Higher Engineering Mathematics

INVERSE LAPLACE TRANSFORMS 641

K

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 66.1. How-
ever, by using partial fractions,
2 s− 3
s(s−3)

≡

1

s

+

1

s− 3

which may be inverted as

1 +e^3 tfrom (i) and (iii) of Table 64.1.
Partial fractions are discussed in Chapter 3, and a
summary of the forms of partial fractions is given in
Table 3.1 on page 18.

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 66.1

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 66.1

Problem 9. Determine

L−^1

{

5 s^2 + 8 s− 1

(s+3)(s^2 +1)

}

5 s^2 + 8 s− 1

(s+3)(s^2 +1)

≡

A

s+ 3

+

Bs+C

(s^2 +1)

≡

A(s^2 +1)+(Bs+C)(s+3)

(s+3)(s^2 +1)

Hence 5s^2 + 8 s− 1 ≡A(s^2 +1)+(Bs+C)(s+3).

Whens=−3, 20= 10 A, from which,A=2.

Equatings^2 terms gives: 5=A+B, from which,

B=3, sinceA=2.

Equatingsterms gives: 8= 3 B+C, from which,

C=−1, sinceB=3.