Higher Engineering Mathematics

628 LAPLACE TRANSFORMS

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 421),

=

[

e−s(∞)

s^2 +a^2

(asina(∞)−scosa(∞))

−

e^0

s^2 +a^2

(asin 0−scos 0)

]

=

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 64.1.

Table 64.1 Elementary standard Laplace transforms

Function Laplace transforms

f(t) L{f(t)}=

∫∞

0 e

−stf(t)dt

(i) 1

1

s

(ii) k

k

s

(iii) eat

1

s−a

(iv) sinat

a

s^2 +a^2

(v) cosat

s

s^2 +a^2

(vi) t

1

s^2

(vii) t^2

2!

s^3

(viii) tn(n=1, 2, 3,...)

n!

sn+^1

(ix) coshat

s

s^2 −a^2

(x) sinhat

a

s^2 −a^2