Higher Engineering Mathematics, Sixth Edition

596 Higher Engineering Mathematics

9. (a)

s+ 1

s^2 + 2 s+ 10

(b)

3

s^2 + 6 s+ 13

[

(a)e−tcos3t (b)

3

2

e−^3 tsin2t

]

10. (a)

2 (s− 3 )

s^2 − 6 s+ 13

(b)

7

s^2 − 8 s+ 12

[

(a)2e^3 tcos2t (b)

7

2

e^4 tsinh2t

]

11. (a)

2 s+ 5

s^2 + 4 s− 5

(b)

3 s+ 2

s^2 − 8 s+ 25

⎡

⎢

⎣

(a)2e−^2 tcosh 3t+

1

3

e−^2 tsinh3t

(b)3e^4 tcos3t+

14

3

e^4 tsin3t

⎤

⎥

⎦

63.3 Inverse Laplace transforms using

partial fractions

Sometimes the function whose inverse is required is not

recognisable as a standard type, such as those listed in

Table 63.1. In such cases it may be possible, by using

partial fractions, to resolve the function into simpler

fractions which may be inverted on sight. For example,

the function,

F(s)=

2 s− 3

s(s− 3 )

cannot be inverted on sight from Table 63.1. However,

by using partial fractions,

2 s− 3

s(s− 3 )

≡

1

s

+

1

s− 3

which

maybeinvertedas1+e^3 tfrom(i)and(iii)of Table61.1.

Partial fractions are discussed in Chapter 2, and a sum-

maryoftheformsofpartial fractionsisgiveninTable2.1

on page 13.

Problem 7. DetermineL−^1

{

4 s− 5

s^2 −s− 2

}

4 s− 5

s^2 −s− 2

≡

4 s− 5

(s− 2 )(s+ 1 )

≡

A

(s− 2 )

+

B

(s+ 1 )

≡

A(s+ 1 )+B(s− 2 )

(s− 2 )(s+ 1 )

Hence 4s− 5 ≡A(s+ 1 )+B(s− 2 ).

Whens= 2 , 3 = 3 A,from which,A=1.

Whens=− 1 ,− 9 =− 3 B,from which,B=3.

HenceL−^1

{

4 s− 5

s^2 −s− 2

}

≡L−^1

{

1

s− 2

+

3

s+ 1

}

=L−^1

{

1

s− 2

}

+L−^1

{

3

s+ 1

}

=e^2 t+ 3 e−t,from (iii) of Table 63.1

Problem 8. FindL−^1

{

3 s^3 +s^2 + 12 s+ 2

(s− 3 )(s+ 1 )^3

}

3 s^3 +s^2 + 12 s+ 2

(s− 3 )(s+ 1 )^3

≡

A

s− 3

+

B

s+ 1

+

C

(s+ 1 )^2

+

D

(s+ 1 )^3

≡

(

A(s+ 1 )^3 +B(s− 3 )(s+ 1 )^2

+C(s− 3 )(s+ 1 )+D(s− 3 )

)

(s− 3 )(s+ 1 )^3

Hence

3 s^3 +s^2 + 12 s+ 2 ≡A(s+ 1 )^3 +B(s− 3 )(s+ 1 )^2

+C(s− 3 )(s+ 1 )+D(s− 3 )

Whens= 3 , 128 = 64 A, from which,A=2.

Whens=− 1 ,− 12 =− 4 D, from which,D=3.

Equatings^3 terms gives: 3=A+B, from which,B=1.

Equating constant terms gives:

2 =A− 3 B− 3 C− 3 D,

i.e. 2 = 2 − 3 − 3 C− 9 ,

from which, 3C=−12 andC=− 4

Hence

L−^1

{

3 s^3 +s^2 + 12 s+ 2

(s− 3 )(s+ 1 )^3

}

≡L−^1

{

2

s− 3

+

1

s+ 1

−

4

(s+ 1 )^2

+

3

(s+ 1 )^3

}

=2e^3 t+e−t−4e−tt+

3

2

e−tt^2 ,

from (iii) and (xi) of Table 63.1