Inverse Laplace transforms 595
(a)L−^1
{
3
s^2 − 4 s+ 13}
=L−^1{
3
(s− 2 )^2 + 32}=e^2 tsin3t,
from (xii) of Table 63.1(b)L−^1
{
2 (s+ 1 )
s^2 + 2 s+ 10}
=L−^1{
2 (s+ 1 )
(s+ 1 )^2 + 32}=2e−tcos3t,
from (xiii) of Table 63.1Problem 6. Determine(a)L−^1{
5
s^2 + 2 s− 3}(b)L−^1{
4 s− 3
s^2 − 4 s− 5}(a)L−^1
{
5
s^2 + 2 s− 3}
=L−^1{
5
(s+ 1 )^2 − 22}=L−^1⎧
⎪⎨⎪⎩5
2( 2 )
(s+ 1 )^2 − 22⎫
⎪⎬⎪⎭=5
2e−tsinh2t,from (xiv) of Table 63.1(b) L−^1
{
4 s− 3
s^2 − 4 s− 5}
=L−^1{
4 s− 3
(s− 2 )^2 − 32}=L−^1{
4 (s− 2 )+ 5
(s− 2 )^2 − 32}=L−^1{
4 (s− 2 )
(s− 2 )^2 − 32}+L−^1{
5
(s− 2 )^2 − 32}=4e^2 tcosh3t+L−^1⎧
⎪⎨⎪⎩5
3( 3 )(s− 2 )^2 − 32⎫
⎪⎬⎪⎭from (xv) of Table 63.1=4e^2 tcosh 3t+5
3e^2 tsinh3t,from (xiv) of Table 63.1Now try the following exerciseExercise 223 Further problemson inverse
Laplace transformsof simple functions
Determine the inverse Laplace transforms of the
following:- (a)
7
s(b)2
s− 5[(a) 7 (b) 2e^5 t]- (a)
3
2 s+ 1(b)2 s
s^2 + 4
[
(a)3
2e−1
2 t (b)2cos2t]- (a)
1
s^2 + 25(b)4
s^2 + 9
[
(a)1
5sin5t (b)4
3sin3t]- (a)
5 s
2 s^2 + 18(b)6
s^2
[
(a)5
2cos3t (b) 6 t]- (a)
5
s^3(b)8
s^4[
(a)5
2t^2 (b)4
3t^3]- (a)
3 s
1
2s^2 − 8(b)7
s^2 − 16
[
(a)6cosh4t (b)7
4sinh4t]- (a)
15
3 s^2 − 27(b)4
(s− 1 )^3
[
(a)5
3sinh3t (b)2ett^2]- (a)
1
(s+ 2 )^4(b)3
(s− 3 )^5
[
(a)1
6e−^2 tt^3 (b)1
8e^3 tt^4]