Maybe my mistakes are in some simple calculations, sorry if that is the case.
I want to solve the following ODE by Laplace transform:
$$\begin{cases} y''+2y'+2y=t\delta_{\pi}(t)+u_{4}(t) \\ y(0)=1\,,y'(0)=1. \end{cases} $$ Calling $\mathcal{L}\{y(t)\}=F(s),$ I will use that:
- $\mathcal{L}\{y'(t)\}=s\mathcal{L}\{y(t)\}-y(0)=sF(s)-1$;
- $\mathcal{L}\{y''(t)\}=s^{2}\mathcal{L}\{y(t)\}-sy(0)-y'(0)=s^2-s-1;$
- $\displaystyle\mathcal{L}\{u_{4}(t)\}=\frac{e^{-4s}}{s};$
- $\displaystyle\mathcal{L}\{t\delta_{\pi}(t)\}=(-1)^{1}(\mathcal{L}\{\delta_{\pi}(t)\})'=-\frac{d}{ds}\left(\frac{e^{-\pi s}}{s}\right)=\frac{e^{-\pi s}(s\pi+1)}{s^{2}};$
So, I have
$$s^2F(s)-s-1+2sF(s)-2+2F(s)=\frac{e^{-\pi s}(s\pi+1)}{s^2}+\frac{e^{-4s}}{s} \\ \implies F(s)=\frac{e^{-\pi s}(s\pi+1)+se^{-4s}+s^3+3s^2}{s^2(s^2+2s+2)}. $$
Did i go wrong somewhere? If I didn't, what can I do?
you may simplify a little $$\implies F(s)=\frac{e^{-\pi s}(s\pi+1)+se^{-4s}+s^3+3s^2}{s^2(s^2+2s+2)}.$$ $$\implies F_1(s)=\frac{s+3}{(s^2+2s+2)}=\frac{s+1}{((s+1)^2+1)}+\frac{2}{(s+1)^2+1}$$ you have a cosine function and a sine function $$\mathcal{L^{-1}}(F_1(s))=e^{-t}(\cos (t)+2\sin(t))$$ $$\implies F_2(s)=\frac{e^{-4s}}{s(s^2+2s+2)}.$$ $$\implies F_2(s)=e^{-4s}(\frac{1}{s((s+1)^2+1)})=\frac {e^{-4s}}2 \left (\frac{1}{s}-\frac {(s+2)}{(s+1)^2+1)} \right )$$ $$ \mathcal{L^{-1}} F_2(s)=\frac {u_4(t)}2 \left (1-e^{-(t-4)}(\cos (t-4)+\sin(t-4)) \right )$$ Try fraction decomposition method for the last expression.