UNIT 5

Laplace
Transforms
5.1 INTRODUCTION
The Laplace transform is a powerful mathematical technique useful to the
engineers and scientists, as it enables them to solve linear differential equations
with given initial conditions by using algebraic methods. The Laplace transform
technique can also be used to solve systems of differential equations, partial
differential equations and integral equations. Starting with the definition of Laplace
transform, we shall discuss below the properties of Laplace transforms and derive
the transforms of some functions which usually occur in the solution of linear
differential equations.

5.1.1 Definition ∞

∫e
− st
If f (t) is a function of t defined for all t ≥ 0, f (t ) dt is defined as the Laplace
0
transform of f(t), provided the integral exists.
Clearly the integral is a function of the parameters s. This function of s is denoted
as f (s) or F(s) or f(s). Unless we have to deal with the Laplace transforms of more
than one function, we shall denote the Laplace transform of f (t) as f(s). Sometimes
the letter ‘p’ is used in the place of s.
The Laplace transform of f (t) is also denoted as L{f (t)}, where L is called the
Laplace transform operator.
∞

Thus L { f (t )} = φ ( s ) = ∫ e−st f (t ) dt.

0

The operation of multiplying f (t) by e−st and integrating the product with respect
to t between 0 and • is called Laplace transformation.
The function f (t) is called the Laplace inverse transform of f(s) and is denoted
by L−1{φ (s)}.

Thus f (t) = L−1{f (s)}, when L {f (t)} = f (s)

5.4 Mathematics II

Note

1. The parameter s used in the definition of Laplace transform is a real or com-

plex number, but we shall assume it to be a real positive number sufficiently
large to ensure the existence of the integral that defines the Laplace trans-
form.
2. Laplace transforms of all functions do not exist. For example, L (tan t) and
2
L (et ) do not exist. We give below the sufficient conditions (without proof)
for the existence of Laplace transform of a function f (t):
Conditions for the existence of Laplace transform If the function f (t) defined
for t ≥ 0 is

(i) piecewise continuous in every finite interval in the range t ≥ 0, and

(ii) of the exponential order, then L{ f (t)} exists.

Note
1. A function f (t) is said to be piecewise continuous in the finite interval a ≤ t ≤
b, if the interval can be divided into a finite number of sub-intervals such that
(i) f (t) is continuous at every point inside each of the sub-intervals and
(ii) f (t) has finite limits as t approaches the end points of each sub-interval
from the interior of the sub-interval.
2. A function f (t) is said to be of the exponential order, if | f (t)| ≤ M eαt, for all t ≥ 0 and
some constants M and α or equivalently, if lim{e−αt f (t )} = a finite quantity.
t →∞

Most of the functions that represent physical quantities and that we encounter in
differential equations satisfy the conditions stated above and hence may be assumed
to have Laplace transforms.

5.2 LINEARITY PROPERTY OF LAPLACE AND

INVERSE LAPLACE TRANSFORMS
L{k1 f1(t) ± k2 f2(t)} = k1L{f1(t)} ± k2 L{ f2(t)},
where k1 and k2 are constants.
∞
Proof: L{k1 f1 (t ) ± k2 f 2 (t )}= ∫ {k1 f1 (t ) ± k2 f 2 (t )}e−st dt
0
∞ ∞

= k1 ∫ f1 (t ) ⋅ e−st dt ± k2 ∫ f 2 (t ) ⋅ e−st dt
0 0

= k1 ⋅ L{f1 (t )} ± k2 ⋅ L{ f 2 (t )}.
Thus L is a linear operator.
As a particular case of the property, we get
L{k f (t)} = kL{f(t)}, where k is a constant.
 Laplace Transforms 5.5

If we take L{f1(t)} = φ1(s) and L{ f2(t)} = φ2(s),

the above property can be written in the following form.
L{k1 f1(t) ± k2 f2(t)} = k1φ1(s) ± k2φ2(s).
∴ L−1 {k1φ1(s) ± k2 φ2(s)} = k1∙ f1(t) ± k2∙ f2(t)
= k1∙L−1{φ1(s) ± k2∙L−1{φ2(s)}
−1
Thus L is also a linear operator
As a particular case of this property, we get
L−1{kf(s)} = k L−1{f (s)}, where k is a constant.

Note
1. L{f1(t) . f2(t)} ≠ L{f1(t)} . L{f2(t)} and
L−1{φ (s) . φ (s)} ≠ L−1(φ (s)} × L−1{φ (s)}
1 2 1 2

2. Generalising the linearity properties,

 n
  n

we get (i) L
∑ kr f r (t ) = ∑ kr ⋅ L{ f r (t )}

 

r = 1  r =1

 n  n

(ii) L−1 ∑ kr φr ( s )
 = ∑ kr ⋅ L {φr ( s )}
−1

 

 r =1  r =1
Using (i), we can find Laplace transform of a function which can be expressed as
a linear combination of elementary functions whose transforms are known.
Using (ii), we can find inverse Laplace transform of a function which can be
expressed as a linear combination of elementary functions whose inverse transforms
are known.

5.3 LAPLACE TRANSFORMS OF SOME

ELEMENTARY FUNCTIONS
k
1. L{k}= , s > 0, where k is a constant,
s
∞

L{k}= ∫ k e−st dt ,by definition.

0

∞
 e−st 
=k  
 −s 
 0

k
= (0 −1)[ e−st → 0 as t → ∞, if s > 0]
−s
k.
=
s
5.6 Mathematics II

1
In particular, L(0) = 0 and L(1) =
s
1
 
∴ L−1 
  =1.

s
 

− at 1
2. L{e } = , where a is a constant,
s+a
∞

L{e−at } = ∫ e−st ⋅ e−at dt

0

∞
∞
 e−(s + a )t 
=∫ e −(s + a )t
dt =  
 −(s + a ) 
0  0

1
= (0 −1) , if (s + a) > 0
−(s + a )

1
= , if s > −a.
s+a

 1 
Inverting, we get L−1   = e−at .
 s + a 

1
3. L{e at } = , where a is a constant, if s −a > 0 or s > a.
s−a
Changing a to − a in (2), this result follows. The corresponding inverse result is

 1 
L−1  = e at
 s − a 

(n + 1)
4. L(t n ) = , if s > 0 and n > − 1.
sn + 1
∞

L(t ) = ∫ e−st ⋅ t n dt
n

∞
 x d x
n

= ∫ e− x   ⋅ , on putting st = x
 s  s
0

∞
1
∫e
−x
= x n dx
sn + 1 0
 Laplace Transforms 5.7

(n + 1) , if s > 0 and n + 1 > 0

=
sn + 1
[by definition of Gamma function]
In particular, if n is a positive integer,

(n + 1) = n!

n!
∴ L(t n ) = , if s > 0 and n is a positive integer.
sn + 1
 n! 
 
Inverting, we get L−1  n + 1  = t n or

s 
 

 1 
  1
L−1  n + 1  = t n

s 
 
 n!

−1 
1 
 1
Changing n to n − 1, we get L  n = t n− 1 , if n is a positive integer.


s 

 (n − 1)!
1
  1 n− 1
If n > 0, then L−1  n  = t


s 
 (n)
1 1
In particular, L(t ) = or L−1  2  = t .
s2  s 
a
5. L(sin at ) =
s2 + a2
∞

L(sin at ) = ∫ e−st sin at dt

0

∞
 e−st 
= 2 ( − s sin at − a cos at )
 s + a2 
 0

s ∞ a ∞
=−
s2 + a2
( e−st sin at)0 − s 2 + a 2 (e−st cos at )0

a
=
s2 + a2
[ e−st sin at and e−st cos at tend to zero at t → ∞, if s > 0]
 a 
Inverting this result we get L−1  2 2
 = sin at .
 s + a 
5.8 Mathematics II

s
6. L(cos at ) =
s2 + a2
∞

L(cos at ) = ∫ e−st cos at dt

0

∞
 e−st 
= 2 ( − s cos at + a sin at )
 s + a2 
 0
s ∞ a ∞
2 (
=− 2 e−st cos at ) + 2 2 (
e−st sin at )
s +a 0 s +a 0

s
= , as per the results stated above.
s + a2 2

 s 
Inverting the above result we get L−1  2 2
 = cos at .
 s + a 
Aliter
L(cos at + i sin at) = L(eiat)
1
= , by result (3).
s − ia
s + ia
=
s2 + a2
s a
i.e., L(cos at ) + iL (sin at ) = +i 2 , by linearity property.
s +a
2 2
s + a2
s
Equating the real parts, we get L(cos at ) = .
s2 + a2
a
Equating the imaginary parts, we get L(sin at ) =
s + a2
2

a
7. L(sinh at ) =
s − a2
2

1 
L(sinh at ) = L  (e at − e−at )
 2 

1
= L (e at ) − L (e−at ) , by linearity property.
2 
1  1 1 
=  −  , if s > a and s > − a.
2  s − a s + a 
a
= , if s > |a|
s − a2
2
 Laplace Transforms 5.9

Aliter
L(sinh at) = − iL(sin i at) [ ∴ sin iθ = i sinh θ]
ia
= −i ⋅ , by result (5)
s +i a
2 2 2

a
=
s −a
2 2

 a 
 
Inverting the above result we get L−1  2 2
= sinh at.


 s − a 


s
8. L(cosh at ) =
s −a2 2

1 
L(cosh at ) = L  (e a t + e−a t )
 2 

1
=  L (e a t ) + L (e−a t ), by linearity property,
2  

1 1 1 
=  +  , if s > |a|.
2  s − a s + a 


s
= , if s > |a|.
s − a2
2

Aliter
L(cosh at) = L(cos iat) [ cos iq = cosh q]
s
= , by result (6)
s + i2a2
2

s
=
s − a2
2

 s 
 
Inverting the above result we get L−1  2 2
= cosh at.

s −a 
 


5.4 LAPLACE TRANSFORMS OF SOME SPECIAL

FUNCTIONS

5.4.1 Definition
0, when t < a

The function f (t ) = 
 , is called Heavyside’s unit step

1, when t > a, where a ≥ 0

function and is denoted by ua (t) or u (t − a)
5.10 Mathematics II

0, when t < 0


In particular, u0 (t ) = 


1, when t > 0

∞

Now L{ua (t )} = ∫ e−s t ua (t ) dt

0
a ∞

= ∫ e−s t ua (t ) dt + ∫ e−s t ua (t ) dt
0 a

= ∫ e−s t dt , by the definition of ua (t)

a

∞
 e−s t  e−a s
=  −S  = S , assuming that s > 0.
 
a

1
In particular, L{u0 (t )} = , which is the same as L(1).
S

 e−as 
 
Inverting the above result, we get L−1 
 
 = ua (t ) .

 s 

 
5.4.2 Definition
lim { f (t )} , where f (t) is defined by
h→0



 1 , when a − h ≤ t ≤ a + h
f (t ) = 
h 2 2 is called Unit Impulse



0 , otherwise

Function or Dirace Delta Function and is denoted by δa(t) or δ (t − a).

Now L{δa (t )} = L  lim { f (t )} ,where f(t) is taken as given in the definition
 h→0 

= lim L{ f (t )}
h→ 0

= lim ∫ e−s t f (t ) dt
h→ 0
0

h
a+
2
1
= lim
h→ 0 ∫ e−s t ⋅
h
dt
h
a−
2

 a+ 
h
 1  e−s t  2 
= lim  ⋅   
h → 0  h  −s   h 
 a− 
2
 Laplace Transforms 5.11

 −s a − h  − s a +  
 h
 e  2  − e  2  
= lim  

h→ 0 sh
 
 
 s h/2 − s h / 2 
−e
= e−a s lim  
e

h→ 0
 sh 
  sh  
 2 sinh   
  2  
= e−a s lim  
h→ 0  sh 
 
 

  sh  
 s cosh   
  2  
= e−a s lim   , by L’ Hospital’s rule.
h→ 0  s 
 
 

 sh 
= e−a s lim  cosh  = e−a s
h→ 0  2 


Aliter

1  
f (t ) =  u h (t ) − u h (t ) , since
h  a− 2 a+
2 

h
u h (t ) = 1, when t > a − and u h (t ) = 0 ,
a− a+
2 2 2

h
when t<a+ and hence u h (t ) − u h (t ) = 1
2 a− a+
2 2

h h
when a− <t <a +
2 2
    
∴ L{f (t )} =
1L 
 u ( t )


 − L


u (t )



h  a− h
  a+
h

  2 
 
 2 

 − a − ⋅⋅ s
 h 
− a + ⋅⋅ s 
 h 

1  e  2  e  2  
=  − 
h s s 
 
 sh 
2 sinh  
 2 
= e−a s
sh
5.12 Mathematics II

  sh  
 2 sinh   
  2  
∴ L {δa (t )} = lim L { f (t )} = e−as ⋅ lim  
h→ 0 h→ 0  sh 
 
 
= e−as
Inverting the above result, we get
L−1{e−as}= δa(t). When a → 0, L−1 {1} = δ (t).

5.5 PROPERTIES OF LAPLACE TRANSFORMS

1. Change of Scale Property

1  s  
  t 
If L{ f(t)} = φ (s), then L{ f (at )}= φ   and L 
 f   = a φ (as ) .
 
a a   a 

Proof
∞

By definition, φ(s ) = L{ f (t )}= ∫ e−st f (t ) dt (1)

0

and L{ f (at )}= ∫ e−st f (at ) dt [ f(at) is a function of t]

0

∞ x
−s d x , putting x = at and making necessary changes.
=∫ e a
f (x)
0
a

∞
1
∫e
−(s / a ) x
= ⋅ f (x) dx (2)
a 0

∞
1
∫e
−(s / a ) t
= ⋅ f (t ) dt , changing the dummy variable x as t.
a 0

Now, comparing (1) and (2), we note that the integral in (2) is the same as the
 
integral in (1) except that ‘s’ in integral in (1) is replaced by  s  in the integral
 a 
in (2).
∴ When the integral in (1) is equal to φ (s), that in (2) is equal to φ (s/a).
1
Thus L{ f (at )} = φ ( s a ) (3)
a

Changing a to 1 in (3) or proceeding as in the proof given above, we have

a
L{ f ( t a )} = a φ (as ) .
 Laplace Transforms 5.13

2. First Shifting Property

If L{ f (t)} = φ (s), then L{e−at f (t)} = φ (s + a)
and L{eat f (t)} = φ (s − a).

Proof
∞

By definition, L{ f (t )} = ∫ e−st f (t ) dt = φ (s ) (1)

0
∞

and L{ e−at f (t )} = ∫ e−st [e−at f (t )]dt

0

[ e−at f (t) is a function of t]

∞

= ∫ e−( s + a ) t f (t ) dt (2)
0

= φ (s + a ) , comparing the integrals in (1) and (2)

Changing a to − a in the above result,
we get L{e at f (t )}=φ (s − a ) .

Note
1. The above property can be rewritten as a working rule (formula) in the fol-
lowing way:
L{e−at f (t )}=φ (s + a )
=[φ (s )]s→ s + a
= L{ f (t )}s→ s + a
‘s → s + a’ means that s is replaced by (s + a).
Thus, to find the Laplace transform of the product of two factors, one of
which is e−at, we ignore e−at, find the Laplace transform of the other factor as
a function of s and change s into (s + a) in it.
Similarly,
L{eat f (t)} = L{f (t)s → s −a
2. The above property can be stated in terms of the inverse Laplace operator as
follows:
−1
If L {φ (s )}= f (t ) , then L−1 {φ (s + a )}= e−at ⋅ f (t )
From this form of the property, we get the following working rule:
L−1 {φ (s + a )}= e−at ⋅ L−1 {φ (s )}
This means that if we wish to find the Laplace inverse transform of a function
that can be identified as a function of (s + a), we have to find the Laplace
inverse transform of the corresponding function of s and multiply it with e−at.
Similarly,
L−1 {φ (s − a )}= e at L−1 {φ (s )} .
5.14 Mathematics II

3. The above property is called so, as it concerns shifting on the s-axis by a (or
− a), i.e., replacing s by s + a (or s − a).
The second shifting property, that follows, concerns shifting on the t-axis by
−a i.e., replacing t by t − a.

3. Second Shifting Property

If L{ f (t)} = φ (s), then L{ f (t − a)ua(t)} = e−asφ (s),
where a is a positive constant and ua(t) is the unit step function.

Proof
∞

By definition, L{ f (t − a )ua (t )}= ∫ e−st f (t − a )ua (t )dt

0
a ∞

= ∫ e−st f (t − a )ua (t ) dt + ∫ e−st f (t − a )ua (t ) dt

0 a
∞

= ∫ e−st f (t − a ) dt , by the definition of ua(t).

a
∞

= ∫ e−s (x+ a ) f (x) dx , putting t − a = x and effecting consequent changes

0
∞

= e−as ∫ e−st f (t ) dt , changing the dummy variable x as t.

0

− as
=e φ (s )

Note
1. Rewriting the above property, we get the following working rule:
L{f (t − a) ua(t)} = e−as L{f (t)}
2. The above property can be stated in terms of the inverse Laplace operator as
follows:
If L−1{φ (s)} = f (t), then L−1{e−as φ (s)} = f (t − a) ua(t).
From this form of the property, we get the following working rule.
L {e−as φ (s)} = L−1 {φ (s)}t → t − a · ua(t).
−1

Thus if we wish to find the Laplace inverse transform of the product of two factors,
one of which is e−as, we ignore e−as, find the Laplace inverse transform of the other
factor as a function of t, replace t by (t − a) in it and multiply by ua(t).

WORKED EXAMPLE 5(a)

Example 5.1 Find the Laplace transforms of the following functions:

 Laplace Transforms 5.15

(t −1)2 ,
 for t >1
(i) f (t ) = 


 for 0 < t < 1
 0,

e k t , for 0 < t < a
(ii) f (t ) = 



 0, for t > a

 π

sin ωt , for 0 < t <

(iii) f (t )= 

ω

 π
 0, for t >


 ω

t, for 0 < t < 4
(iv) f (t ) = 


5,
 for t > 4
∞

L{f (t )}= ∫ e−st f (t ) dt

0

1 ∞
(i) L{f (t )}= ∫ e−st ⋅ odt + ∫ (t −1) e−st dt
2

0 1

∞
 2 e
− st   e−st   e−st 
= (t −1) ⋅  − 2 (t −1) 2  + 2  3 
 −s   s   − s 
 1
2 −s 2 −s
=0+ e = 3e
s3 s
Aliter
 0, for 0 < t <1
(t −1) u1 (t ) =
2

 (t −1)
2
for t >1

Thus f (t) = (t − 1)2 u1 (t)

∴ L{f (t)} = e−s · L(t2), by the second shifting property.

2 −s
= e
s3

a ∞

(ii) L{f (t )}= ∫ e−st ⋅ e kt dt + ∫ e−st ⋅ 0 dt

0 a

 e−(s − k )t 
a

=   = 1 {1− e−a (s − k ) }

 −( s − k ) 0 s − k
5.16 Mathematics II

Aliter
e k t , for 0 < t < a
Consider e kt {1− ua (t )}= 
 0, for t > a
Thus f (t) = e k t − e k t ua (t)

= e k t −e k (t − a) + ka · ua (t)
∴ L{f (t)} = L(e k t) −e ak · L{e k (t − a) · ua(t)}

1
= − e ak ⋅ e−as ⋅L{e kt } , by the second shifting property
s−k

1 1
−e ( ) ⋅
−a s − k
=
s−k s−k
1
{1− e ( ) }.
−a s − k
=
s−k

π ω
(iii) L{f (t )}= ∫ e−st sin ωt dt
0

 e−st 
π ω

= 2 − s sin ωt − ω cos ωt )
 s + ω2 ( 
 0
ω
= 2
s +ω2
(1+ e−π s ω )

4 ∞

(iv) L{f (t )}= ∫ te−st dt + ∫ 5e−st dt.

0 4
∞
  e−st   e−st 
4
   e−st 
 
= t ⋅     
− 2  + 5 − s 
  − s   s  0  4
4 1 1 5
=− e−4 s − 2 e−4 s + 2 + e−4 s
s s s s
1 −4 s 1 −4 s 1
= e − 2e + 2
s s s

Aliter
t, for 0 < t < 4
t{1− u4 (t )} + 5u4 (t ) = 
5, for t > 4

Thus f (t) = t − (t − 5) u4(t)

= t − (t − 4) u4(t) + u4(t)
 Laplace Transforms 5.17

∴ L (f (t )}= L(t ) − e−4 s ⋅ L(t ) + L{u4 (t )}

1 1 −4 s 1 −4 s
= − e + e .
s2 s2 s
Example 5.2 Find the Laplace transforms of the following functions:
1+ 2 t cos t
(i) , (ii) sin t , (iii) .
t t

1+ 2 t 

(i) L
  −1 2
 = L(t ) + 2 L(t )
12


 t 


(1 2) (3 2)
= 12
+2
s s3 2
1
2⋅
π
π ∵ (1 2)= π
= + 2  and (n +1)= n (n)
s s s 
π  1 
= 1+ .
s  s 

( t) +( t)
3 5

(ii) sin t = t− −....∞

3! 5!
1 1
= t1 2 − t 3 2 + t 5 2 −....∞
3! 5!

∴ ( ) 1 1
L sin t = L(t1 2 ) − L (t 3 2 ) + L(t 5 2 ) −....∞
3! 5!
(3 2) 1 (5 2) 1 (7 2)
= − + −....∞
s3 2 3! s 5 2 5! s 7 2

1 1 1 3 1 1 15 3 1 1 
=  (1 2)− ⋅ ⋅ (1 2)⋅ + ⋅ ⋅ (1 2)⋅ 2 −....∞
32
s 2  3! 2 2 s 5! 2 2 2 s 

π  11 1 1 
2

= 3 2 1− ⋅  + ⋅  −......∞

2 s  1!  4 s  2!  4 s  

π
= 3 2 e−1 4 s
2s
 
( ) ( )
2 4
cos t 1  t t 
(iii) = 1− + −.....∞
t t  2! 4! 

 
1 1
= t −1 2 − t1 2 + t 3 2 −.....∞
2! 4!
5.18 Mathematics II

 cos t 
∴ L   = L(t −1/ 2 ) − 1 L (t1 2 ) + 1 L (t 3 2 )−.....∞
 t  2! 4!

(1 2) 1 (3 2) 1 (5 2)
= − + −.....∞
s1 2 2! s 3 2 4! s 5 2
 
= 1− 1 + 1 3 ⋅1 2 −.....∞
π
 2! 2s 4! 2 ⋅ 2 s
s 
 

π  1  1  1  1 
2 
= − ⋅  + ⋅  − 
1   . . ...∞
s  1!  4 s  2!  4s  

π −1 4 s
= e
s

Example 5.3 Find the Laplace transforms of the following functions:

(i) (t3 + 3e2t − 5 sin 3t)e−t (ii) (1 + te−t)3
−2t 3
(iii) e cosh 2t (iv) cosh at cos at
t 3
(v) sinh sin t
2 2
(i) L{(t3 + 3e2t − 5 sin 3t)e−t}
= L(t3 + 3e2t − 5 sin 3t)s → s + 1, by first shifting property.
 3! 3 3 
=  + −5⋅ 2 
 s 4 s − 2 s + 9 s → s +1
6 3 15
= + − 2
( s +1)
4
( s −1) s + 2 s +10

(ii) L(1 + te−t)3 = L(1 + 3t e−t + 3t2 e−2t + t3 e−3t)

= L(1) + 3L(t)s → s + 1 + 3L(t2)s → s + 2 + L(t3)s → s + 3
1 3 6 6
= + + +
s (s +1) 2 (s + 2)3 (s + 3) 4

 −2 t 3 
 −2t  e + e  
2t
−2t
(iii) L(e cosh 2t ) = L e ⋅ 
3
 

  2  

 
1
= L{e−2t (e6t + 3e 2t + 3e−2t + e−6t )}
8
1
= L{e 4t + 3 + 3e−4t + e−8t }
8
1 1 3 3 1 
=  + + + 
8  s − 4 s s + 4 s + 8 
 Laplace Transforms 5.19

  e a t + e−a t 
 
L (cosh at cos at )= L   cos at 
(iv)    
  

 2 
1
=  L (cos at )s → s − a + L (cos at )s → s + a 
2 
 s −a s +a 
=  
1
+
2  ( s − a )2 + a 2 ( s + a )2 + a 2 
 
1  s −a s +a 

= + 2
2  s + 2a − 2as s + 2a + 2as 
 2 2 2

 
1  (s − a ) ( s + 2a + 2as ) + (s + a ) ( s + 2a − 2as )
2 2 2 2

=  
2 ( s 2
+ 2 a )
2 2
− 4 a 2 2
s 
 
s3
= .
s 4 + 4a 4

t 3
(v) L(sinh sin t)
2 2

 et 2 − e−t 2  3 
= L   sin t

   2 

 2  

 
1  3   3  
=  L sin t − L sin t 
2   2 s→s − 1  2 s→s + 1 
 2 2

 
=  
1 3 2 3 2
− 
2  ( s −1 2) + 3 4 ( s +1 2)2 + 3 4 
2
 
3  1 1 
= − 
4  s 2 − s +1 s 2 + s +1
3 s
= ⋅ 4
2 s + s 2 +1

Example 5.4 Find the Laplace transforms of the following functions:

(i) eat cos (bt + c) (ii) e−2t cos2 3t (iii) et sin3 2t
(iv) e−t sin 2t cos 3t (v) e3t sin 2t sin t
(i) L{eatcos (bt + c)}= L{cos (bt + c)}s → s − a
= L{cos c cos bt − sin c sin bt}s→ s − a
(s − a ) cos c b sin c
= −
(s − a ) 2 + b 2 (s − a ) 2 + b 2
5.20 Mathematics II

(ii) L{e−2t cos2 3t}= L(cos2 3t)s → s + 2

1
 

= L  (1+ cos 6t )


2 

s → s + 2
1 1 s+2 
=  + 

2  s + 2 ( s + 2) + 36 
2
 
(iii) L{et sin3 2t} = L(sin3 2t)s → s − 1

3 1 

= L  sin 2t − sin 6t 


 4 4 

s → s −1
3 2 1 6 
=  ⋅ 2 − ⋅ 2 
 4 s + 4 4 s + 36 s → s −1
3 1 1 
=  2 − 2 
2  s − 2 s + 5 s − 2 s + 37 

(iv) L{e−t sin 2t cos 3t} = L(sin 2t cos 3t)s → s + 1


1 

= L  (sin 5t − sin t )

2
 
 s → s +1

1 5 1 
=  2 − 2 
2  s + 25 s +1 s → s +1
1 
= 
5 1
 2 − 2 
2 s + 2 s + 26 s + 2 s + 2 

(v) L{e3t sin 2t sin t} = L(sin 2t sin t)s → s − 3
 1 

= L  (cos t − cos 3t )
 2 s→s−3


1 s  s  
=   2 − 2 
2 
 s +1 s + 9 

s→s −3
 s −3
1 s −3  
= 
 2 − 2 

2

 s − 6 s +10 s − 6 s +18 



Example 5.5 Find the Laplace transforms of the following functions:

(i) (t − 1)2 · u1(t) (ii) sin t · uπ(t) (iii) e−3t · u2(t)
(i) L{(t − 1)2 u1(t)} = e−s · L{t2}, by the second shifting property
2 −s
= e
s3
(ii) L{sint · uπ(t)} = L{sin (t − π + π) · uπ(t)}
= −L{sin (t − π) · uπ(t)}
 Laplace Transforms 5.21

= − e−sp L(sin t)
e −sπ
=−
s 2 +1

(iii) L{e−3t u2(t)} = L{e−3(t−2)−6 ⋅ u2(t)}

−(2 s + 6)
e
= e−6 ⋅ e−2 s ⋅ L {e−3t }=
s +3
Example 5.6 Find the Laplace transforms of the following functions:
(i) t sin at (ii) t cos at (iii) te−4t sin 3t
(i) L (t sin at) = L{t × I.P. of e iat}
= I.P. of L{t e iat}
= I.P. of L(t)s → s − ia
( s + ia )
2
1
= I.P. of = I.P. of
( s − ia ) (s 2 + a 2 )
2 2

 2 
 s − a2 2 as 
= I.P. of  +i 
 ( s 2 + a 2 )2 (s 2 + a 2 ) 
2

2 as
=
(s + a 2 )
2 2

(ii) L(t cos at) = R.P. of L{te iat}

s2 − a2
=
(s 2 + a 2 )
2

(iii) L(t e−4t sin 3t) = I.P. of L {t e −4t e i3t}

= I.P. of L {t · e−(4 − i3)t}

= I.P. of L(t)s → (s + 4 − i3)

( s + 4 + i3)
2
1
= I.P. of = I.P. of
( s + 4 − i3) {(s + 4) + 9}
2 2 2

= I.P. of
{(s + 4) −9} + i 6(s + 4)2

{(s + 4) + 9} 2 2

6 ( s + 4)
=
{(s + 4) + 9}
2 2
5.22 Mathematics II

Example 5.7
 t
(i) Assuming L(sin t) and L(cos t), find the Laplace transforms of Lsin  and
 2 
L(cos 3t).
2s s 2 −1
(ii) Given that L(t sin t )= and L(t cost )= , find L(t sin at) and
(s 2 +1) (s 2 +1)
2 2

 t
Lt cos  .
 a

1
(i) L (sin t )=
s +12

 t 1
∴ L sin = 2 ⋅
 2  (2s ) +1
2

   t  
∵ L 

 f   = a ⋅ L { f (t )}s→as , by the charge of sccale property
  
   a 
 
2
=
4 s 2 +1

s
L (cos t )=
s +12

1 s/3
∴ L (cos3t )= ⋅
3 ( s / 3)2 +1

 1 
∵ L { f (a t )} = L { f (t )} s , by the charge of scale property
 s→ 
a a

s
= .
s +9
2

2s
(ii) Given that L (t sin t )=
(s +1)
2 2

1
∴ L (t sin at ) = L(at sin at )
a
1 1 2s/ a
= ⋅ ⋅ , by change of scale property⋅
a a ( s 2 / a 2 +1)2

2 as
=
(s 2 + a 2 )
2
 Laplace Transforms 5.23

s 2 −1
Given that L (t cos t )=
(s 2 +1)
2

 t t t
∴ L t cos  = a ⋅ L  cos 
 a  
 a a
(as ) −1
2

=a⋅a⋅ , by change of scale property

(as )2 +1
2

 
1
a4 s2 − a2 s2 −2
= or a .
(a 2 s 2 +1)
2
 2 1 
2
 s + 
 a 2 
Example 5.8 Find the inverse Laplace transforms of the following functions:
1 e−2 s s e−s (3a − 4s )
(i) e−s (ii) (iii) (iv) e−bs
( s + 2) (2s − 3) s2 + a2
5/ 2
( s − 3)
3 5

( s + 4)
(v) e−4 s
s2 − 4
(i) From the second shifting property, we have
L−1 {e−as φ (s)} = L−1 {φ (s)t → t − a ⋅ ua(t)

 
  
∴ L−1  −s 1  −1   1 
e ⋅ 5/ 2 
=L  5/ 2 
⋅ u1 (t ) (1)

 ( s + 2 ) 
 
 ( s + 2 ) 


 
 
 
t → t −1

 

Now L−1 
1  −2 t −1   1  
 5/ 2 
=e ⋅ L  5/ 2 
   
 ( s + 2)  
 

 s
 

[ L−1 {φ (s + 2) = e−2t L−1 {φ (s)}, by the first shifting property.]
  
1  −1  1 1 n−1 
= e−2t ⋅ ⋅t3/ 2 ∵L  n = t 
(5 / 2) 
  s (n) 
1
= e−2t ⋅ ⋅t3/ 2 ∵ (n)= (n −1) (n −1)
3 1
⋅ ⋅ (1/ 2)  
2 2
3
4
= t 2 e−2t ∵ (1/ 2)= π 
3 π  
(2)
Inserting (2) in (1), we have

 

 1  4
(t −1) e−2(t −1) ⋅ u1 (t )
3/ 2
L−1 e−s 5/ 2 
=
 
 ( s + 2) 

  3 π

5.24 Mathematics II

0,
 when t < 1


or = 4

 (t −1)3/2 ⋅ e−2 (t −1) , when t > 1


3 π


 e−2 s   = L−1 
  1 

(ii) L−1 
 3

 
3
⋅u2 (t ) (1)


 ( 2 s − 3) 

 

 ( 2 s − 3) 

t→ t −2



Now L−1 

1  = 1 L−1 

 
 1 


3  3


 ( 2 s − 3) 

 8 

 (s − 3 2 ) 


1 2 t −1 
3
 
1 
= e L  3
8 
s 
 

1 3t 1 1 3t
= e2 ⋅ t2 = e2 t2 (2)
8 2! 16

Using (2) in (1), we get


 e−2 s   3
 = 1 e 2 (t −2 ) (t − 2) 2 ⋅ u (t )
L−1 
 3

 (2 s − 3) 
  16

2

 s e−s 
  
 = L−1  s  
(iii) L−1 
 5

 
5
⋅ u1 (t ) (1)


 (s − 3) 

 

 (s − 3) 

t → t −1


 s   = L−1 
  (s − 3) + 3
Now L−1 
 5

 
5 


 (s − 3) 

 

 (s − 3 ) 


 s + 3 
= e3t L−1 
 5  


 s  

1
 3
= e3t L−1  4 + 5 


 s s 



 1 1 

= e 3t  t 3 + 3 ⋅ t 4 
 3!!

 4! 

1 3t 3
= e (4t + 3t 4 ) (2)
24
Using (2) in (1), we have

 s  1 3(t −1)

L−1  e−s 
= e {4(t −1)3 + 3(t −1) 4 }⋅ u1 (t )
 (s − 3)5 

 24

 (3a − 4s ) −bs  
 3a − 4 s 
(iv) L−1 
 2

e  = L−1  2 2

⋅ ub (t ) (1)


 s + a 2


 

 s + a 

t → t −b
 Laplace Transforms 5.25

 3a − 4 s 
   a 
   s 

Now L−1  2 2
= 3L−1  2 2
− 4 L−1  2 2


 s + a 

 

 s + a 

 
 s + a 

= 3 sin at − 4 cos at (2)
Using (2) in (1), we have
 (3a − 4s ) −bs 
 
L−1 
 2 e   = [3 sin a (t − b) − 4 cos a (t − b)]ub (t )

 s +a

2



 ( s + 4) −4 s 
   s+4 
 
(v) L−1  2 e  = L−1  2  ⋅ u4 (t ) (1)


 s − 4 

 

 s − 4 

t → t −4

 s+4 
   s 
−1 
  2 
−1 
Now L−1 
 2 
= L   2 
+ 2 L   2 

 s − 4
 
 
 s − 4
 
 
 s − 4 

= cosh 2t + 2 sinh 2t (2)
Using (2) in (1), we have

 ( s + 4 ) −4 s 

L−1 
 2 e   ={cosh 2(t − 4) + 2 sinh 2(t − 4)}⋅ u4 (t )


 s − 4 


Example 5.9 Find the inverse Laplace transforms of the following functions:
e−s e−2 s e−s e−3 s
(i) (ii) (iii) (iv)
( s − 2) ( s + 3) s ( s 2 +1)
2
s ( s 2 + 4) s + 4 s + 13
2

( s +1)e−π s
(v)
s 2 + 2s + 5


−1 
 e−s 

 
−1  1 

(i) L  = L   
 ⋅ u1 (t ) (1)
 
 ( s − 2) ( s + 3) 
  
 ( s − 2) ( s + 3) 
 t→ t −1


 1 
 1
Now to find L−1   
 , we resolve into partial fractions

 
 (s − 2) (s + 3) 
 (s − 2) (s + 3)
and then use the linearity property of L−1 operator.
1 A B
Let = +
( s − 2) ( s + 3) s − 2 s + 3
Then A (s + 3) + B (s − 2) = 1
1 1
By the usual procedure, we get A= , B =−
5 5

 
  1 5
1 −1  1 5 
∴ L−1 
 
= L   − 

( )( )

 s − 2 s + 3  

 s − 2 s + 3 


5.26 Mathematics II

1  1  1 −1  1 
= L−1  − L  
5  s − 2  5  s + 3 

1
= (e 2 t − e−3 t ) (2)
5

Putting (2) in (1), we have


 e−s 
 1 2 (t −1)
L−1 
 
 = {e − e−3(t −1) }⋅ u1 (t )


 ( s − 2 ) ( s + 3) 

 5


 e−2 s   
−1  1 

(ii) L−1 
 2 2 
= L   2 2 
 ⋅ u2 (t ) (1)


 s ( s + 1) 

 

 s ( s + 1) 

t → t −2
1 1 1 1 1 1
Now = = − = 2− 2
s ( s +1) u (u +1) u u +1 s
2 2
s +1

 1 
 −1  1 
 1 
∴ L−1 
 2 2 
 = L  2 − L  2 
−1
 
 s ( s +1) 
  s   s +1

= t − sin t (2)
Inserting (2) in (1), we get

 e−2 s  
L−1  2 2 
 ={(t − 2) − sin (t − 2} u2 (t )
 s ( s +1) 


 e−s 
−1   
−1 


(iii) L  
= L  
1
 2  2  ⋅ u1 (t ) (1)


 s ( s + 4 ) 

 

 s ( s + 4 ) 

t →t −1

1 A B s +C
Let = + 2
s ( s 2 + 4) s s +4

Then A(s2 + 4) + s (Bs + c) = 1

1 1
∴ A = , B =− and C = 0
4 4
1 1 

 
  s 
1 −1  4 4 
∴ L−1 
 2 
= L  − 2 

 s ( s + 4) 
 
  s s + 4
 
 
1
= (1− cos 2t ) (2)
4
 Laplace Transforms 5.27

Using (2) in (1), we have

 e−s 
  1
L−1 
 2 
 = {1− cos 2 (t −1)}⋅ u1 (t )
 
 s ( s + 4) 
  4

 e−3 s 
 
−1  
L−1  
= L 
1
(iv)  2  2  ⋅ u3 (t ) (1)
 s + 4 s +13

 
 
 s + 4 s +13t →t −3



 1 
 
−1  1 
Now L−1 
 2 
= L   2
 
 s + 4 s +13
  
 ( s + 2) + 3 

2


 1  
= e−2t ⋅ L−1 
 2  , by the shifting property
2


 s + 3 


1  3 
 
= e−2t L−1 
 2 
2
3 

 s + 3 


1
= e−2t sin 3t (2)
3
Using (2) in (1), we get

 e−3 s 
 1 −2 (t − 3)
L−1 
 2 
= e ⋅ sin 3(t − 3) ⋅ u3 (t )

 s + 4 s +13
  3


−1  ( s +1) e
−π s 
 −1 
 s +1  
(v) L   2 
= L   2 
 ⋅ uπ (t ) (1)

 s + 2s + 5
 
 
 s + 2s + 5
 
t →t −π

 s +1 
   ( s +1) 
−1 

Now L−1 
 2 
= L   
2


 s + 2 s + 5 

 

 ( s + 1) 2
+ 2 


 s  
= e− t ⋅ L−1 
 2  , by the shifting property
2

s +2 
 

= e−t cos 2t (2)
Using (2) in (1), we have

 ( s +1) e−π s 

L−1 
 2 
= e
−( t − π )
⋅ cos 2(t − π ) ⋅ uπ (t )

 s + 2s + 5
 

Example 5.10 Find the inverse Laplace transforms of the following functions:

s2 + s − 2 2 s 2 + 5s + 2 s
(i) (ii) (iii)
s ( s + 3) ( s − 2) ( s − 2) 4 ( s + 1) 2 ( s 2 + 1)

1 s
(iv) (v) .
s ( s + 1) ( s 2 + 9)
2 2
( s + 1) ( s + 4) ( s 2 + 9)
2 2
5.28 Mathematics II

 s2 + s − 2 
 
(i) To find L−1 
 
 , we resolve the given function of s into partial


 s ( s + 3) ( s − 2 ) 


fractions and then use the linearity property of L−1 operator.

s2 + s − 2 A B C
Let = + +
s ( s + 3) ( s − 2) s s + 3 s − 2

We find, by the usual procedure, that

1 4 2
A= , B = and C = .
3 15 5

 s2 + s − 2 
  
−1 1 / 3 4 / 15 2 / 5 
∴ L−1 
 
 = L  + +

 s ( s + 3) ( s − 2) 
 
  s s + 3 s − 2 

1 4 2
= + e−3 t + e 2 t
3 15 5

2 s2 + 5 s + 2
(ii) To resolve into partial fractions, we put s − 2 = x, so that
( s − 2) 4

s = x + 2.
2 s 2 + 5 s + 2 2 ( x + 2) 2 + 5 ( x + 2) + 2
Then =
( s − 2) 4 x4

2 x 2 +13 x + 20
=
x4
2 13 20
= 2+ 3+ 4
x x x
2 13 20
= + +
( s − 2) 2 ( s − 2) 3 ( s − 2 ) 4

 2 s 2 + 5 s + 2 
  1 
−1 
 
 +13L−1  1    1 
−1 
∴ L−1 
 = 2L   2

 3
+ 20 L   

 ( s − 2) 

4
  
 ( s − 2) 
  
 ( s − 2) 
   ( s − 2) 4 

 1  −1  1  −1  1 
= 2 e ⋅ L−1 
2t 2t
  2t
 
 s 2  +13 e ⋅ L  s 3  + 20 e ⋅ L  s 4 
2t  13 20 
= e  2⋅ t + t 2 + t 3 
 2! 3! 
2t  13 2 10 3 
= e 2 t + t + t 
 2 3 
 Laplace Transforms 5.29

1 2t
= e (12t + 39t 2 + 20t 3 )
6
s A B C s+D
(iii) Let = + + 2
(s + 1) 2 ( s 2 + 1) s + 1 (s + 1) 2 s +1

∴ A(s + 1) (s2 + 1) + B(s2 + 1) + (Cs + D) (s + 1)2 = s

By the usual procedure, we find that
1 1
A = 0; B = − ; C = 0 and D =
2 2

 

 s  
−1  − 1 / 2 1 / 2 
∴ L−1 
  = L 
 + 2 
 (s + 1) (s + 1) 
 (s + 1) s + 1
2 2 2


 

 
1 1 1  1 
= − e−t ⋅ L−1  2  + L−1  2
2 
 s  2  s + 1
t 1
= − e−t + sin t
2 2
1
(iv) Since 2 2 is a function of s2, we put s2 = u, we resolve
s ( s + 1)( s 2 + 9)
1
into partial fractions and then replace u by s2.
u (u + 1) (u + 9)
1 A B C
Now let = + +
u (u + 1) (u + 9) u u + 1 u + 9
1 1 1
By the usual procedure, we find that A = , B = − and C =
9 8 72

 

 1  −1 1 / 9 1/ 8 1 / 72 
∴ L−1 
 2 2 = L 
  2 − 2 + 2 

 s ( s + 1)( s + 9)
2
 

 s s + 1 s + 9 

 

1 1 1
= t − sin t + sin 3t
9 8 216
s
(v) To resolve 2 into partial fractions, we first resolve
(s +1)(s + 4)(s 2 + 9)
2

1
into partial fractions as shown in (iv).
(s 2 +1)(s 2 + 4)(s 2 + 9)
1 1
Thus =
(s 2 +1)(s 2 + 4)(s 2 + 9) (u + 1) (u + 4) (u + 9)

A B C
= + + , say.
u +1 u + 4 u + 9
5.30 Mathematics II

1 1 1
We find that A = , B =− and C = .
24 15 40
1 1/ 24 1/15 1/ 40
∴ = − +
(s 2 +1)(s 2 + 4)(s 2 + 9) s 2 +1 s 2 + 4 s 2 + 9

s 1 s 1 s 1 s
∴ = ⋅ 2 − ⋅ 2 + ⋅ 2
(s 2
+ 1)( s + 4)( s + 9)
2 2
24 s + 1 15 s + 4 40 s + 9

 
 1 −1  s  1 −1  s 
∴ L−1  
 s 
= − L  2 +
 2  s + 1 15  s + 4 
L  2
(

 s + 1)( s + 4)( s + 9) 24
2 2 
 

1 −1  s 
 s + 9 
L  2
40
1 1 1
= cos t − cos 2t + cos 3t
24 15 40
Example 5.11 Find the inverse Laplace transforms of the following functions:
14 s + 10 2s3 + 4s 2 − s +1
(i) (ii)
49 s 2 + 28s + 13 s 2 ( s 2 − s + 2)
1 1
(iii) (iv)
s3 − a3 s4 + 4
s
(v)
s 4 + s 2 +1

 5 

 s+ 

 14 s + 10 
 14 −1  7 
(i) L−1   = L  
 49 s 2 + 28s + 13

 
 49 
 4 13 
 s 2
+ s + 


 7 49 

 


 5 

 s + 
2 −1  7 
= L  2
    3  
2
7 
  s +  +   
2





 7   7   


 
  


  s +  +
2 3 

2 −1    7 7  
= L  2
7 
     
2

  s + 2  +  3   
  7   7  


 



 

 3  

2  s+ 
2 7 −1 
− t
7 
= e ⋅L  2
7 
   
s 2 +   
3

 


 7 
 


 Laplace Transforms 5.31

2 − 72 t  3 3 
= e cos t + sin t 
7 
 7 7 

2s3 + 4s 2 − s +1 A B Cs + D
(ii) Let = + 2+ 2
s 2 (s 2 − s + 2) s s s −s+2

∴ As(s2 − s + 2) + B(s2 − s + 2) + (Cs + D)s2 = 2s3 + 4s2 − s + 1.

By the usual procedure, we find that

1 1 9 13
A= , B= ,C = and D =
4 2 4 4

 9 13 
 3
  s+

 2 s + 4 s 2 − s + 1
−1 
 1 −1  1  1 −1  1  −1  4 4 
∴ L  2 2  = − L   + L  2  + L  2 

 s ( s − s + 2)   4 s 2 s   s − s + 2 

 
  
 


 

 9  1  35 

  s −  + 
1 t  4  2 8 
= − + + L−1  2
4 2 
  1   7  
2

 
 s −  +   

 2  2  

 

 

 5 7 7 

 ⋅ 
1 t 9 s 4 2 
= − + + et / 2 L−1  ⋅ + 2
4 2 4  7 
2
 7  

 s +  
2
s +   
2

  2   2  
 
9 7 
= − + + et / 2  cos
1 t 7 5 7
t+ sin t
4 2  4 2 4 2 

1 1
(iii) =
s 3 − a 3 (s − a ) ( s 2 + as + a 2 )

1 A Bs + C
∴ Let = + 2
s −a
3 3
s − a s + as + a 2

∴ A(s2 + as + a2) + (s − a)(Bs + C) = 1

By the usual procedure, we find that

1 1 2
A= ,B =− 2 and C = − .
3a 2 3a 3a
5.32 Mathematics II

 1
 
− 2 s − 2 
 1 

−1  1 −1  1  −1   3a 3a
a 
∴ L  3 3
= 2 L   + L  2 2 


 s − a 

 3a  s−a  
 s + as + a 

 

 


 

 

 
1 1  s + 2 a
= e at − 2 L−1  2
3a 2 3a 
  
2  3a  
 a
 s +  +    




 2   2  

 

 3 
 s + a 
1 1 − at2 −1   2 
= e − 2 e ⋅L 
at
2
3a 2 3a 
  3a  

 s 2 +   

  2  

 
 at 
3 
=
1  e at − e− 2 

cos
3
at + 3 sin at 
3a 2   2 2 
 


1 1 1
(iv) = 4 =
s + 4 ( s + 4 s + 4) − 4 s
4 2 2
( s 2 + 2) − ( 2 s ) 2
2

1
=
(s 2
+ 2 s + 2)( s 2 − 2 s + 2)

1 As + B Cs + D
∴Let = 2 + 2
s + 4 s + 2s + 2 s − 2s + 2
4

∴ (As + B) (s2 − 2s + 2) + (Cs + D)(s2 + 2s + 2) = 1

By the usual procedure, we get
A = 1/8, B = 1/4, C = −1/8 and D = 1/4

 1
 1    1
 1 

 8s+ 4   
 8 s − 4 
 1 

−1  −1   −1 
∴ L  4 = L  2 − L  2 

 s + 4
 
 
 s + 2s + 2 
 
 s − 2 s + 2 

 
 
 

 
 
 

1  (s + 1) + 1 
  1 −1  (s −1) −1 
= L−1 
  − L 
 
8 
 (s + 1) + 1

2

 8 
 (s −1) + 1

2

1  t −1  s −1 
 s +1 

=  e−t ⋅ L−1  2 − e ⋅ L  2 

8 
 s + 1
 
  s + 1
 Laplace Transforms 5.33

1
=  e−t (cos t + sin t ) − et (cos t − sin t )
8
1  et + e−t   et − e−t 


=   sint −  cos t 
4  2   2  
1
= (sin t cosh t − cos t sinh t )
4
s s s
(v) = =
s 4 + s 2 + 1 ( s 4 + 2 s 2 + 1) − s 2 ( s 2 + 1)2 − s 2

s
=
(s 2 + s +1)(s 2 − s +1)
s As + B Cs + D
∴Let = 2 + 2
s + s +1 s + s +1 s − s +1
4 2

∴ (As + B) (s2 − s + 1) + (Cs + D) (s2 + s + 1) = s

By the usual procedure, we get
−1 1
A = 0, B = , C = 0 and D = .
2 2

 s  1 −1 
  1  −1 −1 
 1 
∴ L−1  4 = L  2  L  2 
 
 s + s + 1

2
 2  
 s − s + 1
  2 
 s + s + 1

   

   
  
1 −1   1 
 1 
 1 
= L  2
− L−1  2
2       2    
  s + 1  +  3  
2 2

  s − 1  +  3   
  2   2  




 2   2   

 
 
  
 
 
  
  
1  t / 2 −1   1 
 
 1 
= e L   − e−t / 2
⋅ L−1
 2

2   3     3  
2
 
   
 s +   
  s +   
2

2
 2  
 
  2      

 
 
1 2   t/2 3 
= ⋅ 
(e − e ) sin
−t / 2
t
2 3 
 2 
2 3 t
= sin nh .
t sin
3 2 2 .
Example 5.12

 
 
 

−1 s 
 t 
−1  s 
(i) If L  2
= sin t , find L  2
   s + a ) 
(
s + 1) 
(
 2 2  2 2
 
  
5.34 Mathematics II

 2

 s +4 

  2 
−1  s + 1 
(ii) Given that L−1 
 2 
2
= t cosh 2 t , find L  2 
 s − 4)   s −1 2 
(

 

 ( ) 

 
  
 1  1 1
(iii) Given that L−1 
 
2
= (sin 2 t − 2 t cos 2t ), find L −1 
 2
.
  
( s + 9) 
(

 s + 4) 
2

16 2
  
(i) By change of scale property,
1 s
L{ f (at )}= φ  
a  a 
∴ L−1 {f (s/a)} = a L−1{f (s)}t → at (1)

 

−1

 s 
 = t sin t
Given L  2
 s + 1) 
(
 2
 2
 


 


 


 

−1  s / a  = a ⋅ at sin at , by
∴ L  2
(1)

 s   2
 + 1 
2

 

  
 
 a
2
 


 

3 −1 
 s 
 a2
i.e., a L   = t sin at
 s2 + a2 ) 
(
2
  2

 


 
 s  t
∴ L−1   2 = sin at .
 2 2


 ( s + a ) 

2 a

(ii) By change of scale property,

L{ f (t/a)} = a f (as)
1
∴ L−1{φ (as )}= L−1 (φ (s )}t→t / a (2)
a

 

 2
−1  s + 4 

Given L   = t cosh 2t
 s − 4) 
(
2


2


 

 ( 2 s ) 2
+ 4 

 = 1 ⋅ t cosh t , by
∴ L−1 
 2
(2)

[(2 s ) − 4] 

2
 2 2


 2 

1 −1   s +1   t
i.e., L   = cosh t
 
(
s −1) 
2
4  2 4
 

 2
 
 s +1  
∴ L−1  2 2
 = t cosh t .
 


 ( s − 1) 


 Laplace Transforms 5.35


 

 1 
 = 1 (sin 2t − 2t cos 2t )
(iii) Given L−1 
 2 2
 s + 4) 
(
  16
 


 


 


 
 
 
−1 
 1 
 3 −1 
 1 
∴ L  2
= L  2
, by (2)
   


 2 s  2
  + 4 
  2 

 ( s 2
+ 4 ) t → 3 t

  3    2

 
 
 

 

81 −1  1 
 = 3 (sin 3t − 3t cos 3t )
i.e. L  2 2
 s + 9) 
(
16   32
 


 



−1 1 
 = 1 (sin 3t − 3t cos 3t )
∴ L  2
 s + 9) 
(
 2
 54
 


EXERCISE 5(a)

Part A
(Short Answer Questions)
1. Define Laplace transform.
2. State the conditions for the existence of Laplace transform of a function.
3. Give two examples for a function for which Laplace transform does not exist.
4. State the change of scale property in Laplace transformation.
5. State the first shifting property in Laplace transformation.
6. State the second shifting property in Laplace transformation.
7. Find the Laplace transform of unit step function.
8. Find the Laplace transforms of unit impulse function.

e 2t , for 0< t < 1
9. Find L{ f (t)}, if f (t ) = 


0, for t > 1

0, for0 < t < 1



10. Find L{ f (t)}, if f (t ) = 
t , for 1 < t < 2


0, for t > 2


sin 2t , for 0< t < π

11. Find L{ f (t)}, if f (t ) = 


0,
 for t > π

cos t , for 0< t < 2π


12. Find L{ f (t)}, if f (t ) = 


0,
 for t > 2π
5.36 Mathematics II


 2π

0, for 0< t <
 3
13. Find L{ f (t)}, if f (t )= 

  2 π  2 π

cos t −  , for t >
 
 
 3 3

sin t , for 0< t < π
14. Find L{ f (t)}, if f (t )= 


t ,
 for t > π
 1 

( )
15. Find L t and L  
 π t 
16. Find L{(t − 3) u3(t)} and L{u1(t) sin p (t − 1)}.
17. Find L{t2u2 (t)}
Find the Laplace transforms of the following functions:
18. (at + b)3 19. sin (wt + q)
20. sin2 3t 21. cos3 2t
22. sin 2t cos t 23. cos 3t cos 2t
24. sinh3 t 25. cosh2 2t
−2 t  2 
26. (t + 1)2 e−t 27. e cos 3t − sin 3t 
 3 
t 1 
28. e cosh 2t + sinh 2t  29. sin t sinh t
 2 
30. t2 cosh t 31. e3(t + 2)
Find the Laplace inverse transforms of the following functions:
32. e−as/s2 (a > 0). 33. (e−2s − e−3s)/s
e−2 s se−s
34. 35.
s −3 s2 + 9
1 + e−π s
36.
s 2 +1
Find f (t) if L{ f (t)} is given by the following functions:
1 s 2 + 2s + 3
37. 38.
(s + 1)3/2 s3
1 s
39. 40.
(2s − 3) ( s − 2)
4 5

2s + 3 s+6
41. 42.
s2 + 4 s2 − 9
1 1
43. 44.
s (s + a ) s 2 + 2 s +5
s −3
45.
s 2 − 6 s + 10
 Laplace Transforms 5.37

Part B
Find the Laplace transforms of the following functions:
47. e (1+ 2at ) / t
at
46. e3t(2t + 3)3
−t 3
48. e sinh t 49. sin at cosh at − cos at sinh at
 cos 2t 
3

50. (et sin t)2 51.  t 

 e 
52. e−kt sin (wt + q) 53. e−t sin 3t cos t
54. et cost cos 2t cos 3t 55. e−2t sin 2t sin 3t sin 4t
56. t cos 2t 57. te−t cos t
58. t e2t sin 3t
4s
59. Given that L(t sin 2t ) = , find L(t sin t).
(s + 4)
2 2

s2 −9
60. Given that L(t cos 3t ) = , find L(t cos 2t).
( s 2 + 9)
2

Find the Laplace inverse transforms of the following function:

s 2 +1 4 s 2 − 3s + 5
61. s 3 + 3s 2 + 2 s 62.
( s +1)( s 2 − 3s + 2)
s 2 − 3s + 5 7 s −11
63. 64.
( s +1)
3
( s +1)( s − 2)
2

1 s 4 − 8s 2 + 31
65. 66.
(s 2 + s)
2
(s 2 +1)(s 2 + 4)(s 2 + 9)
2s 2s − 9
67. 68.
(s +1)( s + 2)( s + 3) s + 6 s + 34
2 2 2 2

s +1 ls + m
69. 70.
s + s +1
2
as + bs + c
2

3s 2 −16 s + 26 1
71. 72.
s ( s + 4 s +13)
2
s 3 +1

1 s
73. 74.
s + 4a 4
4
s +4
4

s2 4s3
75. 76. 4 s 4 +1
s 4 + 64

s 2 +1
77.
s + s 2 +1
4
5.38 Mathematics II


   
 s   1


78. If L  
−1 −1  s 
 2 2
= t sinh t , find L  2
   2 2
( s −1)  (
s − a ) 
 2 
 
  

 
  
 1  1 −1  1 
79. If L−1 
 2 
2
= (sin 3t − 3t cos 3t ), find L  2  .
   s +1 2 


 ( s + 9 ) 


54
 ( ) 


 
  

 s −1

 t −1 
 s 
80. Given that L  2
= sin 3t , find L  2
 s + 9)   s + 4 
( ( ) 
 2
 6 2
 


5.6 LAPLACE TRANSFORM OF PERIODIC FUNCTIONS

5.6.1 Definition
A function f (t) is said to be a periodic function, if there exists a constant P (> 0) such
that f (t + P) = f (t), for all values of t. Now f (t + 2P) = f (t + P + P) = f (t + P) = f (t), for
all t. In general, f (t + nP) = f (t), for all t, when n is an integer (positive or negative).
P is called the period of the function.
Unlike other functions whose Laplace transforms are expressed in terms of an
integral over the semi-infinite interval 0 ≤ t < ∞, the Laplace transform of a periodic
function f (t) with period P can be expressed in terms of the integral of e−st f (t) over
the finite interval (0, P), as established in the following theorem.
Theorem
If f (t) is a piecewise continuous periodic function with period P, then
P
1
L { f (t )} =
1− e− Ps ∫0
⋅ e−st f (t ) dt.

Proof:
∞

By definition, L { f (t )} = ∫ e f (t ) dt
− st

0
P ∞

= ∫ e−st f (t ) dt + ∫ e−st f (t ) dt (1)

0 P

In the second integral in (1), put t = x + P, ∴ dt = dx and the limits for x become 0 and ∞.
∞ ∞
− s( x + P )
∫e f (t ) dt = ∫ e
− st
∴ f (x + P) dx
P 0
∞
[ f(x + P) = f(x)]
= e−sP ⋅ ∫ e−sx f ( x) dx
0
 Laplace Transforms 5.39

= e−sP ∫ e−st f (t ) dt , on changing the dummy variable x to t.

0

= e−sP ⋅ L { f (t )} (2)

By putting (2) in (1), we have

P

L { f (t )} = ∫ e−st f (t ) dt + e−sP ⋅ L { f (t )}
0

P
∴ (1− e− Ps ) L { f (t )} = ∫ e−st f (t ) dt
0

P
∴ 1
L { f (t )} = ∫e
− st
f (t ) dt
1− e− Ps 0

5.7 DERIVATIVES AND INTEGRALS OF TRANSFORMS

The following two theorems, in which we differentiate and integrate the transform
1 
function f (s) = L{f (t)} with respect to s, will help us to find L{t f (t)} and L  f (t )
 t 
respectively. Repeated differentiation and integration of f (s) will enable us to find
 1 

L{tn f (t)} and L  n f (t )
 , where n is a positive integer.
 t 

Theorem
If L{f (t)} = f (s), then L{t f (t)} = − f' (s).
Proof :
Given: L{f (t)} = f (s)
∞
i.e. (1)
∫e
− st
f (t ) dt = φ (s )
0

Differentiating both sides of (1) with respect to s,

∞
d d
∫
ds 0
e−st f (t ) dt = φ (s )
ds
(2)

Assuming that the conditions for interchanging the two operations of integration
with respect to t and differentiation with respect to s in (2) are satisfied, we have
∞
d
∫ ds {e } f (t ) dt = φ ′(s)
− st

0
5.40 Mathematics II

∫ −t e f (t ) dt = φ ′(s )
− st
i.e.
0

∞
i.e.
∫ e−st [t f (t )]dt = −φ ′(s )
0

i.e. L{t f (t )} = −φ ′(s )

Corollary
Differentiating both sides of (1) n times with respect to s, we get
dn
L{t n f (t )} = ( −1) n φ (s ) or ( −1) nφ (n ) (s ) .
ds n
Note
1. The above theorem can be rewritten as a working rule in the following manner
d
L{t f (t )} = − φ (s )
ds
d
= − L{ f (t )}
ds
Thus, to find the Laplace transform of the product of two factors, one of which is ‘t’,
we ignore ‘t’ and find the Laplace transform of the other factor as a function of s;
then we differentiate this function of s with respect to s and multiply by (− 1).
Extending the above rule,

d2
L{t 2 f (t )} = ( −1) 2 L{ f (t )} and in general
ds 2

dn
L{t n f (t )} = ( −1) nL{ f (t )} .
ds n
2. The above theorem can be stated in terms of the inverse Laplace operator as
follows:
If L−1{φ (s)} = f (t),

then L−1{φ ′(s)} = −t f (t).

From this form of the theorem, we get the following working rule:

1
L−1{ φ (s )} = − L−1{ φ ′(s )}
t

This rule is applied when the inverse transform of the derivative of the given function
can be found out easily. In particular, the inverse transforms of functions of s that
contain logarithmic functions and inverse tangent and cotangent functions can be
found by the application of this rule.
 Laplace Transforms 5.41

Theorem
  ∞
  1 
If L{ f (t )} = φ (s) , then L  1 f (t ) = ∫ φ (s ) ds , provided lim  f (t ) exists.


t 


t →0  t 
s

Proof:
Given: L{ f (t)} = φ (s)
∞

∫e
− st
i.e. f (t ) dt = φ (s ) (1)
0

Integrating both sides of (1) with respect to s between the limits s and ∞ , we have
∞ ∞  ∞

∫  ∫ ef (t ) dt  ds = ∫ φ (s ) ds
− st
(2)
s  0  s
Assuming that the conditions for the change of order of integration in the double
integral on the left side of (2) are satisfied, we have
∞ ∞  ∞
 e−st ds  f (t ) dt = φ (s ) ds
∫  ∫ 

∫
0  s s

s =∞
∞
 e−st  ∞
 
i.e.
∫  −t 
f (t ) dt = ∫ φ (s ) ds
0   s=s s
∞ ∞
1
∫ − t (0 − e ) f (t ) dt = ∫ φ (s) ds ,
− st
i.e.
0 s

assuming that s > 0

∞ ∞
 f (t ) 
i.e. ∫ e−st 
 t 
 dt = ∫ φ (s ) ds
0 s

 ∞
 f (t ) 

i.e. L  = ∫ φ (s ) ds

 t 
 
 s

Corollary

1 
 1  1 

L  2 f (t ) = L   f (t )

t
 

  t 

t 


∞ ∞ 
= ∫  ∫ φ (s ) d s  ds
s  s 
∞ ∞

=∫ ∫ φ (s)ds ds
s s

Generalising this result, we get

1
  ∞

∞ ∞

L  n f (t ) = ∫ ∫ …∫ φ (s) (ds)
n


t 

 s s s
5.42 Mathematics II

Note
1. The above theorem can be rewritten as a working rule as given below:
∞

1 

L  f (t ) = ∫ L{ f (t )}d s


t 

 s
1
Thus, to find the Laplace transform of the product of two factors, one of which is ,
t
1
we ignore , find the Laplace transform of the other factor as a function of s and
t
integrate this function of s with respect to s between the limits s and ∞ .
Extending the above rule. We get;
∞ ∞

1 

L  2 f (t ) = ∫ ∫ L{ f (t )}ds ds and in general

t
 
 s
 s

1
  ∞

∞ ∞

L  n f (t ) = ∫ ∫ …∫ L{ f (t )}(ds) .
n


t 

 s s s

2. The above theorem can be stated in terms of the inverse Laplace operator as
follows:
If L−1{ φ (s )} = f (t ) ,

∞  1
then L−1  ∫ φ (s ) ds  = f (t ) .
 s  t
From this form of the theorem, we get the following working rule:
∞ 
L { φ (s )} = t ⋅ L  ∫ φ (s ) d s 
−1 −1

 s 

This rule is applied when the inverse transform of the integral of the given function
with respect to s between the limits s and ∞ can be found out easily.
In particular, the inverse transforms of proper rational functions whose numerators
are first degree expressions in s and denominators are squares of second degree
expressions in s can be found by applying this rule

WORKED EXAMPLE 5(b)

Example 5.1 Find the Laplace transform of the “saw-tooth wave” function f (t)
which is periodic with period 1 and defined as f (t) = kt, in 0 < t < 1.
The graph of f (t) is shown in Fig. 5.1 below. If the period of the function f (t) is
k
P, the function will be defined as f (t ) = t in 0 < t < P.
P
 Laplace Transforms 5.43

f (t)

t
0 1 2 3
Fig. 5.1
By the formula for the Laplace transform of a periodic function f (t) with period P,
P
1
∫e
− st
L{f (t )} = f (t ) dt
1− e− Ps 0

∴ For the given function,

1
1
∫ kt e
− st
L{f (t )} = dt
1− e−s 0

  e−st  − st 
1

=
k t   −1⋅  e 
1− e−s   −s    s 2 
  0
k  e − s
e − s
1
= −s 
− − 2 + 2
1− e  s s s 

k  (1− e ) e−s 
 −s 
=  − 
1− e−s  s 2 s 
 
−s
k ke
= 2−
s s (1− e−s )

Example 5.2 Find the Laplace transform of the “square wave” function f (t)
defined by
f (t) = k in 0 ≤ t ≤ a

= − k in a ≤ t ≤ 2a

and f (t +2a) = f (t) for all t.

f (t + 2a) = f (t) means that f (t) is periodic with period 2a. The graph of the function
is shown in Fig. 5.2.
For a periodic function f (t) with period P,

P
1
∫e
− st
L{f (t )} = f (t ) dt
1− e− Ps 0
5.44 Mathematics II

f (t)
k

t
O a 2a 3a 4a

–k

Fig. 5.2
∴ For the given function;

1 a 2a 
 k e−st dt + ( − k ) e−st dt 
L{f (t )} =
1− e−2 as  ∫
 0
∫ 

a

 e−st   e−st  
a 2 a

=
k   −   
   
1− e−2aas  −s 0  −s a 
k
= [1 − e−as − e−as + e−2 as ]
s (1− e−2 as )
k (1 − e−as ) 2
=
s (1 − e−as ) (1 + e−as )
k (1 − e−as ) k (e as / 2 − e−as / 2 )
= =
s (1 + e−as ) s (e as / 2 + e−as / 2 )
k  as 
= tanh  
s  2 

Example 5.3 Find the Laplace transform of “triangular wave function f (t) whose
graph is given below in Fig. 5.3.
y = [= f (t)]

A
a

B
t
O a 2a 3a 4a 5a
Fig. 5.3
From the graph it is obvious that f (t) is periodic with period 2a.
Let us find the value of f (t) in 0 ≤ t ≤ 2a, by finding the equations of the lines OA
and AB.
 Laplace Transforms 5.45

OA passes through the origin and has a slope 1.

∴ Equation of OA is y = t, in 0 ≤ t ≤ a
AB passes through the point B(2a, 0) and has a slope − 1.

∴ Equation of AB is y − 0 = (− 1) (t − 2a)
or y = 2a − t in a ≤ t ≤ 2a.

Thus the definition of f(t) [= y] can be taken as

f(t) = t, in 0 ≤ t ≤ a

= 2a − t, in a ≤ t ≤ 2a
and f(t + 2a) = f(a).
2a
1
∫e
− st
Now L{f (t )} = f (t ) dt
1− e−2 a s 0

1 a 2a 
 te−st dt + (2a − t )e−st dt 
=
1− e−2 a s ∫ ∫ 
 0 a 
   e−st  
   
a 2a
1  e−s t   e−st  
  e−st 
=  
 − 1⋅  2   + (2a − t )     
1− e−2 a s
t 
 

 −s   s     −s  + 1⋅  s 2  
 
 
0  a 
1  a −a s e − a s −2 a s − a s 
= − e − 2 + 12 + e 2 + a e−a s − e 2 
1− e−2 a s  s s s s s s 

1 − 2e−a s + e−2 a s (1 − e−a s ) 2
= =
s 2 (1 − e−2 a s ) s 2 (1 − e−a s ) (1 + e−a s )
1 (1 − e−a s ) 1  e a s / 2 − e− a s / 2 
= 2 = 2  a s / 2 
s (1 + e ) s  e
−a s
+ e− a s / 2 
1  as 
= tanh  
s 2  2 

Example 5.4 Find the Laplace transform of the “half-sine wave rectifier” function
f (t) whose graph is given in Fig. 5.4.

f (t)

π t
O π/ω 2π/ω 3π/ω 4π/ω
2ω

Fig. 5.4
5.46 Mathematics II

From the graph, it is obvious that f(t) is a periodic function with period 2π/ω. The
π 
graph of f (t) in 0 ≤ t ≤ π/ω is a sine curve that passes through (0, 0),  , a and
 2ω 
 π 
 , 0
 ω 
∴ The definition of f (t) is given by
f (t) = a sin ω t, in 0 ≤ t ≤ π/ ω
= 0, in π/ ω ≤ t ≤ 2π/ ω
 2π 
and f t +  = f (t ) .
 ω
2π ω
1
Now L { f (t )}=
1− e−2 π s ω ∫ e−st f (t ) dt
0
π ω
a
∫e
− st
= sin ωt dt
1− e−2 π s ω 0

 e−st 
π ω
a
= ⋅
−2 π s ω  2
( − s sin ωt − ω cos ωt 
1− e 
 s +ω
2
0
a  ωe−π s ω + ω 
=
(s + ω ) (1− e−2 π s ω ) 
2 2 

ω a (1+ e−π s ω ) ωa
= =
(s + ω ) (1− e
2 2 −2 π s ω
) (s + ω ) (1− e−π s ω )
2 2

Example 5.5 Find the Laplace transform of the “full-sine wave rectifier” function
f (t), defined as
f (t) = |sin ω t|, t ≥ 0
We note that f (t + π/ ω) = |sin ω (t + π/ ω)|
= |sin ω t|
= f (t)
∴ f (t) is periodic with period π/ ω.
Also f (t) is always positive. The graph of f (t) is the sine curve as shown in
Fig. 5.5.

t
O π/ω 2π/ω 3π/ω

Fig. 5.5
 Laplace Transforms 5.47

π ω
1
∫e
− st
Now L { f (t )}= | sin ω t | dt
1− e−π s ω 0
π ω
1
∫e
− st
= sin ω t dt [ ∵sin ωt > 0 in 0 ≤ t ≤ π ω ]
1− e−π s ω 0

 e−st 
π ω
1  2
= ( − s sin ωt − ω cos ωt )
1− e−π s ω  s +ω 2 
 0
1 ω 1+ e−π s ω 
−π s ω (
= ωe−π s ω + ω )= 2  
(s + ω ) (1− e
2 2
) s + ω 2 1− e−π s ω 

ω  eπ s 2 ω + e−π s 2 ω 
=   , on integration and simplification
s 2 + ω 2  eπ s 2 ω − e−π s 2 ω 

ω  πs 
= coth  
s +ω
2 2  2ω 

Example 5.6 Find the Laplace transforms of the following functions:

(i) t cosh3 t (ii) t cos 2t cos t (iii) t sin3 t (iv) (t sin at)2
  et + e−t 3 
 
  
(i) L{ t cosh 3
t } = L t   

  2   

 

1
= L {t (e3t + 3et + 3e−t + e−3t )} (1)
8

1 d
=− L (e3t + 3et + 3e−t + e−3t )
8 ds
1 d 
 1 3 3 1 
=−  + + + 
8 ds 
 s − 3 s −1 s +1 s + 3
1
 1 3 3 1 
= 
 + + + 
8
 (s − 3)

2
(s −1) 2
(s +1) 2
(s + 3) 2 

Note After getting step (1), we could have applied the first shifting property and
got the same result.

t 

(ii) L(t cos 2t cos t ) = L  (cos 3t + cos t )


 2 


1 d 
= − L (cos 3t +ccos t )

2  ds 
1 d  s s 
=−  2 + 2 
2 ds  s + 9 s +1
5.48 Mathematics II

1  s 2 + 9 − 2 s 2 s 2 +1− 2 s 2 
=−  2 +
2  (s + 9) 2 (s 2 +1) 2 
1  s2 −9 s 2 −1 
=  2 +
2  (s + 9) 2 (s 2 +1) 2 
3
 1 

(iii) L(t sin 3 t ) = L  sin t − sin 3t 


4 4 


1 d
=− {L (3 sin t − sin 3t )}
4 ds
1 d  3 3 
=−  2 − 2 
4 ds  s +1 s + 9 
3 2s 2 s 
=− − 2 + 2 
4 (s +1)

2
(s + 9) 2 
3 
 1 1 
= s
− 2 − 2 2
2 
 (s +1)

2
(s + 9) 

 1− cos 2at 

(iv) L {(t sin at ) 2 }= L t 2  
  2 

2
1 d
= ( −1) 2 2 {L(1− cos 2at )}
2 ds
1 d2  
1 − s 


= 2  2
2 ds  
 s s + 4a 
2


1 d 
 1 s 2 − 4a 2 
= − 2 + 2 
2 ds 
 s
 (s + 4a 2 ) 2 
1  2 (s 2 + 4a 2 ) 2 ⋅ 2 s − (s 2 − 4a 2 ) ⋅ 2 (s 2 + 4a 2 ) ⋅ 2 s 
=  3+
2  s (s 2 + 4a 2 ) 4 

1  2 2 s (12a − s ) 
2 2
=  3+ 2
2  s (s + 4a 2 )3 
1 s (12a 2 − s 2 )
= +
s 3 (s 2 + 4a 2 )3

Example 5.7 Use Laplace transforms to evaluate the following;

∞ ∞

∫ te−2t sin 3t dt ∫ te
−3 t
(i) (ii) cos 2t dt
0 0

∫e
− st
(i) (t sin 3t ) dt = L(t sin 3t ) (by definition) (1)
0
 Laplace Transforms 5.49

d
Now L (t sin 3t ) =− L (sin 3t )
ds
d  3 
=−  2 
ds  s + 9 
6s
= (2)
( s 2 + 9) 2

Putting (2) in (1), we have

∞
6s
∫ te
− st
sin 3t dt = ,s > 0 (3)
0
( s + 9) 2
2

Putting s = 2 in (3), we get

∞
12
∫ te
−2 t
sin 3t dt = .
0
169

∫e
− st
(ii) (t cos 2 t ) dt = L(t cos 2t ) , by definition (1)
0

Now d
L(t cos 2t ) =− L(cos 2t )
ds
d  s 
=−  2 
d s  s + 4 
s2 − 4
= (2)
( s 2 + 4) 2

Inserting (2) in (1), we have

∞
s2 − 4
∫ te
− st
cos 2t dt = ,s > 0 (3)
0
( s 2 + 4) 2

Putting s = 3 in (3), we get

∞
5
∫ te
−3 t
cos 2t dt = .
0
169

Example 5.8 Find the Laplace transforms of the following functions:

(i) te−4t sin 3t; (ii) t cosh t cos t; (iii) te−2t sinh 3t (iv) t 2e−t cos t.
(i) L {te−4t sin 3t}= [ L(t sin 3t )] s → s + 4 (1)
(by the first shifting property)
5.50 Mathematics II

d
Now L(t sin 3t ) =− L(sin 3t )
ds
d  3 
=−  2 
d s  s + 9 
6s
= (2)
( s + 9) 2
2

Using (2) in (1), we have

 6s 
L {te−4t sin 3t}=  2 
 (s + 9) 2 
 s→s+4
6 (s + 4)
= 2
(s + 8s + 25) 2

Note The same problem has been solved by using an alternative method in
Worked Example (6) in Section 5(a).
t
 

(ii) L {t cosh t cos t}= L  (et + e−t ) cos t 


2 


1
= [ L(t cos t ) s → s −1 + L (t cos t ) s → s +1 ] (1)
2
d  s 
Now L (t cos t ) =−  
ds  s 2 +1
s 2 −1 (2)
=
( s 2 +1) 2
Using (2) in (1), we have
1  ( s −1) 2 −1 ( s +1) 2 −1 
L (t cosh t cos t ) =  2 + 2
2  ( s − 2 s + 2) 2
( s + 2 s + 2) 2 
1  s2 − 2 s s 2 + 2 s 
=  2 + 2
2  ( s − 2 s + 2) 2
( s + 2 s + 2) 2 

(iii) L{t e−2t sinh 3t}= L{t sinh 3 t}s → s + 2 (1)

d
Now L (t sinh 3t ) =− L (sinh 3t )
ds
d  3 
=−  2 
ds  s − 9 
6s
= (2)
( s − 9) 2
2

Using (2) in (1), we have

 Laplace Transforms 5.51

6 ( s + 2) 6 ( s + 2)
L {te−2t sinh 3t}= =
{( s + 2) 2 − 9}2 ( s −1) 2 ( s + 5) 2

Aliter

 1 

L {te−2t sinh 3t}= L te−2t ⋅ (e3t − e−3t )


 2 


1
= L { tet − te−5t }
2
1 1 1  
=   − 
2
2
 ( s − 2 ) 2
( s + 5) 


1

 12 s + 24  
 6 ( s + 2)
=  2
=
2

 ( s −1) 2
( s + 5) 

 ( s − 1 ) 2 ( s + 5) 2

(iv) L{t2 e−t cos t} = [L (t2 cos t)]s → s + 1 (1)

d2
Now L (t 2 cos t ) = (−1) 2 L (cos t )
ds 2
d2  s 
= 2  2 
ds  s +1
d 
 1− s 2  

=  2 2

ds 
 ( s + 1 ) 


2 ( s 3 − 3s ) (2)
=
( s 2 +1)3
Using (2) in (1), we have
2 {( s +1)3 − 3 ( s +1)}
L {t 2 e−t cos t}=
( s 2 + 2 s + 2) 3

Example 5.9 Find the inverse Laplace transforms of the following functions:
 a s +a
2 2

(i) log 1−  (ii) log  
 s   s + b
2 2


s +1
2  s −1
(iii) log (iv) s log   + k (k is a constant)
s ( s +1)  s +1

1
(i) L−1 { φ ( s )}=− L−1 { φ ′ ( s )} (1)
t
 a  s − a  1 −1  d  s − a 
∴ L−1 log 1−  = L−1 log   =−  log  
 s 
L
 s   s  t  ds 
5.52 Mathematics II

1 d
=− L−1 {log (s − a ) − log s}
t ds
1  1 1
=− L−1  − 
t  s − a s 
1 1
=− (e at −1) = (1− e at )
t t

(ii) By rule (1),

 s 2 + a 2  1 d
L−1 log 2 2
 =− L−1 [log(s 2 + a 2 ) − log(s 2 + b 2 )]
 s + b  t ds
1   2s 2s  
=− L−1 
 2 − 2 
2
t 

 s + a 2
s + b 


2
= (cos bt − cos at )
t

(iii) By rule (1),

 s 2 +1 
L−1 log   =−1 L−1 d [log (s 2 +1) − log s − log(s +1)]]
 s (s +1)  t ds
 
1   2s 1 1 
=− L−1 
 2 − − 

t  
 s + 1 s s + 1 


1
=− (2 cos t −1− e−t )
t
1
= (1+ e−t − 2 cos t )
t

(iv) By rule (1),

  s −1  1 d
L−1  s log  + k  =− L−1 [s log(s −1) − s log (s +1)] + L−1 (k )

  
 s +1  t ds
1  s s 
=− L−1  + log(s −1) − − log(s +1) + k δ(t )
 
t  s −1 s +1 
1  s s  1 −1
=− L−1  − − L [log(s −1) − log(s +1)] + k δ(t )
 
t  s −1 s +1 t
1  2 s  1  1  1 1 
=− L−1  2  − − L−1  − + k δ(t )
    
t  s −1 t  t   s −1 s +1
 Laplace Transforms 5.53

Note
 −1  s 
[ L   and L−1  s  do not exist, as s and s are improper rational
  s −1   s +1  s−1 s +1

 s s  2s ,
functions. Hence we have simplified  −  as 2 which is a proper
 s −1 s +1   s −1
rational function. ]

2 1
=− cosh t + 2 (et − e−t ) + k δ (t )
t t
2 2
= 2 sinh t − cosh t + k δ (t )
t t
Example 5.10 Find the inverse Laplace transforms of the following functions:
−1  s + a 

(i) cot–1 (as) (ii) tan 
 b 
 2   2 
(iii) cot −1   (iv) tan −1  2 
 s +1  s 

1
(i) L−1 { φ (s )} = − L−1{ φ ′ (s )} (1)
t

1 d 
∴ L−1{cot −1 (as )} = − L−1  cot −1 (as )
t  
d s 
1  −a 
= − L−1  
t  1+ a s
2 2 
a  1 
= L−1  2 2

t 1+ a s 
1  1/ a
 

 = 1 sin t .
= L−1 
 2 2
t 
 s + (1/ a ) 
  t
 a


 −1  s + a  1 −1  d −1  s + a 

(ii) L−1 
tan  
 = − L  tan   


  b  
 t d s  b  

1  1/ b 
= − L−1  
  
2

 1 +  s + a 
t

  b  
 
1  b 
= − L−1  
 
 (s + a ) + b
2 2
t 
5.54 Mathematics II

1
= − e−at sin bt
t .

 −1  2   1 d  2 
(iii) L−1 
cot   = − L−1  cot −1   .

  s + 1 
 t  d s  s + 1 
   
  




 
1   1 
 
2 
=− L−1  − − 
t 


4  (s +1) 2 
 1+  


 (s +1) 2  
1  2 
=− L−1  
 
 (s +1) + 4 
2
t
1
=− e−t sin 2t.
t


 −1  2 
 1 −1  d −1  2 

(iv) L−1 
tan  2 
 =− L  tan  2 


  s 
 t d s  s 
 
 
1 −1  1  −4 
=− L  . 3 
t 1+ 4  s 
 
 s4 
4  s 
= L−1  4 (2)
t  s + 4
 
s s
Consider 4 =
s + 4 (s 2 + 2) 2 − (2 s ) 2
s
= 2
(s − 2 s + 2) (s 2 + 2 s + 2)
1 1 1 
=  2 − 2 ,
4  s − 2 s + 2 s + 2 s + 2 

by resolving into partial fractions
1 1 1 
=  − 
4  (s −1) +1 (s +1) +1
 2 2

 s  1 t
∴ L−1  4  =  e sin t − e−t sin t 
 s + 4  4  
1
= sin t sinh t (3)
2
Using (3) in (2), we have
 −1  2 
  2
L−1 
tan  2 
 = sin t sinh t


  s  
 t
 Laplace Transforms 5.55

Example 5.11 Find the Laplace transforms of the following functions:

sinh t e−at − e−bt e at −cosbt 2 sin 2t sin t

(i) ; (ii) ; (iii) ; (iv) ;
t t t t
 sin t 
2

(v) 
 t 
.

 ∞
 f (t ) 

(i) L  = ∫ L {f (t )}ds (1)

 t 
 
 s

 ∞
 sinh t 

∴ L  = ∫ L (sinh t ) ds

 t 
 
 s
∞
1
=∫ ds
s
s −1
2

∞
1  s −1
=  log  
 2  s +1 s
1  s −1  1  s −1 
=  log  
 − log  
 2  s +1  s → ∞ 2  s +1 
  1 
 1− 
1    s +1
=  log  s 
1
+ log  
2  1  2  s −1 
 1+ 
 s  s→∞
1 1  s +1 1  s +1
= log 1 + log   = log  
 s −1
2 2 
 s −1 2

 e−at − ebt 
  ∞

(ii) L
 
 = ∫ L (e − e ) d s , by rule
− at −bt
(1)


 t 

 s

∞
 1 1 
= ∫  − d s
 s + a s + b 
s

  s + a   s + a 
=  log   − log  

  s + b  s → ∞  s + b 
  a 
 1+   s + b 
 
=  log  s  + log  
 
 1+ b   s + a 
  
 s  s → ∞
5.56 Mathematics II

 s + b 
= log1+ log  
 s + a 
 s + b 
= log  
 s + a 


 e at −cosbt 

∞

(iii) L
 
 ∫ L (e − cos bt ) ds , by rule (1)
= at


 t 

 s
∞
 1 s 
= ∫  −  ds
 s − a s 2 + b 2 
s

∞
 1 
=  log ( s − a ) − log ( s 2 + b 2 )
 2  s

    
  s − a   s − a 
=  log   − log  
  s 2 + b 2   s 2 + b 2 
  s →∞  

    s 2 + b 2 
  1− a /s   
=  log   + log 
 2 2 
  1+ b /s   s − a 
  s →∞  

s2 + b2
= log 1 + log
( s − a )2

1 
 s2 + b2 

= log 
 
2
2 
( s−a ) 
 

∞

 2 sin 2t sin t 

(iv) L 
 
 = ∫ L(2 sin 2t sin t ) ds, by rule (1)


 t 

 s
∞

= ∫ L( cos t − cos3t ) ds
s
∞
 s s 
= ∫  2 − ds
 s +1 s 2 + 9 
s

∞
1  s 2 +1 
=  log  2 
 2  s + 9  s

1   s 2 +1  1  s 2 +1 

= log  
 − log 
2   s 2 + 9  s → ∞ 2  s 2 + 9 
 Laplace Transforms 5.57

  1 
  1+ 2 
1   s  1  s 2 + 9 

=  log   + log 
2  1 + 9/ s 2  2  s 2 +1 
   

  s→∞
1 1  s 2 + 9 
= log1 + log  2 
2 2  s +1 
1  s 2 + 9 
= log  2 
2  s +1 

 ∞
 f (t ) 

∞

(v) L 2  = ∫ ∫ L{ f (t )} ds ds (2)

 t 
 
 s s


 sin 2 t 

∞ ∞

∴ L
 2  = ∫ ∫ L (sin
2
t ) ds d s

 t 
 
 s s

∞ ∞
1− cos 2t 
 
=∫ ∫ L  2
 ds d s



s s

1
∞ ∞
1 s 
=
2 ∫ ∫  s − s 2
 ds ds
+ 4 
s s
∞
 s2 + 4
log  2  ds,
1 1
= ∫
2 s 2  s 
by putting a = 0 and b = 2 in (iii) above.
 ∞ 
1   s 2 + 4   2 s 2  
∞
 

= s log  2  − ∫ s  2
 −  ds
4   s   s + 4 s  

 
s s

by integrating by parts.
s  s 2 + 4   s2 + 4 1 ∞ 8
=  log  2  − log  2 + ∫ 2
s
ds
 4  s  s →∞ 4  s  4 s s + 4

s  s 2   −1 ∞
s 
= L + log  2  − cot  , say
4  s + 4   2 s

s  s 2  s
= L + log  2  + cot −1   (3)
4  s + 4   2 

 4
s/4

Now L = log 1+ 2 

 s 
5.58 Mathematics II

1
 s 2 /4  s
  4  
= log 1+ 2  
 s  

 1
 
  s 
s 2 /4 
 4 
lim ( L) = log  lim 1+ 2   
s →∞
 s →∞  s   


 
 
= log (e ) = log 1 = 0
0
(4)
Using (4) in (3), we have
 sin 2 t  s  s 2  s
L  2  = log  2  + cot −1   .
 t  4   s + 4   2 

Example 5.12 Use Laplace transforms to evaluate the following:

∞ ∞
e−t sin 3 t sin 2 t
(i) ∫ t
dt (ii) ∫ t et
dt
0 0
∞
 cos at −cos bt  ∞
 e−2t − e−4t 
(iii) ∫  t
d t

(iv) ∫  t
 d t .

0 0

∞
e−t sin 3 t ∞  sin 3 t  
∫ d t =  ∫ e−st   d t 
 t  
(i)
0
t  0  s =1
 sin 3 t 
= L   (1)
 t 
s =1

 sin 3 t  ∞
Now L   = ∫ L(sin 3 t ) ds
 t 
s
∞
3
=∫ ds
s s + ( 3 )2
2

∞
  1   s 
=  3 −  cot −1  
  3   3 
 s
 s 
= 0 + cot −1   (2)
 3 

Using (2) in (1), we have

∞
e−1 sin 3t  1  π
∫ d t = cot −1   = .
s
t  3  3
 Laplace Transforms 5.59

∞
sin 2 t
∞
 2 
−t  sin t 
(ii) ∫ t et
d t = ∫  t  dt
e
0 0

∞  sin 2 t  
=  ∫ e−st   dt 
 0  t   s =1
 sin 2 t 
= L   (1)
 t 
s =1

 sin 2 t  1− cos 2t 

Now L   = L 
 2 t 

 t 
∞
1
=
2 ∫ L(1− cos 2t ) ds
s

1
∞
1 s 
=
2 ∫  s − s 2
d s
+ 4 
s
∞
  
1  s 
=  log  
2   s 2 + 4 
 s
   
1 s 
2
1  s 
= log 2  − log 
2  s + 4  2  s 2 + 4 
s→ ∞  
 s 2 + 4 
1  
= log
1  1 
2  + log  
2  1 + 4 /s s → ∞ 2  s 
 s 2 + 4 
1 1  
= log 1+ log  
2 2  s
 
 s 2 + 4 
1  
= log  

2 s (2)


Using (2) in (1), we have

∞
sin 2 t 1 1
∫ t et
dt = log 5 = log 5
2 4
0

∞
 cos at − cos bt  ∞ 
  e−st  cos at − cos bt dt 
(iii) ∫  t



dt = ∫
 0
 t 
0 s = 0
 
 cos at − cos bt 
=L  (1)


 t s = 0
5.60 Mathematics II

∞

 cos at − cosbt 

Now L  = ∫ L(cos at − cos bt ) ds


 t 

 s
∞
 s s 
= ∫  2 − 2  ds
 s + a
s
2
s + b 2 
∞
1  s 2 + a 2 
=  log  2 2

 2  s + b  s
1  s 2 + a 2  1  s 2 + a 2 
=  log  2   − log  
 2  s + b 2  2  s 2 + b 2 
 s →∞
s 2 + b2
= log (2)
s2 + a2
Using (2) in (1), we have
∞
 cos at − cos bt  b
∫   dt = log  
0
t  a

∞
 e−2t − e−4t  ∞  −2 t −4 t  
  e−st  e − e  dtt 
(iv) ∫  t   ∫  t  

 d t = 
0 0  s=0
 e−2t − e−4t 
= L   (1)
 t 
s=0

 e−2t − e−4t  ∞
Now L   = ∫ L (e−2t − e−4t ) ds
 t 
s
∞
 1 1 
= ∫  −  ds
 s + 2 s + 4 
s
∞
  s + 2 
=  log  
+ 
  s 4  s
  s + 2   s + 2 
=  log   − log  

 
  s + 4  s →∞  s + 4 
 s + 4 
= log   (2)
 s + 2 

Using (2) in (1), we have

∞
 e−2t − e−4t 
∫  t
 dt = log 2

0
 Laplace Transforms 5.61

Example 5.13 Find the inverse Laplace transforms of the following functions:
s s
(i) (ii)
(s + a 2 ) 2
2
(s − 4) 2
2

4(s −1) s2 −3
(iii) (iv)
(s 2 − 2 s + 5) 2 (s 2 + 4 s + 5) 2

s +1
(v)
(s + 2 s − 8) 2
2

(i) L−1{φ (s )} = t ⋅ L−1 ∫ φ (s ) ds (1)

s


 s 

∞
s
∴ L−1 
 2  = t ⋅ L−1
2 2 ∫ ds

 
 (s + a ) 
 (s + a 2 ) 2
2
s

∞
1 dx
= t L−1 ∫ 2 x2
, on putting s2 + a2 = x
s2 + a2

∞
t −1  1 
= L − 
2  x s2 + a2

t −1  1 
= L  2 2

2  s + a 
t
= sin at.
2a


 s 

∞
s
(ii) L−1 
 2  = t ⋅ L−1
2 ∫ ds

 
 (s − 4) 
 (s − 4) 2
2
s

t  1 
= L−1  2  , as in (i) above.
2  s − 4 
t
= sinh 2t
4
Note The inverse transform in this case can also found out by resolving the
given function into partial fractions.

 
 4(s −1) 
−1

 
−1  s −1 
(iii) L  2 2
= 4L  2
 (s − 2 s + 5) 

 
  (s −1) + 2  
2 2
 
  


 s 

= 4et L−1 
 2  , by the first shifting property
2 2


 (s + 2 ) 


5.62 Mathematics II

t
= 4et sin 2t , as in problem (i)
4
= tet sin 2t.

 s2 −3 
 = L−1  (s + 4 s + 5) − (4 s + 8) 
 2
(iv) L−1 
 2 2
  
 (s + 4 s + 5) 
 
  (s 2 + 4 s + 5) 2 


 1 
 
−1  s+2 
= L−1 
 2 
− 4 L   2 2

 s + 4s + 5
 
 
 (s + 4 s + 5) 


 1 
  s+2 
= L−1 
 
 − 4 L 
−1

2

 (s + 2) +1

2


2
 {(s + 2) +1} 
t
= e−2t sin t − 4e−2t sin t , as in problem (i)
2
= e−2t (1−2t) sin t.


 s +1 
 
 = L−1  s +1 

(v) L−1 
 2 2

 2 2

 
 (s + 2 s − 8) 
 
[(s +1) − 3 ] 

2



 s 

= e−t ⋅ L−1 
 2 
2 2
,


 (s − 3 ) 


t
= e−t ⋅ sinh 3t , proceeding as in problem (ii)
6

t
= e−t sinh 3t
6

EXERCISE 5(b)

Part A
(Short Answer Questions)
1. State the formula for the Laplace transform of a periodic function.
2. Find the Laplace transform of f(t) = t, in 0 < t < 1 if f (t + 1) = f(t)
3. State the relation between the Laplace transforms of f(t) and t · f(t).
4. State the relation between the inverse Laplace transforms of φ (s) and φ′ (s).
1
5. State the relation between the Laplace transforms of f(t) and f (t ) .
t
6. State the relation between the inverse Laplace transform of φ (s) and its
integral.
Find the Laplace transforms of the following functions:
1
7. t sin at 8. t cos at 9. sin kt − kt cos kt
2a
 Laplace Transforms 5.63

t
10. sin kt + kt cos kt 11. (1− cos at )
a2
1
12. cos kt − kt sin kt .
2
Find the inverse Laplace transforms of the following functions:
 s + 1  s + 1  a
13. log   14. log  15. log 1+ 
 s −1  s   s
 s + a   s   s 2 + 1 
16. log  17. log  18. log  2 
 s + b   s −1  s + 4 
a
19. cot−1 s 20. tan −1
s
Find the Laplace transforms of the following functions:
sin at 1− e−t 1− et
21. 22. 23.
t t t

1− cos at sin 2 t
24. 25.
t t

Part B
Find the Laplace transforms of the following periodic functions:
1
26. f (t ) = E , in 0 ≤ t <
E
1 2π
= 0, in ≤ t <
E n
 2π 
given that f t +  = f (t )
 n
T
27. f (t ) = E , in 0 ≤ t <
2
= − E, in T/2 ≤ t < T
given that f (t + T) = f (t)
28. f (t) = et, in 0 < t < 2π and f (t + 2π) = f (t)
t 
29. f (t ) = sin   , in 0 < t < 2 π and f (t + 2π) = f (t)
 2
30. f (t) = |cos ωt|, t ≥ 0
 π/ω π / 2ω
 Hint: f (t )is periodic with period ( π ω ) and e−st | cos ωt| dt =


∫ ∫ e−st cos ωt dt
0 0

π/ω 
+ ∫ e−st ( − cos ωt ) dt 
π / 2ω 
5.64 Mathematics II

31. f(t) = t, in 0 < t < π

= 0, in π < t < 2π,
given that f(t + 2π) = f(t)
32. f(t) = sin t, in 0 < t < π
= 0, in π < t < 2π,
given that f(t +2π) = f(t).
p
33. f (t ) = 0, in 0 < t <
w
p 2p
= - sin w t , in < t < ,
w w
Ê 2p ˆ
given that f Át +
Ë ˜ = f (t ) .
w¯
34. f(t) = t, in 0 < t < π
= 2π − t, in π < t < 2π,
given that f(t + 2π) = f(t).
35. f(t) = t, in 0 < t < π
= π − t, in π < t < 2π,
given that f(t + 2π) = f(t).
Find the Laplace transforms of the following functions:
36. t sinh3 t 37. t cos3 2t 38. t sin 3t sin 5t
39. t sin 5t cos t 40. (t cos 2t)2 41. t2 sin t cos 2t
42. t2 e−2t sin 3t 43. te3t cos 4t 44. t2 e−3t cosh 2t
45. t sinh 2t sin 3t
Find the inverse Laplace transforms of the following functions:

 a2  s2 + a2 ( s − 2)2
46. log 1 + 2  47. log 48. log
 s  ( s + b)
2
s 2 +1

 s − a  −1  1  −1  s + 2 

49. s log  +a 50. tan   51. tan 
 s + a   2s   3 

−1  a  
52. cot  53. tan-1 (s2)
 s + b 
Find the values of the following integrals, using Laplace transforms:
∞ ∞ ∞
 e−t − e−3t 
∫te 
−2 t
∫t e−t sin t dt ∫  dt
2
54. cos 2t dt 55. 56.  t 
0 0 0
 Laplace Transforms 5.65

∞ ∞ ∞
(1− cost ) e−t  e−at − cosbt  e− 2t sint sinht
57. ∫ dt 58. ∫   dt 59. ∫ dt
0
t 0
 t  0
t
Find the Laplace transforms of the following functions:
1− e−t  sin2t 
2
1− cosat
60. 61. 62.  
t t  t 

sin 3t sint 1− cost

63. 64.
t t2
Find the Laplace inverse transforms of the following functions:
s s s−2
65. 66. 67.
(s 2 + 1) 2 (s 2 − a 2 ) 2 (s 2 − 4 s + 5) 2
(s − a ) 2 s 2 + 8s + 16 s+4
68. 69. 70.
(s 2 − a 2 ) 2 (s 2 + 6 s + 10) 2 (s + 8s + 15) 2
2

5.8 LAPLACE TRANSFORMS OF DERIVATIVES

AND INTEGRALS
In the following two theorems we find the Laplace transforms of the derivatives and
integrals of a function f(t) in terms of the Laplace transform of f(t). These results will
be used in solving differential and integral equations using Laplace transforms.
Theorem
If f(t) is continuous in t ≥ 0, f ′(t) is piecewise continuous in every finite interval in
the range t ≥ 0 and f(t) and f ′(t) are of the exponential order, then
L{f ′(t)} = sL{f(t)} − f(0)

Proof:
The given conditions ensure the existence of the Laplace transforms of f(t) and f ′(t).
∞
By definition, L{f ′(t )} = ∫ e−st f ′(t ) dt
0
∞

= ∫ e−st d [f (t )]
0

∞ ∞

=  e−st ⋅ f (t ) − ∫ (−s) e−st f (t ) dt ,on integration by parts.

 0
0

− st
= li [e f (t )] − f (0) + s ⋅ L{f (t )}
t →∞

= 0 − f(0) + sL {f(t)} [ f(t) is of the exponential order]

=sL {f(t)} − f(0) (1)

5.66 Mathematics II

Corollary 1
In result (1) if we replace f (t) by f ′(t) we get
L{f″ (t)}= sL {f ′ (t)} − f ′(0)
= s [sL {f(t)} − f (0)] −f ′(0), again by (1)
= s2 L{f (t)}−s f (0) −f ′(0) (2)
Note
1. Result (2) holds good, if f(t) and f′(t) are continuous in t ≥ 0,f″ (t) is piecewise
continuous in every finite interval in the range t ≥ 0 and f(t), f ′ (t) and f″ (t) are
of the exponential order.

Corollary 2
Repeated application of (1) gives the following result:
L{ f(n)(t)} = sn L{f(t)}−s n−1 f (0) −sn−2 f ′ (0) −  − f (n−1) (0) (3)
Note
2. Result (3) holds good, if f(t) and its first (n − 1) derivatives are continuous in
t ≥ 0, f(n) (t) is piecewise continuous in every finite interval in the range t ≥ 0
and f(t), f ′(t), … , f (n) (t) are of the exponential order.
3. If we take L { f (t)}= f (s), result (1) becomes

L{ f ′ (t)} = sf (s) − f (0) (4)

If we further assume that f (0) = 0, the result becomes
L{ f ′ (t)}= sf (s) (5)
In terms of the inverse Laplace operator, (5) becomes
L−1 {sf (s)}= f ′ (t) (6)
From result (6), we get the following working rule:
d
L−1{sφ (s ) = L−1{φ (s ) , provided that
dt
f(0) = L−1 {f (s)}t = 0 = 0.
Thus, to find the inverse transform of the product of two factors, one of
which is ‘s’, we ignore ‘s’, and find the inverse transform of the other factor;
we call it f (t), verify that f (0) = 0 and get f ′(t), which is the required inverse
transform.
4. In a similar manner, from result (2) we get
d 2 −1
L−1{s 2φ (s )} = L {φ (s )} , provided that
dt 2
f(0) = 0 and f ′(0) = 0, where
f(t) = L−1 {f (s)}.
Theorem
If f(t) is piecewise continuous in every finite interval in the range t ≥ 0 and is of the
exponential order, then
 Laplace Transforms 5.67

t  1
L  ∫ f (t ) d t  = L{ f (t )}
 0  s
.

Proof:
t

g (t ) = ∫ f (t ) dt
Let
0

∴ g′(t) = f (t)
Under the given conditions, it can be shown that the Laplace transforms of both f(t)
and g(t) exist.
Now by the previous theorem,
L{g′(t)}=sL{g(t)} − g(0)
t  0
s⋅L  ∫ f (t ) dt  − ∫ f (t ) dt = L{ f (t )}
i.e.,
 0  0

t  1
∴ L  ∫ f (t ) d t  = L{ f (t )} (1)
 0  s

Corollary
t t  1
L  ∫ ∫ f (t ) dt dt  = 2 L{ f (t )} , as explained below.
 0 0  s

Let ∫ f (t ) dt = g (t )
.
0

Then, by result (1) above,

t
1
L ∫ g (t ) dt = L{ g (t )}
0
s

t t  1 t
 f (t ) dt dt  = L ∫ f (t )dt
i.e., L ∫ ∫
 0 0  s 0

1 1
= ⋅ L{ f (t )} , again by (l)
s s
1
= L{ f (t )} (2)
s2
Generalising (2), we get
t t t  1
L  ∫ ∫ ∫ f (t ) (dt )  = s n L{f (t )}
 n
(3)
 0 0 0 
5.68 Mathematics II

Note
1. If we put L{ f (t)} = f (s), result (1) becomes
t  1
L  ∫ f (t ) dt  = φ (s ) (4)
 0  s

Result (4) can be expressed, in terms of L−1 operator, as


1 

t
(5)
L−1  φ (s ) = ∫ f (t ) dt

s
 
 0

From (5), we get the following rule:

1 

t

L−1  φ (s ) = ∫ L−1{φ (s )} dt .


s 

 0

Thus, to find the inverse Laplace transform of the product of two factors, one of
1 1
which is , we ignore , find the inverse transform of the other factor and integrate
s s
it with respect to t between the limits 0 and t.
2. In a similar manner, from (2) above, we get
1
 
t t

L−1  2 φ (s ) = ∫ ∫L
−1
{φ (s )} dt dt .


s 
0 0

t  1 1
0

L  ∫ f (t ) dt  = L{ f (t )} + ∫ f (t ) dt
3.  a  s s a
t

g (t ) = ∫ f (t ) dt and g ′(t ) = f (t ),
If we let
a

we get L{g ′(t )} = sL{g (t )} − g (0)

t  0
i.e., L{ f (t )}= sL  ∫ f (t ) dt  − ∫ f (t ) dt
 a  a

t  1 1
0

or L  ∫ f (t ) dt  = L{ f (t )} + ∫ f (t ) dt .
 a  s s a

5.9 INITIAL AND FINAL VALUE THEOREMS

We shall now consider two results, which are derived by applying the theorem on
Laplace transform of the derivative of a function.
 Laplace Transforms 5.69

The first result, known as the initial value theorem, gives a relation between
lim[ f (t )] and lim[sφ (s )], where φ (s ) = L { f (t )}
t →0 s→∞

The second result, known as the final value theorem, gives a relation between
lim[ f (t )] and lim[sφ (s )] .
t →∞ s→0

5.9.1 Initial Value Theorem

If the Laplace transforms of f (t) and f ′(t) exist and L{ f(t)} = f (s), then
lim[f (t )] = lim[sφ (s )]
t →0 s→∞

Proof:
We know that L{f ′(t)} = sf (s) − f (0)
∴ sf (s) = L{f ′(t)} + f (0)
∞

= ∫ e−st f ′(t ) dt + f (0)

0

∫e f ′(t ) dt + f (0),
− st
∴ lim [sφ (s ) = lim
s→∞ s→∞
0

= ∫ lim {e−st f ′(t )} dt + f (0),

s →∞
0

assuming that the conditions for the interchange of the operations of integration
and taking limit hold.
i.e. lim [sφ (s )] = 0 + f (0)
s →∞

= lim [ f (t )].
t →0

5.9.2 Final Value Theorem

If the Laplace transforms of f(t) and f ′(t) exist and L{f (t)} = f (s), then
lim [ f (t )] = lim [sφ (s )] , provided all the singularities of {sf (s)] are in the left
t →∞ s →0

half plane Rl(s)< 0.

Proof:
We know that L{f ′(t)} = sf (s) − f (0)
∴ sf (s) = L{ f ′(t)}+f (0)
∞

= ∫ e−st f ′(t ) dt + f (0)

0
5.70 Mathematics II

∴ lim [sφ (s )] = lim ∫ e−st f ′(t )dt + f (0)

s→0 s→0
0

= ∫ lim{e−st f ′(t )}dt + f (0) , assuming that the conditions for

s→0
0
the interchange of the operations of integration and taking limit hold.
∞

i.e. lim [sφ (s )] = ∫ f ′(t )dt + f (0)

s→0
0

= [ f (t )]∞
0 + f ( 0)

= lim[ f (t )] − f (0)+ f (0)

t →∞

Thus lim [f (t )] = lim [sφ (s )]

t →∞ s →0

5.10 THE CONVOLUTION

Another result, which is of considerable practical importance, is the convolution

theorem that enables us to find the inverse Laplace transform of the product of f (s )
and g (s ) in terms of the inverse transforms of f (s ) and g (s ) .
Definition The convolution or convolution integral of two function f (t) and g (t),
defined in t ≥ 0, is defined as the integral
t

∫ f (u ) g (t − u ) du
0

It is denoted as f (t) * g (t) or (f * g) (t)

t

i.e. f (t ) * g (t ) = ∫ f (u ) g (t − u ) du
0

= ∫ f (t − u ) g [t − (t − u )] du , on using the result

0

t t

∫ φ (u ) du = ∫ φ (t − u ) du
0 0
t

= ∫ g (u ) f (t − u ) du
0

= g (t )* f (t ).
 Laplace Transforms 5.71

Thus the convolution product is commutative.

5.10.1 Convolution Theorem

If f (t) and g (t) are Laplace transformable,
then L{f (t) * g(t)} = L{f (t)}∙L{g (t)}

Proof:
∞

By definition, L{f (t )* g (t )} = ∫ e−st {f (t ) * g (t )} dt ,

0
∞ t 
= ∫ e−st  ∫ f (u ) g (t − u )du  dt,
0  0 
by the definition of convolution.
∞ t

=∫ ∫e
− st
f (u ) g (t − u )du dt (1)
0 0

The region of integration for the double integral (1) is bounded by the lines u = 0,
u = t, t = 0 and t = ∞ and is shown in the Fig. 5.6.

u
t
=

t=∞
u

t=0
(u, u)
(∞, u)

t
u=0
Fig. 5.6
Changing the order of integration in (1), we get,
∞ ∞

L{f (t ) * g (t )} = ∫ ∫e
− st
f (u ) g (t − u ) dt du (2)
0 u

In the inner integral in (2), on putting t − u = v and making the consequent changes,
we get,
∞ ∞

L{f (t ) * g (t )} = ∫ ∫e
− s (u + v )
f (u ) g (v) dv du
0 0
∞ ∞ 
= ∫ e−su f (u )  ∫ e−sv g (v)dv  du
0  0 
5.72 Mathematics II

∞ ∞

= ∫ e−su f (u )du ⋅ ∫ e−sv g (v)dv

0 0

∞ ∞

= ∫ e−st f (t )dt ⋅ ∫ e−st g (t )dt ,

0 0

on changing the dummy variables u and v.

= L{f (t)}·L{g (t)}
Note If L{ f (t ) } = f (s ) and L{g (t )} = g (s ) , the convolution theorem can be put as

L{f (t ) * g (t )} = f (s ) ⋅ g (s ) (3)
In terms of the inverse Laplace operator, result (3) can be written in the following
way.
L−1{f (s ) ⋅ g (s )} = f (t ) * g (t )

= ∫ f (u ) g (t − u ) du (4)
0

Result (4) means that the inverse Laplace transform of the ordinary product of two
functions of s is equal to the convolution product of the inverses of the individual
functions.

WORKED EXAMPLE 5(c)

Example 5.1 Using the Laplace transforms of derivatives, find the Laplace
transforms of
(i) e−at (ii) sin at
(iii) cos2 t (iv) t n (n is a positive integer)

(i) L{f ′(t)} = sL{f(t)} − f (0) (1)

Putting f (t) = e−at in (1), we get

L (−ae−at) = sL (e−at) − 1
i.e. (s + a) L(e−at) = 1
1
∴ L(e−at ) =
s+a
(ii) L{f″(t)} = s2 L{f(t)} − sf (0) − f ′(0) (2)

Putting f (t) = sin at in (2), we get

L (−a2 sin at) = s2L (sin at) − s × 0 − a
 Laplace Transforms 5.73

i.e. (s2 + a2) L (sin at) = a

a
∴ L(sin at ) =
s + a22

(iii) Putting f(t) = cos2 t in (1), we get

L (− 2 cos t sin t) = sL (cos2 t) − 1
i.e. s·L (cos2 t) = 1 − L (sin 2t)
2
= 1−
s +4
2

s2 + 2
=
s2 + 4
s2 + 2
∴ L(cos 2 t ) =
s (s 2 + 4)

(iv) L{f (n ) (t )} = s n L{f (t )} − s n−1 f (0) − s n−2 f ′(0) − f (n−1) (0) (3)

Putting f (t) = tn in (3) and nothing that f (n) (t) = n!

and f (0) = f ′(0) =  = f (n−1) (0) = 0 , we get

L{n!} = sn L (tn)

i.e. n!L (1) = sn L (tn)

1
i.e. n! = s n L(t n )
s

n!
∴ L(t n ) =
s n+1
t  1 
Example 5.2 Find the Laplace transform of and hence find L  .
π  πt 


 t
 1 1 (3 / 2)
L
 = L(t1/ 2 ) = = 3/ 2
 
 π
  π π s

1
(1 / 2)
=
1 1
⋅ 2 3/ 2 = 3/ 2 ( (1/2) = π )
π s 2s
In the result L{f ′(t)} = sL {f(t)} − f (0), we put
t
f (t ) = , we get
π
 1 
  1
L
 
 = s ⋅ 3/ 2 − 0

 2 πt 
 
 2s
5.74 Mathematics II

1
=
2 s

 1   1
∴ L
 
= .


 π t 

 s

Aliter
 1 
  1
 

L
 
 = L ⋅ t / π

 
 πt 
 

t 


∞

= ∫ L( t / π ) ds
s
∞
1
=∫ ds
s
2s 3/ 2
∞
 1  1
= −  = .
 
s s s

Example 5.3 Using the Laplace transforms of the derivatives, find

(i) L(t cos at) and hence L(sin at – at cos at) and L(cos at – at sin at)
1
(ii) L(t sinh at) and hence L (sinh at + at cosh at) and L(cosh at + at sinh at )
2

(i) L { f ′′(t )} = s 2 L { f (t )} − sf (0)-f ′(0) (1)

Put f(t) = t cos at in (1)

Then f ′(t) = cos at − at sin at and f ′(0) = 1

f ″(t) = − a2 t cos at − 2a sin at.
∴ L{− a2 t cos at − 2a sin at} = s2 L {t cos at} − 1
∴
[ f(0) = 0] and f ′(0) = 1
i.e. (s2 + a2) L(t cos at) = 1 − 2a L(sin at)

2a 2
= 1−
s + a2
2

s2 − a2
= 2
s + a2

s2 − a2
∴ L(t cos at ) =
(s 2 + a 2 ) 2
a a (s 2 − a 2 )
Now L(sin at − at cos at ) = −
s 2 + a 2 (s 2 + a 2 ) 2
 Laplace Transforms 5.75

a {(s 2 +a 2 ) − (s 2 − a 2 )}
=
(s 2 + a 2 ) 2
2a 3
=
(s 2 + a 2 ) 2
Taking f (t) = t cos at in the result
L{f ′(t)} = sL{f(t)} − f(0), we get
L{cos at − at sin at} = sL (t cos at) − 0

s (s 2 − a 2 )
=
(s 2 + a 2 ) 2

(ii) Put f (t) = t sinh at in (1).

Then f ′(t) = sinh at + at cosh at and f ′(0) = 0
f ′′(t) = a2t sinh at + 2a cosh at
∴ L{a2t sinh at + 2a cosh at}
=s2L(t sinh at) [ f(0) = 0 = f ′(0)]
i.e. (s2 − a2) L(t sinh at) = 2a L(cosh at)
2as
=
s − a2
2

2as
∴ L(t sinh at ) =
(s 2 − a 2 ) 2
In the result L{f ′(t)} = sL {f(t)} − f(0), if we put f(t) = t sinh at, we
have
2as 2
L(sinh at + at cosh at )= [ f(0) = 0]
(s 2 − a 2 ) 2

1
Now L(cosh at + at sinh at )
2

a
= L(cosh at ) + L(t sinh at)
2
s a 2as
= 2 + ⋅ 2
s −a 2
2 (s − a 2 ) 2
s (s 2 − a 2 ) + a 2 s s3
= = 2 .
(s − a )
2 2 2
(s − a 2 ) 2

Example 5.4 Find the inverse Laplace transforms of the following functions:

s s2
(i) (ii)
(s + 2) 4 (s − 2)3
5.76 Mathematics II

s s
(iii) (iv)
s + 4s + 5
2
(s + 2)(s + 3)
s
(v)
(s 2 + 1)(s 2 + 4)
d −1
(i) L−1 {sφ(s)} = L {φ(s)} , provided L−1{f (s)} vanishes at t = 0 (1)
dt
 s 
 
 , let us first find f (t ) = L−1 
 1 
To find L−1 
 4

 4
and then apply

 (s + 2)  
 
  (s + 2) 

rule (1)
1
Now f (t ) = e−2t L−1  4 
 s 

1 3 1 3 −2t
= e−2t t = te
3! 6

We note that f(0) = 0


 s   = d  1 t 3e−2t 

∴ By (1), L−1 
 4  


 (s + 2 ) 
 dt  6

1
= ( − 2t 3 e−2t + 3t 2 e−2t )
6
1 2 −2 t
= t e (3 − 2t ).
6

d 2 −1
(ii) L−1 {s 2φ(s )} = L {φ(s )} , provided
dt 2
f(0) = 0 f ′(0) = 0, where f(t) = L−1 {f (s)}. (2)
 s 2 
 
 1 
To find L−1 
 3
, we shall find f (t ) = L-1 
 3
and then apply rule (2).
  
 (s − 2) 
 (s − 2) 
 
1
Now f (t ) = e 2t L−1  3 
 s 

1 2 1 2 2t
= e 2t ⋅ t = t e
2! 2
1 2 2t
f ′(t ) = (2t e + 2te 2t )
2
We note that f(0) = 0 and f ′(0) = 0

−1 
 s2  
 d2   1 2 2t 

∴ By (2), L   = 2 
t e 


 (s − 2) 3


 d t 

 2 


 Laplace Transforms 5.77

d 2
= [(t + t )e 2t ]
dt
= 2(t 2 + t )e 2t + (2t + 1)e 2t
= (2t 2 + 4t + 1)e 2t


 s 

(iii) To find L−1  2 ,

 s + 4s + 5
 


 1 

we shall find f (t )=L−1  2  and then apply the rule (1).
 
 s + 4s + 5
 

 1 

Now f (t ) = L−1 
 



 (s + 2 ) 2
+ 1


= e−2t sin t
and f(0) = 0.

 s 
 d −2 t
∴ By (l), L−1  2  = (e sin t )


 s + 4 s + 5 

 dt
= e−2t cos t − 2e−2t sin t
= e−2t (cos t − 2 sin t)

 s 
 
−1  1 
(iv) To find L−1 
 
 , we shall find f (t ) = L   and

 (s + 2) (s + 3) 
 
  (s + 2) (s + 3) 
then apply the rule (1).

 1 
Now f (t ) = L−1 
 

 (s + 2) (s + 3) 


 1
 1 
= L−1  − ,


 s + 2 s + 3
by resolving the function into partial fractions.
= e−2t − e−3t
and f(0) = 0.


 s 
 d −2t
∴Βy (1), L−1 
  −3t
 = (e − e )


 (s + 2 ) (s + 3) 

 dt
= 3e−3t − 2e−2t

 s 
 
−1  1 
(v) To find L−1 
 2 
 , we shall find f (t ) = L  2 

 (s + 1) (s + 4) 

2

  (s + 1) (s + 4) 
2

and then apply the rule (1).

5.78 Mathematics II

 1/ 3
 1/ 3 
Now f (t )=L−1  2 − 2 ,


 s + 1 s + 4 


by resolving the function into partial fractions.
1 1
= sin t − sin 2t
3 6
and f(0) = 0

 s 
 d 1 1 
∴ By (1), L−1 
 2 
 =  sin t − sin 2t 


 (s + 1) (s 2
+ 4 ) 

 d t 

 3 6 

1
= (cos t − cos 2t )
3

Note We have solved the problems in the above example by using the working
rule derived from the theorem on Laplace transforms of derivatives. They can be
solved by elementary methods, such as partial fraction methods, discussed in Section
5(a) also.
Example 5.5 Find the inverse Laplace transforms of the following functions.

s2 s3
(i) (ii)
(s 2 + a 2 ) 2 (s + a 2 ) 2
2

(s + 1) 2 s2
(iii) (iv)
(s 2 + 2 s + 5) 2 (s 2 − 4) 2

(s − 3) 2
(v)
(s 2 − 6 s + 5) 2

 s 

(i) Let f (t ) = L−1 
 2 
2 2

 (s + a ) 
 

t
=sin at [Refer to Worked Example (13) (i) in Section 5(b)]
2a
We note that f (0) = 0
d −1
Now L−1{sφ (s )} = L {φ(s )} , provided f (0) = 0,
dt
where f(t) = L−1 {φ (s)} (1)

 s2  
 s 

∴ L−1 
 2 2 2
= L−1 
s ⋅ 2 
2 2


 (s + a ) 
 

 (s + a ) 


d  t 
=  sin at  , by rule (1)
dt  2a 
 Laplace Transforms 5.79

1
= (sin at + at cos at ) (2)
2a
 s2 
(ii) Let f (t ) = L−1  2 
 (s + a ) 
2 2
 
1
= (sin at + at cos at ) , by (2)
2a
we note that f(0) = 0

 s3 
 = L−1 
  s2 
Now L−1 
 2 2 2

s ⋅ 2 2 2

 (s + a ) 
 
 
 (s + a ) 

d1 
=  (sin at + at cos at ) , by rule (l)
dt  2a 
1
= (2 cos at − at sin at ) .
2
 (s + 1) 2 
   (s + 1) 2 
(iii) L−1 
 2 
 = L−1
 
  2 2
 (s + 2 s + 5)   {(s + 1) + 2 } 
2 2
 


 s2 

= e−t L−1 
 2 
2 2  , by the first shifting property


 (s + 2 ) 



1
= e−t (sin 2t + 2t cos 2t ) , by (2)
4

 s 

(iv) Let f (t ) = L−1 
 2 
2


 (s − 4 ) 


t
= sinh 2t
4
[Refer to Worked Example (13) (ii) in Section 5(b)]
We note that f(0) = 0.

 s2 
  
 = L−1  s 
Now L−1 
 2 2

s ⋅ 2 2

 (s − 4) 
 
 
 (s − 4) 



d −1  s 
= L 
 2 2
, by rule (1)
dt 
 (s − 4) 

d  t 
=  sinh 2t 
dt  4 
1
= (sinh 2t + 2t cosh 2t ) (3)
4
5.80 Mathematics II

 (s − 3) 2  
 (s − 3) 2 

(v) L−1  2  = L−1 

 (s − 6 s + 5) 2  2
  
{(s − 3) − 4} 

2


 s2 
= e3t L−1  2 
 (s − 4) 2  , by the first shifting property.
 
1
= e3t (sinh 2t + 2t cosh 2t ) , by (3).
4
Example 5.6 Find the Laplace transforms of the following functions:
t t

(i) ∫ te−4t sin 3t dt (ii) e−4t ∫ t sin 3t dt

0 0

t
e−t sin t
t
(iii) t ∫ e−4t sin 3t dt (iv) ∫ dt
0 0
t
t t
sin t 1
(v) e−t ∫
t
dt (vi) ∫
t 0
e−t sin t dt
0

t  1
(i) L  ∫ f (t ) dt  = L{ f (t )} (1),
 0  s
by the theorem on Laplace transform of integral
t 
∴ L  ∫ t e−4t sin 3t dt 
 0 

1
= L{te−4t sin 3t}
s
6(s + 4)
=
s (s + 8s + 25) 2
2

[Refer to Worked Example 8(i) in Section 5(b)].

 t   t 
(ii) L  e−4t ∫ t sin 3t dt  =  L ∫ t sin 3t dt  (2), by the first shifting property
 0   0  s → s + 4

t
1
Now L ∫ t sin 3t dt = L(t sin 3t ) , by rule (l)
0
s

1 d 
= − L(sin 3t )
s  ds 

1 d  3 
= − ⋅  2 
s ds  s + 9 
 Laplace Transforms 5.81

1 −3× 2 s 6
=− × 2 = (3)
s (s + 9) 2 (s 2 + 9) 2

Using (3) in (2), we get

 t  6
L  e−4t ∫ t sin3t dt  =
 {(s + 4) + 9}
2 2
 0

6
= 2
(s + 8s + 25) 2
 t 
(iii) L t ⋅ ∫ e−4t sin 3t dt 
 0 

d  
t

ds  ∫0
−4 t
=− L e sin 3t d t  (4)

t
1
Now L ∫ e−4t sin 3t dt = L(e−4t sin 3t ) , by (1)
0
s

1
= [L(sin 3t )]s → s + 4
s
1 3
= ⋅
s (s + 4) 2 + 9
3
= 3 (5)
s + 8s 2 + 25s
Using (5) in (4), we get
 t  d 3 
L t ∫ e−4t sin 3t dt  = −  3 
 0  
 s + 8s + 25s 
ds  2

3(3s 2 + 16s + 25)

= 3
(s + 8s + 25s ) 2
3(3s 2 + 16s + 25)
=
s 2 (s 2 + 8s + 25) 2

e−t sin t 1  e−t sin t 

t

(iv) L∫ dt = L   , by (1) (6)

0
t s  t 

 e−t sin t  ∞
Now L   = ∫ L(e−t sin t ) ds
 t  s
∞
ds
=∫
s
(s + 1) 2 + 1
5.82 Mathematics II

= {− cot −1 (s + 1)}∞
s
(7)
= cot −1 (s + 1)
Using (7) in (6), we get
e−t sin t
t
1
L∫ dt = cot −1 (s + 1)
0
t s

 sin t   sin t 
t t

(v) L  e−t ∫ dt  =  L ∫ dt  (8)

 0
t   0 t  s → s +1

1  sin t 
t
sin t
Now L∫ dt = L   , by (1)
0
t s  t 
∞
1
s ∫s
= L(sin t ) ds

∞
1 ds 1
= ∫ = cot −1 s
s s s +1 s
2
(9)

Using (9) in (8), we get

 t
sin t  1
L  e−t ∫ dt  = cot −1 (s + 1)
 0
t  s + 1
1 t  ∞ t 
(vi) L  ∫ e−t sin t dt  = ∫ L  ∫ e−t sin t dt  ds (10)
 t 0  s  0 
t
1
Now L ∫ e−t sin t dt = L{e−t sin t} , by (1)
0
s

1 1
= ⋅ (11)
s (s + 1) 2 + 1
Using (11) in (10), we get
1 t  ∞ ds
L  ∫ e−t sin t dt  = ∫
t
 0  s
s (s 2
+ 2 s + 2)
∞
1 1 s+2 
=∫  − 2  ds ,
s
2  s s + 2 s + 2 

on resolving the integrand into partial fractions.

1
∞
1 s +1 1 
= ∫  − −  ds
2 s  s (s + 1) + 1 (s + 1) + 1
2 2
 
 Laplace Transforms 5.83

∞
1 1 
=  log s − log {(s + 1) 2 + 1} + cot −1 (s + 1)
2  2  s
∞
1   
=  log 
s  + cot −1 (s + 1)
2   s 2 + 2 s + 2  
  s
1  s + 2 s + 2  1 −1
2
= log   − cot (s + 1).
4 
 s2  2
Example 5.7 Find the inverse Laplace transforms of the following functions:

1 54 1
(i) ; (ii) ; (iii) .
s (s + 2)3 s 3 (s − 3) s (s 2 + 4 s + 5)

1 1  s + 1  5s − 2
(iv) ; (v)  ; (vi) ;
s (s + a 2 )
2 2
s 2  s 2 + 1 s (s −1)(s + 2)
2

1
(vii)
(s + 2) (s 2 + 4 s + 13)

Note All the problems in this example may be solved by resolving the given
functions into partial fractions and applying elementary methods. However we shall
solve them by applying the following working rule and its extensions.

1
  t

L−1  φ(s ) = ∫ L−1 (φ (s ) dt (1)


s 

 0

 1  t
 1 
 
(i) L−1   L−1   dt , by (1)
 s (s + 2)3  ∫
=  3
  

 (s + 2 ) 


0

t
1
= ∫ e−2t L−1  3  dt
 s 
0
t
1 2
= ∫ e−2t t dt
0
2

1   e−2t   e−2t 

t
 e−2t 
= t 2   − 2t   + 2  
2   −2   4   −8 
 0

by Bernoulli’s formula.
1 t2 t 1 1 
= −e−2t  + +  + 
2   2 2 4  4 
1
= [1 − (2t 2 + 2t + 1)ee−2t ]
8
5.84 Mathematics II

 54   
 = 54 ∫ ∫ ∫ L−1  1  dt dt dt , by the extension of rule (1).
t t t
(ii) L−1  3
 s (s − 3)  
 s − 3 
  0 0 0

t t t

= 54 ∫ ∫∫ e 3 t dt d t d t
0 0 0
t
t t
 e3t 
= 54 ∫ ∫  3  dt dt
0 0 0
t t

=18∫ ∫ (e
3t
−1) dt dt
0 0

 e 3t 
t t

=18∫  − t  dt
 3 
0 0
t

= 6 ∫ (e3t − 3t −1) dt
0

 e3t 3t 2 
t

= 6  − − t 
 3 2 
0

= 2e3t − 9t 2 − 6t − 2.

Aliter
We can avoid the multiple integration by using the following alternative method.
 
 54   54 
L 
−1
 =L 
−1

 s 3 (s − 3)   
 ( s − 3 + 3) (s − 3) 
3
 

 1 
= 54e3t L−1   3
,

 s (s + 3) 


by the first shifting property.

t
 1 

= 54e3t ⋅ ∫ L−1 
 3
dt , by rule (1)

 (s + 3) 

0

t

 1 
= 54e3t ∫ e−3t ⋅ L−1  3  dt

 s 

0
t
1 2 −3t
= 54e3t ∫ t e dt
0
2
 Laplace Transforms 5.85

  e−3t  −3 t 
t
− 2t  e  + 2  e 
−3 t
= 27e t 2 
3t
 
 9   −27 
  −3   0
−3 t
= e [2 − e (9t + 6t + 2)]
3t 2

= 2e3t − 9t 2 − 6t − 2



 1 


t

−1  1 
 ∫L 
−1
(iii) L  2 =  2  dt ,

 s (s + 4 s + 5 
 
 0 
 s + 4 s + 5 

by rule (1).
t

 1 
= ∫ L−1 
  dt

 (s + 2) +1

2
0
t

= ∫ e−2t sint dt
0

 −e−2t 
t

= (2sin t + cos t )
 5 
 0
1
= [1− e−2t (2 sin t + cos t )]
5


 1 

t t
 1 
(iv) L−1 
 2 2 2 
 = ∫ ∫L
−1  
 s 2 + a 2  dt dtt
,

 s (s + a ) 
 
 0 0

by the extension of rule (1).

t t
1
=∫ ∫ a sin at dt dt
0 0

1  −cosat 
t t

a ∫0  a 0
=   dt

t
1
=
a2 ∫ (1− cos at ) dt
0

1  sin at 
t

= t − 
a 2  a 0
1
= (at − sin at ).
a3

 1  s +1 
  t t
 s +1 
L−1   =
 2  2  
 ∫ ∫L
(v) 
−1

  s 2 +1 dt dt ,
 s  s +1
  0 0

by the extension of rule (1).

5.86 Mathematics II

t t

=∫ ∫ (cos t + sin t ) dt dt
0 0
t

= ∫ (sin t − cos t )t0 d t

0
t

= ∫ (sin t − cos t +1) d t

0

= ( − cos t − sin t + t )t0

= 1+ t − cos t − sin t.


 5s − 2 

t t

 5 s − 2 
(vi) L−1 
 2 
 ∫
= ∫L
−1 
  dt dt ,

 s (s −1) (s + 2) 
 
 0 
 (s −1) (s + 2) 

0

by the extension of rule (1).

t t
 1 4 
=∫ 
∫L
−1
+  dt dt ,
 s −1 s + 2 
0 0

by resolving the function into partial fractions.

t t

=∫ ∫ (e + 4e
t −2 t
) dt dt
0 0
t

= ∫ (et − 2e−2t )t0 dt

0
t

= ∫ (et − 2e−2t +1) dt

0

= (et + e−2t + t )t0 = et + e−2t + t − 2

 1 
(vii) L−1  
 (s + 2) (s + 4 s +13) 
2

 1 
= L−1  
 (s + 2){(s + 2) 2 + 9}
 1 
= e−2t ⋅ L−1  2  , by the first shifting property.
 s (s + 9) 
t
 1 

= e−2t ∫L
−1
 dt , by rule (1),
 s 2 + 9 
0

t
1 −2 t
=
3
e ∫ sin 3t dt
0
 Laplace Transforms 5.87

1 −2t  −cos 3t 
t

= e  
3  3 0
1
= e−2t (1− cos 3t )
9

Example 5.8 Find the inverse Laplace transforms of the following functions:
1
1 1
(i) ; (ii) ; (iii) (s 2 + 2 s + 5) 2 ;
(s + a 2 ) 2
2
s (s + a 2 ) 2
2

1 1 1
(iv) ; (v) ; (vi) .
(s − 4) 2
2
s (s − 4) 2
2
(s − 2 s − 3) 2
2


 1 
 = L−1  1⋅
 s 
(i) L−1 
 2 2 2



 (s + a ) 

 
 s (s 2
+ a 2 2
) 
t

 s 
= ∫ L−1   2 2 2
d t ,as

 (s + a ) 

0

1
  t

L−1  φ (s ) = ∫ L−1 { φ (s )} d t (1)


s 

 0
t
t
=∫ sin at d t
0
2a

[Refer to Worked Example 13(i) in Section 5(b)]

1   −cos at   −sin at 

t

= t   − 
2 a   a   a 2  0
1
= 3 (siin at − at cos at ) (2)
2a

 1  1 s 
(ii) L−1  2  = L−1  2 ⋅ 2 
 s (s + a 2 ) 2   s (s + a 2 ) 2 
t t

 s 
=∫ ∫ L−1  2 2 2
dt dt ,

 (ss + a ) 

0 0

by the extension of rule (1)

t t
t
=∫ ∫ 2 a sin at dt dt
0 0
t
1
=∫ (sin at − at cos at ) d t , by (2)
0
2 a3
5.88 Mathematics II

 − cos at   sin at   − cos at 

t
1
= 3  − a t   −  
2a  a
   a   a 2 
0

1
= (−2 cos at − at sin at )t0
2a 4
1
= 4 (2 − 2 cos at − at sin at ).
2a

 

 1 
 
−1  1 

−1
L  2  =L 
2 2
(iii)

 ( s + 2 s + 5) 
 
 
 ( s +1) 2 + 4 
 
  

 1 
= e−t ⋅ L−1  2 2

 ( s + 4) 

1
= e−t (sin 2t − 2t cos 2t ), by problem (i)
16

 1  −1 
 1 s 

(iv) L−1 
 2 = L 
2  ⋅ 2 
2


 ( s − 4 ) 

 

 s ( s − 4 ) 


t

 s 

= ∫ L−1   2  dt , by rule (1)
2


 ( s − 4 ) 


0
t
t
=∫ sinh 2t dt
0
4
[Refer to Worked Example 13 (ii) in Section 5(b)]

1   cosh 2t   sinh 2t 

t

= t  − 
4   2   4  0

1
= (2t cosh 2t − sinh 2t ) (3)
16

 1 
 = L−1 
  1 s 
(v) L−1 
 2 2

 2⋅ 2 2

 s ( s − 4) 
 
 
 s ( s − 4) 

t t

 s 
= ∫ ∫ L−1   2 2
dt dt ,

 ( s − 4) 

0 0

by the extension of rule (1).

t t
t
=∫ ∫ 4 sinh 2t dt dt
0 0
t
1
=∫ (2t cosh 2t − sinh 2t ) dt , by (3)
0
16
 Laplace Transforms 5.89

  sinh 2t cosh 2t  cosh 2t 

t
1 2 t  − 
= −
16  
 2 4  2  0
1
= nh 2t − cosh 2t ]t0
[t sin
16
1
= (1+ t sinh 2t − cosh 2t )
16
 

 1 
 −1  1 
(vi) L−1 
 2 
 = L  

 ( s − 2 s − 3)   ( s −1) 2 − 4 
{ }
2 2
 

 
 1 
= et ⋅ L−1  2 ,
 ( s − 4) 2 
 
by the first shifting property.
1 t
= e (2t cosh 2t − sinh 2t ), by problem (iv).
16
Example 5.9
(i) Verify the initial and final value theorems when (a) f(t) = (t + 2)2 e−t; (b)

 1 

f (t ) = L−1 
 
2


 s ( s + 2 ) 


(ii) If L(e−t cos2 t) = f (s), find lim [ sφ ( s )] and lim [ sφ ( s )] .
s →0 s →∞

1
(iii) If L{ f (t )} = , find lim [ f (t )] and lim [ f (t )] .
s ( s + 1)( s + 2) t →0 t →∞

(i) (a) f(t) = (t2 + 4t + 4)e−t

2 4 4
∴ φ ( s ) = L{ f (t )} = + +
( s + 1) 3
( s + 1) 2
s +1
2s 4s 4s
∴ sφ ( s ) = + +
( s + 1) 3
( s + 1) 2
s +1
Now lim [ f (t )] = 4 and slim [ sφ ( s )] = 0 + 0 + 4 = 4
t →0 →∞

Hence the initial value theorem is verified.

Also lim [ f (t )] = 0 and lim [ s φ ( s )] = 0
t →∞ s→0

Hence the final value theorem is verified.


 1 

t
 1 
(i) (b) f (t ) = L−1 
 
2
= ∫ L−1   dt

 s ( s + 2) 
 
 0  ( s + 2) 2 
t

= ∫ te−21 dt
0
5.90 Mathematics II

  e−2t   e−2t 
t

= t   −  

  
  −2   4  0
1
= (1− 2t e−2t − e−2t )
4
1
sφ ( s ) =
( s + 2) 2

Now lim [ f (t )] = 0 = lim [ sφ ( s )]

t →0 s →∞

1
and lim [ f (t )] = = lim [ sφ ( s )]
t →∞ 4 s→0
Hence the initial and final value theorems are verified.
(ii) L(e−t cos2 t) = f (s)
i.e., f (t) = e−t cos2 t
By the final value theorem,
lim [ sφ ( s )] = lim [e−t cos 2 t ] = 0
s→0 t →∞

By the initial value theorem,

lim [ sφ ( s )] = lim [e−t cos 2 t ] = 1
s →∞ t →0

1
(iii) L{ f (t )} =
s ( s +1)( s + 2)

1
∴ sφ ( s ) =
( s +1)( s + 2)
By the initial value theorem,
lim [ f (t )] = lim [ sφ ( s )] = 0
t →0 s →∞

By the final value theorem,

1
lim [ f (t )] = lim [ sφ ( s )] =
t →∞ s→0 2

Example 5.10 Use convolution theorem to evaluate the following

t t

(i) ∫ u 2 e−a (t − u ) du (ii) ∫ sin u cos (t − u ) du

0 0

t t

∫u e ∫
2 −a ( t − u )
(i) du is of the form f (u ) g (t − u ) du
0 0
 Laplace Transforms 5.91

where f(t) = t2 and g(t) = e−at

t

∫u e−a (t − u ) du = (t 2 ) *(e−at )
2
i.e.
0

∴ By convolution theorem,
t 
L  ∫ u 2 e−a (t − u ) du  = L(t ) 2 ⋅ L(e−at )
 0 
2 1
= 3⋅
s s+a

t

 2 

∫u
2
e−a (t − u ) du = L−1 
 3 



 s ( s + a ) 


0


 2 
= e−at ⋅ L−1 
 3

 s ( s − a ) 

t

 2 
= e−at ∫ L−1   3
dt
0  ( s − a ) 


t

= e−at ∫t
2
e at dt
0

 e at e at 
t
e at
= e−at t 2 − 2t 2 + 2 3 
 a a a  0

 t 2 e at 2t e at 2e at 2
= e−at  − 2 + 3 − 3 
 a a a a 
1 2 2
= {a t − 2at + 2 − 2e−at }
a3

t t

(ii) ∫ sin u cos (t − u ) du is of the form ∫ f (u ) g (t − u ) du ,

0 0

where f(t) = sin t and g(t) = cos t

t

i.e. ∫ sin u cos (t − u ) du = (sin t ) * (cos t )

0

∴ By convolution theorem,

t 
L  ∫ sin u cos (t − u ) du  = L(sin t ) ⋅ L(cos t )
 0 
5.92 Mathematics II

s
=
( s 2 + 1) 2
t


 s 


∫ sin u cos (t − u ) du = L
−1
∴  2 2

 ( s + 1) 
 

0

t
= sin t , by Worked Example 13(i) of Section 5(b).
2
Example 5.11 Use convolution theorem to find the inverse Laplace transforms of
the following functions:
1 s s2
(i) (ii) (iii)
( s +1) ( s + 2) (s 2 + a 2 )2 (s 2 + a 2 ) (s 2 +b2 )

4 s2 + s
(iv) (v)
( s + 2 s + 5) 2
2
( s 2 +1) ( s 2 + 2 s + 2)
 1
 1    1 * −1  1 
(i) L−1 
 ⋅ 
 = L 
−1
 L   , by convolution theorem


 s +1 s + 2 

 
 s +1   s + 2 

= e−t * e−2t
t

= ∫ e−u ⋅ e−2 (t − u ) du
0
t

∫e
−2 t
=e u
du
0

= e−2t (et −1) = e−t − e−2t


 s 
 = L−1 
  1 s  
(ii) L−1 
 2 2 2

 2 ⋅ 2 
2


 ( s + a ) 

 

 s + a 2
s + a 


 1  −1  s 
= L−1  2 2
 * L  2 
 s + a 2  , by convolution theorem
 s + a 
1 
=  sin at  * (cos at )
 a 
t
1
=∫ sin au cos a (t − u ) du
0
a
t
1
=
2a ∫ [sin at + sin (2au − at )] du
0

 cos( 2au − at ) 
t
1
=  (sin at ) u − 
2a  2a  0
 Laplace Transforms 5.93

1  1 
= t sin at − (cos at − cos at )

2a  2a 
1
= t sin at.
2a


 s2 
 
 = L−1  s s 
(iii) L−1 
 2 2 

 2 ⋅ 2 2


2 2

 ( s + a )( s + b ) 
 
 s + a s + b 

2

 s
 
 
 * L−1  s 
= L−1 
 2 2 

 2 2
, by convollution theorem

 (s + a ) 
 
 
 s + b 

= ( cos at ) * ( cos bt )
t

= ∫ cos au ⋅ cos b(t − u ) du

0
t
1
2 ∫0
= {cos [(a − b) u + bt ] + cos[(a + b) u − bt ]} du

1  1 
t
1
= sin{(a − b) u + bt + sin{(a + b) u − bt}

2  a −b a +b 
0
1  1 1 
= (sin at − sin bt ) + (sin at + sin bt )
2  a − b a +b 

1  1 1   1
 1  
=  + 
 sin at + +  sin bt 
2  a − b a + b   a + b a − b  

1  2a 2b 
= sin at − 2 sin bt 
2  a − b
2 2
a −b 2 

1
= 2 (a sin at − b sin bt ).
a −b2

   

−1 4   2 2 
(iv) L   = L−1  2 ⋅ 2 
 ( s 2 + 2 s + 5)   ( s + 2 s + 5) ( s + 2 s + 5)
2
 


 2 
 
−1  2 
= L−1 
 
* L   ,

 ( s +1) + 4 

2

 
 ( s +1) + 4 

2

by convolution theorem.
−t −t
= (e sin 2t )*(e sin 2t )
t

= ∫ e−u sin 2u ⋅ e−(t − u ) ⋅ sin 2(t − u ) du

0
5.94 Mathematics II

t
1
= e−t ∫ [cos (4u − 2t ) − cos 2t ] dv
2 0

 sin (4u − 2t ) 
t
1
= e−t  − (cos 2t ) ⋅ u 
2  4  0
1 1 
= e−t  (sin 2t + sin 2t ) − t cos 2t 
2 
4 
1
= e−t (sin 2t − 2t cos 2t ).
4
   s +1
s2 + s s 
(v) L−1  2  = L−1
  2 ⋅ 2
 ( + )( 2
+ + )  s + 2 s + 2 s +1 
s 1 s 2 s 2  
 s +1 
   s 
= L−1 
 
 * L  2 
−1

 ( s +1) +1

2

  s +1
−t
= (e cos t )*(cos t )
t

= ∫ e−u cos u cos (t − u ) du

0
t
1
= ∫
2 0
e−u [cos t + cos (2u − t )] du

1 1 1
= cos t (− e−u ) + ⋅  e−u {− cos (2u − t ) + 2 sin (2u − t )}
t t

2 0 2 5 0

1 1
= cos t (1− e−t ) +  e−t (2 sin t − cos t ) + (2 sin t + cos t )
2 10
1 −t 1
= e (sin t − 3 cos t ) + (sin t + 3 cos t ).
5 5

EXERCISE 5(c)

Part A
(Short Answer Questions)

1. State the relation between the Laplace transforms of f (t) and f ′(t). Under
what conditions does this relation hold good?
2. Express L−1 {sφ (s)} interms of L−1 {φ (s)}. State the condition for the validity
of your answer.
t t 
3. Express L  ∫ ∫ f (t ) dt dt  in terms of L{f(t)}.
 0 0 
 Laplace Transforms 5.95

 1 
4. State the relation between L−1 {f (s)} and L−1  2 φ (s ) .
 s 
5. State the initial value theorem in Laplace transforms.
6. State the final value theorem in Laplace transforms.
7. Define the convolution product of two functions and prove that it is com-
mutative.
8. Verify whether 1 * g (t) = g (t), when g (t) = t.
9. State convolution theorem in Laplace transforms.
Using the Laplace transforms of the derivatives find the Laplace transforms
of the following functions:
10. e at 11. cos a t 12. sin2 t
Find the inverse Laplace transforms of the following functions:
s s2
13. 14.
(s + 2)3 (s −1)3

s s
15. 16.
(s − a ) 2 + b 2 (s +1) (s + 2)
Find the Laplace transforms of the following functions:
t t t
sin t 1− et 1− 2 cos t
17. ∫ t
dt 18. ∫ t
dt 19. ∫ t
dt
0 0 0

t t t

20. ∫ t e− t d t 21. ∫ e− t sin t d t 22. ∫t sin t d t

0 0 0

Find the inverse Laplace transforms of the following functions:

1 1
23. 24.
s (s + a ) s 2 (s +1)

1 1
25. 26.
s (s 2 − a 2 ) s (s 2 +1)

s +3
27. If L { f (t )} = , find lim ( f (t )} and lim{ f (t )} .
(s +1) (s + 2) t →0 t →∞

−1 1 −t −2 t
28. If L { φ (s )}= (1− 2e + e ) , find lim
s →0
(s φ (s )} and lim {s φ (s )}.
2 s →∞

t n−1
29. Show that 1 * 1 * 1 * ......* 1 (n times) = , where * denotes Convolu-
tion. (n −1)!
t
1
30. If L { f (t )} = , evaluate ∫ f (u ) f (t − u ) du .
s 2 +1 0
5.96 Mathematics II

Use convolution theorem to find the inverse Laplace transforms of the

following functions:
1 1
31. 32.
(s −1) (s + 3) (s −1) 2
1 1
33. 34.
s (s + a 2 )
2
s 2 (s +1)

35. 2 / (s +1) (s2 + 1).

Part B
36. Using the Laplace transforms of the derivatives find L (t sin a t) and hence
find L (2 cos at − a t sin a t) and L (sin a t + a t cos a t).
37. Using the Laplace transforms of the derivatives, find L (t cosh a t) and
hence find L (sinh a t + a t cosh a t) and L (a t cosh a t − sinh a t).

 1 
 
−1  
38. Find L−1   s
  and hence find L  .
 
 ( s + a) (s + b) 
  
 ( s + a) (s + b) 

−1  1 
  s2 
39. Find L   
 and hence find L−1  .

 ( s −1) (s − 2) (s − 3) 
 
  ( s −1) (s − 2) (s − 3) 

 
  
 1  −1  s 
40. Find L−1 
 2 
 and hence find L 
 2 
( )( )  ( )( ) 

 s + a 2
s 2
+ b 2   s + a 2
s 2
+ b 2 


 

 s2 
and L−1 
 2 
.



 ( s + a 2
) ( s 2
+ b 2
) 




 

 
41. Given that L−1   = t sin 2 t , find
s
 2
 s + 4) 
(
 2
 4
 


 
 
 
  
 s2 , L−1 
 3
 and L−1  (s + 3)
 2
L−1 
 2
 s
 2  2

    
( s + 6 s + 13) 
(
s + 4) 
(
 2
 s + 4) 
2 2
 
    

 


 −1s 
 = t sinh at
42. Given that L  2 , find
 


 ( s 2
− a 2
) 


2a


 
 
 

    and  s + a 
2
s2 s3
L−1 

,
2
L−1 
 2
L−1   .
     s (s + 2a ) 
( s − a ) 
  ( − ) 
2 2 2 2
  s a 
   
 1 
  1  s 
−1 
43. Given that L−1 
 4 
 = (sin t cosh t − cos t sinh t ), find L  4 ,
 
 s + 4
  4  s + 4 
 Laplace Transforms 5.97

 s2 
   s3 
−1 

L−1 
 4 
 and L   4 
.


 s + 4 

 

 s + 4 


Find the Laplace transforms of the following functions:
t t

∫te ∫ t sin t dt
t t
44. sin t dt 45 . e
0 0

e−2t sin 3t
t t

46. t ∫ et sin t dt 47. ∫ t

dt
0 0

t t
sin 3t 1
t ∫0
48. e−2t ∫ dt 49. e−2t sin 3t dt
0
t
Find the inverse Laplace transforms of the following functions:
1  s −1   
50.    Hint: Consider the function as s2−1 
s 2  s + 1 
 s (s + s ) 
4s + 7 1
51. 52.
s 2 (2s + 3) (3s + 5) s (s 2 + 6 s + 25)
1 1  s − 2  1
53. 54.   55.
(s + 1) (s + 2 s + 2) 2
s 2  s 2 + 4  (s + 9) 2
2

1 1 1
56. 57. 58.
s (s + 9) 2 2
(s + 6s + 10) 2
2
(s − a 2 ) 2
2

1 1
59. 60.
s (s 2 − a 2 ) 2 (s 2 + 4 s ) 2
Verify the initial and final value theorems when

−1  1 
61. f (t) = (2t + 3)2 e−4t 62. f (t ) = L  3
 s (s + 4) 
63. Use convolution theorem to evaluate
t

∫e
−u
sin (t − u ) du
0

64. Evaluate ∫ cos a u cosh a (t − u ) d u , using convolution theorem.

0

Use convolution theorem to find the inverse of the following functions:

s s2
65. 66.
(s + 4) (s 2 + 9)
2
(s + a 2 ) 2
2

1 10
67. 68.
(s + 4) 22
(s + 1) (s 2 + 4)
5.98 Mathematics II

1 1
69. 70.
s (s + 1)3
2
s4 + 4

5.11 SOLUTIONS OF DIFFERENTIAL

AND INTEGRAL EQUATIONS

As mentioned in the beginning, Laplace transform technique can be used to solve

differential (both ordinary and partial) and integral equations. We shall apply this
method to solve only ordinary linear differential equations with constant coefficients
and a few integral and intergo-differential equations. The advantage of this method
is that it gives the particular solution directly. This means that there is no need
to first find the general solution and then evaluate the arbitrary constants as in
the classical approach.

5.11.1 Procedure
1. We take the Laplace transforms of both sides of the given differential equa-
tion in y (t), simultaneously using the given initial conditions. This gives an
algebraic equation in y (s ) = L{y (t )} .

Note L{y (n ) (t )} = s n y (s ) − s n −1 y (0) − s n − 2 y ′ (0) ........ y (n −1) (0).

2. We solve the algebraic equation to get y (s ) as a function of s.
−1
3. Finally we take L {y (s )} to get y(t). The various methods we have dis-
−1
cussed in the previous sections will enable us to find L {y (s )} .
The procedure is illustrated in the worked examples given below:

WORKED EXAMPLE 5(d)

Example 5.1 Using Laplace transform, solve the following equation

di
L + Ri = E e−a t ; i (0) = 0, where L, R, E and a are constants.
dt
Taking Laplace transforms of both sides of the given equation, we get,
L ⋅ L{i ′(t )} + RL{i (t )} = EL {e−at }

E
i.e., L{s i (s ) − i (0)} + Ri (s ) = , where i (s ) = L{i (t )}
s+a

E
i.e., (Ls + R) i (s ) =
s+a

E
∴ i (s ) =
(s + a ) (Ls + R )
 Laplace Transforms 5.99

 1   1  
 

    
    
 R − a L   a L − R  
= E  + 

 s+a s+R L  

 


 

Taking inverse Laplace transforms
 −1  1   1 
i (t ) =
E L  



 − L−1 

 

R − aL   s + a
 
 
 s + R L 

E
= (e−at − e−Rt L )
R−aL

Example 5.2 Solve y″ − 4y′ + 8y = e 2 t, y (0) = 2 and y′ (0) = − 2.

Taking Laplace transforms of both sides of the given equation, we get
1
[s 2 y (s ) − s y (0) − y ′ (0)]− 4 [s y (s ) − y (0)] + 8 y (s ) =
s−2

1
i.e., (s 2 − 4 s + 8) y (s ) = + (2 s −10)
s−2

1 2 s −10
∴ y (s ) = +
(s − 2) ( s 2 − 4 s + 8) s2 − 4s +8
A Bs + C 2 s −10
= + 2 + 2
s − 2 s − 4s +8 s − 4s +8
1 1 1
− s+
= 4 + 4 2 + 2 s −10
s − 2 s2 − 4s +8 s2 − 4s +8
1 7 19
s−
= 4 + 4 2
s − 2 s2 − 4s +8
1 7
(s − 2) − 6
= 4 +4
s − 2 (s − 2) 2 + 4
 7 
 s − 6 
1 −1  1  2t −1  4 
y= L  + e L  2 
4  s − 2   s + 4 
 
 
1  7 
= e 2 t + e 2 t  cos 2t − 3 sin 2t 
4 4 
1
= e 2 t (1+ 7 cos 2t −12 sin 2t )
4
5.100 Mathematics II

Example 5.3 Solve y″ − 2y′ + y = (t + 1)2, y (0) = 4 and y′ (0) = −2.

Taking Laplace transforms of both sides of the given equation, we get,
[s 2 y (s ) − sy (0) − y ′(0)]− 2 [s y (s ) − y (0)]+ y (s ) = L(t 2 + 2 t + 1)
2 2 1
i.e., (s 2 − 2 s + 1) y (s ) − 4 s + 10 = + +
s3 s 2 s
4 s −10 1 2 2
i.e., y (s ) = + + 2 + 3
(s −1) 2
s (s −1) 2
s (s −1) 2
s (s −1) 2
4 6 1 2 2
= − + + 2 + 3
s −1 (s −1) 2
s (s −1) 2
s (s −1) 2
s (s −1) 2
t t t t t t

∴ y = 4e − 6 te + ∫ t e dt + 2 ∫ ∫ t et d t d t + 2 ∫ ∫∫te
t t t t
dt dt dt
0 0 0 0 0 0
t t t

= 4et − 6 t et + (t et − et + 1) + 2 ∫ (t et − et + 1) d t + 2 ∫ ∫ (t e − e
t t
+ 1) d t d t
0 0 0
t

= 3et − 5t et + 1 + 2 (t et − 2et + t + 2) + 2 ∫ (t et − 2et + t + 2) d t

0

t2
= −et − 3 t et + 2 t + 5 + 2 (t et − 3 et + + 2t + 3)
2
= − 7et − t et + t 2 + 6 t + 11

Example 5.4 Solve y″ + 4 y = sin wt, y (0) = 0 and y′(0) = 0.

Taking Laplace transforms of both sides of the equation, we get
ω
[s 2 y (s ) − sy (0) − y ′(0)]+ 4 y (s ) =
s + ω2
2

ω
∴ y (s ) = (1)
(s 2 + 4)(s 2 + ω 2 )
ω  1 1 
=  2 − 2 
ω − 4 s + 4 s + ω 2 
2

Inverting, we have,
1  ω 
y=  sin 2t − sin ωt  , if w ≠ 2.
ω 2 − 4  2 

If w = 2, from (1), we have

2
y (s ) =
(s + 4) 2
2


 1 

∴ y = 2 L−1 
 2 
2

 (s + 4) 
 

 Laplace Transforms 5.101

1
= (sin 2t − 2t cos 2t )
8
[Refer to Worked Example 8(i) in Section 5(c).
Example 5.5 Solve the equation y″ + y′ − 2y = 3 cos 3t −11 sin 3t, y (0) = 0
and y′(0) = 6.
Taking Laplace transforms and using the giving initial conditions, we get
3s 33
(s 2 + s − 2) y (s ) = − 2 +6
s +9 s +9
2

6 s 2 + 3s + 21
=
s2 + 9
6 s 2 + 3s + 21
∴ y (s ) = ,
(s 2
+ 9) (s + 2) (s −1)
As + B C D
= + +
s 2 + 9 s + 2 s −1
3 1 1
= 2 − + by the usual procedure
s + 9 s + 2 s −1
∴ y = sin 3t − e−2t + et.
Example 5.6 Solve the equation (D2 + 4D + 13) y = e−t sin t, y = 0 and
d
D y = 0 at t = 0, where D ≡ .
dt
Taking Laplace transforms and using the given initial conditions, we get

1
(s 2 + 4 s + 13) y (s ) =
s +2s+2
2

1
∴ y (s ) =
(s 2 + 2s + 2)(s 2 + 4s +13)
As + B Cs + D
= +
s 2 + 2 s + 2 s 2 + 4 s + 13
1  −2 s + 7 2s − 3 
=  2 + ,
85  s + 2 s + 2 s 2 + 4 s + 13 

on finding the constants A, B, C, D by the usual procedure.

1  −2(s + 1) + 9 2(s + 2) − 7 
= +
85  (s + 1) 2 + 1 (s + 2) 2 + 9 

1  −t  
 e { − 2 cos t + 9 sin t}+ e−2t 2 cos 3t − sin 3t 
7
∴ y=

85  
 3 
5.102 Mathematics II

Example 5.7 Solve the equation (D2 + 6D + 9) x = 6 t2 e−3t, x = 0 and Dx

= 0 at t = 0.
Taking Laplace transforms and using the initial conditions, use get,
12
(s 2 + 6 s + 9) x (s ) =
(s + 3)3
12
∴ x (s ) =
(s + 3)5
12
∴ x = e−3t ⋅ L−1 .
s5
1
= t 4 e−3t
2
Example 5.8 Solve the equation y″ + 9y = cos 2t, y (0) = 1 and y (p/2) = −1.
Note In all the problems discussed above the values of y and y’ at t = 0 were
given. Hence they are called initial conditions. In fact, the differential equation with
such initial conditions is called an initial value problem.
But in this problem, the value of y at t = 0 and t = p/2 are given. Such conditions
are called boundary conditions and the differential equation itself is called a boundary
value problem.
As y′(0) is not given, it will be assumed as a constant, which will be evaluated
towards the end using the condition y (p/2) = −1.
Taking Laplace transforms, we get
s
s 2 y (s ) − s y (0) − y ′(0) + 9 y (s ) = 2
s +4

s
i.e. (s 2 + 9) y (s ) = + s + A , where A = y′(0).
s +4
2

s s A
∴ y (s ) = + +
(s 2 + 4)(s 2 + 9) s2 + 9 s2 + 9
 s
1 s   s A
=  2 − 2 + 2 + 2
5

 s + 4 s + 9 

 s + 9 s +9

1 4 A
∴ y = cos 2t + cos 3t + sin 3t
5 5 3
π
Given y   = −1
 2 

1 A 12
i.e. −1 = − − ∴ A =
5 3 5
1 4 4
∴ y = cos 2 t + cos 3 t + sin 3 t .
5 5 5
 Laplace Transforms 5.103

Example 5.9 Find the general solution of the following equation

y″ − 2ky′ + k2y = f(t).
Taking Laplace transforms, we get
[s 2 y (s ) − sy (0) − y ′(0)] − 2 k [s y (s ) − y (0)]+ k 2 y (s ) = f (s )

i.e. (s 2 − 2ks + k 2 ) y (s ) = sy (0) +[y ′(0) − 2 ky (0)]+ f (s )

i.e. (s − k ) 2 y (s ) = As + B + f (s )

[As y (0) and y′(0) are not given, they are assumed as arbitrary contants]

As B f (s )
∴ y (s ) = + +
(s − k ) 2
(s − k ) 2
(s − k ) 2
A (s − k ) + (Ak + B ) f (s )
= +
(s − k ) 2
(s − k ) 2
C1 C2 f (s )
= + + , where C1 = A and C2 = Ak + B
s − k (s − k ) 2 (s − k ) 2


 1 
∴ y = C1 e kt + C2 t e kt + L−1 
 f (s ) ⋅ 


 (s − k ) 2 
i.e. y = (C1 + C2t) ekt + f (t) * t ekt
t

i.e. y = (C1 +C2 t ) e + ∫ f (t − u ) ue ku du .

kt

Example 5.10 Solve the equation (D3 + D) x = 2, x = 3, D x = 1 and D2x = −2

at t = 0.
Taking Laplace transforms and using the initial conditions, we get
2
(s 3 + s ) y (s ) = 3s 2 + s + 1 +
s

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

t t t

∴ y = 3 cos t + sin t + ∫ sin t dt + 2 ∫ ∫ sin t dt dt

0 0 0
t

= 3 cos t + sin t + (1− cos t ) + 2 ∫ (1− cos t ) dt

0

= 2 cos t + sin t + 1 + 2 (t − sin t )

= 2 cos t − sin t + 1 + 2 t .
5.104 Mathematics II

d 4 y d3 y dy d 2 y d3 y
Example 5.11 Solve the equation − = 0 , y = = 2 and = =1
d x 4 d x3 dx d x 2 d x3
at x = 0.

Note Change in the independent variable from t to x makes no difference in the

procedure.)
Taking Laplace transforms and using the initial conditions, we get

(s 4 − s 3 )y (s ) = 2 s 3 − s

2 s 2 −1 A B C
∴ y ( s )= = + 2+ .
s ( s −1) s s
2
s −1

1 1 1
i.e. y ( s )= + 2 +
s s s −1
∴ y = 1 + t + et
dy
Example 5.12 Solve the simultaneous differential equation + 2 x = sin 2t
dt

dy
and + 2 x = cos 2t , x (0) = 1, y (0) = 0.
dt
Taking Laplace transforms of both sides of the given equation and using the given
initial conditions, we get
2
sy ( s ) + 2 x ( s ) = (1)
s +4
2

s
and sx ( s ) − 2 y ( s ) = +1 (2)
s +4
2

Solving (1) and (2), we have

1 s 2
x ( s )= + 2 and y ( s )=− 2
s +4 s +4
2
s +4

1
∴ x = sin 2t + cos 2t and y = − sin 2 t.
2

Example 5.13 Solve the simultaneous equations

2x′ − y′ + 3x = 2t and x′ + 2y′ − 2x − y = t2 − t, x (0) = 1 and y (0) = 1.
Taking Laplace transforms of both the equations, we get
2
2[ sx ( s ) −1]− [ sy ( s ) −1] + 3 x ( s ) = and
s2
 Laplace Transforms 5.105

2 1
[ sx ( s ) −1] + 2[ s y ( s ) −1]− 2 x ( s ) − y ( s ) = −
s3 s 2
2
i.e. (2 s + 3) x ( s ) − sy ( s ) = +1 (1)
s2
2 1
and ( s − 2) x ( s ) + (2 s −1) y ( s ) = − +3 (2)
s3 s 2
Solving (1) and (2), we have
3 5s −1
x ( s )= +
s ( s +1)(5s − 3) ( s +1)(5s − 3)
 −1 3 8 25 8   3 4 5 4 
=  + +  + 
 + 
 s s +1 5s − 3   s +1 5s − 3 

1 98 78
=− + +
s s +1 s − 3 5

9 7
∴ x =−1+ e−t + e3 5t (3)
8 8
Eliminating y′ from the given equations,
we get 5x′ + 4x − y = t2 + 3t

∴ y = 5x′ + 4x − t2 − 3t
 9 21 3 t   7 3t 
= 5 − e−t + e 5  + 4 −1+ e−t + e 5 
9
 8 40   8 8 
− t2 − 3t, on using (3)
3
9 49 t
i.e. y =− e−t + e − t 2 − 3t − 4 .
5
8 8
Example 5.14 Solve the simultaneous equations
Dx + Dy = t and D2x − y = e−t; nx = 3,
Dx = − 2 and y = 0 at t = 0.
Taking Laplace transformed of both the equations, we get
1
sx ( s ) − 3 + sy ( s ) = and
s2
1
s 2 x ( s ) − 3s + 2 − y ( s ) =
s +1
1 3
i.e. x (s) + y ( s) = + (1)
s3 s
1
and s 2 x ( s) − y ( s) = + 3s − 2 (2)
s +1
5.106 Mathematics II

Solving (1) and (2), we have

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

1  1 1 
 − s
 2 2 
3 1 3s 2
= 2 + 2 + + 3 2 + 2 − 2
s +1 s +1 s ( s +1) s ( s +1) s +1 s +1
2

1 3 5
s
3 1
= 2 − 2 2 + 22 + + 3 2
s +1 s +1 s +1 s ( s +1) s ( s +1)
2

t t t t
1 3 5
∴ x = e−t − sin t + cos t + 3∫ sin t dt + ∫ ∫ ∫ sin t dt dt dt
2 2 2 0 0 0 0
2
1 3 5 t
= e−t − sin t + cos t + 3 (1− cos t ) + + cos t −1
2 2 2 2

1 −t 3 1 t2
i.e. x= e − sin t + cos t + + 2 (3)
2. 2 2 2
1 3 1
∴ x ′′ = e−t + sin t − cos t +1 (4)
2 2 2
From the given second equation we have
y = x″ − e−t
1 3 1
=1− e−t + sin t − cos t.
2 2 2

Example 5.15 Solve the simultaneous equations

D2x − Dy = cos t and Dx + D2y = − sin t; x = 1,
Dx = 0, y = 0, Dy = 1 at t = 0.
Taking the Laplace transforms of both equations, we get
s
s 2 x ( s ) − sy ( s ) = +s (1)
s +1
2

1
and sx ( s ) + s 2 y ( s ) = 2 − (2)
s 2 +1

Solving (1) and (2), we have

s 2 s 1
x ( s )= + + −
s +1 s ( s +1) ( s +1) s ( s +1)2
2 2 2 2 2

s 2 s 2 −1
= + +
s 2 +1 s ( s 2 +1) s ( s 2 +1)2
 Laplace Transforms 5.107

1 2
and y ( s )= −
(s2 +1)
2 2
s +1


 
t

  1 
t
−1  1 
∴ x = cos t + 2 ∫ sin t d t + ∫  L  2 − 2 L−1
2
dt
  s +1 s 2 +1) 
0 0 

 ( 
t

= cos t + 2 (1− cos t ) + ∫ (sin t − sin t + t cos t ) dt

0

= 1 + t sin t
1
and y = sin t − 2× (sin t − t cos t )
2
= t cos t.
Example 5.16 Show that the solution of the equation
t
di 1
L + Ri + ∫ i d t = E , i (0) = 0 [where L, R, E are constants] is given by
dt C 0

Ï E - at
Ô w L e sin w t ,if w > 0
2

Ô
Ô E
i = Ì te - at ,if w = 0
Ô L
Ô E - at 2
Ô kL e sinh kt ,if w < 0
Ó
R 1 R2
where a= , w2 = - 2 and k2 = −w2.
2L LC 4 L
Note The given equation is an integro-differential equation, as the unknown
(dependent variable) i occurs within the integral and differential operations.]
Taking Laplace transforms of the given equation, we get
1 E
Lsi ( s ) + Ri ( s ) + i (s) =
Cs s

i.e. (LCs2 + RCs +1) i (s) = EC

EC
∴ i ( s) =
2
LCs + RC s +1
E 1
= ⋅
L s2 + R s + 1
L LC
5.108 Mathematics II

E 1
= ◊
Rˆ Ê 1 R2 ˆ
2
L Ê
ÁË s + 2 L ˜¯ + Á LC - 2 ˜
Ë 4L ¯
E 1
= ◊ ,if w 2 > 0
L (s + a ) 2 + w 2
E - at
∴ i (t ) = ◊ e sin w t
Lw

If ω = 0,

E 1
i (s )= ⋅
L ( s + a)2
E −at
∴ i (t )= te
L
If ω2 < 0 a n d ω2 = − k 2 ,

E 1
i (s ) = ⋅
L (s + a ) 2 − k 2
E −at
∴ i (t ) = e sinh kt .
Lk

Example 5.17 Solve the simultaneous equations

3x′ + 2y′ + 6x =0 and
t

y ′ + y + 3∫ x dt = cos t + 3 sin t , x (0) = 2 and y (0) = − 3.

0

Taking Laplace transforms of both the equations, we get

3[s x (s ) − 2] + 2[s y (s ) + 3] + 6 x (s ) = 0 and
3 s 3
s y (s ) + 3 + y (s ) + x (s ) = 2 +
s s + 1 s 2 +1
i.e. (3s + 6) x (s ) + 2 s y (s ) = 0 (1)

3 s +3
and x (s ) + (s +1) y (s ) = −3 (2)
s s +1 2

Solving (1) and (2) for x (s ), we have

2 s (s + 3)
[3(s + 2) (s +1) − 6] x (s ) =− + 6s
s +1
2

2 1 2
i.e. x (s )=− ⋅ +
3 s +1 s + 32
 Laplace Transforms 5.109

2
∴ x =− sin t + 2e−3t .
3
Solving (1) and (2) for y (s ) , we have

3s 2 + 5s − 2  1 2 s −1
y (s ) = =−  + 2
(s + 3) ( s +1)  
 s + 3 s +1 
2

∴ y = − e−3t − 2 cos t + sin t.

Example 5.18 Solve the integral equation
t
t2
2 ∫0
y (t ) = − u y (t − u ) du

Noting that the integral in the given equation is a convolution type integral and
taking Laplace transforms, we get

1
y (s ) = − L(t ) ⋅ L{ y (t )}
s3
1 1
= 3 − 2 y (s )
s s

1+ s 2  1 1
∴   y (s ) = 3 or y (s ) =
2 
 s  s s (1+ s 2 )
t

∴ y (t ) = ∫ sin t dt
0

= 1 − cos t.

Example 5.19 Solve the integral equation

t

y (t ) = a sin t − 2 ∫ y (u ) cos (t − u ) du.

0

Taking Laplace transforms,

a
y (s ) = − 2 L { y (t )}⋅ L (cos t )
s 2 +1

a 2s
i.e. y (s ) = − ⋅ y (s )
s 2 + 1 s 2 +1

(s +1) 2 a
i.e. y (s ) = 2
s +1
2
s +1
5.110 Mathematics II

a
i.e. y (s ) =
( s +1)
2

∴ y (t ) = a t e−t .
Example 5.20 Solve the integro-differential equation

y ′ (t ) = t + ∫ y (t − u ) cos u d u , y (0) = 4
0

Taking Laplace transforms,

1 s
s y (s ) − 4 = + 2 y (s )
s 2
s +1

 1 
s 1− 2  y (s ) = 2 + 4
1
i.e.
 s +1 s

(s 2 +1)(1+ 4 s 2 )
i.e. y (s ) =
s5
4 5 1
= + 3+ 5
s s s

5 1
∴ y (t ) = 4 + t 2 + t 4 .
2 24

EXERCISE 5(d)

Part A
(Short Answer Questions)
Using Laplace transforms, solve the following equations:
1. x ′ + x = 2 sin t, x (0) = 0
2. x ′ − x = et , x (0) = 0
3. y ′ − y = t, y (0) = 0
4. y ′ + y = 1, y (0) = 0
t

5. y + ∫ y (t ) d t = e−t
0
t

6. x + ∫ x (u ) d u = t + 2t
2

0
 Laplace Transforms 5.111

7. x + ∫ x (t ) d t = cos t + sin t
0

8. x − 2 ∫ x (t ) d t =1
0

t
9. y =1+ 2 ∫ e−2u y (t − u ) d u
0

10. y =1+ ∫ y (u ) sin (t − u ) d u

0

11. f (t ) = cos t + ∫ e f (t − u ) d u
−u

12. y (t ) = t + ∫ sin u y (t − u ) d u
0

Part B
Solve the following differential equations, using Laplace transforms:
13. x″ + 3x′ +2x = 2 (t2 +t + 1), x (0) = 2,x′ (0) = 0
14. y″ − 3y′ − 4y = 2e−t, y (0) = y′ (0)=1
15. x″ + 4x′ + 3x = 10 sin t, x (0) = x′ (0) = 0
16. (D2 + D − 2) y = 20 cos 2t, y = − 1, Dy = 2 at t = 0
17. x″+ 4x′ + 5x = e−2t (cos t − sin t), x (0) = 1, x′ (0) = −3
18. y″ + 2y′ + 2y = 8 et sin t, y (0)= y′ (0) = 0
19. x″ − 2x′ + x = t 2 et, x (0) = 2, x′ (0) = 3
20. y″ + y = t cos 2t, y (0) = y′ (0) = 0
21. x″ + 9x=18t, x (0) = 0, x (π/2) = 0
22. y″ + 4y′ = cos 2t, y (π) = 0, y′ (π) = 0
23. (D2 + a2) x = f (t)
24. (D3 − D) y = 2 cos t, x = 3, Dx = 2, D2x = 1 at t = 0
25. x′″ − 3x″ + 3x′ − x =16 e3t, x (0) = 0, x′ (0) = 4, x″ (0) = 6
26. (D4 − a4) y = 0, y (0) = 1, y′ (0) = y″(0) = y′″ (0) = 0
Solve the following simultaneous equations, using Laplace transforms:
27. x′ − y = et; y′ + x = sin t, given that x (0)= 1 and y (0) = 0
28. x′ − y = sin t; y′ − x = − cos t, given that x = 2 and y = 0 for t = 0
29. x′ + 2 x − y = − 6 t; y′ − 2x + y = − 30 t, given that x = 2 and y = 3 at t = 0
30. Dx + Dy + x − y = 2; D2x + Dx − Dy = cos t, given that x = 0, Dx = 2 and y = 1
at t = 0
31. D2x + y = − 5 cos 2t; D2y + x = 5 cos2t, given that x = Dx = Dy = 1 and
y = − 1 at t = 0
5.112 Mathematics II

32. x ′ + y ′ − x = 2et + e−t ; 2x ′ + y + 3 ∫ y dt = 2et (t + 3) , given that x (0) =

0
− 1 and y (0) =2
Solve the following integral equations, using Laplace transforms:
t

33. x ′ + 3 x + 2 ∫ x dt = t , x (0) = 0
0

34. y ′ + 4 y + 5 ∫ y dt = e , y (0) = 0
−t

35. x ′ + 2 x + ∫ x dt = cos t , x (0) = 1

0

36. y ′ + 4 y + 13 ∫ y dt = 3e sin 3t , y (0) = 3

−2 t

37. x (t ) = 4t − 3 ∫ x (u ) sin (t − u ) du
0

38. y (t ) = e − 2 ∫ y (u ) cos (t − u ) du
−t

t
x(u )
39.
∫ t −u
du = 1 + t + t 2
0

40. ∫ y (u ) y (t − u ) du = 2 y (t ) + t − 2
0

ANSWERS

Exercise 5(a)
2 1
{1 − e−(s − 2 ) }
t
3. tan t ; e 9.
s −2

1 1 2 1
10.  + 2  e−s −  + 2  e−2 s
s s  s s 

2 2
11. (1 − e−π s ) ⋅ 12. (1 − e− 2 π s ) ⋅
s +4
2
s +1
2
 Laplace Transforms 5.113

s 1 π 1
13. ⋅ e−2 π s / 3 14. (1 + e− πs ) +  + 2  e−πs
s2 + 1 s2 + 1  s s 

1 π 1 1 −3 s π
15. ; 16. e ; 2 e−s
2s s s s2 s + π2
2 4 4  −2 s 6a 3 6a 2b 3ab 2 b3
17.  3 + 2 +  e + 3 + 2 +
18.
s s s s4 s s s

1 1 s  
(ω cos θ + s sin θ ) 20.  − 2
1 
36 
19.
s +ω2 2
2  s s +
1  3s s  1  3 1 
 2 − 2   2 − 2 
4  s + 4 s + 36 
21. 22.
2  s + 9 s + 1
1  s s  1  1 1 3 1 
 2 + 2   − + − 
8  s − 3 s − 1 s + 1 s + 3 
23. 24.
2  s + 25 s + 1
1  1 3 2  2 2 1
 + +  + +
s 
25. 26.
4 s − 2 s +2 (s + 1)3 (s + 1) 2 s + 1
s 1

 3 1 
27. 28.  + 
(s + 2) 2 + 9 4
 s − 3 s + 1

2s 1 1
29. 30. +
s +4
4
(s − 1) 3
(s + 1)3
2e3
31. 32. (t − a) ua (t)
2s − 3
33. u2 (t) − u3 (t) 34. e3(t −2). u2 (t)
35. cos 3 (t −1) u1 (t) 36. sin t + sin (t − π) uπ(t)
t −t 1
37. 2 e 38. (2 + 4t + 3t 2 )
π 2
1 3 3t / 2 1 3
39. t e 40. t (2 + t ) e 2t
96 12
3
41. 2 cos 2t + sin 2t 42. cosh 3t + 2 sinh 3t
2
1 1 −t
43. (1 − e−at ) 44. e sin 2t
a 2
8 12 9
45. e3t cos t 46. + +
(s − 3) 3
(s − 3) 2
s −3

π s 1

 1 3 3 1 
47. ⋅ 48.  − + − 
s−a s−a 8
 s − 2 s s + 2 s + 4 

5.114 Mathematics II

4a 3 2
49. 50.
s 4 + 4a 4 (s − 2) ( s 2 − 4 s + 8)


(s + 3) 
 3 1 


51.  + 
4  
 (s − 3) 2
+ 4 (s + 3) 2
+ 36 



w cos q + (s + k ) sin q 3 ( s 2 + 2 s + 9)
52. 53.
(s + k ) 2 w 2 (s 2 + 2s + 5)(s 2 + 2s + 17)

(s − 1) 
 1 1 1 1 
54.  + + + 
4   (s − 1)

2
(s − 1) + 4 (s −1) + 16 (s + 1) + 36 
2 2 2


1
 1 3 5 9 
55.  + + − 
4
 (s + 2)

2
(s + 2) + 9 (s + 2) + 25 (s + 2) + 81
2 2 2

s2 − 4 s (s + 2)
56. 57.
(s + 4) (s + 2 s + 2)
2 2 2
2

6 (s − 2) 2s
58. 59.
(s − 4 s + 13) (s + 1)
2 2 2 2

s2 − 4 1 5
60. 61. − 2et + e− 2t
(s + 4)
2
2 2 2
 9 
62. 2e − t − 3et + 5e2t 63. 1 − 5t + t 2  e−t
 2 
64. − 2e − t + 2e2t + te2t 65. − 2 + t + 2e−t + t e−t
1 1
66. sin t − sin 2t + sin 3t 67. cos t − 2 cos 2 t + cos 3 t
2 3
 
e− t / 2  3 cos
1 3 3
68. e − 3t (2 cos 5t − 3 sin 5t) 69. t + sin t 
3  2 2 

1 − bt / 2 a 
  λt   m b  2a  λt 
70. e ⋅
cos   +  − ⋅ sin  
a 
  2a   l
 2a  λ  2a 

if b2 − 4ac < 0 and = − λ2

1 − bt / 2 a 
  kt   m b  2a  kt 
e ⋅
cosh   +  − ⋅ sinh   ,
  2a   l 2a  k  2a 
a 
 
if b2 − 4ac > 0 and = k2

1 − bt / 2 a 

 m b 
1 +  −  t 
 , if b − 4ac = 0
2
e
a 

  l 2a 
 

 Laplace Transforms 5.115

71. 2 + e − 2t (cos 3t + 2 sin 3t)

1  − t 
t/2  3 3 
72.  e − e cos t − 3 sin t 
3   2 2 

1
73. (sin at cosh at − cos at sinh at)
4a 3
1
74. sin t sinh t
2
1
75. (sinh 2t cos 2t + cosh 2t sin 2t)
4
t t 2 3 t
76. cos cosh 77. sin t cosh
2 2 3 2 2
1 1
78. t sinh at 79. (sin t − t cos t )
2a 2

1
80. t sin 2t
4

Exercise 5(b)

1 1 s
2. − 7.
s (e − 1) (s 2 − a 2 )
2 2
s
s
s2 − a2 2k 3
8. 9.
(s 2 + a 2 ) (s 2 + k 2 )
2 2

2ks 2 3s 2 + a 2
10. 11.
(s 2 + k 2 ) s 2 (s 2 + a 2 )
2 2

s3 2sinht
12. 13.
(s 2
+k )
2 2 t

1− e− t 1− e− at
14. 15.
t t
e− bt − e− at et −1
16. . 17.
t t
2 sin t
18. (cos 2t − cos t ) 19.
t t
sin at
20. 21. cot − 1 (s/a)
t
5.116 Mathematics II

 s + 1  s −1
22. log   23. log  
 s   s 

1  s 2 + a 2  1  s 2 + 4 
log   log  
 s 2   s 2 
24. 25.
2 4

E (1− e−s / E ) E  sT 
26. 27. tanh  
s (1− e −2 π s / n
) s  4 

28. [1 − e−2π(s − 1)]/(s − 1) (1 − e−2πs)

2 Ï
1 Ê ps ˆ¸
coth (πs) Ì s + w cosech Á ˝
Ë 2w ˜¯ ˛
29. 30.
4 s +1
2 2
s +w Ó 2

1 
1 π 

−2 π s  2 (
31. 1− e−π s ) − e−π s 
1− e 

 s s 



32. 1/(s2 + 1) (1 − e−πs) 33. ω/(s2 + ω2) (eπ s/ω − 1)

1  πs 
34. tanh  
s 2
2

1 π −π s 
35.  2 (e−π s −1) + e (e−π s −1) / (1− e−2 π s )
2

 s s 

1 1 3 3 1 
36.  − + − 

8  (s − 3) 2
(s −1) 2
(s + 1) 2
(s + 3) 2 

1  3( s 2 − 4) ( s 2 − 36) 
37. +
4  ( s 2 + 4) ( s 2 + 36) 
2 2

1  ( s 2 − 4) ( 2 )
38. − s − 64 2 
2  ( s 2 + 4) ( s 2 + 64) 
2

 s s  1 s (48 − s 2 )
39. s  +  40. −
 ( s 2 + 36)2
( s 2 + 4) 
2
s 3 ( s 2 + 16)3

9 ( s 2 − 3) 1− 3s 2 18(s 2 + 4 s + 1)
41. + 42.
( s 2 + 9)
3
( s 2 + 1)
3
(s 2 + 4 s + 13)3

s 2 − 6s − 7 1 1
43. 44. +
(s 2 − 6 s + 25) 2 (s + 1) 3
(s + 5)3
 s−2 s+2 
45. 3  2 − 
 (s − 4 s + 13) 2 (s 2 + 4 s + 13) 2 
 Laplace Transforms 5.117

2
46. (1− cos a t )
t
2 −bt 2
47. (e − cos a t ) 48. (cos t − e 2t )
t t

2 2a 1 t 
49. sinh a t − cosh at + a δ (t ) 50. sin  
t2 t t  2

1 −2 t 1 −bt
51. − ⋅ e sin 3t 52. − e sin at
t t
2 t t
53. − sin sinh 54. 0
t 2 2
1
55. 56. log 3
2
1 b
57. log 2 58. log  
2 a
π  s + 1
59. 60. log  
8  s 
1  s 2 + a 2  1  s 2 + 16 
 
 s 2   s 2 
61. log 62. log
2 4
1  s 2 + 16  s
log   64. s log + cot −1 s
63.
4  s 2 + 4  s +1
2

t t
65. sin t 66. sinh at
2 2a
t 2t 1
67. e sin t 68. sin at − t sin a t
2 a
t −4t
69. e−3t(1 + t) sin t 70. e sinh t.
2

Exercise 5(c)

t2 1 s
8. No; 1* t = ≠t 10. 11.
2 s−a s + a2
2

t t 
2
2
12. 13. t (1 − t) e−2t 14. e  + 2 t + 1
s ( s 2 + 4) 2 

1 at
15. e (a sin b t + b cos b t ) 16. 2 e−2t − e−t
b
5.118 Mathematics II

1 1  s −1 1  s 2 + 1
17. cot −1 s 18. log   19. log  
s s  s  s  s 
1 1 2
20. 21. . 22.
s (s + 1) 2 s (s 2 + 2 s + 2) (s 2
+ 1)
2

1 1
23. (1− e−at ). 24. e−at + t − 1 25. (cosh a t −1)
a a2
1
26. 1 − cos t 27. 1;0 28. ; 0 30. sin t
2
1 t 1
31. (e − e−3t ). 32. t et 33. 2 (1− cos a t )
4 a
34. e −t + t − 1 35. sin t − cos t + e −t
2a s 2s3 2a s 2
36. ; ;
(s 2 + a 2 ) (s 2 + a 2 ) (s 2 + a 2 )
2 2 2

s2 + a2 2a s 2 2a 3
37. ; ;
(s 2 − a 2 ) (s 2 − a 2 ) (s 2 − a 2 )
2 2 2

1 1
38. (e−bt − e−at ); (a e−at − b e−bt )
a −b a −b
1 t 1 1 9
39. e − e 2 t + e 3 t ; e t − 4e 2 t + e 3 t
2 2 2 2
1  1 1  1
40.  sin b t − sin a t  ; 2 (cos b t − cos a t );
a 2 − b 2  b a a − b2
1
(a sin at − b sin bt )
a − b2
2

1 t
41. (sin 2 t + 2 t sin 2 t ); (cos 2 t − t sin 2 t ); e−3t sin 2 t
4 4
1 1 t −at
42. (sinh a t + a t cosh a t ); (a t sinh a t + 2 cosh a t ); e sinh at
2a 2 2a
1 1
43. sin t sinh t ; (sin t cosh t + cos t sinh t ); cos t cosh t
2 2
2 (s −1) 2 3s 2 − 4 s + 2
44. 45. 46.
s ( s 2 − 2 s + 2) ( s 2 − 2s + 2) s 2 ( s 2 − 2 s + 2)
2 2 2

1 −1  s + 2  1  s + 2 
47. cot   48. cot −1  
s  3  s+2  3 
3  s 2 + 4 s + 13  6 −1  s + 2 
49. log   − cot  
26 
 s 2
 13  3 
 Laplace Transforms 5.119

50. 2 − t − 2 e−t
4 −3t /2 3 −5t /3 7 73
51. e − e + t−
9 25 15 225
1
52. [4 − e−3t (3 sin 4 t + 4cos 4 t )]
100
1
53. e − t (1 − cos t) 54. (1− 2t − cos 2 t + sin 2 t )
4
1 1
55. (sin 3t − 3 t cos 3 t ) 56. (2 − 2 cos 3 t − 3 t sin 3 t )
54 162
1 −3 t 1
57. e (sin t − t cos t ) 58. (a t cosh a t − sinh a t )
2 2a 3
1
59. (2 − 2 cosh a t + a t sinh a t )
2a 4
1 −2 t
60. e (2 t cosh 2 t − sinh 2 t )
16
1 1
63. (cos t − sin t − e−t ) 64. (sin a t + sinh a t )
2 2a
1 1
65. (cos 2t − cos 5 t ) 66. (sin a t + a t cos a t )
5 2a
1
67. (sin 2 t − 2 t cos 2 t ) 68. 2 e − t + sin 2 t − 2 cos 2 t
16
e−t 2 1
69. (t + 4 t + 6) + t − 3 70. (sin t cosh t − cos t sinh t )
2 4

Exercise 5(d)
1. x = e − t − cos t + sin t 2. x = t e t 3. y = e t − t − 1
4. y = 1 − e −t 5. y = (1 − t)e −t 6. x = 2t
7. x = cos t 8. x = e 2 t 9. y = 1 + 2t
t2 t3
10. y = 1 + 11. f (t) = cos t + sin t 12. y = t +
2 6
1
13. x = t 2 − 2t + 3 − e −2t 14. y = (13e−t −10t e−t + 12e 4t )
25
5 −t 1 − 3 t
15. x = e − e − 2 cos t + sin t
2 2
2 −2 t 4 t
16. y = e + e − 3 cos 2t + sin 2t
3 3
−2 t  3 t 
17. x = e  cos t − sin t + (sin t + cos t )
 2 2 
5.120 Mathematics II

18. y = 2 (sin t cosh t − cos t sinh t)

 t4  5 4 t
19. x =  + t + 2 e t 20. y =− sin t + sin 2 t − cos 2 t
12  9 9 3
1
21. x = 2 t + π sin 3 t 22. y = (t − π ) sin 2 t
4
t
1
23. x = A cos a t + B sin a t + ∫ sin a u ⋅ f (t − u ) d u
a 0
24. y = 3 sinh t + cosh t − sint + 2
1
25. x = 2e 3 t − 5t 2 e t − 2e t 26. y = (cos a t + cosh a t ).
2
1 t
27. x = (e +2sin t + cos t − t cos t );
2
1
y = ( − e−t − sin t + cos t + t sin t )
2
28. x = 2 cosh t; y = 2 sinh t − sin t
29. x = 1 + 2t − 6t 2 + e−3t; y = 4 − 2t − 12t 2 − e −3t
30. x = t + sin t; y = t + cos t
31. x = sin t + cos 2t; y = sin t − cos 2 t
32. x = t e t − e −t; y = e t + e −t
1 1 − t 1 −2 t
34. y =− e + e (cos t + 3 sin t )
−2 t −t
33. x = (1+ e ) − e
2 2 2
1
35. x = {(1− t ) e−t + cos t}
2

−2 t  7 3 
36. y = e  
3 cos 3t − sin 3t + t sin 3t + t cos 3t 


 3 2

3
37. x = t + sin 2 t 38. y (t) = e −t (1− t)2
2
1  −1 2 8 3 2
t + 2t + t 
12
39. x (t )=
π 3 

40. y (t) = 1