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]