Higher Engineering Mathematics, Sixth Edition

Introduction to Laplace transforms 585

=

1

s^2 +a^2

[( 0 )− 1 ( 0 −a)]

=

a

s^2 +a^2

(provideds> 0 )

(b) From equation (1),

L{t^2 }=

∫∞

0

e−stt^2 dt

=

[

t^2 e−st

−s

−

2 te−st

s^2

−

2e−st

s^3

]∞

0

by integration by parts twice,

=

[

( 0 − 0 − 0 )−

(

0 − 0 −

2

s^3

)]

=

2

s^3

(provideds> 0 )

(c) From equation (1),

L{coshat}=L

{

1

2

(eat+e−at)

}

,

from Chapter 5

=

1

2

L{eat}+

1

2

L{e−at},

equations (2) and (3)

=

1

2

(

1

s−a

)

+

1

2

(

1

s−(−a)

)

from (iii) of Table 61.1

=

1

2

[

1

s−a

+

1

s+a

]

=

1

2

[

(s+a)+(s−a)

(s−a)(s+a)

]

=

s

s^2 −a^2

(provideds>a)

Problem 4. Determine the Laplace transforms of:

(a) sin^2 t (b) cosh^23 x.

(a) Since cos2t= 1 −2sin^2 tthen

sin^2 t=

1

2

( 1 −cos2t). Hence,

L{sin^2 t}=L

{

1

2

( 1 −cos2t)

}

=

1

2

L{ 1 }−

1

2

L{cos2t}

=

1

2

(

1

s

)

−

1

2

(

s

s^2 + 22

)

from (i) and (v) of Table 61.1

=

(s^2 + 4 )−s^2

2 s(s^2 + 4 )

=

4

2 s(s^2 + 4 )

=

2

s(s^2 +4)

(b) Since cosh 2x=2cosh^2 x−1then

cosh^2 x=

1

2

( 1 +cosh2x)from Chapter 5.

Hence cosh^23 x=

1

2

( 1 +cosh 6x)

ThusL{cosh^23 x}=L

{

1

2

( 1 +cosh 6x)

}

=

1

2

L{ 1 }+

1

2

L{cosh6x}

=

1

2

(

1

s

)

+

1

2

(

s

s^2 − 62

)

=

2 s^2 − 36

2 s(s^2 − 36 )

=

s^2 − 18

s(s^2 −36)

Problem 5. Find the Laplace transform of

3sin(ωt+α),whereωandαare constants.

Using the compound angle formula for sin(A+B),

from Chapter 17, sin(ωt+α)may be expanded to

(sinωtcosα+cosωtsinα). Hence,

L{3sin(ωt+α)}

=L{ 3 (sinωtcosα+cosωtsinα)}

=3cosαL{sinωt}+3sinαL{cosωt},

sinceαis a constant

=3cosα

(

ω

s^2 +ω^2

)

+3sinα

(

s

s^2 +ω^2

)

from (iv) and (v) of Table 61.1

=

3

(s^2 +ω^2 )

(ωcosα+ssinα)

Now try the following exercise

Exercise 219 Further problems on an

introduction to Laplace transforms

Determine the Laplace transforms in Problems

1to9.