I have a question asking to solve the following with Laplace transformations. I get through most of it, but I'm stumped in the final stage.
$$y''+4y'+29y=e^{-2t}\sin(5t),\quad \ y(0)=5,\ y'(0)=-2.$$
I run through the whole thing until I end up here.
$$(s^2+4s+29)Y=5s+3+\frac{5}{(s+2)^2+25}$$
I can divide and solve for the $5s+3$ bit fine, but the issue is that the part in the very right, the Laplace transformation of $e^{-2t}\sin(5t)$, gains a power of two on the bottom, as $s^2+4s+29$ becomes $(s+2)^2+25$. I have no idea how to inverse Laplace from that.
Hint:
$$ \mathcal{L}\{t\sin at \}(s) = -\frac{d}{ds}\mathcal{L}\{\sin at\}(s) = -\frac{d}{ds} \left( \frac{a}{s^2+a^2} \right) = \frac{2as}{(s^2 + a^2)^2} $$
$$ \mathcal{L}\{t\cos at\}(s) = -\frac{d}{ds}\mathcal{L}\{\cos at\}(s) = - \frac{d}{ds} \left( \frac{s}{s^2+a^2} \right) = \frac{s^2 - a^2}{(s^2 + a^2)^2} $$
This suggests a partial fraction decomposition of the form
$$ \frac{(5s+3)\big((s+2)^2 + 25\big) + 5}{\big((s+2)^2 + 25\big)^2} = A\frac{s+2}{(s+2)^2 + 25} + B\frac{5}{(s+2)^2 + 25} + \\ C\frac{(s+2)^2-25}{\big((s+2)^2+25\big)} + D\frac{10(s+2)}{\big((s+2)^2 + 25\big)^2} $$
For sanity, I suggest writing $u=s+2$