Question

How do you find the Laplace transform of le="9t e^{-t} sin(3t)" alt="9t e^{-t} sin(3t)" align="absmiddle" class="latex-formula">?

Answer 1

If $F(s)$ is the Laplace transform of $f(t)$, then first recall the phase shift property of the transform:

$\mathcal L_s\{e^{ct}f(t)\}=F(s-c)$

In this case, $c=-1$, and $F(s)$ is the transform of $f(t)=9t\sin3t$.

We can easily (if tediously) derive the following result:

$\mathcal L_s\{t\sin(at)\}=\dfrac{2as}{(s^2+a^2)^2}$

so that

$F(s)=\dfrac{9(6s)}{(s^2+9)^2}=\dfrac{54s}{(s^2+9)^2}$

and so

$F(s+1)=\mathcal L_s\left\{9te^{-t}\sin3t\right\}=\dfrac{54(s+1)}{((s+1)^2+9)^2}$
$F(s+1)=\dfrac{54s+54}{s^4+4s^3+24s^2+40s+100}$