c hapterlaplace transform method the laplace transform was developed by the french mathematician by...
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CCCChapterhapterhapterhapter 4 4 4 4LAPLACE TRANSFORM
2
OUTLINE OF CHAPTER 4: OUTLINE OF CHAPTER 4: OUTLINE OF CHAPTER 4: OUTLINE OF CHAPTER 4:
�Definition of Laplace transform �Properties of Laplace transform �Inverse Laplace Transform �Properties of Inverse of Laplace Transform �Application of Laplace transform
Laplace transform methodLaplace transform methodLaplace transform methodLaplace transform method
�The Laplace transform was developed by the French mathematician by the same name (1749-1827) and was widely adapted to engineering problems in the last century.
�Its utility lies in the ability to convert differential equations to algebraic forms that are more easily solved. The notation has become very common in certain areas as a form of engineering “language” for dealing with systems.
Steps Steps Steps Steps involved in using the involved in using the involved in using the involved in using the Laplace Laplace Laplace Laplace transformtransformtransformtransform
SolutionDetermine
inverse transform
Solve equ. by algebra
Transform diff. equ.
to algebraic
form
Differential Equation
4.1: Definition of 4.1: Definition of 4.1: Definition of 4.1: Definition of laplacelaplacelaplacelaplace transformtransformtransformtransform
( )
( )
( ) ( ){ } ( )
0
0
is called Laplace transform of ( ) if the integral exists. We write as
where is interpreted as a
L
et be a function defined for all
n
0. The int
eg
operat
a
or
r lst
st
f
e f t dt
F s
f t t
f t e f dt
t
t
∞ −
∞ −=
≥
=
∫
∫L
L
Example 4.1Example 4.1Example 4.1Example 4.1
( )( )
2
2
By using the definition of Laplace transform, find theLaplace transform of the following functions:a)
b) t
f t t
f t e−=
=
Solution (a):Solution (a):Solution (a):Solution (a):( )( ) ( ){ } { }
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
[ ]
2
2
2
0
2
2 30
2
2 3
2 0 0 0
2 3
3 3
(solve using B.P/T.M)
2 2
2 2
0 2 0 2
2 20 0 0 0 0
st
st st st
s s s
s s s
f t t
F s f t t
t e dt
t e te es s s
e e es s s
e e es s s
s s
∞ −
∞− − −
− ∞ − ∞ − ∞
− − −
=
= =
=
⎡ ⎤= − − −⎢ ⎥⎣ ⎦
⎡ ⎤∞ ∞= − − − −⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤− − −⎢ ⎥⎢ ⎥⎣ ⎦
⎡ ⎤= − − − − − =⎢ ⎥⎣ ⎦
∫
L L
2
2
3
2
2
0
st
st
st
st
t eets
eses
−
−
−
−
+
− −
+
− −
Solution (b):Solution (b):Solution (b):Solution (b):
( )( ) ( ){ } { }
( )
( )
( )( ) ( )( )
[ ]
2
2
2
0
2
0
2
0
2 2 0
2
2 2
1 102 2
t
t
t st
s t
s t
s s
f t e
F s f t e
e e dt
e dt
es
e es s
s s
−
−
∞ − −
∞ − −
∞− −
− − ∞ − −
=
= =
=
=
⎡ ⎤= ⎢ ⎥− −⎣ ⎦
⎡ ⎤ ⎡ ⎤= −⎢ ⎥ ⎢ ⎥− − − −⎣ ⎦ ⎣ ⎦
⎡ ⎤= − =⎢ ⎥− − +⎣ ⎦
∫∫
L L
Example 4.2Example 4.2Example 4.2Example 4.2
( )
Find the Laplace transform of the following function using the definition of Laplace transform
1, 0 22, 2 43, 4
tf t t
t
≤ <⎧⎪= ≤ <⎨⎪ ≥⎩
Solution:Solution:Solution:Solution:( ) ( ){ }
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
2 4
0 2 42 4
0 2 42 4
0 2 4
2 0 4 2 4
1 2 3
2 3
2 3
2 3
st st st
st st st
st st st
s s s s s s
F s f t
e dt e dt e dt
e dt e dt e dt
e e es s s
e e e e e es s s s s s
∞− − −
∞− − −
∞− − −
− − − − − ∞ −
=
= + +
= + +
⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤ ⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + − + −⎢ ⎥ ⎢ ⎥ ⎢⎜ ⎟ ⎜ ⎟ ⎜ ⎟
− − − − − −⎢ ⎥ ⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦ ⎣
∫ ∫ ∫∫ ∫ ∫
L
2 4 2 4
2 4 2 4
2 4
1 2 3 0
1 2 2 3
1
s s s s
s s s s
s s
e e e es s s s s
e e e es s s s se e
s s s
− − − −
− − − −
− −
⎤⎥⎥⎦
⎡ ⎤ ⎡ ⎤ ⎡ ⎤= − + + − + + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦
= − + − + +
= + +
Exercise 4.1Exercise 4.1Exercise 4.1Exercise 4.1
( ) ( )
By using the definition of Laplace transform, find theLaplace transform of the following functions:
cos , where be constantf t at a=
4.2: Table of 4.2: Table of 4.2: Table of 4.2: Table of laplacelaplacelaplacelaplace transform transform transform transform
Exercise 4.2Exercise 4.2Exercise 4.2Exercise 4.2
( )( ) ( )( )( )( )
( )
10
3
3
By using the table of Laplace transforms, findthe Laplace transforms for the given
a) 5 f)
b)
c) cos3
d)
2e) sinh3
t
t
f t
f t f t e
f t e
f t t
f t t
f t t
−= =
=
=
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
4.3: properties of 4.3: properties of 4.3: properties of 4.3: properties of laplacelaplacelaplacelaplace transformtransformtransformtransform
� LINEARITY
� FIRST SHIFT THEOREM
�MULTIPLYING BY THEOREM nt
Linearity theoremLinearity theoremLinearity theoremLinearity theorem
Example 4.3 Example 4.3 Example 4.3 Example 4.3
( )( )( )
3
3
5
Find the Laplace transform for the following functions:a) 2 7
b) 11 6
c) 2cos5 6 9
t
t
f t t t
f t e t
f t t t e−
= − +
= +
= − + −
Solution Solution Solution Solution ::::
( )( ) ( ){ } { }( ) { } { }
3
3
3
3 1 1 1
4 2
a) 2 7
2 7
2 7
3! 1!2 7
12 7
f t t t
F s f t t t
F s t t
s s
s s
+ +
= − +
⇒ = = − +
= − +
⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
−= +
L L
L L
Solution Solution Solution Solution ::::
( )( ) ( ){ } { }
{ } { }
3
3
3
1 1
2
b) 11 6
11 6
11 6
1 1! 11 63
11 6 3
t
t
t
f t e t
F s f t e t
e t
s s
s s
+
= +
⇒ = = +
= +
⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠
= +−
L L
L L
Solution Solution Solution Solution ::::
( )( ) ( ){ } { }
{ } { } { }
( )
5
5
5
2 2 1 1
2 2
c) 2cos5 6 9
2cos5 6 9
2 cos5 6 9
1! 1 2 6 95 5
2 6 9 25 5
t
t
t
f t t t e
F s f t t t e
t t e
ss s s
ss s s
−
−
−
+
= − + −
⇒ = = − + −
= − + −
⎛ ⎞⎛ ⎞ ⎛ ⎞= − + − ⎜ ⎟⎜ ⎟ ⎜ ⎟+ − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= − + −+ +
L L
L L L
First shift theoremFirst shift theoremFirst shift theoremFirst shift theorem
Example 4.4 Example 4.4 Example 4.4 Example 4.4
( )( ) ( )( ) ( )
2 4
3
Find the Laplace transform for the following functions:a)
b) cos 6
c) sinh 2
t
t
t
f t t e
f t e t
f t e t
−
=
=
=
Solution Solution Solution Solution ::::
( )
( ) { }
{ } ( )( )
2 4
22 1 3
2 43
a)2! 2
2 44
t
t
f t t e
F s ts s
t e F ss
+
=
⇒ = = =
= − =−
L
L
Solution Solution Solution Solution ::::
( ) ( )
( ) ( ){ }( ){ } ( )( )
( )
( )
( )
2 2 2
2
2
2
b) cos 6
cos 66 36
cos 6 1
11
1 361
1 361
2 37
t
t
f t e ts sF s t
s se t F s
F ss
ss
ss
s s
−
−
=
⇒ = = =+ +
= − −
= +
+=
+ +
+=
+ +
+=
+ +
L
L
Solution Solution Solution Solution ::::
( ) ( )
( ) ( ){ }( ){ } ( )
( )
3
2 2 2
3
2
2
c) sinh 22 2sinh 2
2 4 sinh 2 3
2 3 42 6 5
t
t
f t e t
F s ts s
e t F s
s
s s
=
⇒ = = =− −
= −
=− −
=− +
L
L
Multiplying by Multiplying by Multiplying by Multiplying by nt
Example 4.5 Example 4.5 Example 4.5 Example 4.5
( )( ) ( )( ) ( )
2
2
Find the Laplace transform for the following functions:a)
b) sin 3
c) cosh
tf t te
f t t t
f t t t
−=
=
=
Solution Solution Solution Solution ::::
( )
( ) { }
{ } ( )
( ) ( )
( )
( )
2
2
12
1 1
2
2
a)1
21 1
2
1 2
1 1 212
t
t
t
f t te
F s esdteds sd sdss
s
−
−
−
−
−
=
⇒ = =+⎡ ⎤= − ⎢ ⎥+⎣ ⎦
⎡ ⎤= − +⎣ ⎦
= − ⋅− +
=+
L
L
Solution Solution Solution Solution ::::
( ) ( )
( ) ( ){ }
( ){ } ( )
( ) ( )
( )
( )
2
12
11 2
22
22
b) sin 33sin 3
33 sin 3 1
3
1 3 2
1 3 1 2 2
6
2
f t t t
F s tsdt tds sd sdss s
ss
−
−
=
⇒ = =+⎡ ⎤= − ⎢ ⎥+⎣ ⎦
⎡ ⎤= − ⋅ +⎢ ⎥⎣ ⎦
= − ⋅ ⋅− + ⋅
=+
L
L
Solution Solution Solution Solution ::::
( ) ( )
( ) ( ){ }
( ){ } ( )
( )( ) ( )( )
( )
2
2
222
2 2
2
22
2
22
c) cosh
cosh1
cosh 11
1 1 2
1
1
1
f t t tsF s t
sd st tds s
s s sdds s
d sds s
=
⇒ = =−⎡ ⎤= − ⎢ ⎥−⎣ ⎦
⎡ ⎤− −⎢ ⎥=⎢ ⎥−⎣ ⎦⎡ ⎤− −⎢ ⎥=⎢ ⎥−⎣ ⎦
L
L
Solution (c0nt.) Solution (c0nt.) Solution (c0nt.) Solution (c0nt.) ::::
( )( ) ( ) ( ) ( )( )
( )( ) ( )( )
( )
( ) ( )
2
22
2 12 2 2
42
2 12 2 2
42
5 3 3
4 32 2
1
1
1 2 1 2 1 2
1
2 1 4 1 1
1
2 4 6 2 6@1 1
d sds s
s s s s s
s
s s s s s
s
s s s s ss s
⎡ ⎤− −⎢ ⎥=⎢ ⎥−⎣ ⎦
− − − − − ⋅ − ⋅=
−
− − + + −=
−
+ − +=
− −
4.4: inverse 4.4: inverse 4.4: inverse 4.4: inverse laplacelaplacelaplacelaplace transform transform transform transform
( )( ){ } ( )
( ) ( )
( ){ } ( )1
If Laplace transform of is
Then, the inverse of Laplace transform of is . The notation for the operation of inverse Laplace transformis
F
f t
f ts
f t F s
F s f t−
=
=
L
L
Example 4.7 Example 4.7 Example 4.7 Example 4.7
( ) ( )
( ) ( )
( ) ( )
( ) ( )
4
2 2
2 2
Find the inverse Laplace transform of the following functions:6 6a) b)
1 1c) d) 9 7
3e) f) 9 25
2g) h) 4 16
F s F ss s
F s F ss s
sF s F ss s
sF s F ss s
= =
= =− +
= =+ +
= =− −
Solution Solution Solution Solution ::::
( )
( )
( )
( )
( )
( )
1
4
1 1 34 3 1
1 9
6a)
6 6
6b)
6 3!
1c)9
19
t
F ss
f ts
F ss
f t ts s
F ss
f t es
−
− −+
−
=
⎧ ⎫⇒ = =⎨ ⎬⎩ ⎭
=
⎧ ⎫ ⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭
=−⎧ ⎫⇒ = =⎨ ⎬−⎩ ⎭
L
L L
L
1 a as
− ⎧ ⎫ =⎨ ⎬⎩ ⎭
L
11
! nnn ts
−+
⎧ ⎫ =⎨ ⎬⎩ ⎭
L
1 1 ates a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
Solution Solution Solution Solution ::::
( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
1 1 7
2
1 12 2 2
2
1 12 2 2
1d)7
1 17 7
3e)93 3 sin 3
9 3
f)25
cos 525 5
t
F ss
f t es s
F ss
f t ts ssF s
ss sf t t
s s
− − −
− −
− −
=+
⎧ ⎫⎪ ⎪⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬+ − −⎩ ⎭ ⎪ ⎪⎩ ⎭
=+⎧ ⎫ ⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬
+ +⎩ ⎭ ⎩ ⎭
=+⎧ ⎫ ⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬
+ +⎩ ⎭ ⎩ ⎭
L L
L L
L L
( )12 2 coss ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
( )12 2 sina ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
1 1 ates a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
Solution Solution Solution Solution ::::
( )
( ) ( )
( )
( ) ( )
2
1 12 2 2
2
1 12 2 2
2g)42 2 sinh 2
4 2
h)16
cosh 416 4
F ss
f t ts ssF s
ss sf t t
s s
− −
− −
=−⎧ ⎫ ⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭
=−⎧ ⎫ ⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭
L L
L L
( )12 2 coshs ats a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
( )12 2 sinha ats a
− ⎧ ⎫− =⎨ ⎬+⎩ ⎭L
Example 4.8 Example 4.8 Example 4.8 Example 4.8
( ) ( )
( ) ( )
( )
3 2
2 2
Find the inverse Laplace transform of the following functions:14 3a) b)
162 42c) d)
16 25 491e)
2 5
F s F ss s
sF s F ss s
F ss
= =+
= =+ −
=+
Solution Solution Solution Solution ::::
( )
( )
( )
( ) ( )
3
1 1 1 23 2 1 2 1
2
1 12 2 2
14a)
14 14 2! 2!7 72!
3b)16
3 3 4 3 sin 416 4 4 4
F ss
f t ts s s
F ss
f t ts s
− − −+ +
− −
=
⎧ ⎫ ⎧ ⎫ ⎧ ⎫⇒ = = = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭ ⎩ ⎭
=+⎧ ⎫ ⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭
L L L
L L
11
! nnn ts
−+
⎧ ⎫ =⎨ ⎬⎩ ⎭
L
( )12 2 sina ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
Solution Solution Solution Solution ::::
( )
( )
2
1 12
2
12
2
2c)16 25
2 22516 25 1616
2 1 5cos16 8 45
4
sF ss
s sf ts s
s ts
− −
−
=+
⎧ ⎫⎪ ⎪⎧ ⎫⇒ = =⎨ ⎬ ⎨ ⎬+⎩ ⎭ ⎪ ⎪+⎩ ⎭
⎧ ⎫⎪ ⎪⎪ ⎪ ⎛ ⎞= =⎨ ⎬ ⎜ ⎟
⎝ ⎠⎛ ⎞⎪ ⎪+ ⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
L L
L
( )12 2 coss ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
Solution Solution Solution Solution ::::
( )
( )
( )
( )
( )
2
1 1 12 2 2 2
51 1 1 2
42d)4942 1 42 742
49 49 7 76sinh 7
1e)2 5
1 1 1 1 1 15 52 5 2 2 22 2
t
F ss
f ts s st
F ss
f t es s s
− − −
−− − −
=−⎧ ⎫ ⎧ ⎫ ⎧ ⎫⇒ = = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬− − −⎩ ⎭ ⎩ ⎭ ⎩ ⎭
=
=+
⎧ ⎫⎧ ⎫⎪ ⎪⎪ ⎪ ⎪ ⎪⎧ ⎫⇒ = = = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬
+ ⎛ ⎞⎩ ⎭ ⎪ ⎪ ⎪ ⎪+ − −⎜ ⎟⎩ ⎭ ⎪ ⎪⎝ ⎠⎩ ⎭
L L L
L L L
1 1 ates a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
( )12 2 sinha ats a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
4.5: properties of inverse 4.5: properties of inverse 4.5: properties of inverse 4.5: properties of inverse laplacelaplacelaplacelaplace transformtransformtransformtransform
� LINEARITY
� FIRST SHIFT THEOREM
� RATIONAL FUNCTION
LinearityLinearityLinearityLinearity
Example 4.9 Example 4.9 Example 4.9 Example 4.9
( ) ( )
( ) ( )
( ) ( )
2 2
2 2
2 2
Find the inverse Laplace transform of the following functions:3 2 7a) b) 16 4
3 6 4 7c) d) 4 16 16
6 6 5e) f) 4 9 9 16
s sF s F ss ss sF s F ss ss sF s F ss s
+ += =
− +− − +
= =+ −− +
= =− +
Solution Solution Solution Solution ::::
( )12 2 coshs ats a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
( )12 2 sinha ats a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
( )
( )
( ) ( )
2
1 1 12 2 2
1 12 2 2 2
3a)16
3 1316 16 16
3 4 4 4 4
3 cosh 4 sinh 44
sF ss
s sf ts s s
ss s
t t
− − −
− −
+=
−+⎧ ⎫ ⎧ ⎫ ⎧ ⎫⇒ = = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬− − −⎩ ⎭ ⎩ ⎭ ⎩ ⎭
⎧ ⎫ ⎧ ⎫= +⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭
= +
L L L
L L
Solution Solution Solution Solution ::::
( )12 2 coss ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
( )12 2 sina ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
( )
( )
( ) ( )
2
1 1 12 2 2
1 12 2 2 2
2 7b)4
2 7 12 74 4 4
7 2 22 2 27 2cos 2 sin 22
sF ss
s sf ts s s
ss s
t t
− − −
− −
+=
++⎧ ⎫ ⎧ ⎫ ⎧ ⎫⇒ = = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬+ + +⎩ ⎭ ⎩ ⎭ ⎩ ⎭
⎧ ⎫ ⎧ ⎫= +⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭
= +
L L L
L L
Solution Solution Solution Solution ::::
( )12 2 coss ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
( )12 2 sina ats a
− ⎧ ⎫ =⎨ ⎬+⎩ ⎭L
( )
( )
( )
( ) ( )
2
1 1 12
2 2
1 12 2
1 12 2 2 2
3 6c)4 16
3 6 3 6 116 164 16 4 44 4
3 6 1 4 4 4 43 6 2 4 2 4 2 23 3 cos 2 sin 24 4
sF ss
s sf ts s s
ss ss
s s
t t
− − −
− −
− −
−=
+⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪−⎧ ⎫⇒ = = −⎨ ⎬ ⎨ ⎬ ⎨ ⎬+⎩ ⎭ ⎪ ⎪ ⎪ ⎪+ +⎩ ⎭ ⎩ ⎭⎧ ⎫ ⎧ ⎫= −⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭⎧ ⎫ ⎧ ⎫= −⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭
= −
L L L
L L
L L
First shift theoremFirst shift theoremFirst shift theoremFirst shift theorem
( )( )
( )( )( )
( )( ) ( )
( )( )( )
( )( )
2 2
2 2 2
2 2
:a) If of the given function contains ' ' ' '
in accordance to term the term found in the deno
8 8
mi
4 4 4 44 8 4 8
nator
Exa8 8
m
8
e:
4
pl
s a a s as as a s a s
Notes s
a
a
s
a
sss s s s
s a
s a
− + −= = +
− + − + − + − +
⇒
− + −= = +
− + − + − + −
− +
−
numeratornumeratornumeratornumerator
( )
( ) ( )
2 2
2 22
2
2 2 22
b) If contains the quadratic term, we need to do the completing the square firs
4 8
3 3
2 1
38 5 8 8 4 21
t
Exam
52
pl 2 2
2
:
2 1
e
s ss bs c
s
b bs
s ss
c
+ +=
+ + ⎛ ⎞ ⎛ ⎞
+
= =+ − + −⎛ ⎞ ⎛ ⎞+ − + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝
+ − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎠
denominatordenominatordenominatordenominator
Example 4.10 Example 4.10 Example 4.10 Example 4.10
( )( )
( )( )
( )
2
2
2
Find the inverse Laplace transform of the following functions:6a)
2 43 6b)
1 52 5c)
4 6
F sssF s
ssF s
s s
−=
− −
−=
− +
+=
+ −
Solution Solution Solution Solution ::::
( )12 2 sinha ats a
− ⎧ ⎫ =⎨ ⎬−⎩ ⎭L
( )( )
( )( )
( )
2
12
2 12
2 12 2
2
6a)2 4
162 4
164
6 22 23 sinh 2
t
t
t
F ss
f ts
es
es
e t
−
−
−
−=
− −
⎧ ⎫⎪ ⎪⇒ = − ⎨ ⎬− −⎪ ⎪⎩ ⎭⎧ ⎫= − ⎨ ⎬−⎩ ⎭⎧ ⎫= − ⎨ ⎬−⎩ ⎭
= −
L
L
L
( ){ } ( )1 22 tF s e f t− − =L
Solution Solution Solution Solution ::::
( )( )
( ) ( )( )
( )( )
( )( )
( )( ) ( )
( )
2
1 1 12 2 2
1 12 2
1 12 2
122
3 6b)1 5
3 1 1 6 3 1 3 6 3 1 31 5 1 5 1 5
1 13 31 5 1 5
13 35 5
35
t t
t
sF ss
s s sf t
s s s
ss s
se es s
ses
− − −
− −
− −
−
−=
− +
⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎡ ⎤− + − − + − − −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎣ ⎦⇒ = = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬− + − + − +⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭
⎧ ⎫ ⎧ ⎫−⎪ ⎪ ⎪ ⎪= −⎨ ⎬ ⎨ ⎬− + − +⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
⎧ ⎫ ⎧ ⎫= −⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭⎧
=+
L L L
L L
L L
L( )
( ) ( )
122
3 55 5
33 cos 5 sin 55
t
t t
es
e t e t
−
⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪−⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ + ⎪⎩ ⎭ ⎩ ⎭
= −
L
( ){ } ( )1 1 tF s e f t− − =L
Solution Solution Solution Solution ::::
( )
( )( )
( )( )
( )( )
( )( )
( )( )
2
1 12 2 2
1 12 2
12
12
2 5c)4 6
2 5 2 54 4 2 1062 2
2 2 2 5 2 2 4 52 10 2 10
2 2 12 10
22
2 10
sF ss s
s sf tss
s ss s
ss
ss
− −
− −
−
−
+=
+ −⎧ ⎫⎪ ⎪ ⎧ ⎫+ +⎪ ⎪ ⎪ ⎪⇒ = =⎨ ⎬ ⎨ ⎬
+ −⎛ ⎞ ⎛ ⎞ ⎪ ⎪⎪ ⎪ ⎩ ⎭+ − −⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭⎧ ⎫ ⎧ ⎫⎡ ⎤+ − + + − +⎪ ⎪ ⎪ ⎪⎣ ⎦= =⎨ ⎬ ⎨ ⎬
+ − + −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎧ ⎫+ +⎪ ⎪= ⎨ ⎬
+ −⎪ ⎪⎩ ⎭⎧ ⎫+⎪= ⎨ ⎬
+ −⎪⎩
L L
L L
L
L( )
12
12 10s
−⎧ ⎫⎪ ⎪ ⎪+ ⎨ ⎬
+ −⎪ ⎪ ⎪⎭ ⎩ ⎭L
( ){ } ( )1 22 tF s e f t− − =L
Solution Solution Solution Solution ::::
( )( ) ( )
( ) ( )( ) ( )
1 12 2
2 1 2 12 2
2 1 2 12 2
2 2
2 2
2 122 10 2 10
1210 10
1 1021010 10
12 cosh 10 sinh 1010
t t
t t
t t
ss s
se es s
se es s
e t e t
− −
− − − −
− − − −
− −
⎧ ⎫ ⎧ ⎫+⎪ ⎪ ⎪ ⎪= +⎨ ⎬ ⎨ ⎬+ − + −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎧ ⎫ ⎧ ⎫= +⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= +⎨ ⎬ ⎨ ⎬⎪ − ⎪ ⎪ − ⎪⎩ ⎭ ⎩ ⎭
= +
L L
L L
L L
Inverse Inverse Inverse Inverse laplacelaplacelaplacelaplace transform of rational transform of rational transform of rational transform of rational functionsfunctionsfunctionsfunctions
Example 4.11 Example 4.11 Example 4.11 Example 4.11
( )( )( )
( ) ( )
( ) ( )( )
2
2
2
2
2
Find the inverse Laplace transform of the following functions:2 1a)
2 4
2 1b) 3
4c)36 6
s sF ss s
sF ss s
sF ss s
− +=
+ +
+=
−
=− −
Solution Solution Solution Solution ::::
( )( )( )
( )( )( ) ( )( )
( )( )
( )
2
2
2
22
2 2
2 1a)2 4
2 12 42 4
2 1 4 2
9Let 2 : 9 88
Let 0 : 1 4 29 71 4 28 4
Let 1: 0 5 3
9 7 10 5 38 4 8
s sF ss s
s s A Bs Cs ss s
s s A s Bs C s
s A A
s A C
C C
s A B C
B B
− +=
+ +
− + +⇒ = +
+ ++ +
− + = + + + +
= − = ⇒∴ =
= = +
⎛ ⎞= + ⇒∴ = −⎜ ⎟⎝ ⎠
= = + +
⎛ ⎞ ⎛ ⎞= + − ⇒∴ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Solution Solution Solution Solution ::::
( )( )
( )( )
( )
2
2 2 22
21 1 1 1
2 22
1 1 12 2 2 2
2 1 9 1 1 7 12 4 8 2 8 4 4 42 4
2 1 9 1 1 7 18 2 8 4 4 42 4
9 1 1 7 28 2 8 2 4 2 2
s s A Bs C ss s s s ss s
s s ss s ss s
ss s s
− − − −
− − −
− + + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ = + = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + + + ++ + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎧ ⎫− +⎪ ⎪ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫= − −⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬+ + ++ + ⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎪ ⎪⎩ ⎭⎧ ⎫ ⎧ ⎫ ⎧ ⎫= − −⎨ ⎬ ⎨ ⎬ ⎨ ⎬+ + +⎩ ⎭ ⎩ ⎭ ⎩ ⎭
L L L L
L L L
( ) ( )29 1 7cos 2 sin 28 8 8
te t t−= − −
Solution Solution Solution Solution ::::
( ) ( )
( )( ) ( )
2
2 2
2
2 1b)3
2 13 3
2 1 3 31Let 0 : 1 33
7Let 3 : 7 99
Let 1: 3 2 21 7 73 2 23 9 9
sF ss s
s A B Cs s s s s
s As s B s Cs
s B B
s C C
s A B C
A A
+=
−
+⇒ = + +
− −
+ = − + − +
= = − ⇒∴ = −
= = ⇒∴ =
= = − − +
⎛ ⎞= − − − + ⇒∴ = −⎜ ⎟⎝ ⎠
Solution Solution Solution Solution ::::
( )
( )
( )
2 2 2
1 1 1 12 2
1 1 11 1
3
2 1 7 1 1 1 7 13 3 9 3 9 3
2 1 7 1 1 1 7 13 9 3 9 3
7 1 1 1! 7 1 9 3 1! 9 3
7 1 7 9 3 9
t
s A B Cs s s s s s s s
ss s s s s
s s s
t e
− − − −
− − −+
+ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⇒ = + + = − − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎧ ⎫+ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫= − − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬− −⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎩ ⎭⎧ ⎫ ⎧ ⎫ ⎧ ⎫= − − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬−⎩ ⎭ ⎩ ⎭ ⎩ ⎭
= − − +
L L L L
L L L
Solution Solution Solution Solution ::::
( ) ( )( )
( )( ) ( )( )( ) ( ) ( )
( ) ( ) ( ) ( ) ( )( )( ) ( ) ( )
( )
2
2
2 2 2
22
2
2 2
22
4c)36 6
4 4 46 6 636 6 6 6
46 66 6 6
4 6 6 6 6Let 6 : 144 12 12Let 6 : 144 144 1Let 0 : 0 36 6 36
0 36 6 12 36 3
sF ss s
s s ss s ss s s s
s A B Cs ss s s
s A s s B s C ss B Bs C Cs A B C
A A
=− −
⇒ = =− + −− − − +
= + +− +− + −
= − + + + + −
= = ⇒∴ == − = ⇒∴ == = − + +
= − + + ⇒∴ =
Solution Solution Solution Solution ::::
( )( ) ( ) ( ) ( )
( ) ( ) ( )
( )( ) ( )
( )
( )
2
22
2
21 1 1 1
22
1 1 12
6 6 12
46 636 6 6
3 12 1 6 66
4 1 1 13 126 636 6 6
1 1 1 3 126 66
1 3 12t t
s A B Cs ss s s
s ss
ss ss s s
s ss
e es
− − − −
− − −
−
⇒ = + +− +− − −
= + +− +−
⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎧ ⎫ ⎧ ⎫= + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬− +− − ⎩ ⎭ ⎩ ⎭−⎪ ⎪⎪ ⎪ ⎩ ⎭⎩ ⎭⎧ ⎫⎪ ⎪⎧ ⎫ ⎧ ⎫= + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬− +⎩ ⎭ ⎩ ⎭−⎪ ⎪⎩ ⎭
⎧ ⎫⎪= + ⎨⎪⎩
L L L L
L L L
L
( )
6
6 6 1 61 1
6 6 6
12 1! 31!
3 12
t
t t t
t t t
e
e e es
e te e
−
− −+
−
⎪+⎬⎪⎭
⎧ ⎫⎪ ⎪= + +⎨ ⎬⎪ ⎪⎩ ⎭
= + +
L
Application of Application of Application of Application of laplacelaplacelaplacelaplace transform transform transform transform
Given a problem of
1st order ODE:
2nd order ODE:
( ) ( )dya by f t ay by f tdx
′+ = ⇒ + =
( ) ( )2
2d y dya b cy f t ay by cy f tdx dx
′′ ′+ + = ⇒ + + =
( ){ } ( )( ){ } ( ) ( )( ){ } ( ) ( ) ( )2
0
0 0
y t Y s
y t sY s y
y t s Y s sy y
=
′ = −
′′ ′= − −
L
L
L
Application of Application of Application of Application of laplacelaplacelaplacelaplace transform transform transform transform
Then solve the given problem using the following method of solution:
Step 1: Transform the left-hand and right-hand side of the differential equation to generate an algebraic equation,
Step 2: Substitute conditions and
Step 3: Solve for
Step 4: Apply inverse Laplace transform,
Step 5: Obtained which is the original solution of the given differential equation
( ).Y s
( )y a b= ( ) .y a c′ =
( ).Y s
( ){ }1 Y s−L
( )y t
Example 4.12 Example 4.12 Example 4.12 Example 4.12
( )
( )( )
2
3
Use the method of Laplace transform to find the solution of the following initial values problems.
a) 6 4 , 0 3
b) , 0 2
c) 4 5, 0 1
t
t
dy y e ydty y te y
y y y
−− = =
′− = = −
′+ = = −
Solution Solution Solution Solution ::::
( )
{ } { } { }( ) ( ) ( )
( ) ( )
( )
( )[ ] ( )( ) ( )
( ) ( )( )
2
2
a) 6 4 , 0 3
6 4Step 1:
Step 2:
Ste
40 62
43 62
Solve for
4 3 2
p 3:
4 10 36 32 2 2
10 32 6
t
t
dy y e ydt
y y e
sY s y Y ss
sY s Y ss
Y s
s sY s ss s s
sY ss s
−
−
− = =
′ − =
− − =+
− − =+
+ + +− = + = =
+ + +
+∴ =
+ −
L L L { } 1ates a
=−
L
( ){ }
( ){ } ( )( )
( )( )( ) ( )
( )( )
( )( )
1
1 1
1 1
Apply :
10 32 6
10 32 6 2 6
10 3 6 21Let 2 : 4 82
7Let 6 : 28 82
10 3 1 1 7
Step 4:
12 6 2 2 2 6
10 3 1 1 72 6 2 2
Y s
sY ss s
s A Bs s s ss A s B s
s A A
s B B
ss s s s
ss s s
−
− −
− −
⎧ ⎫+= ⎨ ⎬+ −⎩ ⎭
+⇒ = +
+ − + −
+ = − + +
= − = − ⇒∴ = −
= = ⇒∴ =
+ ⎛ ⎞ ⎛ ⎞∴ = − +⎜ ⎟ ⎜ ⎟+ − + −⎝ ⎠ ⎝ ⎠
⎧ ⎫+ ⎧ ⎫= − +⎨ ⎬ ⎨ ⎬+ − +⎩ ⎭⎩ ⎭
L
L L
L L 1 12 6s
− ⎧ ⎫⎨ ⎬−⎩ ⎭
L
( )
( )( )
( )
1 1 1
2 6
Obtain :
10 3 1 1 7 12 6 2 2 2 6
1 7
Step 5:
2 2
t t
y t
ss s s s
y t e e
− − −
−
⎧ ⎫+ ⎧ ⎫ ⎧ ⎫= − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬+ − + −⎩ ⎭ ⎩ ⎭⎩ ⎭
= − +
L L L
Solution Solution Solution Solution ::::
( ){ } { } { }( ) ( ) ( )
( ) ( )
( ) ( )( )
( )
( )[ ]( )
( )( ) ( )
( )( ) ( )
3
3
1 1 2
2
2 2
2 2 2
2
2
Step 1:
Step 2:
S
a) , 0 2
1! 103 3
123
Solve for
1 2 31 2 12 171 23
tep 3:
3 3
2 12 173 1
t
t
y y te y
y y te
sY s y Y ss s
sY s Y ss
Y s
s s sY s ss s s
s sY ss s
+
′ − = = −
′ − =
− − = =− −
+ − =−
− − − + −− = − = =
− − −
− + −∴ =
− −
L L L{ }
( ) 1!atn
ntes a +=−
L
( ){ }
( ){ }( ) ( )
( ) ( ) ( )( )( ) ( ) ( )
1
21 1
2
2
2 2
22
Apply :
2 12 173 1
2 12 173 13 1 3
2 12 17 3 1 1 31Let
Step
3 : 1 22
7Let 1: 7 44
Let 0 : 17 3 91 717 3 92 4
4: Y s
s sY ss s
s s A B Cs ss s s
s s A s s B s C s
s B B
s C C
s A B C
A
−
− −⎧ ⎫− + −⎪ ⎪= ⎨ ⎬
− −⎪ ⎪⎩ ⎭− + −
⇒ = + +− −− − −
− + − = − − + − + −
= = ⇒∴ =
= − = ⇒∴ = −
= − = − +
⎛ ⎞− = − + − ⇒∴⎜ ⎟⎝ ⎠
L
L L
14
A = −
( ) ( ) ( )( )
( ) ( ) ( )
2
2 2
21 1 1
2 2
1
3 3 12
6 8 1 1 1 1 174 3 2 13 1 3
Obtain :
6 8 1 1 1 14 3 23 1 3
7 1 4 1
1 1 1 7 4 2 4
Step 5:
t t t
s ss ss s s
y t
s sss s s
s
e e es
− − −
−
−
⎛ ⎞− + − ⎛ ⎞ ⎛ ⎞∴ = − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− −⎝ ⎠ ⎝ ⎠− − −⎝ ⎠
⎧ ⎫ ⎧ ⎫− + −⎪ ⎪ ⎪ ⎪⎧ ⎫= − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬−⎩ ⎭− − −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎧ ⎫− ⎨ ⎬−⎩ ⎭
⎧ ⎫= − + −⎨ ⎬⎩ ⎭
=
L L L
L
L
( )
( )
3 3 11 1
3 3
1 1 1! 74 2 4
1 1 74 2 4
t t t
t t t
e e es
y t e te e
−+
⎧ ⎫⎪ ⎪− + −⎨ ⎬⎪ ⎪⎩ ⎭
∴ = − + −
L
Solution Solution Solution Solution ::::
( ){ } { } { }
( ) ( ) ( )
( ) ( )
( )
( )[ ]
( ) ( )
a) 4 5, 0 1
4 550 4
51 4
Solve f
Step 1:
Step 2:
Step or 5 54 1
3
54
:
y y y
y y
sY s y Y ss
sY s Y ss
Y ssY s s
s ssY s
s s
′ + = = −
′ + =
− + =
+ + =
−+ = − =
−∴ =
+
L L L{ } aa
s=L
( ){ }
( ){ } ( )
( ) ( )( )
( ) ( )
1
1 1
Apply :
54
54 4
5 45Let 0 : 5 44
9Let 4
Step 4:
: 9 44
5 5 1 9 14 4 4 4
Y s
sY ss s
s A Bs s s s
s A s Bs
s A A
s B B
ss s s s
−
− − ⎧ ⎫−= ⎨ ⎬+⎩ ⎭
−⇒ = +
+ +
− = + +
= = ⇒∴ =
= − = − ⇒∴ = −
⎛ ⎞− ⎛ ⎞∴ = − ⎜ ⎟⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
L
L L
( )
( ) ( )
( )
1 1 1
4
Obtain :
5 5 1 9 14 4 4 4
5 9 4 4
Step 5:
t
y t
ss s s s
y t e
− − −
−
⎧ ⎫ ⎧ ⎫− ⎧ ⎫= −⎨ ⎬ ⎨ ⎬ ⎨ ⎬+ +⎩ ⎭⎩ ⎭ ⎩ ⎭
= −
L L L
Example Example Example Example 4.13 4.13 4.13 4.13
( ) ( )
( ) ( )
( ) ( )
2
2
22
2
Use the method of Laplace transform to find the solution of the following initial values problems.
a) 2 8 4, 0 2 and 0 3
b) 4 , 0 2 and 0 1
c) 2 , 0 1 and 0 2
t
t
d y dy y y ydt dtd y y e y ydty y y te y y
−
′+ − = = =
′+ = = =
′′ ′ ′+ + = = = −
Solution Solution Solution Solution ::::
( ) ( )
{ } { } { } { }
( ) ( ) ( ) ( ) ( )( ) ( )
( ) ( )( ) ( )
( ) ( ) ( )
( )
( )
( )
2
2
2
2
2
22
2
2
Step 1:
St
a) 2 8 4, 0 2 and 0 3
2 8 440 0 2 0 8
42 3 2 2 8
42 8 2 7
Solve for
4 4 2 72 8 2 7
4 2 7
ep 2:
Step 3:
d y dy y y ydt dt
y y y
s Y s sy y sY s y Y ss
s Y s s sY s Y ss
s Y s sY s Y s ss
Y s
s sY s s s ss s
s sY ss s
′+ − = = =
′′ ′+ − =
′− − + − − =
− − + − − =
+ − − − =
+ +⎡ ⎤+ − = + + =⎣ ⎦
+ +∴ =
L L L L
( ) ( )( )22 7 4
2 42 8s s
s s ss+ +
=− ++ −
{ } aas
=L
( ){ }
( ){ } ( )( )
( )( )( )( ) ( ) ( )
( )( )
1
21 1
2
2
2
Apply :
2 7 42 4
2 7 42 4 2 4
2 7 4 2 4 4 21Let 0 : 4 82
13Let
Step
2 : 26 1261Let 4 : 8 243
2 7 4
4:
1 1 13 12 4 2 6
Y s
s sY ss s s
s s A B Cs s s s s s
s s A s s Bs s Cs s
s A A
s B B
s C C
s ss s s s
−
− − ⎧ ⎫+ += ⎨ ⎬− +⎩ ⎭
+ +⇒ = + +
− + − +
+ + = − + + + + −
= = − ⇒∴ = −
= = ⇒∴ =
= − = ⇒∴ =
+ + ⎛ ⎞∴ = − +⎜ ⎟− + ⎝ ⎠
L
L L
1 12 3 4s s
⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠
( )
( )( )
( )
21 1 1
1
2 4
Obtain :
2 7 4 1 1 13 12 4 2 6 2
1 1 3 41 13 1 2 6 3
Step 5:
t t
y t
s ss s s s s
s
y t e e
− − −
−
−
⎧ ⎫+ + ⎧ ⎫ ⎧ ⎫= − +⎨ ⎬ ⎨ ⎬ ⎨ ⎬− + −⎩ ⎭ ⎩ ⎭⎩ ⎭⎧ ⎫+ ⎨ ⎬+⎩ ⎭
= − + +
L L L
L
Solution Solution Solution Solution ::::
( ) ( )
{ } { } { }( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( )
( ) ( )( )
( ) ( )( )( )( )
22
2
2
2
2
2
2
2
a) 4 , 0 2 and 0 1
4
10 0 42
1
Step 1:
Step 2:
Step 3
2 1 42
14 2 12
Solve for
1 2 1 214 2 12 2
1 2 1 22 4
:
t
t
d y y e y ydt
y y e
s Y s sy y Y ss
s Y s s Y ss
s Y s Y s ss
Y s
s sY s s s
s ss s
Y ss s
−
−
′+ = = =
′′ + =
′− − + =+
− − + =+
+ − − =+
+ + +⎡ ⎤+ = + + =⎣ ⎦ + +
+ + +∴ =
+ +
L L L { } 1ates a
=−
L
( ){ }
( ){ } ( )( )( )( )
( )( )( )( ) ( )
( )( ) ( ) ( )( )
( )( )
1
1 12
2 2
2
Apply :
1 2 1 22 4
1 2 1 222 4 4
1 2 1 2 4 2
1Let 2 : 1 88
1 5Let 0 : 3 4 2 3 4 28 4
Let 4 : 55 20 4 6
155 2
Step
0 2
4:
48
Y s
s sY s
s s
s s A Bs Css s s
s s A s Bs C s
s A A
s A C C C
s A B C
B
−
− −⎧ ⎫+ + +⎪ ⎪= ⎨ ⎬
+ +⎪ ⎪⎩ ⎭+ + + +
⇒ = +++ + +
+ + + = + + + +
= − = ⇒∴ =
⎛ ⎞= = + ⇒ = + ⇒∴ =⎜ ⎟⎝ ⎠
= = + +
⎛ ⎞= +⎜ ⎟⎝ ⎠
L
L L
( )( )( )( ) 2 22
5 1564 8
1 2 1 2 1 1 15 5 18 2 8 4 4 42 4
B
s s ss s ss s
⎛ ⎞+ ⇒∴ =⎜ ⎟⎝ ⎠
+ + + ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∴ = + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + ++ + ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
( )
( )( )( )( )
( )
( ) ( )
1 1 1 12 22
2 1 12 2 2 2
2
Obtain :
1 2 1 2 1 1 15 5 18 2 8 4 4 42 4
1 15 5 2 = 8 8
Step 5
2 4 2 21 15 5 cos 2 si8 8
:
8
t
t
y t
s s ss s ss s
ses s
y t e t
− − − −
− − −
−
⎧ ⎫+ + +⎪ ⎪ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫= + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬+ + ++ + ⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎪ ⎪⎩ ⎭
⎧ ⎫ ⎧ ⎫+ +⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭
= + +
L L L L
L L
( )n 2t
Solution Solution Solution Solution ::::( ) ( )
{ } { } { } { }( ) ( ) ( ) ( ) ( )( ) ( )
( )
( ) ( ) ( ) ( ) ( )( ) ( )( )
( ) ( )( ) ( )( )
( ) ( ) ( )( )
( )
( )
21 1
22
22
22
2
a) 2 , 0 1 and 0 2
2
1!0 0 2 01
10 0 2
Step 1:
Step 2:
Step 3:
01
12 2 11
121
Solve for
2 1
t
t
y y y te y y
y y y te
s Y s sy y sY s y Y ss
s Y s sy y sY s y Y ss
s Y s s sY s Y ss
s Y s sY s Y s ss
Y s
Y s s s
+
′′ ′ ′+ + = = = −
′′ ′+ + =
′− − + − + =−
′− − + − + =−
− + + − + =−
+ + − =−
⎡ ⎤+ + =⎣ ⎦
L L L L
( )( )
( )
( ) ( )( ) ( )
( )( ) ( )
2
2 2
2 2
2 2 22
1 111 1
1 1 1 11 2 1 1 1
s ss
s s
s s s sY s
s s s s s
+ −+ =
− −
+ − + −∴ = =
− + + − +
{ }( ) 1
!atn
ntes a +=−
L
( ){ }
( ){ } ( )( ) ( )
( )( ) ( ) ( ) ( ) ( ) ( )
( ) ( )( ) ( ) ( ) ( ) ( )
1
21 1
2 2
2
2 2 2 2
2 2 2 2 2
Apply :
1 11 1
1 11 11 1 1 1
1 1 1 1 1 1 1 11Let 1: 1 44
3Let 1: 3 44
Let 0 : 11 314 4
32
Step 4: Y s
s sY s
s s
s s A B C Ds ss s s s
s s A s s B s C s s D s
s B B
s D D
s A B C D
A C
−
− −⎧ ⎫+ −⎪ ⎪= ⎨ ⎬
− +⎪ ⎪⎩ ⎭
+ −⇒ = + + +
− +− + − +
+ − = − + + + + − + + −
= = ⇒∴ =
= − − = ⇒∴ = −
= = − + + +
= − + + −
=
L
L L
(1)A C− + − − − − − − −
( )( ) ( ) ( ) ( )
2
2 2 2 2
Let 2 : 3 9 9 31 33 9 9 34 4
3 9 3 (2)2
1 5Solve (1) and (2), ,4 4
1 1 1 1 1 1 5 1 3 14 1 4 4 1 41 1 1 1
s A B C D
A C
A C
A C
s ss ss s s s
= = + + +
⎛ ⎞= + + −⎜ ⎟⎝ ⎠
= + − − − − − − − −
⇒ ∴ = − =
⎛ ⎞ ⎛ ⎞+ − ⎛ ⎞ ⎛ ⎞∴ = − + + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠− + − +⎝ ⎠ ⎝ ⎠
( )( )
( ) ( ) ( )
( )
21 1 1 1
2 2 2
12
1 12 2
Obtain :
1 1 1 1 1 1 5 14 1 4 4 11 1 1
3 1 4 1
1 1 1 5 3 1 = 4 4 4 41 1
Ste
= 4
4
p 5:
t t t t
t t
y t
s ss ss s s
s
e e e es s
e e
− − − −
−
− − − −
−
⎧ ⎫ ⎧ ⎫+ −⎪ ⎪ ⎪ ⎪⎧ ⎫ ⎧ ⎫= − + +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬− +⎩ ⎭ ⎩ ⎭− + −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎧ ⎫⎪ ⎪− ⎨ ⎬
+⎪ ⎪⎩ ⎭⎧ ⎫ ⎧ ⎫− + + −⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭
− +
L L L L
L
L L
L
( )
1 11 1 1 11! 5 3 1!
4 41 1 5 3 4 4 4 4
t t
t t t t
e es s
y t e te e te
− − −+ +
− −
⎧ ⎫ ⎧ ⎫+ −⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭
= − + + −
L