602 Higher Engineering Mathematics
(i) L
{
d^2 y
dx^2
}
− 3 L
{
dy
dx
}
=L{ 9 }
Hence [s^2 L{y}−sy( 0 )−y′( 0 )]
−3[sL{y}−y( 0 )]=
9
s
(ii) y( 0 )=0andy′( 0 )= 0
Hences^2 L{y}− 3 sL{y}=
9
s
(iii) Rearranging gives:
(s^2 − 3 s)L{y}=
9
s
i.e. L{y}=
9
s(s^2 − 3 s)
=
9
s^2 (s− 3 )
(iv) y=L−^1
{
9
s^2 (s− 3 )
}
9
s^2 (s− 3 )
≡
A
s
+
B
s^2
+
C
s− 3
≡
A(s)(s− 3 )+B(s− 3 )+Cs^2
s^2 (s− 3 )
Hence 9≡A(s)(s− 3 )+B(s− 3 )+Cs^2.
Whens= 0 , 9 =− 3 B, from which,B=−3.
Whens= 3 , 9 = 9 C, from which,C=1.
Equatings^2 terms gives: 0=A+C, from which,
A=−1, sinceC=1. Hence,
L−^1
{
9
s^2 (s− 3 )
}
=L−^1
{
−
1
s
−
3
s^2
+
1
s− 3
}
=− 1 − 3 x+e^3 x,from (i),
(vi) and (iii) of Table 63.1.
i.e.y=e^3 x− 3 x− 1
Problem 4. Use Laplace transforms to solve the
differential equation:
d^2 y
dx^2
− 7
dy
dx
+ 10 y=e^2 x+20, given that when
x=0,y=0and
dy
dx
=−
1
3
Using the procedure:
(i) L
{
d^2 y
dx^2
}
− 7 L
{
dy
dx
}
+ 10 L{y}=L{e^2 x+ 20 }
Hence [s^2 L{y}−sy( 0 )−y′( 0 )]−7[sL{y}
−y( 0 )]+ 10 L{y}=
1
s− 2
+
20
s
(ii) y( 0 )=0andy′( 0 )=−
1
3
Hence s^2 L{y}− 0 −
(
−
1
3
)
− 7 sL{y}+ 0
+ 10 L{y}=
21 s− 40
s(s− 2 )
(iii) (s^2 − 7 s+ 10 )L{y}=
21 s− 40
s(s− 2 )
−
1
3
=
3 ( 21 s− 40 )−s(s− 2 )
3 s(s− 2 )
=
−s^2 + 65 s− 120
3 s(s− 2 )
Hence L{y}=
−s^2 + 65 s− 120
3 s(s− 2 )(s^2 − 7 s+ 10 )
=
1
3
[
−s^2 + 65 s− 120
s(s− 2 )(s− 2 )(s− 5 )
]
=
1
3
[
−s^2 + 65 s− 120
s(s− 5 )(s− 2 )^2
]
(iv) y=
1
3
L−^1
{
−s^2 + 65 s− 120
s(s− 5 )(s− 2 )^2
}
−s^2 + 65 s− 120
s(s− 5 )(s− 2 )^2
≡
A
s
+
B
s− 5
+
C
s− 2
+
D
(s− 2 )^2
≡
(
A(s− 5 )(s− 2 )^2 +B(s)(s− 2 )^2
+C(s)(s− 5 )(s− 2 )+D(s)(s− 5 )
)
s(s− 5 )(s− 2 )^2
Hence
−s^2 + 65 s− 120
≡A(s− 5 )(s− 2 )^2 +B(s)(s− 2 )^2
+C(s)(s− 5 )(s− 2 )+D(s)(s− 5 )