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

21.5 Laplace transform of derivatives and integrals 635
EXERCISES21.4
1 Find the Laplace transformsofthe following
functions:
(a)3t 2 −4
(c)2−t 2 +2t 4
( )
(e) 1 3 sin3t −4cos t
2
(f) 3t 4 e 5t +t
(g)sinh2t +3cosh2t
(h)e −t sin3t +4e −t cos3t
(b) 2sin4t+11−t
(d) 3e 2t +4sint
2 TheLaplacetransform of f (t)is given as
F(s) = 3s2 −1
s 2 +s+1
Find the Laplace transformof
(a)e −t f (t) (b) e 3t f (t) (c) e −t/2 f (t)
3 Given
find
L{f(t)} =
4s
s 2 +1
(a) L{u(t −1)f(t −1)}
(b) L{3u(t −2)f(t −2)}
{
}
(c) L u(t−4) f(t−4)
2
4 Find the final value ofthe followingfunctionsusing
the final value theorem:
(a) f(t) =e −t sint
(b) f(t)=e −t +1
(c) f(t) =e −3t cost +5
Solutions
1 (a)
(c)
(e)
(g)
6
s 3 − 4 s
2
s − 2 s 3 + 48
s 5
1
s 2 +9 − 16s
4s 2 +1
3s+2
s 2 −4
(b)
(d)
(f)
(h)
8
s 2 +16 + 11
s − 1 s 2
3
s −2 + 4
s 2 +1
72
(s −5) 5 + 1 s 2
4s+7
(s+1) 2 +9
2 (a)
(c)
3 (a)
3s 2 +6s+2
s 2 +3s+3
12s 2 +12s −1
4s 2 +8s+7
4se −s
s 2 (b)
+1
(b)
12se −2s
s 2 +1
4 (a)0 (b)1 (c)5
3s 2 −18s +26
s 2 −5s+7
(c)
2se −4s
s 2 +1
21.5 LAPLACETRANSFORMOFDERIVATIVES
ANDINTEGRALS
In later sections we shall use the Laplace transform to solve differential equations. In
order to do this we need to be able to find the Laplace transform of derivatives of functions.Let
f (t)beafunctionoft,and f ′ and f ′′ thefirstandsecondderivativesof f.The
Laplace transform of f (t)isF(s). Then
L{f ′ } =sF(s) − f(0)
L{f ′′ } =s 2 F(s)−sf(0)−f ′ (0)
where f(0) and f ′ (0) are the initial values of f and f ′ . The general case for the
Laplace transform ofannthderivative is
L{f (n) } =s n F(s) −s n−1 f(0) −s n−2 f ′ (0)−···−f (n−1) (0)