Question

(with steps please) Find the inverse Laplace transform, f(t), of the function: s/((s^2+9)*(s^2+16))

Answer 1

Explanation:

Given is a Laplace transform of f(t) as

`F(s) = (s)/((s^2+9)(s^2+16))`

Let

`(s)/((s^2+9)(s^2+16))=(As+B)/((s^2+9))+(Cs+D)/((s^2+16))`

Solving we get
`\:s=s^3\left(A+C\right)+s^2\left(B+D\right)+s\left(16A+9C\right)+\left(16B+9D\right)`

`A+C=0:  B+D=0: 16A+9C=1 and 16B+9D=0`

Solving B=0=D

`A=(1)/(7) \\C=(-1)/(7)`

Hence we have

f(t) = inverse Laplace of

`(1)/(7)[ (s)/((s^2+9))-(s)/((s^2+16))]`

=
`(1)/(7) (cos 3t-cos 4t)`