Question

y''+2y'+2y=delta(t-2)+u_5(t); y(0)=0,y'(0)=0

Answer 1

Take the Laplace transform of both sides:
\[\begin{align*}
y''+2y'+2y&=\delta(t-2)+u(t-5)\\\\
\mathcal{L}\{\cdots\}&=\mathcal{L}\{\cdots\}\\\\
\mathcal{L}\{y''\}+2\mathcal{L}\{y'\}+2\mathcal{L}\{y\}&=\mathcal{L}\{\delta(t-2)\}+\mathcal{L}\{u(t-5)\}\\\\
\left(s^2\mathcal{L}\{y\}-sy(0)-y'(0)\right)\quad\quad\\+2\left(s\mathcal{L}\{y\}-y(0)\right)+2\mathcal{L}\{y\}&=e^{-2s}+\frac{e^{-5s}}{s}\\\\

\left(s^2+2s+2\right)\mathcal{L}\{y\}&=e^{-2s}+\frac{e^{-5s}}{s}\\\\

\mathcal{L}\{y\}&=\frac{e^{-2s}}{s^2+2s+2}+\frac{e^{-5s}}{s(s^2+2s+2)}
\end{align*}\]
Take the inverse transform. Give it a try, I'm not going to write every single step.