definition 3.1 (laplace transform) · 13 s ss. ll=−=−>0. chapter 3 : laplace transform ......
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CHAPTER 3 : LAPLACE TRANSFORM
1
The Laplace transform of a function is defined by )(tf
( )( ) ( ) ( )0
stf t F s e f t∞ −= = ∫L dt
Example 3.1.1
Example 3.1.1:
Using definition of Laplace Transform, find if )(sF ( )f t = a , a is constant.
Solution:
( )
[ ]
0
0
0
0
0
( )
1
0 1 ( ) , 0
st
st
st
st
s s
F s e f t dt
e a dt
aes
aesa e es
a aa ss s
∞ −
∞ −
− ∞
∞−
− ∞ −
=
=
=−
⎡ ⎤= − ⎣ ⎦
⎡ ⎤= − −⎣ ⎦
= − − ∴ = >
∫∫
L
Definition 3.1 (Laplace Transform)
Remember !!!
( ) ( )1 31 and 3 , ss s
= − = −L L 0>
CHAPTER 3 : LAPLACE TRANSFORM
2
Example 3.1.2:
Find if , a is constant )(sF atetf =)(
Solution:
( )( )
( )
( )
( ) [ ] ( )
0
0
0
1 10 1 , 0 ,
st at
s a t
s a t
at
F s e e dt
e dt
es a
s a s a es a s a s a
∞ −
∞ − −
∞− −
=
=
=− −
= − − − > = > ∴ =− −
∫∫
L 1−
Example 3.1.3:
Find ( )sin atL
Solution:
( )
( )
( )
0
2 20
0
2 2 2 2
sin sin
sin cos
0 sin 0 cos 0 ,
st
st
at e at dt
e s at a ats a
e as a ss a s a
∞ −
∞−
=
⎡ ⎤= − −⎢ ⎥+⎣ ⎦
⎡ ⎤= − − − = >⎢ ⎥+ +⎣ ⎦
∫L
0
( )2 1 , 22
te ss
− = > −+
L
( ) 2
3sin 39
ts
=+
L
CHAPTER 3 : LAPLACE TRANSFORM
3
Example 3.1.4:
Find ( )( )f tL if
( )
⎪⎩
⎪⎨
⎧
<<<<<
=te
tt
tft 10,
105,050,2
4
Solution:
( )( )
( )
( )( )
( )( )
( )
5 10 4
0 5 105 4
0 10
5 4
0 10
10 45 0
10 45
2 0
2
24
2 04
2 2 , 44
st st st t
s tst
s tst
ss
ss
F s e dt e dt e e
e dt e dt
e es s
e e es s s
e e ss s s
∞− − −
∞ −−
∞−−
−− −
− −−
= + +
= +
⎡ ⎤⎡ ⎤⎢ ⎥= +⎢ ⎥
− −⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦⎡ ⎤⎡ ⎤
= − + −⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎣ ⎦
= − + >−
∫ ∫ ∫∫ ∫
dt
CHAPTER 3 : LAPLACE TRANSFORM
4
Elementary Laplace Transform
( )( ) ( )f t F s=L
( )f t
( )F s Condition on s
a
as
0s >
, 0,1,2,nt n = ... 1
!n
ns + 0s >
ate 1
s a− s a>
sin at 2 2
as a+ 0s >
cosat 2 2
ss a+ 0s >
sinh at 2 2
as a− s a>
cosh at 2 2
ss a− s a>
Note:
This table will enable us to obtain the transforms of many other functions.
CHAPTER 3 : LAPLACE TRANSFORM
5
Let f, f1, f2 be a functions whose Laplace Transforms exist for s > α and c be a constant. Then for s > α,
( ) ( ) ( )1 2 1 2f f f f± = ±L L L
( ) ( )cf c f=L L
Example 3.2.1:
Find ( )41 5 6 sin 2te t+ −L
Solution:
( ) ( ) ( ) ( )
( ) ( )
4 4
4
2
1 5 6 sin 2 1 5 6 sin 2
1 5 6 sin 2
1 5 124 4
t t
t
e t e t
e ts
s s s
+ − = + −
= + −
= + −− +
L L L L
L L
Exercises 3.2.1:
1. Show ( ) 2 2sinh aats a
=−
(
L using the linearity property.
2. Find )2sin 2 .t
( )3 cosh .te t
L
3. Find L
Definition 3.2 (Linearity Property of Laplace Transform)
CHAPTER 3 : LAPLACE TRANSFORM
6
If , ( ) ( )( )F s f t= L
then ( )( ) ( )ate f t F s a= −L
Example 3.3.1:
Find ( )2te tL
( )2 ,a f t t= =
.
Solution
( ) ( )2 2te t F s∴ = −L
( ) ( )( ) ( )where 2
1F s f t ts
= = =L L
( ) ( )( )
2
21 22
te t F ss
∴ = − =−
L
Exercises 3.3.1:
Find ( )4 coste tL .
Definition 3.3 (First Shifting Property)
CHAPTER 3 : LAPLACE TRANSFORM
7
If ( )( ) ( ) ,f t F s=L
( )( )
then
( 1) , 1,2,3,...n
n nn
d Ft f t nds
= − =L
Example 3.4.1:
Find . ( )sin 6t tL
Solution:
( ) ( ) 2
61 , sin 6 ,36
n f t t F ss
= = =+
( )
( )( ) ( )( )
( ) ( )
1
2
22
2 22 2
sin 6 ( 1)
36 0 6 2
36
12 12
36 36
dFt tds
s s
s
s s
s s
= −
+ −= −
+
−= − =
+ +
L
Exercises 3.4.1:
1. Find L . ( )tt e
(2. Find )sin 2te t t−L .
Definition 3.4 (Derivatives of the Laplace Transform)
CHAPTER 3 : LAPLACE TRANSFORM
8
Graph of
( )F t
where 0, 0 and 1,
Example 3.5.1:
( ) ( )0 , 4
Draw 41 , 4
tf t H t
t<⎧
= − = ⎨ ≥⎩
t
1
0 4
( )F t
Exercises 3.5.1:
1. Draw
2. Draw 2
3. Draw 2
Definition 3.5 (Heaviside Unit Step Function)
0, 01,
The function is defined by
CHAPTER 3 : LAPLACE TRANSFORM
9
Effect of the Unit Step Function
1) ( ) ( ) ( )3 6H t H t= − − −f t
( )F t
2) ( )0 , 0
, 0t
f t tt t
<⎧= = ⎨ ≥⎩
( )F t
3) ( ) ( )0 , 2
2, 2
tf t t H t
t t<⎧
= ⋅ − = ⎨ ≥⎩
( )F t
4) ( ) ( ) ( )0 , 2
2 22 , 2
tf t t H t
t t<⎧
= − − = ⎨ − ≥⎩
( )F t
CHAPTER 3 : LAPLACE TRANSFORM
10
Definition: Step Function A step function is a piecewise the form continuous function of
, 0 , ,
,
The step functions can be expressed into the unit step functions forms.
Given a step function , 0,
The Heaviside un it step function is
0, 0 ,
0, 0
1,
Example 3.5.2:
Express the following function rms o functions. s in te f unit step
4, 0 2, 2 40, 4
Solution:
4
4 4 2 0
4 4 2 4
CHAPTER 3 : LAPLACE TRANSFORM
11
Example 3.5.3:
Express the f e ns form. ollowing unit step functions into th step functio
1 1 2 4 Solution:
1 1 2 4
1
1 1
1 1 2
2
1, 0 2
2, 2 42, 4
Laplace Transform of H( t – a )
, 0
Theorem: Laplace Transform of Unit Step Functions
Inverse Laplace Transform of Unit Step Functions
CHAPTER 3 : LAPLACE TRANSFORM
12
Example 3.5.4:
Find ( ){ }f tL if ( ) ( ) ( )2 4f t H t H t= − − − .
Solution:
a) Using the Laplace Transform Table
( ){ } ( ) ( ){ }( ){ } ( ){ }
2 4
2 4
2 4s s
f t H t H t
H t H t
e es s
− −
= − − −
= − − −
= −
L L
L L
b) Using the Laplace Transform definition ‐ Change the unit step function into the step function form
H t 2 H t 4 0 1 2 11 2 1 3
0; 1 1; 1 0.
2
4
Hence 0, 0 21, 2 40, 4
U e Laplace transform definition sing th
· 0 · 1 · 0
CHAPTER 3 : LAPLACE TRANSFORM
13
Laplace Transform of H( t – a ). F( t – a)
Example 3.5.5:
Find ( ) ( ){ }24 4t H t− −L .
Solution
( ) ( )23
2!4, ,c f t t F ss
= = =
( ) ( ){ } ( )2 4 43
2!4 4 s st H t F s e es
− −− − = =L
( ) ( ){ }4
23
24 4set H t
s
−
∴ − − =L
Exercises 3.5.2:
1. Find . ( ) ( ){ }sin 3 3t H t− −L
2. Find . ( ) ( ){ }5 5te H t− −L
3. Find cos 2 2 2
Theorem: Second‐shift Property
Inverse Laplace Transform with Second‐shift Property
CHAPTER 3 : LAPLACE TRANSFORM
14
Example 3.5.6
Find the function whose transform is
4se2s
−
.
Solution
The numerator corresponds to ase−
where 4a = and therefore indicate
( )4H t − .
Then ( ) { } ( )2
1 F s t f t ts
= = ∴L = ( ) ( )4
12 4 4
se t H ts
−− ⎧ ⎫
∴ = − −⎨ ⎬⎩ ⎭
L
Exercises 3.5.3:
1. A function ( )f t is defined by ( )4 , 0 22 3 ,
tf t
t t 2< <⎧
= ⎨ − >⎩
Sketch the graph of the function, expressing the function ( )f t in unit
step form and determine its Laplace transforms.
2. Write the following ( )f t in terms of unit step functions and determine
the Laplace transforms.
( ) , 0 22 1 , 2
te tf t
t t
−⎧ < <= ⎨
− ≥⎩
( ){ }
( )2 12
2
1 3 2Answer:1 1
ss ef t e
s s s s
− +− ⎛ ⎞= + + −⎜ ⎟+ +⎝ ⎠
L
Note: Remember that in writing the final result ( )f t is replaced by ( )f t c− .
CHAPTER 3 : LAPLACE TRANSFORM
15
3. A function ( )f t is defined by
( )6 , 0 18 2 , 1 34 , 3
tf t t t
t
< <⎧⎪= − < <⎨⎪ >⎩
Sketch the graph and find the Laplace transform of the function.
( ){ }2 3
2 2
6 2 2 2Answer:3s s se e ef t
s s s s
− −
= − + +L−
<
4. Given
( )2
0 , 0 2, 2 5
, 5t
tf t t t
e t
⎧ < <⎪= <⎨⎪ >⎩
Find the Laplace transform of the function.
( ){ }2 10 5 5 2 5
2 2
2 5Answer:2
s s s se e e e e ef ts s s s s
− − − −⋅= + − + −
−L
s−
5. Determine the function ( )f t for which,
( ){ }2
2 2
3 4 5s se ef ts s s
− −
= − +L .
Find its inverse transform and sketch the graph of ( )f t .
( )3 , 0 1
Answer: 7 4 , 1 23 , 2
tf t t t
t t
< <⎧⎪= − < <⎨⎪ − >⎩
CHAPTER 3 : LAPLACE TRANSFORM
16
Example 3.6.1:
( ){ }6 6 ast a eδ −⋅ − =L
Example 3.6.2:
Given , 0 35, 3
( ) ( ){ } ( )( ) ( ){ } ( )
3 3
2 2
3 3 5
2 2 2
s s
s s
f t t f e e
f t t f e e
δ
δ
− −
− −
⋅ − = =
⋅ − = =
L
L
Definition 3.6 (Dirac Delta Function)
∞,0,
The Dirac Delta function is defined by
1
And
THE R M the Dirac Delta Function O E : Laplace Transform for For 0, and the inverse Laplace Transform is
For 0,
CHAPTER 3 : LAPLACE TRANSFORM
17
In many technological problems, we are dealing with forms of
mechanical vibrations or electrical oscillations and the necessity to
express such periodic functions in Laplace transforms soon arises. Let
( )f t be a periodic function of period T i.e. ( ) ( )f t f t T= + , . 0T ≠
Example 3.7.1:
Show that ( ) sin 2f t tπ= is a periodic function.
Solution:
( )( ) ( )( ) ( )
( )
sin 2
sin 2
sin 2 sin 2 = sin 2 cos 2 cos 2 sin 2cos 2 1, sin 2 0 1,2,3,...
f t t
f t T t T
f t f t T
t t Tt T t
T T T
π
π
π πTπ π π
π π
=
+ = +
= +
= +
+= = ⇒ =
π
Exercises 3.7.1:
1. Show that ( ) 2f t t= is not a periodic function.
2. Sketch the following periodic function.
( )1 , 0 1
1 , 1 2t
f tt
< <⎧= ⎨− < <⎩
( ) ( )2f t f t= +
Definition 3.7 (Periodic Function)
CHAPTER 3 : LAPLACE TRANSFORM
18
Example 3.7.2:
Sketch the following periodic function and find its Laplace transform.
( )3 , 0 20 , 2 4
tf t
t< <⎧
= ⎨ < <⎩
( ) ( )4f t f t= +
Solution:
( )F t
The expression for
( ){ } ( )
( )
4
4 0
2 4
4 0 2
2
11
1 3 01
31
sts
st sts
s
f t e f t dte
e dt e de
s e
−−
− −−
−
=−
⎡ ⎤= ⋅ +⎢ ⎥⎣ ⎦−
=+
∫
∫ ∫
L
t⋅
f (t)
Laplace Transform of Periodic Function
If ( )f t is a periodic function of period T,
then ( ){ } ( )0
1 ; 01
T stsTf t e f t dt s
e−
−= >− ∫L .
CHAPTER 3 : LAPLACE TRANSFORM
19
Exercises 3.7.2:
Sketch the following periodic function and find its Laplace transform.
( ), 0 1
1 , 1 2t t
f tt
< <⎧= ⎨ < <⎩
( ) ( )2f t f t= +
CHAPTER 3 : LAPLACE TRANSFORM
20
Definition 3.8 (Inverse Laplace Transform)
Recall that,
Notation for Laplace Transform; ( )( ) ( )f t F=L s
So, the inverse form; ( )( ) ( )1 F s f t− =L
Linearity property of Inverse Laplace Transform
If ( )( ) ( )1 F s f t− =L and ( )( ) ( )1 G s g t− =L
with α and β is a constants, then
( ) ( )( ) ( )( ) ( )( )1 1
( ) ( )
F s G s F s G s
f t g t
α β α β
α β
− −± = ±
= ±
L L 1−L
First shifting property for Inverse Laplace Transform
If ( )( ) ( )1 F s f t− =L with α as constant, then
( )( ) ( )1 atF s a e f t− − =L
or we can write as . ( )( ) ( )( )1 1atF s a e F s− −− =L L
CHAPTER 3 : LAPLACE TRANSFORM
21
Example 3.8.1:
( ) ( )2 2
4 4 3 4 sin 39 3 9 3
F s fs s
⎛ ⎞= = ⇒ =⎜ ⎟+ +⎝ ⎠t t
Example 3.8.2:
( ) ( ) 45 5
1 1 4! 14! 24
F s f t ts s
⎛ ⎞= = ⇒ =⎜ ⎟⎝ ⎠
Example 3.8.3:
( )( )4
61
F ss
=−
By shifting property, ( ) ( )1 , 1G s a G s a− = − = .
( )
( ) ( )4 4
3 3
6 3!
t
G ss sg t t f t e t
= =
⇒ = ⇒ =
Exercises 3.8.1:
Determine
1. ( )222 9s
−⎧ ⎫⎪ ⎪⎨ ⎬
− +⎪ ⎪⎩ ⎭
1L
2. ( )33
2 5s−⎧ ⎫⎪ ⎪⎨ ⎬
+⎪ ⎪⎩ ⎭
1L
CHAPTER 3 : LAPLACE TRANSFORM
22
Example 3.8.4:
-1 -1
2 2 2 cos525 5
s s ts s
⎧ ⎫ ⎧ ⎫= =⎨ ⎬ ⎨ ⎬+ +⎩ ⎭ ⎩ ⎭L L
We can write down the corresponding function in t, provided we can
recognize it from our table of transforms.
But, what about -1
2
3 16
ss s
+⎧ ⎫⎨ ⎬− −⎩ ⎭
L ?
Solution:
( )( )2
3 1 3 16 2
s ss s s s
+ +=
− − + −3 = 1 2
2 3s s+
+ −
-1 12
3 1 1 26 2 3
⎫sL Ls s s s
−+⎧ ⎫ ⎧∴ = +⎨ ⎬ ⎨− − + −⎩ ⎭ ⎩⎬⎭
from table: = tt ee 32 2+−
The two simpler functions of 12s +
and 23s −
are called the partial
fractions of 2
3 16
ss s
+− − .
Therefore, u need to know partial fractions!!
( ) ( )2
2
3 16 2 3
3 2 33: 5 10 2
2 : 5 5 1
3 1 1 26 2 3
s A Bs s s s
1A s B s ss B Bs A A
ss s s s
+= +
− − + −− + + = +
= = ⇒ == − − = − ⇒ =
+= +
− − + −
CHAPTER 3 : LAPLACE TRANSFORM
23
Example 3.8.5:
Determine 2
5 112
ss s
− +⎧ ⎫⎨ − −⎩1L ⎬
⎭. Answer: 3 43 2t te e− +
( ) ( )2
2
3 42
5 112 3 4
4 3 5 13: 7 14 2; 4 : 7 21 3
5 1 2 312 3 45 1 2 3 2 3
12 3 4t t
s A Bs s s sA s B s ss A A s B B
ss s s s
s e es s s s
− − −
+= +
− − + −− + + = +
= − − = − ⇒ = = = ⇒ =+
= +− − + −
+⎧ ⎫ ⎧ ⎫= + = +⎨ ⎬ ⎨ ⎬− − + −⎩ ⎭ ⎩ ⎭1 1L L
Partial Fractions
There are few types of denominator that u should know:
1) A linear factor gives a partial fraction (s a+ ) As a+ where A is a
constant to be determined.
2) A repeated factor ( 2)s a+ gives ( )2A B
s a s a+
+ + .
3) Similarly ( gives )3s a+ ( ) ( )2 3A B C
s a s a s a+ +
+ + + .
4) A quadratic factor ( )2s ps q+ + gives 2
Ps Qs ps q
++ + .
5) Repeated quadratic factors ( )22s ps q+ + gives
( )22 2
Ps Q Rs Ts ps q s ps q
+ ++
+ + + + .
CHAPTER 3 : LAPLACE TRANSFORM
24
Exercises 3.8.2:
Determine
1. ( )( )2
2
4 5 61 4
s ss s
−⎧ ⎫− +⎪ ⎪⎨ ⎬
+ +⎪ ⎪⎩ ⎭
1L 3 cos 2 3sin 2e t t+ −. Answer: t−
2. ( ){ } for F s−1L ( )102
12 ++
+=
ssssF
3. ( ){ } ( ) 2
2 3 for 2
sF s F ss s
− −=
+ −1L
4. ( )32 3
4s
s−⎧ ⎫+⎪ ⎪⎨
+⎪⎩ ⎭
1L ⎬⎪ Answer:
4 25( ) 22
tf t e t t− ⎛ ⎞∴ = −⎜ ⎟⎝ ⎠
CHAPTER 3 : LAPLACE TRANSFORM
25
Example 3.8.6:
Find the inverse Laplace transform for ( )2
14s s + using Convolution
theorem.
Solution:
Let ( ) 1F ss
= and ( ) 2
14
G ss
=+ , then
( ) ( ) 11, sin 22
f t g t t= = .
( ) ( ) ( )11 , sin 22
f u g t u t u⇒ = − = − .
( ) ( ) ( )
( )
( )
[ ]
12 0
0
1 11 sin 224
cos 212 2
1 cos 2 0 cos 241 1 cos 24
t
t
t u dus s
t u
t
t
−⎧ ⎫⎪ ⎪ = −⎨ ⎬
+⎪ ⎪⎩ ⎭
− −⎡ ⎤= ⎢ ⎥−⎣ ⎦
= −⎡ ⎤⎣ ⎦
= −
∫L
Convolution Theorem
( ) ( ){ } ( ) ( )1
0
tF s G s f u g t u du− = −∫L
CHAPTER 3 : LAPLACE TRANSFORM
26
Exercises 3.8.3:
Find ( )2
22 4
s
s−⎧ ⎫⎪ ⎪⎨ ⎬
+⎪ ⎪⎩ ⎭
1L using Convolution theorem.
Find ( )22
1
1s−⎧ ⎫⎪ ⎪⎨
+⎪⎩ ⎭
1L ⎬⎪using Convolution theorem.
CHAPTER 3 : LAPLACE TRANSFORM
27
To solve a differential equation by Laplace transforms, we go through
four distinct stages.
(a) Rewrite the equation in terms of Laplace transforms.
(b) Insert the given initial conditions.
(c) Rearrange the equation algebraically to give the transform of the
solution.
(d) Determine the inverse transform to obtain the particular solution.
Transforms of Derivatives
{ }If ( ) ( ), theny t Y s=L
( ){ } 0( )y t sY s′ y= −L
{ } 20 0( ) ( )y t s Y s sy y′′ ′= − −L
( ){ } 3 20 0( )y t s Y s s y sy y′′′ ′ ′′0= − − −L
( ) ( ){ } ( )11 20 0 0
( )n nn n ny t s Y s s y s y y −− − ′= − − − −L
Definition 3.9 (Solution of Differential equations by Laplace Transforms)
CHAPTER 3 : LAPLACE TRANSFORM
28
Solution of 1st Order Differential equations
Example 3.9.1:
Solve the equation 2dy y− = 4dt , given that at 0, 1t y= = .
Solution:
(a) Rewrite the equation in Laplace transforms using the last notation
{ } { } { }2 4 2dy dyy ydt dt
⎧ ⎫ ⎧ ⎫− = ⇒ − =⎨ ⎬ ⎨ ⎬⎩ ⎭ ⎩ ⎭L L L L 4L
We have
{ }
{ }
{ }
0( ) ( ) ,
( ) ( ),44
dy y t sY s ydt
y t Y s
s
⎧ ⎫ ′= = −⎨ ⎬⎩ ⎭
=
=
L L
L
L
.
Then the equation becomes
( )04( ) 2 ( )sY s y Y ss
⇒ − − = .
(b) Insert the initial condition that at 0, 1t y= = i.e. 0 1y = .
( ) 4( ) 1 2 ( )sY s Y ss
⇒ − − =
(c) Now we rearrange this to give an expression for ( )Y s
( ) ( )4 4 42 ( ) 1 ( )
2s ss Y s Y s
s s s s+ +
⇒ − = + = ∴ =−
CHAPTER 3 : LAPLACE TRANSFORM
29
(d) Finally we take inverse transforms to obtain x
( )( ) ( )4 2
42 2
s A Bs s s s
s A s B s∴ + = − +
3B
+= +
− −
i) Let 2 6 2s B= ⇒ = ∴ =
ii) Let ( )0 4 2s A A 2= ⇒ = − ∴ = −
( )4 3( )2 2
sY s 2s s s s
+∴ = =
− −−
Therefore, taking inverse transforms
{ } ( )1 1
1
2
4( ) ( )2
3 22
( ) 3 2t
sy t Y ss s
s s
y t e
− −
−
⎧ ⎫+⎪ ⎪= = ⎨ ⎬−⎪ ⎪⎩ ⎭
⎧ ⎫= −⎨ ⎬−⎩ ⎭
∴ = −
L L
L
Example 3.9.2:
Solve the equation 2 44 2 t tdx x e e
dt− = + given that at 0, 0t x= = .
Solution:
{ }2 44 2 t tdx x e edt
⎧ ⎫− = +⎨ ⎬⎩ ⎭L L
( )0
2 1( ) 4 ( )2 4
sX s x X ss s
⇒ − − = +− − .
CHAPTER 3 : LAPLACE TRANSFORM
30
0, 0t x= =
( ) 2 1( ) 0 4 ( )2 4
sX s X ss s
⇒ − − = +− −
( )
( )( ) ( )2
2 14 ( )2 4
2 1( )2 4 4
can use partial fraction can transformdirectly fromthe table
s X ss s
X ss s s
⇒ − = +− −
∴ = +− − −
Partial Fraction
( )( )2 1
2 4 2 4 2 4A B
s s s s s s1−
= + = +− − − − − −
( )( ) ( ) ( )
( )
1 12 2
2 4 4
4 2
2 1 1 1 1( ) 2 4 2 44 4
( )1
t t t
t t
x ts s s ss s
x t e e tee t e
− −⎧ ⎫ ⎧ −⎪ ⎪ ⎪= + = + +⎨ ⎬ ⎨− − − −− −⎪ ⎪ ⎪⎩ ⎭ ⎩
⎫⎪⎬⎪⎭
∴ = − + +
= + −
L L
Exercises 3.9.1:
Solve the equation 32 10 tdx x e
dt+ = given that at 0, 6t x= = .
CHAPTER 3 : LAPLACE TRANSFORM
31
Solution of 2nd Order Differential equations
The method is, in effect, the same as before, going through the same
four distinct stages.
Example 3.9.3:
Solve the equation
23
2 3 2 2 td y dy y edt dt
− + = given that ( )0y = 5 and
( )0 7y′ = .
Solution:
(a) We rewrite the equation in terms of its transforms, remembering
that
{ } ( )y Y s=L
{ } 0( ) ( )y t sY s′ = −L y
{ } 20 0( ) ( )y t s Y s sy y′′ ′= − −L
The equation becomes
⇒( ) ( )
( )
20 0 0
2( ) 3 ( ) 2 ( )3
t y ty ty⎧ ⎫⎪ ⎪⎧ ⎫ ⎛ ⎞⎧ ⎫ ⎛ ⎞⎪ ⎪⎪ ⎪ ⎨ ⎬⎜ ⎟⎨ ⎬⎜ ⎟⎨ ⎬ ⎝ ⎠⎪ ⎪⎩ ⎭⎝ ⎠⎪ ⎪⎪ ⎪⎩ ⎭ ⎩ ⎭
′′′ LLL
s Y s sy y sY s y Y ss
′− − − − + =−
(b) Insert the initial conditions. In this case 0 5y = and . 0 7y′ =
⇒ ( )2 2( ) 5 7 3 ( ) 5 2 ( )3
s Y s s sY s Y ss
− − − − + =−
CHAPTER 3 : LAPLACE TRANSFORM
32
(c) Rearrange to obtain ( )Y s
( )
( )( )
( )( )( ) ( )( )
2 23 2 ( ) 5 7 153
2 21 2 ( ) 5 7 15 5 83 3
2 5 Y(s)3 1 2 1 2
s s Y s ss
s s Y s s ss s
ss s s s s
− + − − + =−
− − = + + − = + −− −
−= +
8− − − − −
(d) Now for partial fractions
( )( )( ) ( )( )
( ) ( ) ( ) ( ) ( ) ( ) ( )
2 5 83 1 2 1 21 1 2 3 2 4 1
3 1 2 1 2 3
ss s s s s
s s s s s s s 1
−+
− − − − −
−= + + + + = +
− − − − − − −
Therefore, taking inverse transforms
( )( )( ) ( )( ) ( ) ( )1 1
3
2 5 8 4( )3 1 2 1 2 3 1
( ) 4 t t
sy ts s s s s s s
y t e e
− −⎧ ⎫ ⎧−⎪ ⎪ ⎪= + =⎨ ⎬ ⎨− − − − − − −⎪ ⎪ ⎪⎩ ⎭ ⎩
∴ = +
L L1 ⎫⎪+ ⎬
⎪⎭
Exercises 3.9.2:
1. Solve the equation 2
2 4 24cos 2d x x tdt
− = given that at and
.
( )0x = 3
( )0 4x′ =
2. Solve the boundary value problem equation
( ) ( )3 4 cosh , 5y y t t yδ′− = − = 0.
CHAPTER 3 : LAPLACE TRANSFORM
33
REFERENCES:
1. Nagle, Saff and Snider, Fundamentals Of Differential Equations 5th Edition, Addison Wesley Longman; 2000.
2. Alan Jeffrey, Advanced Engineering Mathematics, Academic Press; 2002.
3. Abd. Wahid Md. Raji & Mohd Nor Mohamad, Persamaan Pembeza Biasa; Jabatan Matematik; UTM;Skudai 2002.
4. Glynn James et al., Advanced Modern Engineering Mathematics, Addison Wesley; 1993.
5. Boyce and di Prima, Elementary Differential Equations and Boundary Value Problems, 4th edition, John Wiley and Sons; 1986.
6. K. A. Stroud, Advanced Engineering Mathematics; MacMillan Ltd.; London; 1996.
7. Normah Maan, Halijah Osman et al., Differential Equations Module, Jabatan Matematik; UTM; Skudai 2007.
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