Answer:
Step-by-step explanation:
Given is a Laplace transform of f(t) as
[tex]F(s) = \frac{s}{(s^2+9)(s^2+16)}[/tex]
Let
[tex]\frac{s}{(s^2+9)(s^2+16)}=\frac{As+B}{(s^2+9)}+\frac{Cs+D}{(s^2+16)}[/tex]
Solving we get [tex]\:s=s^3\left(A+C\right)+s^2\left(B+D\right)+s\left(16A+9C\right)+\left(16B+9D\right)[/tex]
[tex]A+C=0: B+D=0: 16A+9C=1 and 16B+9D=0[/tex]
Solving B=0=D
[tex]A=\frac{1}{7} \\C=\frac{-1}{7}[/tex]
Hence we have
f(t) = inverse Laplace of
[tex]\frac{1}{7}[ \frac{s}{(s^2+9)}-\frac{s}{(s^2+16)}][/tex]
=[tex]\frac{1}{7} (cos 3t-cos 4t)[/tex]
