Question

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

Answer 1

Answer:

Step-by-step explanation:

Given is a Laplace transform of f(t) as

$F(s) = \frac{s}{(s^2+9)(s^2+16)}$

Let

$\frac{s}{(s^2+9)(s^2+16)}=\frac{As+B}{(s^2+9)}+\frac{Cs+D}{(s^2+16)}$

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

$A=\frac{1}{7} \\C=\frac{-1}{7}$

Hence we have

f(t) = inverse Laplace of

$\frac{1}{7}[ \frac{s}{(s^2+9)}-\frac{s}{(s^2+16)}]$

=$\frac{1}{7} (cos 3t-cos 4t)$