Finding the inverse Laplace transform of $Y(s)=\frac{1-e^{-{\pi}s}}{(s^2+1)(s^2+4)}+\frac{s+1}{s^2+4}$

Question

I've encountered an IVP, and I need to solve it by applying the Laplace transform. The problem is:

$$y^{''}+4y=g(t), y(0)=y^{'}(0)=1$$

where $g(t)=\sin(t)$ for $0\leq{t}\leq\pi$ and $g(t)=0$ for $\pi\leq{t}$

Now, the Laplace transform is:

LHS: $L\{y^{''}\}+L\{4y\}=s^2Y(s)-sy(0)-y^{'}(0)+4Y(s)=(s^2+4)Y(s)-s-1$

RHS: $L\{g(t)\}=L\{\sin(t)-\sin(t)u(t-\pi)\}=\frac{1-e^{-{\pi}s}}{s^2+1}$

Then I get:

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

Now I'm struggling with finding the inverse transform of $Y(s)$. Can anyone help me with this probelm. Thank you in advance!

Answer 1

Rewrite $Y$ into $$Y(s)=\frac{1-e^{-{\pi}s}+(s+1)(s^2+1)}{(s^2+4)(s^2+1)}$$

We can expand the contour of the inverse laplace transform with a semicircle. Note that the poles of $Y$ are all of order $1$. By the residue theorem it follows that $$\mathcal L^{-1}\{Y(s)\}(t)=\sum_{j\in\{i,-i,2i,-2i\}}\operatorname{Res}(Y(s)e^{st};j).$$

Answer 2

This is mechanized in CASes. E.g. the command of Mathematica does the job:

InverseLaplaceTransform[(1 - Exp[-Pi*s])/(s^2 + 1)/(s^2 + 4) + (s + 
 1)/(s^2 + 4), s, t]

$ -\frac{1}{3} \theta (t-\pi ) (\sin (t) (-\cos (t))-\sin (t))+\cos(2 t)+\frac{1}{2} \sin (2 t)+\frac{1}{3} (\sin (t)-\sin (t) \cos (t))$