Determine the Laplace Transform of $te^{-3t}\sin(2t)$

Question

As the question says, I want to find the Laplace Transform of $te^{-3t}\sin(2t)$. It seems like I can just use the formula here right? So $L=\frac{2}{(s+3)^2+4}$, using the formula and definition of a Laplace Transform.

Answer 1

$ \newcommand{\L}[1]{\mathcal{L} \left\{ #1 \right\}(s)} \newcommand{\dv}{\frac{\mathrm{d}}{\mathrm{d}s}} $ Not quite. You seem to be finding just $\L{ e^{-3t} \sin(2t) }$, which, while necessary in the derivation of the answer, will need to be differentiated and multiplied by $-1$ to get there.

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In more detail:

Using $(30)$ from here with $n=1$ and $f(t) = e^{-3t} \sin(2t)$ yields

$$\L{te^{-3t} \sin(2t)} = -\dv \L{ e^{-3t} \sin(2t) }$$

From here, then we apply identity $(21)$ from that source with $a=-3$ and $b=2$ to obtain

$$\L{ e^{-3t} \sin(2t) } = \frac{2}{(s+3)^2 + 4}$$

Then we need to take the derivative of this, and get

$$\dv \L{ e^{-3t} \sin(2t) } = \dv \frac{2}{(s+3)^2 + 4} = \frac{-4(s+3)}{((s+3)^2 + 4)^2}$$

and consequently

$$\L{te^{-3t} \sin(2t)} = \frac{ 4(s+3)}{((s+3)^2 + 4)^2}$$

Answer 2

Let $F(t)=\sin 2t$, the Laplace transform is $$ f(s) =\mathcal{L} \{ F(t) \} =\frac{2}{s^2+4} $$ Then use respectively the two Laplace transform properties $\mathcal{L} \{ e^{at} F(t) \} =f(s-a)$ and $\mathcal{L} \{ t F(t) \} = - \frac{d}{ds} f(s)$.

The LT of $G(t)=e^{-3t} F(t)$ is $$ g(s) =\frac{2}{(s+3)^2+4} $$ and the required LT is $$ h(s) = - \frac{d}{ds} g(s) = \frac{4(s+3)}{[(s+3)^2+4]^2} $$