Production Engineering - Laplace Transformation

Ekeeda – Production Engineering
UNIT – IV Laplace Transformations Class 1
Section I
Introduction
The knowledge of Laplace Transformations has in recent years became an
essential part of Mathematical background required of engineers and
scientists.
This is because the transform methods provide an easy and effective means
for the solution of many problems arising in engineering.
This subject originated from the operational methods applied by the English
engineer Oliver Heaviside (1850 – 1925) to problems in electrical
engineering.
Unfortunately, Heaviside’s treatment was unsystematic and lacked rigour,
which was placed on sound mathematical footing by Bromwich and carson
during 1916 – 1917.
It was found that Heaviside’s operational calculus is best introduced by
means of a particular type of definite integrals called “Laplace Transforms”.
The method of Laplace Transforms has the advantage of directly giving the
solution of differential equations with given boundary values without the
necessary of first finding the general solution and then evaluating from it
the arbitrary constants.
Moreover, the ready tables of Laplace Transforms reduce the problems of
solving differential equations to mere algebraic manipulation.
Definition: Integral transform
Let K(s, t) be a function of two variables s and t where s is a parameter [s 
R or C] independent of t. Then the function f(s) defined by an Integral which
is convergent. i.e., 

−
= dttFtsKsf )(),()( is called the Integral Transform of
the function F(t) and is denoted by [T{F(t)], K(s, t) is kernel of the
transformation.

If kernel K(s, t) is defined as







=
−
0
0
1
0,0
),(
te
t
ort
tsK
st
then


−
=
0
)()( dttFesf st
is called “Laplace
Transform” of the function F(t) and is also denoted by L{ F(t) } or )(sf .
 L { F(t) } = 

−
=
0
)()( dttFesf st
= )(sf
Definition: Laplace Transformation
P f(P)
f is real valued /
complex valued function
Domain Range
Let F(t) be a real or complex valued function defined on [0, ). Then the
function f(s) defined by 

−
=
0
)()( dttFesf st
is called Laplace Transformation
of F(t) if the integral exists and f(s) = L{ F(t) }.
Note: If L { F(t) } = )(sf  F(t) = L-1 { )(sf }. Then F(t) is called the inverse
Laplace Transform of )(sf which transforms F(t) into )(sf is called “The
Laplace Transforms Operator”.
Linearity Property of Laplace Transformation:
A transformation T is said to be linear if  a1, a2 constants and F1(t), F2(t) be
any functions of F. Then L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) } or
F1(t), F2(t), F3(t) there exists a1, a2, a3 constants such that
L {a1 F1(t) + a2 F2(t) – a3 F3(t)} = a1 L{ F1(t) + a2 L{ F2(t) } – a3 L{ F3(t)}.
Theorem: Laplace transformation is a linear transformation i.e.,  a1, a2
constants L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) }.

Proof: By definition, 

−
=
0
)()(=}F(t)L{ dttFesf st
 L.H.S. = L{ a1 F1(t) + a2 F2(t) }
= 

−
0
2211 dt](t)Fa+(t)Fa[st
e
= 

−
0
11 dt(t)F.a st
e + 

−
0
22 dt(t)Fa st
e
= a1 L{ F1(t) + a2 L{ F2(t) }
= R.H.S.
 L{ a1 F1(t) + a2 F2(t) } = a1 L{ F1(t) + a2 L{ F2(t) }
Hence Laplace Transformation is a linear Transformation.
This completes the proof of the theorem.
Definition: Piece-wise or sectionally continuous
A function F(t) is said to be piece-wise or sectionally continuous on a closed
interval a  t  b, if it is defined on that interval can be divided into finite
number of sub-intervals in each of which F(t) is continuous and has finite
left limit and right hand limits.
i.e., Lim F(t) = Lim F(t) = Lim F(t) = finite F say  a  t  b.
t → - 0 t → + 0 t → 0
therefore, F is continuous.
Geometrically:
F(t)
0 a t1 t2 t3 b t
I1 I2 I3 I4

Figure. piece-wise or sectionally continuous.
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
Definition: Functions of an Exponential order
A function F(t) is said to be an exponential order  as t tends to  if there
exist a positive real number M a number  and a finite number t0 such that
| F(t) | < M e t or | e-t F(t) | < M,  t  t0.
Note: If a function F(t) is of an exponential order . It is also of  such that
 > .
Definition: A function of class A
A function which is piece-wise continuous or sectionally continuous on
every finite interval in the range t  0 and is of an exponential order , as t
→  is known as “A function of class A”.

Theorem: Existence of Laplace Transformation
If F(t) is a function which is piecewise or sectionally continuous on every
interval (finite) in the range t  0 and satisfies | F(t) |  M. eat,  t  0 where
a and M are constants. Then the Laplace Transform exists for every s > a.
Proof:
-  0 t t0 t1 
|---------------0-------→|------I1--------→|---------- I2---------------→|
By definition, 

−
=
0
)()(=}F(t)L{ dttFesf st
= 
−
0
0
)(
t
st
dttFe + 

−
0
)(
t
st
dttFe ....................... (1)

−
0
0
)(
t
st
dttFe exists and piece-wise continuous in interval 0  t  t0 so that


−

−

00
)()(
t
st
t
st
dttFedttFe


−

0
..
t
atst
dteMe (since, |F(t)  M eat)
= 

−−
0
)(
.
t
tas
dteM (since, s > a implies s – a > 0  0)
= 

−−
−0
0)(
.
)(t
tas
dte
as
M
0)(
0
.
)(
)( tas
t
st
e
as
M
dttFe −−

−
−
 for any s > a
If t0 is too large then R.H.S. is too small or infinite small therefore L{ F(t) }
exists for s > a.

UNIT–IV Table of General Properties of Laplace Transform Class 3
F(s) = 

−
0
)( dttfe st
S No. Name Laplace Transform Inverse Laplace Transform
01. Definition L{f(t)} = f(s) L-1{f(s)} = f(t)
02. Linearity af1(t) +b f2(t) a f1(s)+bf2(s)
03. Change of scale f(at) )(
1
a
s
f
a
04. First shifting Th. eat f(t) f(s-a)
05. second shifting u(t-a) = {




−
at
atatf
,0
,)(
e-as f(s)
06. Derivative
(multiply by s) f (t) sf(s)-f(0)
07. Second derivative
(multiply by s2) f (t) s2f(s) – s f(0) - f (0)
08. n th derivative
(multiply by sn) f (n)(t) snf(s) – sn-1 f(0) - sn-2 f (0).....– f n-1(0)
09.Integral(division by s) 
t
duuf
0
)(
s
sf )(
10. Multiple integral  −
−−

tntt
duuf
n
nut
duuf
000
)(
)!1(
1)(
)(...... n
s
sf )(
(division by sn)
11. Multiply by t - t f(t) f (s)
12. Multiply by t2 t2 f(t) f (s)
13. Multiply by tn (-1)n tn f(t) f n(s)
14. Division by t
t
tf )(


1
)( duuf
15. Convolution f(t) * g(t) =  −
t
duutguf
0
)()(

=  −
t
duugutf
0
)()( f(s)*g(s)=L(f * g)
= L-1(f(s) g(s)}
16. f-periodic with period p f(t) = f(t + p) 
−
−
−
p
su
sp
duufe
e 0
)(
1
1
Table of some Important Laplace Transforms By Dr N V Nagendram
________________________________________________________________
S No. Laplace Transform Inverse Laplace Transform
01. L{f(t)} = f(s) L-1{f(s)} = f(t)
02. 1
s
1
03. t 2
1
s
04. t2
3
!2
s
05. tn n = 0,1,2,3,... 1
!
+n
s
n
06. eat
as −
1
07. e-at
as +
1
08.
)!1(
et at1)-(n
−n n
as )(
1
−
n = 1,2,3,.....
09.
)(
et at1)-(k
k k
as )(
1
−
k = 1,2,3,.....
10. sin at 22
as
a
+
11. cos at 22
as
s
+
12. ebt sin at 22
)( abs
a
+−
13. ebt cos at 22
)( abs
bs
+−
−
14. sinh at 22
as
a
−
15. cosh at 22
as
s
−
16. ebt sinh at 22
)( abs
a
−−

17. ebt cosh at 22
)( abs
bs
−−
−
18. u(t - a)
s
e as−
19. f(t – a). u(t – a) e-as f(s)
20.
ab
ee atbt
−
−
))((
1
bsas −−
,a  b
Table of some Important Laplace Transforms By Dr N V Nagendram
________________________________________________________________
S No. Laplace Transform Inverse Laplace Transform
21.
ab
aebe atbt
−
−
ba
absas
s

−−
,
))((
22. 3
2
cossin
a
atatat −
222
)(
1
as +
23.
a
att
2
sin
222
)( as
s
+
24.
a
atatat
2
cossin +
222
2
)( as
s
+
25. atatat sin
2
1
cos − 222
3
)( as
s
+
26. t cos at 222
22
)( as
as
+
−
27. 3
2
sinhcos
a
atathat −
222
)(
1
as −
28.
a
att
2
sinh
222
)( as
s
−
29. (sinh at + at cosh at)/ 2a 222
2
)( as
s
−
30. (cosh at + 1/2at sinh at) 222
3
)( as
s
−
31. t cosh at 222
22
)( as
as
−
+
32.
a
att
2
sin2
322
22
)(
3
as
as
−
−
33.
2
cos2
att
322
23
)(
3
as
sas
+
−

34.
6
cos3
att
422
4224
)(
6
as
asas
+
+−
35.
a
att
24
sin3
422
23
)( as
sas
+
−
36.
t
ee atbt
−
as
bs
Log
−
−
37.
t
tsin
s
1
tan 1−
38.  (1) – log t
s
sLog
UNIT–IV Laplace Transformations and its applications Class 4
Definition: Laplace Transform of Periodic function
A function f(t) is said to be a periodic function of period T > 0 if
F(t) = f(T + t) = f(2T + t) = f(3T + t ) = ..... = f(nT + t).
Sin t , cos t are periodic functions of period 2.
The Laplace transform of a piecewise periodic function f(t) with period p is
L{ f(t) } = 
−
−
−
s
st
st
dttfe
e 0
)(.
1
1
; s > 0
Note: (1) = , (
2
1
) =  ,(
2
3
) =(
2
1
+1) =
2
1
(
2
1
) =
2
1
 ,  ,(
2
5
) =(
2
3
+1)
=
2
3
(
2
3
) =
2
3
2
1
 and (- 1
2
3
+ ) =(-
2
1
) = -2. 

UNIT–IV Problems Vs Solutions on Laplace transformations Class 5
Problem Solution
01. L{ 3t - 5} s
s
sf 5
3
)( 2
−=
02. L{ 2t3 – 6t + 8 }
sss
sf
8612
)( 24
+−=
03. L { 6 sin 2t – 5 cos 2t }
4
512
4
5
4
12
)( 222
+
−
=
+
−
+
=
s
s
s
s
s
sf
04. L { 3 cosh 5t – 4 sinh 5t }
25
203
25
20
25
3
)( 222
−
−
=
−
−
−
=
s
s
ss
s
sf
05. L{ (t2 + 1)2 } 5
24 24
4)(
s
sssf ++=
06. L[ cos2 t }
16
22
)( 2
+
+=
s
s
s
sf
07. L{ 3t4-2t3 + 4 e-3t – 2 sin 5t + 3 cos 2t }
4
3
5
10
3
41272
)( 2245
+
+
+
−
+
+−=
s
s
ssss
sf
08. L{ sin2 at }
)4(
2
)( 22
2
ass
a
sf
+
=
09. L{ 2 e3t – e-3t }
9
9
)( 2
−
+
=
s
s
sf
10. L{ sin 5t + cos 3t }
925
5
)( 22
+
+
+
=
s
s
s
sf
11. L{ e-2t – e-3t }
3
1
2
1
)(
+
−
+
=
ss
sf
12. L{ F(t) } if F(t) =





10
10
t
tet
( ))1(
1
1
1
)( −−
−
−
= s
e
s
sf

13. L{ F(t) } if F(t) =





24
200
t
t
s
e
sf
s2
.4)(
−
=
14. L{ F(t) } if F(t) =





53
50
t
tet
( ) ss
e
s
e
s
sf 5)1(5 3
1
1
1
)( −−−
+−
−
=
15. L{ F(t) } if F(t) =





2
201
tt
t
2
22
1
)(
s
e
s
e
s
sf
ss −−
++=
16. L{ F(t) } if F(t) =





45
40
t
tt
2
4
2
)1(1
)(
s
es
s
sf
s−
−
−=
17. L{ F(t) } if F(t) =
t
1
s
sf
1
)( =
18. L{ }
1
3






−
t
t 







++−= −
2
1
2
1
2
3
2
5
81263
4
)(
ssss
sf

19. L{ e2t + 4 t3 – 2 sin 3t + 3 cos 3t } 





+
−
++
−
=
9
)2(324
2
1
)( 24
s
s
ss
sf
20. L{ 1 + 2 t +3 t } 







++= 3
23
1
)(
sss
sf

21. L{ cosh at – cos at } 





−
= 44
2
2
)(
as
sa
sf
22. L{ cos (at+b) } 





+
−
= 22
sincos
)(
as
babs
sf
23. L{ (sint – cos t)2 }
)4(
42
)( 2
2
+
+−
=
ss
ss
sf
24. L{ sin 2t cos 3t }
)25)(1(
)5(2
)( 22
2
++
−
=
ss
s
sf
25. L{ sin at sin bt } 





−+++
= 222222
])([])([
2
)(
basbas
abs
sf

UNIT–IV Problems Vs Solutions on Laplace transformations Class 6
Section II
. First shifting / Translation Lemma
. Second shifting / Translation Lemma
. Change of scale property
. Multiplication by tn
. Problems Vs Solutions

Section II
Lemma: First shifting / Translation Lemma.
If L ( F(t) ) = )(sf where s >  then L ( eat F(t) ) = )( asf − , s >  + a or If
)(sf is a Laplace transformation of F(t) then )( asf − is a Laplace
transformation of eat F(t).
Proof: Y
)( asf − )(sf
O L ( eat F(t) ) L ( F(t) )
X
Figure

Given )(sf is a Laplace transformation. Then )(sf = dttFe ts
)(
0


−
= L { F(t) }
So, )( asf − = dttFe tas
)(
0
)(


−−
= dttFee tsta
)(
0


−
= eat dttFe ts
)(
0


−
= eat )(sf
 )( asf − = eat )(sf since, )(sf = dttFe ts
)(
0


−
= L { F(t) }
 )( asf − is a Laplace transformation of eat L{ F(t) }.
Lemma: Second Translation / shifting Lemma.
If L { F(t) } = )(sf and G(t) =




−
at
atatF
0
)(
then, L { G(t) } = e-a s
)(sf .
Proof:
− a 
|------------------------ t < a ------------→|0------|--- t > a -----→|
Given )(sf = dttFe ts
)(
0


−
= L { F(t) } and given G(t) =




−
at
atatF
0
)(
So that, )(sf = dttFe ts
)(
0


−
= +
−
dte
a
ts
0.
0
dtatFe
a
ts
)( −

−
Put t – a = x implies dt = dx , since da/dx = 0 implies a = c.
= dtatFe
a
ts
)( −

−
= dxxFe
a
axs
)()(


+−
= e-a s dxxFe
a
xs
)(

−
= e – a s dttFe
a
ts
)(

−
This completes the proof of lemma.
Note: for every x  (a, ) such that x  (0, ) and for every t  (0, ), a = 0

 dxxFee
a
xssa
)(.

−−
= dxxFee xssa
)(.
0


−−
For any t, e-as
 L { G(t) } = e-as L { F(t) }
)(sf e-as = L { G(t) } since, dxxFee
a
xssa
)(.

−−
= dttFe
a
ts
)(

−
 L { G(t) } = e-as )(sf
 L { G(t) } = e-as L { F(t) }
Lemma: Change of scale property
If L { F(t) } = )(sf then L { F(at) } =
a
1






a
s
f .
Proof: Given )(sf = dttFe ts
)(
0


−
= L { F(t) }
Let us consider L { F(at) } = dtatFe ts
)(
0


−
Put at = x implies t = x/a so dt = (1/a) dx
L { F(at) } = dtatFe ts
)(
0


−
=
a
1
dxxFe a
x
s
)(
0

 





−
=
a
1
dxxFe a
s
x
)(
0

 





−
=
a
1
dttFe a
s
t
)(
0

 





−
[since,  =
b
a
b
a
dttfdxxf )()( ]

=
a
1






a
s
f [since, )(sf = dttFe ts
)(
0


−
]
 L { F(at) } =
a
1






a
s
f or L { F(at) } =
a
1
dttFe a
s
t
)(
0

 





−
This completes the proof of the lemma.
Lemma: Multiplication by tn
If L { F(t) } = )(sf then L { tn F(t) } = (-1)n
n
n
ds
d
})({ sf for n =
1,2,3,………………
We have )(sf = dttFe ts
)(
0


−
On differentiation w.r.t. s,








−
dttFe
ds
d ts
)(
0
=
ds
d
{ )(sf }
By Leibnitz principle for differentiation under integral sign,
dttFe
ds
d ts
)()(
0


−
=
ds
d
{ )(sf }  dttFet ts
)(
0


−
− =
ds
d
{ )(sf }
 dttFet ts
)(
0


−
= −
ds
d
{ )(sf }
for n = m, dttFte mts
])([
0


−
= (−1)m
m
m
ds
d
{ )(sf }
This completes the proof of lemma on multiplication by tn.
Lemma: Division by t

If L { F(t) } = )(sf then L {
t
1
F(t) } = 

S
dssf )( provided integral exists.
Proof: since )(sf = dttFe ts
)(
0


−
On integrating both sides w.r.t. s from s →  we get,


S
dssf )( = dsdttFe ts
S








−

)(
0
= dtdsetF ts
S


−

0
)( = dtdsetF ts
S
 

−







0
)( here, since t is independent
of s
= dt
t
e
tF
S
ts

 −






0
)( = dt
t
tF
e ts


−
0
)(
= L {
t
1
F(t) } by definition of
Laplace transformation.
 L {
t
1
F(t) } = 

S
dssf )(
Section II
Problems Vs Solutions
Problem #1: Find L { t sin at }
Problem #2: Find L { t2 sin at }
Problem #3: Find L { t3 e-3t }
Problem #4: Find L { t e-t sin 3t }

Problem #5: Find L





 −
t
et
)1(
Problem #6: Find L 




 −
t
btCosatCos )(
Problem #7: Find 

−
0
2
dttSinet t
and evaluate at s = 2
*Problem #8: Find 

0
dt
t
mtSin
Problem #9: Find L








0
dt
t
tSinet
Problem #1: Find L { t sin at }
Solution: Since L { sin at } = 22
as
a
+
L { t sin at } = −
ds
d
{ 22
as
a
+
} = −
( )222
.2
as
as
+
 L { t sin at } =
( )222
.2
as
as
+
−
is required solution.
Problem #2: Find L { t2 sin at }
Solution: Since L { sin at } = 22
as
a
+

L { t2 sin at } = ( −1)2
2
2
ds
d
{ 22
as
a
+
}
=
( ) 







+
−
222
.2
as
as
ds
d
= −
( ) ( )
( ) 







+
+−+
422
22222
2.2.2.2
as
sasasasa
=
( ) 







+
−−−−+
422
53244234
24428.8
as
aassassasa
=
( ) 







+
−−−−+
422
53244234
24428.8
as
aassassasa
=
( ) 







+
−+
422
5324
246
as
aasas
=
( ) 







+
−+
422
5324
246
as
aasas
= −
( ) 







+
−+
322
232
822
as
asaas
= −
( ) 







+
−+
322
232
822
as
asaas
=
( ) 







+
−
322
32
26
as
aas
L { t2 sin at } = =
( ) 







+
−
222
22
)3(2
as
asa
Problem #3: Find L { t3 e-3t }
Solution: since, L { e-3t } = )3(
1
+s
L { t3 e-at } = ( −1)3
3
3
ds
d






+ 3
1
s
= - 2
2
ds
d
( ) 







+
−
2
3
1
s
= -
ds
d
( ) 







+
−−
3
3
2.1
s
= -
ds
d
( ) 







+
3
3
2
s
= -
( ) 







+
−
4
3
3.2
s

= ( ) 







+
4
3
6
s
L { t3 e-at } = ( ) 







+
4
3
6
s is required solution.
Problem #4: Find L { t e-t Sin 3t }
Solution: since, L { Sin 3t } = 22
3
3
+s
And L { t Sin 3t } = -
ds
d






+ 22
3
3
s
=
( )222
3
6
+s
s
So, L { t e-t Sin 3t } =
 222
3)1(
)1(6
++
+
s
s
L { t e-t Sin 3t } =
 222
3)1(
)1(6
++
+
s
s
Problem #5: Find L





 −
t
et
)1(
Solution: since L { 1- et } = s
1
−
1
1
−s
= )(sf
Now L





 −
t
et
)1(
= 

S
dssf )( = 







−
−
S
ds
ss 1
11
=  
−− SsLogsLog )1(
=













− S
s
s
Log
1
= − 







−
s
Log
11
1

= 




 −
s
s
Log
1
L





 −
t
et
)1(
= 




 −
s
s
Log
1
Problem #6: Find L 




 −
t
btCosatCos )(
Solution : since L { Cos at – Cos bt } = 22
as
s
+
− 22
bs
s
+
L 




 −
t
btCosatCos )(
= 

S
dssf )( = 







+
−
+S
ds
bs
s
as
s
2222
=







+−+
S
asLogasLog )(
2
1
)(
2
1 2222
= 0 −
)(
)(
2
1
22
22
bs
as
Log
+
+
=
)(
)(
22
22
bs
as
Log
+
+
 L 




 −
t
btCosatCos )(
=
)(
)(
22
22
bs
as
Log
+
+

−
0
2
dttSinet t
and evaluate at s = 2
Solution : Let 

−
0
2
dttSinet t
= 

−
0
2
)( dttSinte t
= - 1. 





+1
1
2
sds
d
= 





+1
2
2
s
s
at s = 2 

−
0
2
dttSinet t
= 25
4
 

−
0
2
dttSinet t
= 





+1
2
2
s
s
and s = 2 value is 25
4
Is required solution.

0
dt
t
mtSin
Solution: L { Sin mt } = 





+ 22
ms
m

Now 

0
dt
t
mtSin
=
−

=





+ S
S
msds
ms
m
)]/(tan[ 1
22
 

0
dt
t
mtSin
= 





− −
)(tan
2
1
m
s
as s → 0
= 2

if m > 0
= − 2

if m < 0
 

0
dt
t
mtSin
= 





− −
)(tan
2
1
m
s
Problem #9: Find L








0
dt
t
tSinet
Solution: since L { t
tSin
} = sCotsTans
s
ds
S
S
111
22
2
)](tan[
1
−−−

=−==





+

 L { t
tSin
et
} =L { et Cot-1s} = Cot-1 (s – 1) [ by shifting lemma ]
 L








0
dt
t
tSinet
= s
1
.Cot-1 (s − 1). Hence the solution.
Section II
S
NO
F(t) L { F(t) } Solution
01 Sin t Cos t L{Sin t Cos t}
= ½ L{ 2 sin t cos t}
= ½ L { Sin 2t }
22
2
1
+s
, for s>|2|

02 Cosh2 2t L{Cosh2 2t }
= ½. L{1+Cosh 4t} )4(
8
22
2
−
−
ss
s
,s > 0
03 Sinh at – Sin at L{Sinh at – Sin at}
44
3
2
as
a
−
, for s>|a|
04 Sin (at + b) L{ Sin (at+b) }
=L{Sin at cos b + Cos at
Sin b}
22
as
bSinsbCosa
+
+
, for
s>|a|
05 Cos( t +  ) L { Cos( t +  ) }
,22


+
−
s
SinCoss
for
s>||
06 Cos3 3t = 1/4 cos3t + 3/4 cos t L{ Cos3 3t }
= L{ 1/4 cos3t + 3/4 cos t} ( )( ) |9|,
93
)63(
2222
3

++
+
s
ss
ss
07 Cos3 t L{ Cos3 t }
( )( ) |3|,
31
)7(
2222
3

++
+
s
ss
ss
08 Sin3 2t
(Sin 3t = 3 sin t – 4 sin3 t)
L { Sin3 2t }
( )( ) |6|,
62
4
2222

++
s
ss
s
09 Sin 2t Sin 3t L { Sin 2t Sin 3t }
=½ L{ Cos t - Cos 5t } ( )( ) |5|,
51
12
2222

++
s
ss
s
10 Cos 5t Cos 2t L { Cos 5t Cos 2t }
=½ L{ Cos 7t + Cos 3t } ( )( ) |7|,
73
)29(
2222
3

++
+
s
ss
ss
11
F(t) =

51{
502


t
tt L { F(t) }
0,
5
9
)1(
2 55
2
−− −−
see
s
ss
12
F(t) =

22{
211{
100



t
t
t L { F(t) }
)(
1 2 ss
ee
s
−−
+
13
F(t) =



20{
20


t
ttCos L { F(t) }
)1(
1
2
2
−
+
− s
e
s
s 
14 e-t[3 cos 5t – 4 sin 5t] L{e-t[3 cos 5t – 4 sin 5t] }
( )
0,
51
73
22

++
−
s
s
s

15 e2t(3 sinh 2t – 5 cosh 2t) L{ e2t(3 sinh 2t – 5 cosh 2t)}
( )
0,
22
516
22

−−
−
s
s
s
16
!)1(
. 1
−
−−
n
te nta
L {
!)1(
. 1
−
−−
n
te nta
} as
as n
−
+
,
)(
1
17 e-t cos2 t L (e-t cos2 t } =
L { e-t
2
)2cos1( t+
} ( )
0,
)1](21[
532
22
2

+++
++
s
ss
ss
18 e-at Sin bt L { e-at Sin bt }
0,
)( 22

−+
s
bas
b
19 Cosh at – Cos at L { Cosh at – Cos at }
0,
]4[
2
44
3

+
s
as
s
20 ( 1 + t e-t )3 L { ( 1 + t e-t )3 }
0
,
)3(
6
)2(
6
)1(
31
432

+
+
+
+
+
+
s
ssss
21 t e-4t Sin 3t L [t e-4t Sin 3t }

0,
3)4[()4(
3
222

+++
s
ss
22 (t-1)3[u(t – 1) L { (t-1)3 [u(t – 1) }
6 4
1
s
e s−
23 eat [ u( t – a ) } L [eat [ u( t – a ) }
)1( +
−
s
e sa
24 e-2t {1- u(t – 1) } L { e-2t } – L { e-2t (1- u(t – 1) )}
)2(
1 )2(
+
− +−
s
e s
25
If L { F(t) } = )(sf then L { F(t/a) } = a. )( asf − Change of scale property.
26 If L [ F(t) } = L[3e2t sin t – 4 e2t
cos 4t }=
( ) 0,
204
420
2

+−
−
s
ss
s
L[ F(3t) } =
( ) 0,
18012
)15(4
2

+−
−
s
ss
s
27
If L {
t
tSin
}= Tan-1 





s
1
L {
t
atSin
}= Tan-1 





s
a
28 F(t) =







−
4
3
0
4
3
)
4
3
(


t
ttCos
L { F(t) }=
)1( 2
3
+





 −
s
e
s

29
F(t) =







tt
ttCos
sin
0)( L { F(t) }
)1( 2
)1(
+
−−−
s
Se SS
30 F(t)= t2 + at + b L { F(t) }
,
2
23
s
b
s
a
s
++
31 F(t) = t3 + 5 Cos t L { F(t) }
1
56
24
+
+
s
s
s
32. Evaluate 

−
0
3
dttSinet t
[Ans. L{ 

−
0
3
dttSinet t
}= 30
5

33. Evaluate 

−
0
2
dttCoset t
[Ans. L{ 

−
0
2
dttCoset t
}= 25
3

34. Evaluate 

−
0
4
dttSinet t
[Ans. L{ 

−
0
4
dttSinet t
}= )172(
)1(8
2
++
+
sss
s

35. Evaluate 
−
t
dt
t
btatCos
0
cos
[Ans. L{ 
−
t
dt
t
btatCos
0
cos
}= 





+
+
)(
)(
2
1
22
22
as
bs
Log
s 
36. Evaluate 
 −−
−
0
3
dt
t
ee tt
[Ans. L{ 
 −−
−
0
3
dt
t
ee tt
}=
( )3Log 
36. Evaluate 
 −
0
2
dt
t
tSine t
[Ans. L{ 
 −
0
2
dt
t
tSine t
}=
( )
4
5Log

37. Evaluate 
−
t
t
dtCoste
0
[Ans. L{ 
−
t
t
dtCoste
0
}= )22(
)1(1
2
++
+
sss
s

UNIT – IV Inverse Laplace Transformations Class 9
Section III

Having found Laplace Transformation of a new functions let us now
determine the inverse Laplace Transformations of given functions of S.
We have seen L { F(t) } is an algebraic function which is rational. Hence to
find inverse laplace transforms, we have to express the given function of S
into partial fractions which will, then to recognize as one of the following
standard forms:
Sl.
NO
Inverse Laplace Function Solution
1
L-1 { s
1
}
1
2
L-1 { as −
1
}
eat
3
L-1 { n
s
1
} ,
!)1(
1
−
−
n
tn
n=1,2,3..
4
L-1 { n
as )(
1
− } eat ,
!)1(
1
−
−
n
tn
n=1,2,3..
5
L-1 { 22
1
as + } atSin
a
1
6
L-1 { 22
as
s
+ }
atCos
7
L-1 { 22
1
as − } athSin
a
1
8
L-1 { 22
)(
1
bas +− } btCose
b
at1
9
L-1 { 22
as
s
− }
Cosh at

10
L-1 { 22
)( bas
as
+−
−
}
eat Cos bt
11
L-1 {
( )222
as
s
+
} atSint
a2
1
12
L-1 {
( )222
1
as +
} )(
2
1
3
atCosatatSin
a
−
Note: Reader is strongly advised to commit these results to memory.

Section III
Problem #1: Evaluate L-1





 +−
3
2
43
s
ss






+−
+
43
2
2
ss
s






−+−
+−
6116
562
23
2
sss
ss






+−
+
)2()1(
54
2
ss
s






++−
+
)52()1(
35
2
sss
s






+ )4( 44
as
s





 −
5
22
2
)2(3
s
s






−
−
+
+
−
22
9
184
254
52
s
s
s
s






++ )4)3( 2
s
s






−+
+
82
23
2
ss
s






−−
+
32
73
2
ss
s






 +−
3
2
43
s
ss
Solution: L-1





 +−
3
2
43
s
ss
= L-1






+− 333
2
43
ss
s
s
s
= L-1






3
2
s
s
+ L-1






− 3
3
s
s
+ L-1






3
4
s
= L-1






s
1
− L-1






2
3
s + L-1






3
4
s
= 1 – 3 t + 4 !2
2
t
 L-1





 +−
3
2
43
s
ss
= 1 – 3 t + 4 !2
2
t






+−
+
43
2
2
ss
s
Solution: L-1






+−
+
43
2
2
ss
s
= L-1






+−
+−
22
3)2(
42
s
s
= L-1






+−
+
+−
−
2222
3)2(
4
3)2(
2
ss
s
= L-1






+−
−
22
3)2(
2
s
s
+ L-1






+− 22
3)2(
4
s
= tCose t
3
3
4 2
+ tCose t
32
 L-1






+−
+
43
2
2
ss
s
= tCose t
3
3
4 2
+ tCose t
32







−+−
+−
6116
562
23
2
sss
ss
Solution: L-1






−+−
+−
6116
562
23
2
sss
ss
By using synthetic division method, we can get factors as
S3 s2 s c
S= - 1 1 - 6 11 - 6
0 1 -5 6
1 -5 6 0
S = 2 0 2 - 6
1 - 3 0
 6116 23
−+− sss =(S – 1) (S – 2) (S – 3) factor so by partial fractions






−+−
+−
6116
562
23
2
sss
ss
= 321 −
+
−
+
− s
C
s
B
s
A
on solving We get A = ½, B = -1, C = 5/2
L-1






−+−
+−
6116
562
23
2
sss
ss
= L-1






− )1(2
1
s - L-1






− )2(
1
s + L-1






− )3(2
5
s
= ½ et - e-2t + 5/2 e3t
 L-1






−+−
+−
6116
562
23
2
sss
ss
= ½ et - e-2t + 5/2 e3t is required solution.







+−
+
)2()1(
54
2
ss
s
Solution: To find L-1






+−
+
)2()1(
54
2
ss
s
by using partial fractions
4s+5 = A( s – 1 ) ( s + 2 ) + B ( s + 2 )
Put s = 1  9 = 3B  B = 3 and co efficient of s2, 0 = A – 1/3  A = 1/3
 L-1






+−
+
)2()1(
54
2
ss
s
= L-1






+
−
−
+
− )2(3
1
)1(3
1
3
)1(3
1
2
sss
= L-1






− )1(3
1
s +L-1






− 2
)1(
1
3
s
− L-1






+ )2(3
1
s
= 1/3 et + 3 t et – 1/3 e-2t
 L-1






+−
+
)2()1(
54
2
ss
s
= 1/3 et + 3 t et – 1/3 e-2t is required solution.






++−
+
)52()1(
35
2
sss
s
Solution: L-1






++−
+
)52()1(
35
2
sss
s
)52()1(
35
2
++−
+
sss
s
=
)1( −s
A
+
)52( 2
++
+
ss
CBs
5s + 3 = A )52( 2
++ ss +(s − 1)( CBs + )
On solving we get, s = 1  8 = 8A  A = 1
s = 0  3 = 5A − C  C = 2
 L-1






++−
+
)52()1(
35
2
sss
s
= L-1






++
+
++
−
− )52(
2
)52()1(
1
22
ssss
s
s
= L-1






− )1(
1
s
−L-1






++ )52( 2
ss
s
+ L-1






++ )52(
2
2
ss
= et - e- t Cos 2t + 3/2 e-t Sin 2t
 L-1






++−
+
)52()1(
35
2
sss
s
= = et - e- t Cos 2t + 3/2 e-t Sin 2t







+ )4( 44
as
s






+ )4( 44
as
s
For that by known formula, s4 + 4a4 = (s2+2a2)2-(2as)2=(s2+2a2+2as)(s2+2a2-
2as)
)4( 44
as
s
+
=
)22()22( 2222
asas
dCs
asas
bAs
−+
+
+
++
+
By partial fractions
 s = (As + b) )22( 22
asas −+ + (Cs + d) )22( 22
asas ++
Co efficient of s3  A – C = 0
Co efficient of s2  B + D = 0
Co efficient of s  − A + C = 0 and on solving, B =
a4
1−
; D – B =
a2
1
so D =
a4
1
 L-1






+ )4( 44
as
s
= a4
1−
L-1






++ ))(
1
22
aas + a4
1
L-1






+− ))(
1
22
aas
 L-1






+ )4( 44
as
s
= a4
1−
e-a t sin at + a4
1
ea t sin at is required solution.





 −
5
22
2
)2(3
s
s
Solution: L-1





 −
5
22
2
)2(3
s
s
= L-1






s2
3
−6 L-1






3
1
s + 3 L-1






5
1
s
= 2
3
−6 !2
2
t
+ !4
4
t
 L-1





 −
5
22
2
)2(3
s
s
= 2
3
−6 !2
2
t
+ !4
4
t







−
−
+
+
−
22
9
184
254
52
s
s
s
s
Solution: L-1






−
−
+
+
−
22
9
184
254
52
s
s
s
s
= L-1
( ) ( ) 

































+
−



 +
2
22
2
2
5
1
4
5
2
54
2
ss
s
+L-1
  

















−
+
−
− 2222
3
3.1
3
18
3
4
ss
s
= ( )tSinhtCoshtSintCos 3634
2
5
4
5
2
5
2
1
+−+





−
= tSinhtCoshtSintCos 3634
2
5
4
5
2
5
2
1
+−−
 L-1






−
−
+
+
−
22
9
184
254
52
s
s
s
s
= tSinhtCoshtSintCos 3634
2
5
4
5
2
5
2
1
+−−






++ )4)3( 2
s
s
Solution: L-1






++ )4)3( 2
s
s
= L-1






++
−+
)4)3(
33
2
s
s
= L-1






++
+
)2)3(
3
22
s
s
− L-1






++ )2)3(
3
22
s
[Since, L-1 tCose
s
s
LandtCos
as
s t
2
)2)3(
2
)(
3
22
1
22
−−







+−







+

= e-3t Cos 2t - 3/2 e-3t Sin 2t
 L-1






++ )4)3( 2
s
s
= e-3t Cos 2t - 3/2 e-3t Sin 2t is required solution.







−+ 82
3
2
ss
s
Solution: L-1






−+ 82
3
2
ss
s
= L-1






−+
−+
22
3)1(
113
s
s
= L-1






−+
+
22
3)1(
13
s
s
− L-1






−+ 22
3)1(
1
s
= 3 e-t Cosh 3t − 3. 3
1
.Sinh 3t.e-t
= 3 e-t 




 + −
2
33 tt
ee
− 




 − −
2
33 tt
ee
e-t
= 




 −
−
+ −−
22
33 3342 tttt
eeee
e-t
= 2
33 4422 tttt
eeee −−
++−
=
tt
ee 42
2 −
+
 L-1






−+ 82
3
2
ss
s
=
tt
ee 42
2 −
+ is required solution.







−−
+
32
73
2
ss
s
Solution: L-1






−−
+
32
73
2
ss
s
= L-1






−−
++−
22
2)1(
73)1(3
s
s
= L-1






−−
+
−−
−
2222
2)1(
10
2)1(
)1(3
ss
s
= L-1






−−
−
22
2)1(
)1(3
s
s
+ 10 L-1






−− 22
2)1(
1
s
= 3 et 6 Sinh 2t + 5 L-1






−− 22
2)1(
2
s
= 3 et 




 + −
2
22 tt
ee
+ 5 Sinh 2t et
= 3 




 + −
2
3 tt
ee
+ 5 




 − −
2
3 tt
ee
 L-1






−−
+
32
73
2
ss
s
=
tt
ee −
−3
4 is required solution.

Section III






−+
−+
)2)(3(
22
sss
ss






+−−
+−
65)(7(
1310
2
2
sss
ss






− 22
)1(s
s






+−
+
22
)2()1(
12
ss
s






+− )4)(2( 2
ss
s






++ )1()1( 22
ss
s






− 44
3
as
s






− 33
1
as






++
+
)3)(1(
6
222
2
ss
s






++
−
134
32
2
ss
s






++
+
22
)54(
2
ss
s







+++
+
22)(1(
5
22
2
sss
s






++ 124
ss
s






+
−
44
22
4
2(
as
asa






− 3
2
)2s
s






− 3
2
)2(s
s






+−
+
)134
3
2
ss
s






+ )(
1
22
ass






+ 3
)(
1
ass






+ 222
)as
s






− 222
)(
1
as






−
+
)1(
1
s
s
Log






+
+
)1(
12
ss
s
Log











−
2
1 2
s
Tan












−
2
1 s
Cot
Section III






−+
−+
)2)(3(
22
sss
ss






−+
−+
)2)(3(
22
sss
ss
By partial fractions, L-1






−+
−+
)2)(3(
22
sss
ss
= L-1






s
A
+ L-1






+ )3(s
B
+ L-1






− )2(s
C
We get A = 3
1
; B =
15
4
; C =
5
2
L-1






−+
−+
)2)(3(
22
sss
ss
= L-1






s3
1
+ L-1






+ )3(15
4
s + L-1






− )2(5
2
s
= 3
1
L-1






s
1
+ 15
4
L-1






+ )3(
1
s + 5
2
L-1






− )2(
1
s
= 3
1
+ 15
4
e-3t + 5
2
e2t
 L-1






−+
−+
)2)(3(
22
sss
ss
= 3
1
+ 15
4
e-3t + 5
2
e2t is required solution.







+−−
+−
65)(7(
1310
2
2
sss
ss
Solution: L-1






+−−
+−
65)(7(
1310
2
2
sss
ss
= L-1






−−−
−−−
)2)(3)(7(
8)3)(7(
sss
ss
By partial fractions )2)(3)(7(
8)3)(7(
−−−
−−−
sss
ss
=
)2()3()7( −
+
−
+
− s
C
s
B
s
A
We get, A = 5
2
− ; B = 2 ; C =
5
3
−
L-1






−−−
−−−
)2)(3)(7(
8)3)(7(
sss
ss
= L-1






−
−
)7(5
2
s + 2 L-1






− )3(
1
s 5
3
− L-1






− )2(
1
s
L-1






−−−
−−−
)2)(3)(7(
8)3)(7(
sss
ss
= 5
2
− L-1






− )7(
1
s + 2 L-1






− )3(
1
s 5
3
− L-1






− )2(
1
s
= 5
2
− e7t + 2 e3t
5
3
− e2t
 L-1






+−−
+−
65)(7(
1310
2
2
sss
ss
== 5
2
− e7t + 2 e3t
5
3
− e2t is required solution.







− 22
)1(s
s
Solution: L-1






− 22
)1(s
s
= L-1






+−
−−+
22
22
)1()1(4
)1()1(
ss
ss
= L-1






+−
+
22
2
)1()1(4
)1(
ss
s
− L-1






+−
−
22
2
)1()1(4
)1(
ss
s
= L-1






− 2
)1(4
1
s − L-1






+ 2
)1(4
1
s
= 4
1
L-1






− 2
)1(
1
s − 4
1
L-1






+ 2
)1(
1
s
= 4
1
t et -
4
1
t e-t
 L-1






− 22
)1(s
s
= 4
1
t et -
4
1
t e-t is required solution.







+−
+
22
)2()1(
12
ss
s
Solution: L-1






+−
+
22
)2()1(
12
ss
s
= L-1






+−
−++
22
)2()1(
)1()2(
ss
ss
= L-1






+−
+
22
)2()1(
)2(
ss
s
+ L-1






+−
−
22
)2()1(
)1(
ss
s
= L-1






+− )2()1(
1
2
ss + L-1






+− 2
)2)(1(
1
ss = L-1






+−
−−+
)2()1(3
)1()2(
2
ss
ss
+ L-1






+−
−−+
2
)2)(1(3
)1()2(
ss
ss
= L-1






+−
+
)2()1(3
)2(
2
ss
s
-L-1






+−
−
)2()1(3
)1(
2
ss
s
+ L-1






+−
+
2
)2)(1(3
)2(
ss
s
-L-
1






+−
−
2
)2)(1(3
)1(
ss
s
= 3
1
L-1






− 2
)1(
1
s - 3
1
L-1






+− )2)(1(
1
ss + 3
1
L-1






+− )2)(1(
1
ss - 3
1
L-1






+ 2
)2(
1
s
= 3
1
L-1






− 2
)1(
1
s - 3
1
L-1






+− )2)(1(
1
ss + 3
1
L-1






+− )2)(1(
1
ss - 3
1
L-1






+ 2
)2(
1
s
= 3
1
t et -
3
1
t e-2t
 L-1






+−
+
22
)2()1(
12
ss
s
= = 3
1
t et -
3
1
t e-2t is required solution.







+− )4)(2( 2
ss
s






+− )4)(3( 2
ss
s
By partial fractions )4)(3( 2
+− ss
s
=
)2()3( 22
+
+
+
− s
CBs
s
A
Put s = 3  A = 13
3
Put s = 0  C = 13
4
Put s = 1  B = 13
3
−
L-1






+− )4)(3( 2
ss
s
= L-1






+
+
+
− )2()3( 22
s
CBs
s
A
= L-1






+
+−
+
− )2(13
43
)3(13
3
22
s
s
s
= L-1






− )3(13
3
s − L-1






+ )2(13
3
22
s
s
+ L-1






+ )2(13
4
22
s
= L-1






− )3(13
3
s − L-1






+ )2(13
3
22
s
s
+ L-1






+ )2(13
4
22
s
=13
3
L-1






− )3(
1
s − 13
3
L-1






+ )2( 22
s
s
+ 13
4
L-1






+ )2(
1
22
s
=13
3
e3t − 13
3
Cos 2t + 13
2
Sin 2t
 L-1






+− )4)(2( 2
ss
s
=13
3
e3t − 13
3
Cos 2t + 13
2
Sin 2t is required solution.







++ )1()1( 22
ss
s
Solution: L-1






++ )1()1( 22
ss
s
= L-1






++
+−+
)1()1(2
)1()1(
22
22
ss
ss
= L-1






++
+
−
++
+
)1()1(2
)1(
)1()1(2
)1(
22
2
22
2
ss
s
ss
s
= L-1






++
+
)1()1(2
)1(
22
2
ss
s
− L-1






++
+
)1()1(2
)1(
22
2
ss
s
= L-1






+ )1(2
1
2
s − L-1






+ 2
)1(2
1
s
= L-1






+ )1(2
1
2
s − L-1






+ 2
)1(2
1
s
= 2
1
Sin t -
2
1
t e-t
 L-1






++ )1()1( 22
ss
s
= 2
1
Sin t -
2
1
t e-t is required solution.







− 44
3
as
s
Solution: L-1






− 44
3
as
s
= L-1






−+
−++
))((2
)()[(
2222
2222
asas
asass
= L-1






−+
−
+
−+
+
))((2
)[(
))((2
)[(
2222
22
2222
22
asas
ass
asas
ass
= L-1






−+
+
))((2
)[(
2222
22
asas
ass
+ L-1






−+
−
))((2
)[(
2222
22
asas
ass
= L-1






− )(2 22
as
s
+ L-1






+ )(2 22
as
s
= 2
1
Cosh at + 2
1
Cos at
 L-1






− 44
3
as
s
= 2
1
Cosh at + 2
1
Cos at is required solution.







− 33
1
as
Solution: L-1






− 33
1
as = L-1






−+− )(3)(
1
3
asasas = L-1






+−− ]3))[((
1
2
asasas
L-1






+−− ]3))[((
1
2
asasas = L-1






+−
+
+
− ]3)[()( 2
asas
CBs
as
A
by partial fractions
Here 1= A[(s-a)2+3as] + (Bs + c) (s-a) ; Put s = a A = 2
3
1
a
; Put s = 0 C =
a3
2−
; Put s = 1 B = 2
)1(3
1
aa −
L-1






− 33
1
as = L-1






+−
+
+
− ]3)[()( 2
asas
CBs
as
A
= L-1






+−
−
+−−
+
− ]3))[(3(
2
]3)][()1(3[3)(
1
2222
asasaasasaa
s
aas
= L-1






− 2
3)(
1
aas + L-1






+−− ]3)][()1(3[ 22
asasaa
s
- L-1






+− ]3))[(3(
2
2
asasa
=
at
e
a2
3
1
+ 2
)1(3
1
aa − L-1






+
−
+
+
22
)()( as
a
as
as
− a3
2
L-1






+
−
+
+
22
)()( as
s
as
as
=
at
e
a2
3
1
+ 2
)1(3
1
aa − L-1






+
+
2
)( as
as
− 2
)1(3
1
aa − L-1






+ 2
)( as
a
− a3
2
L-1






+
+
2
)( as
as
+ a3
2
L-1






+ 2
)( as
s
=
at
e
a2
3
1
+ 2
)1(3
1
aa − e-at − aa )1(3
1
−
t e-at −
a3
2
e-at +
a3
2
t e-at Cos at
 L-1






− 33
1
as =
at
e
a2
3
1
+ 2
)1(3
1
aa − e-at − aa )1(3
1
−
t e-at −
a3
2
e-at +
a3
2
t e-at Cos at







++
+
)2)(1(
6
222
2
ss
s
Solution: L-1






++
+
)2)(1(
6
222
2
ss
s
= L-1






++
+
++
+++
)2)(1(
1
2
7
)2)(1(
)4()1(
2
1
222222
22
ssss
ss
= L-1






++
+
)2)(1(
)1(
2
1
222
2
ss
s
+ L-1






++
+
)2)(1(
)4(
2
1
222
2
ss
s
+ L-1






++ )2)(1(
7
2
1
222
ss
= L-1






+ )2(
1
2
1
22
s + L-1






+ )1(
1
2
1
2
s + L-1






++ )2)(1(
7
2
1
222
ss
= L-1






+ )2(
1
2
1
22
s + L-1






+ )1(
1
2
1
2
s + L-1






++
+−+
)2)(1(
)]1()2[(
2
7
222
222
ss
ss
= L-1






+ )2(
1
2
1
22
s + L-1






+ )1(
1
2
1
2
s + L-1






++
+
)2)(1(
)]2[(
3.2
7
222
22
ss
s
− L-
1






++
+
)2)(1(
)]1[(
3.2
7
222
22
ss
s
= L-1






+ )2(
1
2
1
22
s + L-1






+ )1(
1
2
1
2
s + L-1






+ )1(
1
6
7
2
s − L-1






+ )2(
1
6
7
22
s
= 4
1
Sin 2t +
2
1
Sin t +
6
7
Sin t -
12
7
Sin 2t
= 4
1
Sin 2t +
2
1
Sin t +
6
7
Sin t -
6
7
Sin 2t
= 3
5
Sin t -
3
1
Sin 2t
= 3
1
[5 Sin t - Sin 2t]

 L-1






++
+
)2)(1(
6
222
2
ss
s
= 3
1
[5 Sin t - Sin 2t] is required solution.






++
−
134
32
2
ss
s
Solution: L-1






++
−
134
32
2
ss
s
= L-1






++
−+
22
3)2(
7)2(2
s
s
= L-1






++
−
++
+
2222
3)2(
7
3)2(
)2(2
ss
s
= L-1






++
+
22
3)2(
)2(2
s
s
− L-1






++ 22
3)2(
7
s
= L-1






++
+
22
3)2(
)2(2
s
s
− L-1






++ 22
3)2(
7
s
= 2 e-2t Cos 3t − 7 e-2t
3
1
sin 3t
 L-1






++
−
134
32
2
ss
s
= 2 e-2t Cos 3t − 7 e-2t
3
1
sin 3t is required solution.






++
+
22
)54(
2
ss
s
Solution: L-1






++
+
22
)54(
2
ss
s
= L-1






++
+
22
1)2(
2
s
s
= e-2t Cos t
 L-1






++
+
22
)54(
2
ss
s
= e-2t Cos t is required solution.







+++
+
)22)(1( 22
2
sss
ss
Solution: L-1






+++
+
)22)(1( 22
2
sss
ss
= L-1






+++
−++++
)22)(1(
3)1()22
22
22
sss
sss
= L-1






+++
−
+++
+
+
+++
++
)22)(1(
3
)22)(1(
)1(
)22)(1(
)22(
2222
2
22
2
ssssss
s
sss
ss
= L-1






+++
++
)22)(1(
)22(
22
2
sss
ss
+ L-1






+++
+
)22)(1(
)1(
22
2
sss
s
- L-1






+++ )22)(1(
3
22
sss
= L-1






+ )1(
1
2
s + L-1






++ ]1)1[(
1
2
s - 3 L-1






+++ )22)(1(
1
22
sss
= Sin t + e-t Sin t - 3 L-1






+++ )22)(1(
1
22
sss
 L-1






+++
+
)22)(1( 22
2
sss
ss
= Sin t + e-t Sin t - 3 L-1






+++ )22)(1(
1
22
sss







++ 124
ss
s
Solution: L-1






++ 124
ss
s
= L-1






















+
−+
2
22
2
3
)
2
1
(
2
1
2
1
s
s
= L-1






















+
−








+
+
2
22
2
22
2
3
)
2
1
(
2
1
2
3
)
2
1
(
2
1
ss
s
=
= L-1






















+
+
2
22
2
3
)
2
1
(
2
1
s
s
− L-1






















+
2
22
2
3
)
2
1
(
2
1
s
= e-1/2t
3
2
Cos 







t
2
3
− 2
1
3
2
e-1/2t Sin 







t
2
3
 L-1






++ 124
ss
s
= e-1/2t
3
2
Cos 







t
2
3
− 2
1
3
2
e-1/2t Sin 







t
2
3







+
−
44
22
4
2(
as
asa
Solution: L-1






+
−
44
22
4
2(
as
asa
= L-1






+
−
+ 44
2
44
2
4
2
4 as
a
as
as
= L-1






+ 44
2
4as
as
−L-1






+ 44
2
4
2
as
a
= a L-1






+ 224
2
)2( as
s
− aL-1






+ 224
)2(
2
as
a
=a Cos 2a2t− a Sin 2a2t
 L-1






+
−
44
22
4
2(
as
asa
=a Cos 2a2t− a Sin 2a2t is required solution.






− 3
2
)2(s
s
Solution: L-1






− 3
2
)2(s
s
= L-1






−
+−+−
3
2
)2(
4)2(4)2(
s
ss
= L-1






−
+
−
−
+
−
−
333
2
)2(
4
)2(
)2(4
)2(
)2(
ss
s
s
s
= L-1






−
−
3
2
)2(
)2(
s
s
+ L-1






−
−
3
)2(
)2(4
s
s
+ L-1






− 3
)2(
4
s
= L-1






− )2(
1
s + L-1






− 2
)2(
4
s + L-1






− 3
)2(
4
s
= e2t + 4 t e2t + 4 t2 e2t

 L-1






− 3
2
)2(s
s
= e2t + 4 t e2t + 4 t2 e2t is required solution.






+−
+
)134
3
2
ss
s
Solution: L-1






+−
+
)134
3
2
ss
s
= L-1






+−
+
)3)2(
3
22
s
s
= L-1






+−
+
+−
−
)3)2(
5
)3)2(
2
2222
ss
s
= L-1






+−
−
)3)2(
2
22
s
s
+L-1






+− )]3)2[(3
5
22
s
= e2t Cos 3t + 5/3 e2t Sin 3t
 L-1






+−
+
)134
3
2
ss
s
= e2t Cos 3t + 5/3 e2t Sin 3t is required solution.






+ )(
1
22
ass
Solution: Since, L-1






+ )(
1
22
as = a
atsin
L-1






+ )(
1
22
ass = 
t
dt
a
at
0
sin
= 2
1
a
( )t
atCos 0− = 2
1
a
( )atCos−1
 L-1






+ )(
1
22
ass = 




 −
2
1
a
atCos







+ 3
)(
1
ass
Solution: L-1






+ 3
)(
1
ass = L-1






+−+ 3
)]()[(
1
asaas = e-at L-1






− 3
)(
1
sas
Here we have, L-1






− )(
1
ass = 
t
at
dte
0
= at
eat
1
− ;
L-1






− )(
1
2
ass = a
1
 −
t
at
dte
0
)1( = 2
1
a
( )1−− ateat
and
L-1






− )(
1
3
ass = 2
1
a  −−
t
at
dtate
0
)1( = 3
1
a 





−−− 1
2
22
at
ta
eat
 L-1






+ 3
)(
1
ass = 3
a
e at−






−−− 1
2
22
at
ta
eat
= 3
1
a 





−−− −−− atatat
eat
ta
ee .
2
1
22
 L-1






+ 3
)(
1
ass = 3
1
a 





−−− −−− atatat
eat
ta
ee .
2
1
22






+ 222
)as
s
Solution: Let F (t) = L-1






+ 222
)as
s

Hence, L 





t
tF )(
= 

+0
222
)
ds
as
s
= −







+ s
as )
1
2
1
22 = 





+ 22
1
2
1
as since, after
applying limits. So, 





t
tF )(
= 





+
−
22
1 1
2
1
as
L = att
a
sin
2
1
 L-1






+ 222
)as
s
= att
a
sin
2
1






− 222
)(
1
as
Solution: L-1






− 222
)(
1
as
= L-1






− 222
)( ass
s
= 
t
dttF
0
)(
= 
t
dt
a
atSint
0
2
= 












 −
−




 −

tt
dt
a
at
a
at
t
a 00
cos
.1
cos
.
2
1
= 





+
−
2
cos
.
2
1
a
atSin
a
at
t
a
=  atCosatSin
a
−3
2
1
 L-1






− 222
)(
1
as =  atCosatSin
a
−3
2
1







−
+
)1(
1
s
s
Log
Solution: Let F (t) = L-1






−
+
)1(
1
s
s
Log
So, t. F(t) = - L-1












−
+
)1(
1
s
s
Log
ds
d
= - L-1












−
+
)1(
1
s
s
Log
ds
d
= - L-1 ( )












−
−+
)1(
1
1
s
LogsLog
ds
d
= - L-1 ( ) 






+1sLog
ds
d
+ L-1












− )1(
1
s
Log
ds
d
= - L-1






+1
1
s + L-1






−1
1
s
= - e-t + et
 t. F(t) = - e-t + et
 F(t) = L-1






−
+
)1(
1
s
s
Log =2 t
tSinh







+
+
)1(
12
ss
s
Log
Solution: t. F(t) = L-1












+
+
−
)1(
12
ss
s
Log
ds
d
= − L-1  






+−−+ )1()1( 2
sLogsLogsLog
ds
d
= − L-1






+ )1( 2
sLog
ds
d
+ L-1  






sLog
ds
d
+ L-1






+ )1(sLog
ds
d
= − L-1






+1
2
2
s
s
+ L-1






s
1
+ L-1






+1
1
s
t. F(t) = − 2 Cos t + 1 + t e-t
F(t) = tCos
t
2−
+ e-t +
t
1
 F(t) = t
1
( )tCoset t
21−+−












−
2
1 2
s
Tan
Solution: t F(t) = L-1


















− −
2
1 2
s
Tan
ds
d
= − L-1

















−
2
1 2
s
Tan
ds
d
= − L-1














−






+
3
2
4
2
2
.
2
1
1
s
s
= L-1
( )





+ 24
2
4
s
s
= L-1
( )





−+ 222
)2()2
4
ss
s
= L-1
( ) 





−+
−+−++
222
22
)2()2
)22()22(
ss
ssss
= L-1
( )





−+++
−+−++
)22)(22
)22()22(
22
22
ssss
ssss
= L-1
( ) ( )





−+++
−+
−
−+++
++
)22)(22
)22(
)22)(22
)22(
22
2
22
2
ssss
ss
ssss
ss
= L-1
( )





−+++
++
)22)(22
)22(
22
2
ssss
ss
− L-1
( )





−+++
−+
)22)(22
)22(
22
2
ssss
ss
= L-1






−+ ss 22
1
2 − L-1






++ ss 22
1
2 = L-1






+− 1)1(
1
2
s − L-1






++ 1)1(
1
2
s
= et Sin t − e- t Sin t = 2 




 − −
2
tt
ee
Sin t = 2 sinh t sin t
t F(t) = 2 Sinh t sin t

 F(t) = 





t
tSintSinh2











−
2
1 s
Cot
Solution: Let t F(t) = L-1











−
2
1 s
Cot
= L-1


















− −
2
1 1 s
Cot
ds
d
= −L-1






+ 22
2
2
s
t F(t) = −Sin 2t
 F(t) = t
tSin 2−

Section III
Convolution theorem:
If L-1{ )(sf } = F(t) and L-1{ )(sg } = G(t) then L-1{ )()( sgsf } =  −
t
duutguf
0
).().( = F * G is
called the convolution or falting of F and G.
u
u= t
t=u t → 
O u = 0 t
Let, (t) =  −
t
duutguf
0
).().(
L{ (t) } = dtduutgufe
t
st






−

−
00
).().( = dtduutgufe
t
st


−
−
0 0
).().(
= dudtutgufe
u
st

 
−
−
0
).().(

= dudtutgeufe utsst
 
 
−−−






−
0 0
)(
).(.)( here put t-u =v then dv = dt
= dudvgveufe svsu
 
 
−−






0 0
..)( = dusgufe su


−
0
)(.)( = )().( sgsf = F * G.
 L-1{ )()( sgsf } =  −
t
duutguf
0
).().( = F * G
Exercises: Try yourself….. By Dr N V Nagendram
Problem #01 Evaluate L-1






+ 222
)( as
s
[Ans. a2
1
t Sin at ]






++ ))(( 2222
2
bsas
s
[Ans. a2
1
t Sin at ]






+ )2(
1
2
Ss [Ans. − )12(
4
1 2
−+−
te t
]






+ 3
)2(
1
Ss [Ans. 8
1
− )
2
1
(
4
1 22
++−
tte t
]






+ )( 222
bsa
s
[Ans. 2
1
a
Cos 





a
b
t ]






+ )(
1
222
ass [Ans. 2
1
a
( t-
a
1
Sin at) ]






+ )1(
1
23
ss [Ans. 





−+ ttCos
t
2
2
]






+ 2
)( as
s
[Ans. e-at(1 – at) ]






+ 222
)(
2
as
as
[Ans. ]






+ 3
2
)( as
s
[Ans. 2
1
e-at ( )2422
+− atta ]











 +
s
s
Log
1
[Ans. t
1
(1 - e-at)
]












+
+
bs
as
Log [Ans. t
1
(e-b t - e-at) ]













++
+
)3)(2(
1
ss
s
Log [Ans. e- t − e-2t − e-3t ]












+
+
)(2
1
22
22
as
bs
Log [Ans. t
1
(Cos at – Cos bt) ]












− 2
2
1
s
a
Log [Ans. t
2
(1 – Cosh at) ]












−
+
2
2
)1(
1
s
s
Log [Ans. t
2
(et – Cos t) ]











−
s
Tan
21
[Ans. t
tSin 2
]
Problem #18 Evaluate L-1 ( ) 11
+−
sCot [Ans. t
tSine t−
]












+
−
1
1
.
s
s
Logs [Ans. )(
2
2
tCohttSin
t
− ]






+ 222
)( as
s
Solution: L-1






+ 222
)( as
s
= L-1






++ )(
1
.
)( 2222
asas
s
[Since, L-1 atCos
as
s
=






+
.
)( 22 and L-1 atSin
aas
1
.
)(
1
22
=






+ ]
By convolution theorem,
L-1






++ )(
1
.
)( 2222
asas
s
=  −
t
duauSin
a
utaCos
0
..
1
)( = a2
1
 −−
t
duatauSinatSin
0
)2([
= a2
1 t
atauCos
a
atSinu 0])2(
2
1
[[ −−
L-1






+ 222
)( as
s
= a2
1 t
atauCos
a
atSinu 0])2(
2
1
[[ −− = a2
1
t Sin at
 L-1






+ 222
)( as
s
= a2
1
t Sin at is required solution.







++ ))(( 2222
2
bsas
s






++ ))(( 2222
2
bsas
s
Since, L-1 atCos
as
s
=






+
.
)( 22 and L-1 btCos
bs
=






+
.
)(
1
22
by convolution theorem,
L-1






++ ))(( 2222
2
bsas
s
= L-1






++ )(
.
)( 2222
bs
s
as
s
=  −
t
dubuCosutaCos
0
.)( = 2
1
 −−++−
t
dubuutaCosbuutaCos
0
]])([])([[
= 2
1






+
−+−
−






−
−−+
ab
atSintabatSin
ab
atSintabatSin ])([])([
= ( )atSinabtSinb
ab
22
)(2
1
22
−
−
= ( )atSinabtSinb
ab
−
− )(
1
22
 L-1






++ ))(( 2222
2
bsas
s
= ( )atSinabtSinb
ab
−
− )(
1
22

UNIT – IV Application of Laplace Transform Class 13
Section III
Application of Laplace Transform to differential equations with
constant coefficients:
Laplace transform is especially suitable to obtain the solution of linear non-
homogeneous ordinary differential equations with constant coefficients,
when all the boundary conditions are specified for the unknown function
and its derivatives at a single point.
Consider the initial value problem )(2
2
trby
dt
dy
a
dt
yd
=++ …………………………..
(1)
y(t = 0) = k0, y1(t=0) = k1 …………………………………………………………………
(2)
Where a,b,k0,k1 are all constants and r(t) is a function of t.
Method of Solution to differential equation (D.E) by Laplace Transform (L.T.):
1. Apply Laplace transform on both sides of the given differential equation
(1), resulting in a subsidiary equation as

[s2Y – s y(0) - y(0)] +a[sY-y(0)]+bY = R(s) …………………………………………. (3)
where Y = L{y(t)} and R(s) = L{r(t)}.
Replace y(0), y(0) using given initial conditions (2).
2. Solve (3) algebraically for Y(s), usually to a sum of partial fractions.
3. Apply inverse Laplace transform to Y(s) obtained in 2. This yields the
solution of O.D.E. (1) satisfying the initial conditions (2) as y(t) = L-1[Y(s)].
Problem #1 Solve the differential equation by using Laplace transformation
y-2y-8y = 0, y(0) = 3, y(0) = 6.
Solution: On applying Laplace transform (s2Y - 3s - 6) – 2( sY – 3 ) - 8Y = 0
On solving, Y(s) =
82
3
2
−− ss
s
=
)2)(4(
3
+− ss
s
By using partial fractions, Y(s) =
)4(
2
−s
+
)2(
1
+s
Applying 1 L.T. y(t) = L-1(Y(s)) = 2 L-1






− )4(
1
s
+L-1






+ )2(
1
s
= 2 e4t + e-2t.
y(t) = 2 e4t + e-2t is required solution.

y+2y+5y = e-t Sin t, y(0) = 0, y(0) = 1.
Solution: Using L.T. On applying Laplace transform we get the given
differential equation as, (s2Y - 0 - 1) + 2( sY – 0 ) = L{ e-t Sin t } = 22
1)1(
1
++s
On solving Y =
)22)(52(
32
22
2
++++
++
ssss
ss
By partial fractions,
)22)(52(
32
22
2
++++
++
ssss
ss
=
)52( 2
++
+
ss
BAs
+
)22( 2
++
+
ss
DCs
322
++ ss = (As + B) ( 222
++ ss )+ (Cs + D ) ( 522
++ ss )
322
++ ss = s3(A + C) + s2 (2A + 2C + B + D ) + s (2A + 5C + 2B + 2D )+ 2B +
5D
Equating coefficients of s on either side,
A + C = 0 ; 2A + 2C + B + D = 1; 2A + 5C + 2B + 2D = 2; 2B + 5D = 3
 A = 0, B =
3
1
, C = 0, D =
3
2
Y(s)=
)52( 2
++
+
ss
BAs
+
)22( 2
++
+
ss
DCs
=
3
1
)2)1(
1
22
++s
+
3
2
)1)1(
1
22
++s
Applying I.T. y(t) = L-1{Y}=
3
1
L-1






++ )2)1(
1
22
s
+
3
2
L-1






++ )1)1(
1
22
s
using first shifting theorem, y(t)=
3
1
e-t L-1






+ )1
1
2
s
+
3
2
e-t L-1






+ 22
2
1
s

y(t)= ]2[
3
tSintSin
e t
+
−
y+n2y = a Sin (nt+2), y(0) = 0, y(0) = 0.
Solution: y+n2y = a Sin (nt+2)
= a [Sin nt Cos 2 + Cos nt Sin 2]
Applying L.T. L{ y} + n2 L { y } = a cos 2. L{Sin nt} + a Sin 2 L{Cos nt}
Using L.T. On applying Laplace transform we get the given differential
equation as,
[s2Y – sy(0) - y(0)] + n2Y = 22
ns
n
+
.a.Cos 2 + 22
ns
n
+
.a. Sin 2
Solving Y, Y(s) = 222
)( ns
n
+
.a. Cos 2 + 222
)( ns
n
+
.a. Sin 2
Applying I.T. we get
y(t) = n. a. Cos 2. L-1






+ 222
)(
1
ns
+ a. Sin 2. L-1






+ 222
)(
1
ns
…… (1) From
I.L.T. tables, we know that 2nd term in R.H.S.

L-1






+ 222
)( ns
s
=
n
ntSint
2
.
……………………………………………………………… (2)
to find first term in R.H.S.
L-1






+ 222
)(
1
ns
=L-1






+ 222
)(
.
1
ns
s
s
= 
t
dtnttSint
n 0
...
2
1
= ].[
2
1
3
ntSinntCosnt
n
+− ….(3)
Thus substituting (2) and (3) in (1), we get
y(t) = a.n. Cos 2. ].[
2
1
3
ntSinntCosnt
n
+− + a Sin 2
n
ntSin
2
= 2
2n
a
[ ntCosCosnt .2.− + Cos 2. Sin nt + nt. Sin 2 . Sin nt]
= 2
2n
a
[ )2.2.(.2. SinntSinCosntCosntCosntSin −− ]
= 2
2n
a
[ )2(.2. +− ntCosntCosntSin ]
 y(t) = 2
2n
a
[ )2(.2. +− ntCosntCosntSin ] is required solution.
Problem #4 Find the general solution of the differential equation by using
Laplace transformation y - 3y +3y - y =t2 et, y(0) = 1, y(0) = 0,y(0) = -
2.
Solution: Since the initial conditions are arbitrary assume y(0) = a, y(0)
=b, y(0) =c.
Then Using L.T. On applying Laplace transform we get the given differential
equation as, (s2Y – as2 – bs - c) - 3( s2Y – as -b ) + 3(sY – a) - Y = 3
)1(
2
−s
Y = 33
2
)1(
2
)1(
)33()3(
−
+
−
+−+−+
ss
cbasabas
By partial fractions Y = 61
3
2
2
3
1
)1(
2
)1()1()1( −
+
−
+
−
+
− ss
c
s
c
s
c
where c1, c2, c3 are
constants depending on a, b, c.
Applying I.T. and using first shifting theorem
y(t) = c1
2
2
t
et + c2 t et + c3 et + t
e
t
60
5

y - 3y +3y - y =t2 et, y(0) = 1, y(0) = 0,y(0) = - 2.
Solution: Applying L.T. to D.E. L{ y - 3y +3y - y} =L{ t2 et }
 L{ y} - 3L{ y} +3 L { y} - L{ y} =L{ t2 et }
[s3Y – s2y(0) - sy(0) - y(0) ] – 3 [s2Y – sy(0) - y(0)] + 3[sY – y(0)] – Y = 3
)1(
2
−s
Using the initial condition y(0) = 1, y(0) = 0,y(0) = - 2, and solving for Y
(s3 – 3a2 + 3s – 1)Y – s2 + 3s – 1= 3
)1(
2
−s
Y = 63
2
)1(
2
)1(
13
−
+
−
+−
ss
ss
= 63
2
)1(
2
)1(
12
−
+
−
−+−
ss
sss
= 63
2
)1(
2
)1(
1)1()1(
−
+
−
−−−−
ss
ss
Y = 632
)1(
2
)1(
1
)1(
1
)1(
1
−
+
−
−
−
−
− ssss
On applying I L.T. we get

y(t) = L-1 { Y } = L-1






− )1(
1
s
− L-1






− 2
)1(
1
s
− L-1






− 3
)1(
1
s
+2 L-1






− 6
)1(
1
s
y(t) = L-1 { Y } = et − t et −
2
2 t
et
+
60
5 t
et
Section III Exercise Try yourself……..
Use L.T. to solve each of the following I.V.P. consisting of a D.E. with I.C.
01. 0=− yy , general solution [Ans. Assume y(0)=0=A=constant;y=Aet]
02. 2)0(,3
==− yeyy t
[Ans. y=(3et +e3t)/2 ]
03. ByAyyy ===+ )0(,)0(,0 [Ans.G.S. y=C + De-t, C=A + B,D= − B ]
04. 2)0(,0)0(,2 ===+ yyeyy t
[Ans.G.S. y=et+ Cos t + Sin t ]
05. 9)0(,2)0(,096 ===+− yyyyy [Ans.G.S. y=(3t + 2)e3t ]
06. 7)0(,0)0(,94 ===+ yytyy [Ans.G.S. y= tSint 2
4
19
4
9
+ ]
07.
1)0(
,0)0(,4107 3
−=
==++ −
y
yeyyy t
[Ans.G.S. y=e-2t-2e-3t+e-5t ]
08.
10)0(
,5)0(,9158 2
=
==+−
y
yteyyy t
[Ans.G.S. y=4e2t+3te2t+3e3t-2e5t ]

09. 0)0(,0)0(,2 ===+ yytCostyy [Ans.G.S. y= tCosttSintSin 2
3
1
9
5
2
4
9
−− ]
10.
0)0(
,0)0(),(2
=
=+=+
y
yntSinayny 
[Ans. y= )([
2 2
 +− ntCosntCosntSin
n
a
]
11.
0)0(
,1)0(,2
=
==+
y
ytSintSinyy
[Ans.G.S. y= tCostSin
t
tCos 3
16
1
416
15
++ ]
12. 0)0(,0)0(,2
===+ −
yytSineyy t
[Ans.y= )(
8
)(
8
1
tCostSin
e
tCostSin
t
++−
−
]
13.
3)0(,0)0(,0)0(
,10254
===
=+++
yyy
tCosyyyy
[Ans.y=−e-2t + 2e-t -2te-t – Cos t +2 Sin t]
14. 0)0(,0)0(,0)0(, ====− yyyeyy t
[
2
)
2
3
2
3
5
2
3
9(
183
1 2
1
tt
t e
tSintCos
e
e −++
−
]
15.
18)(,0)(
,2)0(,0)0(,3016)(
−==
===−
 yy
yytSinyy iv
[Ans.G.S. y= 2(Sin 2t – Sin t ]
Section IV
Application of Laplace Transform to system of simultaneous differential
equations:
Laplace transform can also be used to solve a system or family of m
simultaneous equations in m dependent variables which are functions of the
independent variable t. consider a family of two simultaneous differential
equations in the two dependent variables x, y which are functions of t.
)(165432
2
22
2
1 tRyaxa
dt
dy
a
dt
dx
a
dt
yd
a
dt
xd
a =+++++ ………………………………………..
(1)

)(265432
2
22
2
1 tRybxb
dt
dy
b
dt
dx
b
dt
yd
b
dt
xd
b =+++++ ………………………………………..
(2)
Initial conditions x(0) = c1; y(0) = c2; x(0) = c3; y(0) = c4 …………………………
(3)
Here a1, a2, a3, a4, a5, a6,b1, b2, b3, b4, b5, b6 and c1, c2, c3, c4 are all
constants and R1(t), R2(t) are functions of t.
Method of solution to system of differential equation (D.E.):
I. Apply Laplace transform on both sides of each of the two differential
equation (1) and (2) above. This reduces (1) and (2) to two algebraic
equations in X(s) and Y(s) where X(s) = L{ x(t) } and Y(s) = L{ y(t) }.
a1[s
2X-sx(0) - x(0)] + a2[ s2Y – sy(0) - y(0)] + a3[sX – x(0) ] + a4[sY- y(0) ]+ a5 X + a6 Y =
Q1(s)…..(4)
b1[s
2X-sx(0) - x(0)] + b2[ s2Y – sy(0) - y(0)] + b3[sX – x(0) ] + b4[sY- y(0) ]+ b5 X + b6 Y = Q2(s)…
(5)
use the initial conditions (3) and substitute for x(0) , y(0) , x(0) , y(0).
II. Solve (4) and (5) for X(s) and Y(s).
III. The required solution is obtained by taking the inverse Laplace
Transform of X(s) and Y(s) as
x(t) = L-1{ X(s) } and y(t) = L-1 { Y(s) }.
Problem #1 Solve 314232
2
+=+−− ty
dt
dy
dt
dx
dt
xd
; conditions
dt
dy
x
dt
dx
;13 =+− x(0) =
0; y(0) = 6.5; x(0) = 0;
Solution: Taking Laplace Transformation of the given differential equation
we have
[s2X-sx(0) - x(0)] -3 [sX – x(0) ] - [sY- y(0) ]+ 2Y = 14 2
1
s
+3
s
1
[sX – x(0) ] – 3X + [sY- y(0) ] =
s
1
Use given initial conditions (I.C.) x(0) = 0; y(0) = 6.5=13/2; x(0) = 0;
S(s - 3)X +(2 – s)Y = 2
2
2
13628
s
ss −+
and (s – 3 )X + sY =
s
s
2
132 +
On solving we get Y(s) =
)2(2
2861513
22
23
−+
−−+
sss
sss
and X(s) =
)1)(2)(3(
)67(4
−+−
−
ssss
s

Taking inverse Laplace Transform y(t) = L-1{Y(s)} = L-1






−+
−−+
)2(2
2861513
22
23
sss
sss
= L-1






+
+
−
++
212
s
D
s
C
s
B
s
A
= L-1






+
+
−
−+
)2(2
5
1
157
2
ssss
 y(t) = 7t + 5 – et +
2
5
e-2t
Similarly, x(t) = L-1{X(s)} = L-1






−+−
−
)1)(2)(3(
)67(4
ssss
s
= L-1






+
−
−
−
−
−
)2(
1
)1(2
1
)3(2
12
ssss
= 2−
2
1
et
−
2
1
e3t
− e-2t
 x(t) =2−
2
1
et
−
2
1
e3t
− e-2t
Problem #2 Solve t
eyx
dt
dy
dt
dx −
=−−+2 ; conditionseyx
dt
dy
dt
dx t
;2 =+++ x(0) = 2;
y(0) = 1;
Solution: Taking Laplace Transformation of the given differential equation
we have
2[sX(s) - x(0)] + sY(s)- y(0) –X(s)- Y(s) =
1
1
+s
sX(s) – x(0) + sY(s) - y(0)+ 2 X(s) + Y(s) =
1
1
−s
Use given initial conditions (I.C.) x(0) = 2; y(0) = 1.
(2s - 1)X(s) +(s – 1)Y(s) =
1
65
+
+
s
s
and (s + 2 )X(s) + (s+1)Y(s) =
1
23
−
−
s
s
On solving we get Y(s) =
)1)(1(
1412
222
23
−+
+−−
ss
sss
=
)1(
1
)1(
13
22
−
+
+
−
ss
s

and X(s) =
)1(
82
2
+
+
s
s
Taking inverse Laplace Transform y(t) = L-1{Y(s)} = L-1






−
+
+
−
)1(
1
)1(
13
22
ss
s
= Cos t – 13 Sin t + Sinh t
 y(t) = Cos t – 13 Sin t + Sinh t
Similarly, x(t) = L-1{X(s)} = L-1






+
+
)1(
82
2
s
s
= 2 Cos t + 8 Sin t
 x(t) =2 Cos t + 8 Sin t Is required solution.
Exercise Solve the following system of equations
Try yourself ……..
01. yx
dt
dx
32 −= ; xy
dt
dy
2−= , x(0) = 8, y(0) = 3
[Ans. x(t) = 5 e-t + 3 e4t; y(t) = 5 e-t – 2 e4t]
02. t
ex
dt
dy
dt
xd −
=++ 1532
2
; tSiny
dt
dx
dt
yd
215342
2
=+− ,
x(0) = 35,y(0) = 27; x(0) = -48,y(0) = -55.
[Ans. x(t) = 30 Cos t – 15 Sin 3t + 3 e-t + 2 Cos 2t;

y(t) = 30 Cos t – 60 Sin t - 3 e-t + Sin 2t]
03. t
eyx
dt
xd 2
2
2
32 ++= ; yx
dt
yd
22
2
−−= , x(0) = y(0) = 1; x(0) = y(0) = 0
[Ans. x(t) = ttt
etSintCosee 2
5
2
)219(
10
1
)73(
4
1
+−−+ −
;
y(t) = ttt
etSintCosee 2
5
1
)219(
10
1
)73(
12
1
−−++
− −
]
04. t
ey
dt
dx
+= ; xtSin
dt
dy
−= , x(0) = 1, y(0) = 0
[Ans. x(t) =
2
)2( tCosttSintCoset
−++
; y(t) =
2
)( tSintCosetSint t
−+−
]
05. 123 =++ x
dt
dy
dt
dx
; 034 =++ y
dt
dy
dt
dx
, x(0) = 3, y(0) = 0
[Ans.x(t)=
10
)325( 11/6tt
ee −−
−−
; y(t) =
5
)3( 11/6tt
ee −−
−
]
06. t
e
dt
yd
dt
dx −
=+ 2
2
2 ; 12 =−+ yx
dt
dx
, x(0) = y(0)= y(0) = 0
[Ans. x(t)=1+ e-t – e-at-e-bt; y(t)=1+e-t– be-at− a e-bt ]
Note: a = )22(
2
1
− ; b = )22(
2
1
+
07. tyx
dt
dy
dt
dx
2122 −=+−− ; 022
2
=++ x
dt
dx
dt
xd
, x(0) = 0, x(0) = 0
[Ans. x(t) = 2−2e-t (1+t); y(t) = 2−t-2e-t (1+t)]
08. t
eyx
dt
dy
dt
dx
342 =−++ ; t
eyx
dt
dy
dt
dx
=+++ 22 , x(0) =1, y(0) =0
[Ans. x(t) = e-2t - t et; y(t) =
3
1
et +tet ]
09. t
eyx
dt
dx
836 =+− ; t
exy
dt
dy
42 =−− ,x(0) =-1, y(0) = 0
[Ans. x(t) = -2 et + e4t; y(t)=
3
2−
et +
3
2
e4t]
10. tSiny
dt
dx
= ; tCosx
dt
dy
=+ , x(0) = 2, y(0) = 0
[Ans. x(t)=e-t + et = 2Cosh t; y(t)=Sin t – 2 Sinh t]

Production Engineering - Laplace Transformation

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