Find the inverse Laplace transform of: $\frac{1}{(s^2+a^2)(s^2+b^2)}$

Question

I'm having trouble doing this homework problem because I'm not sure how to deal with the $a$ and $b$. I did it the usual way we were taught - use partial fraction decomposition and then try to solve for the coefficients. When I solved for them, this is the conclusion I came to:

$$ \frac{A}{B}=\frac{-a^2}{b^2} $$

Thus my original problem becomes $$ \frac{-a^2}{(s^2+a^2)}+\frac{b^2}{(s^2+b^2)} $$

But I'm stuck here. I know that the inverse of $\frac{a}{(s^2+a^2)}$ is $sin(at)$ but I'm not sure if I can use that here....

Also, is there another way of solving this without using partial fraction decomposition?

Answer 1

Your partial fractions decomposition is off. You should get $$ \frac{1}{(s^2+a^2)(s^2+b^2)} = \frac{1}{b^2-a^2} \left( \frac{1}{s^2+a^2} - \frac{1}{s^2+b^2} \right), $$ so the inverse Laplace transform is $$ \frac{1}{b^2-a^2} \left( \frac1a \sin at - \frac1b \sin bt \right). $$

Answer 2

As you said, $$\mathcal{L}^{-1}\left(\frac{a}{s^2+a^2}\right)=\sin(at)\tag{1} $$ hence, assuming $a\neq b$ and $a,b\neq 0$: $$\begin{eqnarray*}\mathcal{L}^{-1}\left(\frac{1}{(s^2+a^2)(s^2+b^2)}\right)&=&\frac{1}{b^2-a^2}\cdot\mathcal{L}^{-1}\left(\frac{1}{s^2+a^2}-\frac{1}{s^2+b^2}\right)\\&=&\frac{1}{b^2-a^2}\left(\frac{\sin(at)}{a}-\frac{\sin(bt)}{b}\right)\\&=&\color{red}{\frac{b\sin(at)-a\sin(bt)}{ab(b^2-a^2)}}.\tag{2}\end{eqnarray*}$$

Answer 3

Just note $$\frac{1}{s^2+a^2}-\frac{1}{s^2+b^2}=\frac{b^2-a^2}{(s^2+a^2)(s^2+b^2)}$$ Then, if $a^2\neq b^2$ and $ab\neq 0$, we have \begin{align*} \frac{1}{(s^2+a^2)(s^2+b^2)}&=\frac{\frac{1}{b^2-a^2}}{s^2+a^2}-\frac{\frac{1}{b^2-a^2}}{s^2+b^2}\\ \mathcal{L}^{-1}\left\{\frac{1}{(s^2+a^2)(s^2+b^2)}\right\}&=\frac{1}{a(b^2-a^2)}\sin(at)-\frac{1}{b(b^2-a^2)}\sin(bt) \end{align*}