Inverse Laplace Transform Table, Absolution of Form

Question

Do I need to ensure I don't stray from the transform in the table?

$\frac{-2}{s-1}$ this looks like $-2*\frac{a}{s^2-a^2},$ for $a=1$

Does this yield $-2\sinh(t)$, or should it fit perfectly to another form?

Answer 1

Beware: one has a square ($\frac a{s^2-a^2}$), whereas the other not ($\frac1{s-a}$). It changes everything.

Notice that, under Laplace transform,

$$1 \rightarrow \frac1s$$ $$e^{at}f(t)\rightarrow F(s-a)$$

Hence $\frac1{s-a}$ is the Laplace transform of $e^{at}$.