594 Higher Engineering Mathematics
Hence L
−1
1
s 2 + 9
= L
−1
1
s 2 + 3 2
=
1
3
L
−1
3
s 2 + 3 2
=
1
3
sin 3t
(b) L
−1
5
3s − 1
= L
−1
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
5
3
s −
1
3
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
=
5
3
L
−1
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
1
s −
1
3
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
=
5
3
e
1
3 t
from (iii) of Table 63.1
Problem 2. Find the following inverse Laplace
transforms:
(a) L −1
6
s 3
(b) L −1
3
s 4
(a) From (vii) of Table 63.1, L −1
2
s 3
= t 2
Hence L −1
6
s 3
= 3L −1
2
s 3
= 3t 2 .
(b) From (viii) of Table 63.1, if s is to have a power
of 4 then n = 3.
Thus L −1
3!
s 4
= t 3 i.e. L −1
6
s 4
= t 3
Hence L −1
3
s 4
=
1
2
L −1
6
s 4
=
1
2
t 3 .
Problem 3. Determine
(a) L −1
7s
s 2 + 4
(b) L −1
4s
s 2 − 16
(a) L −1
7s
s 2 + 4
= 7L −1
s
s 2 + 2 2
= 7 cos 2t,
from (v) of Table 63.1
(b) L −1
4s
s 2 − 16
= 4L −1
s
s 2 − 4 2
= 4 cosh 4t,
from (x) of Table 63.1
Problem 4. Find
(a) L −1
3
s 2 − 7
(b) L −1
2
(s − 3) 5
(a) From (ix) of Table 63.1,
L
−1
a
s 2 − a 2
= sinh at
Thus
L
−1
3
s 2 − 7
= 3L
−1
1
s 2 − (
√
7) 2
=
3
√
7
L
−1
√
7
s 2 − (
√
7) 2
=
3
√
7
sinh
√
7t
(b) From (xi) of Table 63.1,
L
−1
n!
(s − a) n+1
= e
at t
n
Thus L
−1
1
(s − a) n+1
=
1
n!
e
at t
n
and comparing with L −1
2
(s − 3) 5
shows that
n = 4 and a = 3.
Hence
L
−1
2
(s − 3) 5
= 2L
−1
1
(s − 3) 5
= 2
1
4!
e
3t t
4
=
1
12
e
3t t
4
Problem 5. Determine
(a) L −1
3
s 2 − 4s + 13
(b) L −1
2(s + 1)
s 2 + 2s + 10