I am trying to solve
$$\mathscr{L}^{-1} \left\{\frac{3}{(s-1)^2} \right\}$$
However, it seems worse than easy. We can write this out as
$$\mathscr{L}^{-1} \left\{\frac{3}{(s-1)}\frac{1}{(s-1)} \right\}, $$ where the factor $\dfrac{1}{s-1}$ recurs, so the answer must have the inverse of this in it, so we can write $$\mathscr{L}^{-1} \left\{\frac{3}{(s-1)}\frac{1}{(s-1)} \right\} = e^t \, f(t).$$
We can then consider the damping rule which says that $\mathscr{L}^{-1} \left\{F(s+b) \right\} =e^{-bt} \, f(t).$
Then, with $b=-1$, we obtain:
$$\mathscr{L}^{-1}\left\{ \frac{3}{(s-1)}\frac{1}{(s-1)} \right\}=F(s-1).$$
But the question now is, what is $F(s)$?
Any ideas how to solve this?
Thanks
If $L\{f(t)\}=F(s)$, then we have $L\{tf(t)\}=-F'(s)$ (this link); and note that:
$$\frac {3}{(s-1)^2}=-\left(\frac {3}{s-1}\right)'.$$
Moreover we have $L\{3 \, e^t\}=\frac {3}{s-1}$; therefore:
$$L^{-1} \left\{ \frac{3}{(s-1)^2} \right\}=3 \, t \, e^t.$$