082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

638 Chapter 21 The Laplace transform
Solutions
1 (a) sY−3
(b) s 2 Y−3s−1
(c) 3s 2 Y −sY −9s
(d) (s 2 +2s+3)Y−3s−7
(e) (3s 2 −s +2)Y −9s
(f) (−4s 2 +5s −3)Y +12s −11
(g) (3s 2 +6s +8)Y −9s −21
(h) (4s 2 −8s +6)Y −12s +20
2 (a) (3s−2)F−6
(b) (3s 2 −s+1)F−6s−7
(c) s 3 F−2s 2 −3s+1
(d) (2s 3 −s 2 +4s−2)F−4s 2 −4s−3
1
3 (b)
(s +1) 2
21.6 INVERSELAPLACETRANSFORMS
As mentioned in Section 21.5, the Laplace transform can be used to solve differential
equations. However, before such an application can be put into practice, we must study
theinverseLaplacetransform.Sofarinthischapterwehavebeengivenfunctionsoftand
foundtheirLaplacetransforms.Wenowconsidertheproblemoffindingafunction f (t),
havingbeengiventheLaplacetransform,F(s).ClearlyTable21.1andthepropertiesof
Laplace transformswill helpus todothis. If L{f (t)} =F(s)wewrite
f(t) = L −1 {F(s)}
and call L −1 the inverse Laplace transform. Like L, L −1 can be shown to be a linear
operator.
Example21.11 Findthe inverse Laplace transformsofthe following:
(a)
2
s 3
(b)
16
s 3
(c)
s
s 2 +1
(d)
1
s 2 +1
(e)
s +1
s 2 +1
Solution (a) We need to find a function of t which has a Laplace transform of
Table 21.1 wesee L −1 { 2
s 3 }
=t 2 .
{ } { } 16 2
(b) L −1 = 8L −1 = 8t 2
s 3 s 3
{ } s
(c) L −1 =cost
s 2 +1
{ } 1
(d) L −1 =sint
s 2 +1
{ } s +1
(e) L −1 s 2 +1
{ } { } s 1
= L −1 + L −1 =cost+sint
s 2 +1 s 2 +1
2
s3. Using
In parts (a), (c) and (d) we obtained the inverse Laplace transform by referring directly
to the table. In (b) and (e) we used the linearity properties of the inverse transform, and
then referredtoTable 21.1.