Higher Engineering Mathematics

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

}