Higher Engineering Mathematics, Sixth Edition

Introduction to Laplace transforms 583

(a) f(t)= 1. From equation (1),

L{ 1 }=

∫∞

0

e−st( 1 )dt=

[

e−st

−s

]∞

0

=−

1

s

[e−s(∞)−e^0 ]=−

1

s

[0−1]

=

1

s

(provideds> 0 )

(b) f(t)=k. From equation (2),

L{k}=kL{ 1 }

HenceL{k}=k

(

1

s

)

=

k

s

,from (a) above.

(c) f(t)=eat(whereais a real constant=0).
From equation (1),

L{eat}=

∫∞

0

e−st(eat)dt=

∫∞

0

e−(s−a)tdt,

from the laws of indices,

=

[

e−(s−a)t

−(s−a)

]∞

0

=

1

−(s−a)

( 0 − 1 )

=

1

s−a

(provided(s−a)> 0 ,i.e.s>a)

(d) f(t)=cosat(where a is a real constant).

From equation (1),

L{cosat}=

∫∞

0

e−stcosatdt

=

[

e−st

s^2 +a^2

(asinat−scosat)

]∞

0

by integration by parts twice (see page 423),

=

[

e−s(∞)

s^2 +a^2

(asina(∞)−scosa(∞))

−

e^0

s^2 +a^2

(asin0−scos0)

]

=

s

s^2 +a^2

(provideds> 0 )

(e) f(t)=t. From equation (1),

L{t}=

∫∞

0

e−sttdt=

[

te−st

−s

−

∫

e−st

−s

dt

]∞

0

=

[

te−st

−s

−

e−st

s^2

]∞

0

by integration by parts,

=

[

∞e−s(∞)

−s

−

e−s(∞)

s^2

]

−

[

0 −

e^0

s^2

]

=( 0 − 0 )−

(

0 −

1

s^2

)

since(∞× 0 )= 0 ,

=

1

s^2

(provideds> 0 )

(f) f(t)=tn(wheren=0, 1, 2, 3, ...).

By a similar method to (e) it may be shown

thatL{t^2 }=

2

s^3

andL{t^3 }=

( 3 )( 2 )

s^4

=

3!

s^4

.These

results can be extended tonbeing any positive

integer.

ThusL{tn}=

n!

sn+^1

provideds> 0 )

(g) f(t)=sinhat. From Chapter 5,

sinhat=

1

2

(eat−e−at). Hence,

L{sinhat}=L

{

1

2

eat−

1

2

e−at

}

=

1

2

L{eat}−

1

2

L{e−at}

from equations (2) and (3),

=

1

2

[

1

s−a

]

−

1

2

[

1

s+a

]

from (c) above,

=

1

2

[

1

s−a

−

1

s+a

]

=

a

s^2 −a^2

(provideds>a)

A list of elementary standard Laplace transforms are

summarized in Table 61.1.

61.5 Worked problems on standard

Laplace transforms

Problem 1. Using a standard list of Laplace

transforms determine the following:

(a)L

{

1 + 2 t−

1

3

t^4

}

(b)L{5e^2 t−3e−t}.