From the Laplace transform table, we have:
$$\mathcal{L}\{t^{n} e^{at} \sinh(bt)\} = \frac{n!}{(s-a)^{n+1}[(s-a)^{2} + b^{2}]}$$
Applying this formula to $t^{3} e^{-4t} \sinh 2t$, we get:
$$\mathcal{L}\{t^{3} e^{-4t} \sinh 2t\} = \frac{3!}{(s+4)^{4}[(s+4)^{2} +
Show more…