Question

can someone please show me how to use the convolution theorem to solve for the inverse laplace transform of s/(s^2+a^2)(s^2+b^2)

Answer 1

First, you do the following:
\[\mathcal{L}_s^{-1}\left[\frac{1}{a^2+s^2}\right](t)=\frac{\sin (a t)}{a}\\

\mathcal{L}_s^{-1}\left[\frac{s}{b^2+s^2}\right](t)=\
cos (b t)
\]

Answer 2

Now you can use that the Inverse Laplace transform of the product is the convolution of
their inverse transform

Answer 3

Then you do the work to find that
\[
\mathcal{L}_s^{-1}\left[\frac{s}{\left(a^2+s^2\right)
\left(b^2+s^2\right)}\right](t)=\int_0^t
\frac{\sin (a s) \cos (b (t-s))}{a} \,
ds=\frac{\cos (b t)-\cos (a t)}{a^2-b^2}
\]

Answer 4

\[
\mathcal{L}_s^{-1}\left[\frac{s}{\left(a^2+s^2\right)
\left(b^2+s^2\right)}\right](t)=\\\int_0^t
\frac{\sin (a s) \cos (b (t-s))}{a} \,
ds=\frac{\cos (b t)-\cos (a t)}{a^2-b^2}
\]

Answer 5

thank you so so so much!!!

Answer 6

YW