Higher Engineering Mathematics, Sixth Edition

Inverse Laplace transforms 597

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.

Equating s terms gives: 8= 3 B+C, from which,
C=−1, sinceB=3.

HenceL−^1

{

5 s^2 + 8 s− 1

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

}

≡L−^1

{

2

s+ 3

+

3 s− 1

s^2 + 1

}

≡L−^1

{

2

s+ 3

}

+L−^1

{

3 s

s^2 + 1

}

−L−^1

{

1

s^2 + 1

}

=2e−^3 t+3cost−sint,

from (iii), (v) and (iv) of Table 63.1

Problem 10. FindL−^1

{

7 s+ 13

s(s^2 + 4 s+ 13 )

}

7 s+ 13

s(s^2 + 4 s+ 13 )

≡

A

s

+

Bs+C

s^2 + 4 s+ 13

≡

A(s^2 + 4 s+ 13 )+(Bs+C)(s)

s(s^2 + 4 s+ 13 )

Hence 7s+ 13 ≡A(s^2 + 4 s+ 13 )+(Bs+C)(s).

Whens= 0 , 13 = 13 A, from which,A=1.

Equating s^2 terms gives: 0=A+B, from which,

B=−1.

Equatingsterms gives: 7= 4 A+C, from which,C=3.

HenceL−^1

{

7 s+ 13

s(s^2 + 4 s+ 13 )

}

≡L−^1

{

1

s

+

−s+ 3

s^2 + 4 s+ 13

}

≡L−^1

{

1

s

}

+L−^1

{

−s+ 3

(s+ 2 )^2 + 32

}

≡L−^1

{

1

s

}

+L−^1

{

−(s+ 2 )+ 5

(s+ 2 )^2 + 32

}

≡L−^1

{

1

s

}

−L−^1

{

s+ 2

(s+ 2 )^2 + 32

}

+L−^1

{

5

(s+ 2 )^2 + 32

}

≡ 1 −e−^2 tcos3t+

5

3

e−^2 tsin3t

from (i), (xiii) and (xii) of Table 63.1

Now try the following exercise

Exercise 224 Further problemson inverse

Laplace transformsusing partial fractions

Use partial fractions to find the inverse Laplace

transforms of the following functions:

1.

11 − 3 s

s^2 + 2 s− 3

[2et−5e−^3 t]

2.

2 s^2 − 9 s− 35

(s+ 1 )(s− 2 )(s+ 3 )

[4e−t−3e^2 t+e−^3 t]

3.

5 s^2 − 2 s− 19

(s+ 3 )(s− 1 )^2

[2e−^3 t+3et−4ett]

4.

3 s^2 + 16 s+ 15

(s+ 3 )^3

[e−^3 t( 3 − 2 t− 3 t^2 )]

5.

7 s^2 + 5 s+ 13

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

[

2cos

√

2 t+

3

√

2

sin

√

2 t+5e−t

]

6.

3 + 6 s+ 4 s^2 − 2 s^3

s^2 (s^2 + 3 )

[2+t+

√

3sin

√

3 t−4cos

√

3 t]