Higher Engineering Mathematics, Sixth Edition

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 )