INVERSE LAPLACE TRANSFORMS 639
K
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 66.1,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 66.1, 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
}
=7 cos 2t,
from (v) of Table 66.1
(b) L−^1
{
4 s
s^2 − 16
}
= 4 L−^1
{
s
s^2 − 42
}
=4 cosh 4t,
from (x) of Table 66.1
Problem 4. Find
(a)L−^1
{
3
s^2 − 7
}
(b)L−^1
{
2
(s−3)^5
}
(a) From (ix) of Table 66.1,
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 66.1,
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=4 anda=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
}
(a)L−^1
{
3
s^2 − 4 s+ 13
}
=L−^1
{
3
(s−2)^2 + 32
}
=e^2 tsin 3t,
from (xii) of Table 66.1
(b)L−^1
{
2(s+1)
s^2 + 2 s+ 10
}
=L−^1
{
2(s+1)
(s+1)^2 + 32
}
=2e−tcos 3t,
from (xiii) of Table 66.1
Problem 6. Determine
(a)L−^1
{
5
s^2 + 2 s− 3
}
(b) L−^1
{
4 s− 3
s^2 − 4 s− 5
}