Higher Engineering Mathematics

(Greg DeLong) #1
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

}
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