Higher Engineering Mathematics, Sixth Edition

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)

⎫

⎪⎪

⎪⎪

⎪⎪

⎬

⎪⎪

⎪⎪

⎪⎪

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(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

1. Derive the Laplace transform of the first
   derivative from the definition of a Laplace
   transform. Hence derive the transform

L{ 1 }=

1

s