Our professor was doing a Laplacian transform example in class.
Original problem:
$$ y''+4y=\sin t $$
He was working on the problem and got to this step:
$$ L^{-1}\left(L(y)\right)=L^{-1}\left(\frac{1}{s^2+1}\cdot\frac{1}{s^2+4}\right) $$
He claimed that this is possible:
$$ (s^2+4)-(s^2+1)=3 $$ $$ \frac{(s^2+4)}{3}-\frac{(s^2+1)}{3}=1 $$
Thus:
$$ L^{-1}(L(y))=L^{-1}\left(\frac{1}{3}\cdot \frac{1}{s^2+1}\right)-L^{-1}\left(\frac{1}{3} \cdot \frac{1}{s^2+4}\right) $$ And finally: $$ y=\frac{\sin t}{3}-\frac{\sin2t}{6} $$
Why does his claim work? What is the reasoning behind why that works? And is that allowed when the two expressions have more than just $2$ terms, such as $s^2-6s+9$ and $2s^2+s-1$?
He wrote that
$$\frac{1}{(s^2+1)(s^2+4)} = \frac{1}{3}\frac{1}{s^2+1} - \frac{1}{3}\frac{1}{s^2+4}$$
It's a classical method of partial fraction decomposition:
https://en.wikipedia.org/wiki/Partial_fraction_decomposition