634 LAPLACE TRANSFORMS
Now try the following exercise.
Exercise 232 Further problems on Laplace
transforms of the form eatf(t)Determine the Laplace transforms of the follow-
ing functions:- (a) 2te^2 t(b)t^2 et
[
(a)
2
(s−2)^2(b)2
(s−1)^3]- (a) 4t^3 e−^2 t(b)
1
2t^4 e−^3 t[
(a)24
(s+2)^4(b)12
(s+3)^5]- (a) etcost(b) 3e^2 tsin 2t
[
(a)
s− 1
s^2 − 2 s+ 2(b)6
s^2 − 4 s+ 8]- (a) 5e−^2 tcos 3t(b) 4e−^5 tsint
[
(a)
5(s+2)
s^2 + 4 s+ 13(b)4
s^2 + 10 s+ 26]- (a) 2etsin^2 t(b)
1
2e^3 tcos^2 t⎡⎢
⎢
⎣(a)1
s− 1−s− 1
s^2 − 2 s+ 5(b)1
4(
1
s− 3+s− 3
s^2 − 6 s+ 13)⎤⎥
⎥
⎦- (a) etsinht(b) 3e^2 tcosh 4t
[
(a)
1
s(s−2)(b)3(s−2)
s^2 − 4 s− 12]- (a) 2e−tsinh 3t(b)
1
4e−^3 tcosh 2t[
(a)6
s^2 + 2 s− 8(b)s+ 3
4(s^2 + 6 s+5)]- (a) 2et( cos 3t−3 sin 3t)
(b) 3e−^2 t( sinh 2t−2 cosh 2t)
[
(a)2(s−10)
s^2 − 2 s+ 10(b)−6(s+1)
s(s+4)]65.3 The Laplace transforms of
derivatives(a) First derivativeLet the first derivative off(t)bef′(t) then, from
equation (1),L{f′(t)}=∫∞0e−stf′(t)dtFrom Chapter 43, when integrating by parts
∫
udv
dtdt=uv−∫
vdu
dtdtWhen evaluating∫∞
0 e−stf′(t)dt,letu=e−standdv
dt=f′(t)from which,
du
dt=−se−standv=∫
f′(t)dt=f(t)Hence∫∞0e−stf′(t)dt=[
e−stf(t)]∞
0 −∫∞0f(t)(−se−st)dt=[0−f(0)]+s∫∞0e−stf(t)dt=−f(0)+sL{f(t)}assuming e−stf(t)→0ast→∞, andf(0) is the
value off(t)att=0. Hence,L{f′(t)}=sL{f(t)}−f(0)or L{
dy
dx}
=sL{y}−y(0)⎫
⎬⎭(3)wherey(0) is the value ofyatx=0.(b) Second derivativeLet the second derivative off(t)bef′′(t), then from
equation (1),L{f′′(t)}=∫∞0e−stf′′(t)dtIntegrating by parts gives:
∫∞0e−stf′′(t)dt=[
e−stf′(t)]∞
0 +s∫∞0e−stf′(t)dt=[0−f′(0)]+sL{f′(t)}