Higher Engineering Mathematics

646 LAPLACE TRANSFORMS

Hencey=L−^1

{

8 s+ 38

2 s^2 + 5 s− 3

}

=L−^1

{

12

2 s− 1

−

2

s+ 3

}

=L−^1

{

12

2

(

s−^12

)

}

−L−^1

{

2

s+ 3

}

Hencey=6e

1

2 x−2e−^3 x, from (iii) of

Table 66.1.

Problem 2. Use Laplace transforms to solve

the differential equation:

d^2 y

dx^2

+ 6

dy

dx

+ 13 y=0, given that when

x=0,y=3 and

dy

dx

=7.

This is the same as Problem 3 of Chapter 50,
page 477. Using the above procedure:

(i)L

{

d^2 x

dy^2

}

+ 6 L

{

dy

dx

}

+ 13 L{y}=L{ 0 }

Hence [s^2 L{y}−sy(0)−y′(0)]

+6[sL{y}−y(0)]+ 13 L{y}=0,

from equations (3) and (4) of Chapter 65.

(ii)y(0)=3 andy′(0)= 7

Thus s^2 L{y}− 3 s− 7 + 6 sL{y}

− 18 + 13 L{y}= 0

(iii) Rearranging gives:

(s^2 + 6 s+13)L{y}= 3 s+ 25

i.e. L{y}=

3 s+ 25

s^2 + 6 s+ 13

(iv)y=L−^1

{

3 s+ 25

s^2 + 6 s+ 13

}

=L−^1

{

3 s+ 25

(s+3)^2 + 22

}

=L−^1

{

3(s+3)+ 16

(s+3)^2 + 22

}

=L−^1

{

3(s+3)

(s+3)^2 + 22

}

+L−^1

{

8(2)

(s+3)^2 + 22

}

=3e−^3 tcos 2t+8e−^3 tsin 2t, from (xiii)

and (xii) of Table 66.1

Hencey=e−^3 t(3 cos 2t+8 sin 2t)

Problem 3. Use Laplace transforms to solve

the differential equation:

d^2 y

dx^2

− 3

dy

dx

=9, given that whenx=0,y= 0

and

dy

dx

=0.

This is the same problem as Problem 2 of Chapter 51,

page 482. Using the procedure:

(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)=0 andy′(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.