ) = L{ y ( t )}
17 y (t )
''
s 2 Y ( s ) − sy ( 0 ) − y ' ( 0 )
18 e at f (t ) F ( s − a)
19 t n f (t ) , n = 1,2 ,... d
( − 1) n F ( s)
ds n
20 µa ( t ) f ( t ) e − as L{ f ( t + a )}
21 µa ( t ) f ( t − a ) e − as L{ f ( t )}

6. BMM 104: ENGINEERING MATHEMATICS I Page 6 of 8
22 f (t ) is periodic with period ρ ρ
∫e
−st
f (t )
0
1 − e −ρs
23 t F ( s )G ( s )
∫ f (τ ) g ( t −τ ) dτ
o
24 t 1
F (s)
∫ f (τ )dτ
o
s
PROBLEM SET: CHAPTER 9
1. Evaluate the following Laplace transform.
(a) {
L e −7 t + e −2 t }
(b) L{e 3t + sin 3t }
(c) L{3 sin 4t − 2 cos 4t }
(d) L{ sin at cos at}
(e) L{5 sinh 3t − 2 cosh 2t}
2. Solve by using First Shift Theorem.
(a) {
L e 2 t sinh t }
(b) L{e 2 t sin 3t cos 3t }
(c) L{e −3t cos( 2t + 4 )}
3. Solve by using Multiplication by t Theorem.
(a) L{t cos at }
(b) L{t 2 e 2 t }
(c) {
L t sin 2 t }
4. Find the inverse of the following Laplace transform.
1
3s( s 2 + 4 )
(a)

7. BMM 104: ENGINEERING MATHEMATICS I Page 7 of 8
2
(b)
s + s −6
2
2s − 1
(c)
s + 4s + 5
2
1
s ( s + 4 s + 13)
(d) 2
1
s( s + 1) ( s 2 + 4 s + 5 )
(e)
5. Solve the following initial value problem.
(a) y '' + 9 y = t 2 ; y ( 0 ) = 1; y ' ( 0 ) = 0
(b) y '' − y = sin t ; y(0 ) = y' (0 ) = 0
(c) y '' + 4 y ' + 8 y = cos 2t ; y ( 0 ) = 2; y ' ( 0 ) = 1
(d) y '' − 2 y ' + y = te t ; y ( 0 ) = 1; y ' ( 0 ) = 0
(e) y '' + 2 y ' + 5 y = e −t sin t ; y ( 0 ) = 1; y ' ( 0 ) = 1
ANSWERS FOR PROBLEM SET: CHAPTER 9
2s + 9
1. (a)
( s + 7 )( s + 2 )
s( s + 3)
(b)
( s − 3) ( s 2 + 9 )
2( 6 − s )
(c)
s 2 + 16
a
(d)
s + 4a 2
2
− 2 s 3 + 15 s 2 + 18 s − 60
(e)
( s 2 − 4 )( s 2 − 9 )
1
2. (a)
( s − 2) 2 − 1
3
(b)
( s − 2 ) 2 + 36
( s + 3) cos 4 − 2 sin 4
(c)
( s + 3) 2 + 4

8. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
s2 − a2
3. (a)
(s 2
+ a2 ) 2
2
(b)
( s − 2) 3
2( 3 s 2 + 4 )
(c)
s 2 ( s 2 + 4)
2
1
4. (a) ( 1 − cos 2t )
12
(b)
5
(e − e −3t )
2 2t
(c) e −2 t ( 2 cos t − 5 sin t )
1  −2t  2 
(d) 1 − e  cos 3t + 3 sin 3t 
13   
1 1 −t 1 −2 t
(e) − e + e ( 3 cos t + sin t )
5 2 10
1 2 1
5. (a) t + ( 83 cos 3t − 2 )
9 81
(b)
1 t
4
(e − e −t ) − 2 sin t OR 2 ( sinh t − sin t )
1 1
1 1 39 −2 t 47 −2 t
(c) cos 2t + sin 2t + e cos 2t + e sin 2t
20 10 20 20
 1 
(d) et 1 − t + t 3 
 6 
1 −t
(e) e ( sin t + sin 2t )
3

9. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
s2 − a2
3. (a)
(s 2
+ a2 ) 2
2
(b)
( s − 2) 3
2( 3 s 2 + 4 )
(c)
s 2 ( s 2 + 4)
2
1
4. (a) ( 1 − cos 2t )
12
(b)
5
(e − e −3t )
2 2t
(c) e −2 t ( 2 cos t − 5 sin t )
1  −2t  2 
(d) 1 − e  cos 3t + 3 sin 3t 
13   
1 1 −t 1 −2 t
(e) − e + e ( 3 cos t + sin t )
5 2 10
1 2 1
5. (a) t + ( 83 cos 3t − 2 )
9 81
(b)
1 t
4
(e − e −t ) − 2 sin t OR 2 ( sinh t − sin t )
1 1
1 1 39 −2 t 47 −2 t
(c) cos 2t + sin 2t + e cos 2t + e sin 2t
20 10 20 20
 1 
(d) et 1 − t + t 3 
 6 
1 −t
(e) e ( sin t + sin 2t )
3

10. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
s2 − a2
3. (a)
(s 2
+ a2 ) 2
2
(b)
( s − 2) 3
2( 3 s 2 + 4 )
(c)
s 2 ( s 2 + 4)
2
1
4. (a) ( 1 − cos 2t )
12
(b)
5
(e − e −3t )
2 2t
(c) e −2 t ( 2 cos t − 5 sin t )
1  −2t  2 
(d) 1 − e  cos 3t + 3 sin 3t 
13   
1 1 −t 1 −2 t
(e) − e + e ( 3 cos t + sin t )
5 2 10
1 2 1
5. (a) t + ( 83 cos 3t − 2 )
9 81
(b)
1 t
4
(e − e −t ) − 2 sin t OR 2 ( sinh t − sin t )
1 1
1 1 39 −2 t 47 −2 t
(c) cos 2t + sin 2t + e cos 2t + e sin 2t
20 10 20 20
 1 
(d) et 1 − t + t 3 
 6 
1 −t
(e) e ( sin t + sin 2t )
3

11. BMM 104: ENGINEERING MATHEMATICS I Page 8 of 8
s2 − a2
3. (a)
(s 2
+ a2 ) 2
2
(b)
( s − 2) 3
2( 3 s 2 + 4 )
(c)
s 2 ( s 2 + 4)
2
1
4. (a) ( 1 − cos 2t )
12
(b)
5
(e − e −3t )
2 2t
(c) e −2 t ( 2 cos t − 5 sin t )
1  −2t  2 
(d) 1 − e  cos 3t + 3 sin 3t 
13   
1 1 −t 1 −2 t
(e) − e + e ( 3 cos t + sin t )
5 2 10
1 2 1
5. (a) t + ( 83 cos 3t − 2 )
9 81
(b)
1 t
4
(e − e −t ) − 2 sin t OR 2 ( sinh t − sin t )
1 1
1 1 39 −2 t 47 −2 t
(c) cos 2t + sin 2t + e cos 2t + e sin 2t
20 10 20 20
 1 
(d) et 1 − t + t 3 
 6 
1 −t
(e) e ( sin t + sin 2t )
3