590 Higher Engineering Mathematics
(b) Second derivative
Let the second derivative off(t)bef′′(t), then from
equation (1),
L{f′′(t)}=
∫∞
0
e−stf′′(t)dt
Integrating by parts gives:
∫∞
0
e−stf′′(t)dt=
[
e−stf′(t)
]∞
0 +s
∫∞
0
e−stf′(t)dt
=[0−f′( 0 )]+sL{f′(t)}
assuming e−stf′(t)→0ast→∞,andf′( 0 )is the
value off′(t)att=0. Hence
{f′′(t)}=−f′( 0 )+s[s(f(t))−f( 0 )], from equa-
tion (3),
i.e.
L{f′′(t)}
=s^2 L{f(t)}−sf(0)−f′(0)
or L
{
d^2 y
dx^2
}
=s^2 L{y}−sy(0)−y′(0)
⎫
⎪⎪
⎪⎪
⎪⎪
⎬
⎪⎪
⎪⎪
⎪⎪
⎭
(4)
wherey′( 0 )is the value of
dy
dx
atx=0.
Equations (3) and (4) are important and are used in
the solution of differential equations (see Chapter 64)
and simultaneous differential equations (Chapter 65).
Problem 5. Use the Laplace transform of the first
derivative to derive:
(a)L{k}=
k
s
(b)L{ 2 t}=
2
s^2
(c)L{e−at}=
1
s+a
From equation (3),L{f′(t)}=sL{f(t)}−f( 0 ).
(a) Letf(t)=k,thenf′(t)=0andf( 0 )=k.
Substituting into equation (3) gives:
L{ 0 }=sL{k}−k
i.e. k=sL{k}
Hence L{k}=
k
s
(b) Let f(t)= 2 tthenf′(t)=2andf( 0 )=0.
Substituting into equation (3) gives:
L{ 2 }=sL{ 2 t}− 0
i.e.
2
s
=sL{ 2 t}
Hence L{ 2 t}=
2
s^2
(c) Letf(t)=e−atthenf′(t)=−ae−atandf( 0 )=1.
Substituting into equation (3) gives:
L{−ae−at}=sL{e−at}− 1
−aL{e−at}=sL{e−at}− 1
1 =sL{e−at}+aL{e−at}
1 =(s+a)L{e−at}
HenceL{e−at}=
1
s+a
Problem 6. Use the Laplace transform of the
second derivative to derive
L{cosat}=
s
s^2 +a^2
From equation (4),
L{f′′(t)}=s^2 L{f(t)}−sf( 0 )−f′( 0 )
Letf(t)=cosat,thenf′(t)=−asinatand
f′′(t)=−a^2 cosat,f( 0 )=1andf′( 0 )= 0
Substituting into equation (4) gives:
L{−a^2 cosat}=s^2 {cosat}−s( 1 )− 0
i.e. −a^2 L{cosat}=s^2 L{cosat}−s
Hence s=(s^2 +a^2 )L{cosat}
from which, L{cosat}=
s
s^2 +a^2
Now try the following exercise
Exercise 221 Further problems on the
Laplace transformsof derivatives
- Derive the Laplace transform of the first
derivative from the definition of a Laplace
transform. Hence derive the transform
L{ 1 }=
1
s