Inverse Laplace transforms 595
(a) L
−1
3
s 2 − 4s + 13
= L
−1
3
(s − 2) 2 + 3 2
= e
2t sin 3t,
from (xii) of Table 63.1
(b) L
−1
2(s + 1)
s 2 + 2s + 10
= L
−1
2(s + 1)
(s + 1) 2 + 3 2
= 2e
−t cos 3t,
from (xiii) of Table 63.1
Problem 6. Determine
(a) L −1
5
s 2 + 2s − 3
(b) L −1
4s − 3
s 2 − 4s − 5
(a) L
−1
5
s 2 + 2s − 3
= L
−1
5
(s + 1) 2 − 2 2
= L
−1
⎧
⎪ ⎨
⎪ ⎩
5
2
(2)
(s + 1) 2 − 2 2
⎫
⎪ ⎬
⎪ ⎭
=
5
2
e
−t sinh 2t,
from (xiv) of Table 63.1
(b) L −1
4s − 3
s 2 − 4s − 5
= L −1
4s − 3
(s − 2) 2 − 3 2
= L
−1
4(s − 2) + 5
(s − 2) 2 − 3 2
= L
−1
4(s − 2)
(s − 2) 2 − 3 2
+ L
−1
5
(s − 2) 2 − 3 2
= 4e
2t cosh 3t + L
−1
⎧
⎪ ⎨
⎪ ⎩
5
3
(3)
(s − 2) 2 − 3 2
⎫
⎪ ⎬
⎪ ⎭
from (xv) of Table 63.1
= 4e
2t cosh 3t +
5
3
e
2t sinh 3t,
from (xiv) of Table 63.1
Now try the following exercise
Exercise 223 Further problems on inverse
Laplace transforms of simple functions
Determine the inverse Laplace transforms of the
following:
1. (a)
7
s
(b)
2
s − 5
[(a) 7 (b) 2e 5t ]
2. (a)
3
2s + 1
(b)
2s
s 2 + 4
(a)
3
2
e
−
1
2 t
(b) 2 cos2t
3. (a)
1
s 2 + 25
(b)
4
s 2 + 9
(a)
1
5
sin 5t (b)
4
3
sin 3t
4. (a)
5s
2s 2 + 18
(b)
6
s 2
(a)
5
2
cos 3t (b) 6t
5. (a)
5
s 3 (b)
8
s 4
(a)
5
2
t 2 (b)
4
3
t 3
6. (a)
3s
1
2
s 2 − 8
(b)
7
s 2 − 16
(a) 6 cosh 4t (b)
7
4
sinh 4t
7. (a)
15
3s 2 − 27
(b)
4
(s − 1) 3
(a)
5
3
sinh 3t (b) 2 e t t 2
8. (a)
1
(s + 2) 4 (b)
3
(s − 3) 5
(a)
1
6
e −2t t 3 (b)
1
8
e 3t t 4