Compute Laplace Transform

Question

$$f(t) = 2t\ e^t\sin(2t)$$

Individually I found the Laplace of $2t\ e^t$ to be $\dfrac{2}{s^2}$ and the Laplace of $\sin(2t)$ to be $\dfrac{1}{1-s} + \dfrac{2}{s^2+4}$

From here I don't know where to go.

Answer 1

If you see it as: $\mathscr{L}\{tf(t)\}$, with $f(t)=e^t \sin(2t)$, you can solve for $f(t)$ with the solution: $$\mathscr{L}\{e^{-at}\sin(bt)\}=\frac{b}{(s+a)^2+b^2}$$ and after apply the property: $$\mathscr{L}\{tf(t)\}=-\frac{dF(s)}{ds}$$ So you obtain: $$F(s)=\frac{2}{(s-1)^2+4}$$ and so: $$-\frac{dF(s)}{ds}=\frac{4(s-1)}{(s^2-2s+5)^2}$$ Finally, the solution is: $$\mathscr{L}\{2te^t \sin(2t)\}=\frac{8(s-1)}{(s^2-2s+5)^2}$$