Inverse Laplace transform of $\frac{40.5}{s(s^2-9)}$ using convolution theorem

Question

Find the inverse Laplace transform of $$\frac{40.5}{s(s^2-9)}$$ using the convolution theorem.

I see how you can solve this using partial fractions, but apparently it's supposed to be easier if you use convolution. I don't see how to apply convolution here because the inverse of $1/s$ is just $1$ which doesn't fit into the integral when I try to do it.

Answer 1

All you need to know is that the ILT of $1/(s^2-9)$ is $(1/3) \sinh{(3 t)}$. The convolution integral...doesn't matter which is a function of $t'$ or $t-t'$. Thus, the ILT is

$$\frac{40.5}{3} \int_0^t dt' \sinh{(3 t')} $$

which I am sure you can handle.

Answer 2

With ${\cal L}(1)=\dfrac{1}{s}$ and ${\cal L}(\sinh3t)=\dfrac{3}{s^2-9}$ then $${\cal L}^{-1}(\dfrac{1}{s(s^2-9)})=\dfrac13\int_0^t\sinh3x\ dx=\dfrac19\cosh3x\Big|_0^t=\dfrac{\cosh3t-1}{9}$$