Use Laplace Transform to solve the following IVP:

Question

I know that this is a somewhat simple problem but I have been having trouble coming up with the little "tricks" that help with Laplace.

The problem is:

$y''+2y' +5y = e^{-t}\sin(2t)$ where $y(0) = 2, y'(0) = -1$

Attempt at Solution

$(s^2+2s+5)Y = \frac{2}{(s+1)^2+4} + (s+2)(2) + 1(-1)$

$Y = \frac{2}{(s^2+2s+5)^2}+\frac{2s+3}{s2+2s+5}$

$Y = \frac{2}{(s^2+2s+5)^2}+2(\frac{s+1}{s^2+2s+5})+\frac{1}{s^2+2s+5}$

$Y = \frac{2}{(s^2+2s+5)^2}+\frac{1}{2}\sin(2t)+2e^{-t}\cos(2t)$

And I am stuck here.

The answer given in the book is as follows:

Y = $\frac{5}{8}e^{-t}\sin(2t)+2e^{-t}\cos(2t)-\frac{1}{4}te^{-t}\cos(2t)$

Any help would be greatly appreciated.

Answer 1

We can take inverse Laplace transform by using the Bromwich integral. That is, \begin{align} \mathcal{L}^{-1}\{Y(s)\} &= \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma -iT}^{\gamma +iT}Y(s)e^{st}ds\\ &= \sum\text{Res} \end{align} where $$ Y(s) = \frac{2}{(s^2 + 2s + 5)^2} + \frac{2s + 3}{s^2 + 2s + 5} $$ Then the poles of $s$ are at $s = -1\pm 2i$. We can then evaluate the integrals by splitting them into two. In the first one, the poles are of order two.

Answer 2

You need the inverse Laplace transform of $$ \dfrac{2}{(s^2 + 2 s + 5)^2}$$ This may help: $$ {\mathcal L} \{t e^{at} \cos(bt)\} = \dfrac{(s-a)^2 - b^2}{((s-a)^2+b^2)^2} = \dfrac{1}{(s-a)^2 + b^2} - \dfrac{2b^2}{((s-a)^2+b^2)^2}$$