91Ó°ÊÓ

The Laplace transform, is an integral transform that converts a function of a real variable usually t, the time domain to a function of a complex variable s.

Given initial value problem,

$y\text{'}\text{'}+2y\text{'}+2y=u(t-2\mathrm{Ï€})-u(t-4\mathrm{Ï€})$

Where$y\left(0\right)=1$ and$y\text{'}\left(0\right)=1$

Taking Laplace transform for the initial value problem

$$\begin{gathered} \mathcal{L}y\text{'}\text{'}\left(s\right)+2\mathcal{L}y\text{'}\left(s\right)+2\mathcal{L}y\left(s\right)=\mathcal{L}\left[u\right(t-2\mathrm{Ï€})-u(t-4\mathrm{Ï€}\left)\right] \\ {s}^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+2s\mathcal{L}y\left(s\right)-2y\left(0\right)+2\mathcal{L}y\left(s\right)=\frac{e^{-2\mathrm{Ï€}s}}{s}-\frac{e^{-4\mathrm{Ï€}s}}{s} \end{gathered}$$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s-1+2s\mathcal{L}y\left(s\right)-2+2\mathcal{L}y\left(s\right)=\frac{e^{-2\mathrm{Ï€}s}}{s}-\frac{e^{-4\mathrm{Ï€}s}}{s} \\ \left(s^2+2s+2\right)\mathcal{L}y\left(s\right)-(s+3)=\frac{e^{-2\mathrm{Ï€}s}}{s}-\frac{e^{-4\mathrm{Ï€}s}}{s} \end{gathered}$$

$$\begin{gathered} \mathcal{L}y\left(s\right)=\frac{s+3}{\left(s^2+2s+2\right)}+\frac{e^{-2\mathrm{Ï€}s}}{s\left(s^2+2s+2\right)}-\frac{e^{-4\mathrm{Ï€}s}}{s\left(s^2+2s+2\right)} \\ =\frac{s+1+2}{{(s+1)}^2+1}+\frac{e^{-2\mathrm{Ï€}s}}{s\left(s^2+2s+2\right)}-\frac{e^{-4\mathrm{Ï€}s}}{s\left(s^2+2s+2\right)} \end{gathered}$$

$\frac{1}{s\left(s^2+2s+2\right)}=\frac{\frac{1}{2}}{s}-\frac{\frac{1}{2}s+1}{s^2+2s+2}$

$\mathcal{L}y\left(s\right)=\frac{s+1}{{(s+1)}^2+1}+\frac{2}{{(s+1)}^2+1}+\frac{\frac{e^{-2\mathrm{Ï€}s}}{2}}{s}-\frac{\frac{e^{-2\mathrm{Ï€}s}}{2}s+e^{-2\mathrm{Ï€}s}}{s^2+2s+2}-\frac{\frac{e^{-4\mathrm{Ï€}s}}{2}}{s}+\frac{\frac{e^{-4\mathrm{Ï€}s}}{2}s+e^{-4\mathrm{Ï€}s}}{s^2+2s+2}$

$$\begin{gathered} y\left(t\right)=e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}u(t-2\mathrm{Ï€})-\frac{1}{2}{e}^{-t+2\mathrm{Ï€}}\cos (t-2\mathrm{Ï€})u(t-2\mathrm{Ï€}) \\ +\frac{1}{2}{e}^{-t+2\mathrm{Ï€}}\sin (t-2\mathrm{Ï€})u(t-2\mathrm{Ï€})-e^{-t+2\mathrm{Ï€}}\sin (t-2\mathrm{Ï€})u(t-2\mathrm{Ï€}) \\ -\frac{1}{2}u(t-4\mathrm{Ï€})+\frac{1}{2}{e}^{-t+4\mathrm{Ï€}}\cos (t-4\mathrm{Ï€})u(t-4\mathrm{Ï€}) \\ -\frac{1}{2}{e}^{-t+4\mathrm{Ï€}}\sin (t-4\mathrm{Ï€})u(t-4\mathrm{Ï€})+e^{-t+4\mathrm{Ï€}}\sin (t-4\mathrm{Ï€})u(t-4\mathrm{Ï€}) \end{gathered}$$

$$\begin{gathered} y\left(t\right)=e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}u(t-2\mathrm{Ï€})-\frac{1}{2}{e}^{-t+2\mathrm{Ï€}}\cos tu(t-2\mathrm{Ï€})-\frac{1}{2}{e}^{-t+2\mathrm{Ï€}}\sin tu(t-2\mathrm{Ï€}) \\ -\frac{1}{2}u(t-4\mathrm{Ï€})+\frac{1}{2}{e}^{-t+4\mathrm{Ï€}}\cos tu(t-4\mathrm{Ï€})+\frac{1}{2}{e}^{-t+4\mathrm{Ï€}}\sin tu(t-4\mathrm{Ï€}) \\ =e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}\left[1-e^{-t+2\mathrm{Ï€}}[\cos t+\sin t]\right]u(t-2\mathrm{Ï€})-\frac{1}{2}\left[1-e^{-t+4\mathrm{Ï€}}[\cos t+\sin t]\right]u(t-4\mathrm{Ï€}) \end{gathered}$$

hence

$y\left(t\right)=e^{-t}\cos t+2e^{-t}\sin t+\frac{1}{2}\left[1-e^{-t+2\mathrm{Ï€}}[\cos t+\sin t]\right]u(t-2\mathrm{Ï€})-\frac{1}{2}\left[1-e^{-t+4\mathrm{Ï€}}[\cos t+\sin t]\right]u(t-4\mathrm{Ï€})$