594 Higher Engineering Mathematics

HenceL−^1

{

1

s^2 + 9

}

=L−^1

{

1

s^2 + 32

}

=

1

3

L−^1

{

3

s^2 + 32

}

=

1

3

sin3t

(b)L−^1

{

5

3 s− 1

}

=L−^1

⎧

⎪⎪

⎨

⎪⎪

⎩

5

3

(

s−

1

3

)

⎫

⎪⎪

⎬

⎪⎪

⎭

=

5

3

L−^1

⎧

⎪⎪

⎨

⎪⎪

⎩

1

(

s−

1

3

)

⎫

⎪⎪

⎬

⎪⎪

⎭

=

5

3

e

(^13) t
from (iii) of Table 63.1
Problem 2. Find the following inverse Laplace
transforms:
(a)L−^1
{
6
s^3
}
(b)L−^1
{
3
s^4
}
(a) From (vii) of Table 63.1,L−^1
{
2
s^3
}
=t^2
HenceL−^1
{
6
s^3
}
= 3 L−^1
{
2
s^3
}
= 3 t^2.
(b) From (viii) of Table 63.1, ifsis to have a power
of 4 thenn=3.
Thus L−^1
{
3!
s^4
}
=t^3 i.e.L−^1
{
6
s^4
}
=t^3
Hence L−^1
{
3
s^4
}

1
2
L−^1
{
6
s^4
}

1
2
t^3.
Problem 3. Determine
(a)L−^1
{
7 s
s^2 + 4
}
(b)L−^1
{
4 s
s^2 − 16
}
(a) L−^1
{
7 s
s^2 + 4
}
= 7 L−^1
{
s
s^2 + 22
}
=7cos2t,
from (v) of Table 63.1
(b)L−^1
{
4 s
s^2 − 16
}
= 4 L−^1
{
s
s^2 − 42
}
=4cosh4t,
from (x) of Table 63.1
Problem 4. Find
(a)L−^1
{
3
s^2 − 7
}
(b)L−^1
{
2
(s− 3 )^5
}
(a) From (ix) of Table 63.1,
L−^1
{
a
s^2 −a^2
}
=sinhat
Thus
L−^1
{
3
s^2 − 7
}
= 3 L−^1
{
1
s^2 −(
√
7 )^2
}

3
√
7
L−^1
{ √
7
s^2 −(
√
7 )^2
}

3
√
7
sinh
√
7 t
(b) From (xi) of Table 63.1,
L−^1
{
n!
(s−a)n+^1
}
=eattn
Thus L−^1
{
1
(s−a)n+^1
}

1
n!
eattn
and comparing withL−^1
{
2
(s− 3 )^5
}
shows that
n=4anda=3.
Hence
L−^1
{
2
(s− 3 )^5
}
= 2 L−^1
{
1
(s− 3 )^5
}
= 2
(
1
4!
e^3 tt^4
)

1
12
e^3 tt^4
Problem 5. Determine
(a)L−^1
{
3
s^2 − 4 s+ 13
}
(b)L−^1
{
2 (s+ 1 )
s^2 + 2 s+ 10
}