How do you convert this Laplace equation to the Time Domain?

Question

If I have a Laplace function:

$V(s) = c * \frac{s}{(s+a)(s+b)}$

Where $c$, $a$, and $b$ are constants, how can I convert that to the time domain?

ie. Where for example if:

$Y(s)=\frac{1}{(s+a)(s+b)}$

$y(t) = \frac{e^{-at}-e^{-bt}}{b-a}$

I have this table of similar transforms: https://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html

But I don't see anything for what to do in the case of an $s$ in the numerator like I've got.

Is it reasonably solvable? If so, how?

Thanks.

Answer 1

$$F(s)=\frac{s}{(s+a)(s+b)} =\frac{A}{s+a}+\frac{B}{s+b}=\frac{(A+B)s+Ab+Ba}{(s+a)(s+b)} \implies A+B=1, Ab+Ba=0$$ $$ \implies A=\frac{a}{a-b}, B=\frac{b}{b-a}$$ Then use $$L e^{at}=\frac{1}{s-a}$$ $$F(s)= A L e^{-at} -B L e^{-bt}=Lf(t)=L \left( \frac{ae^{-at}-be^{-bt}}{a-b} \right)$$ So $$f(t)=\left( \frac{ae^{-at}-be^{-bt}}{a-b} \right)$$