91Ó°ÊÓ

The use of Laplace transformation is to convertdifferential equationsinto algebraic equations. The formula for Laplace transform is

$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{+}\mathbf{∞}}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{·}{\mathit{e}}^{\mathbf{-}\mathbf{s}\mathbf{·}\mathbf{t}}\mathbf{·}\mathit{d}\mathit{t}$

Where, F(s) = Laplace Transform

S is complex number

t = real number >=0

t’ = first derivative of the function f(t)

Given initial value problem

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

Where.$Y\left(0\right)=0andy\text{'}\left(0\right)=0$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+4\mathcal{L}y\text{'}\left(s\right)+4\mathcal{L}y\left(s\right)=\mathcal{L}\left[u\right(t-\mathrm{Ï€})-u(t-2\mathrm{Ï€}\left)\right]$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+2s\mathcal{L}y\left(s\right)-2y\left(0\right)+2\mathcal{L}y\left(s\right)=\frac{e^{-\mathrm{Ï€}s}}{s}-\frac{e^{-2\mathrm{Ï€}s}}{s} \\ {s}^2\mathcal{L}y\left(s\right)-0-0+4s\mathcal{L}y\left(s\right)-0+4\mathcal{L}y\left(s\right)=\frac{e^{-\mathrm{Ï€}s}}{s}-\frac{e^{-2\mathrm{Ï€}s}}{s} \end{gathered}$$

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

Using partial fraction

$\frac{1}{s\left(s^2+1s+1\right)}=\frac{\frac{1}{4}}{s}-\frac{\frac{1}{4}}{s+2}-\frac{\frac{1}{2}}{{(s+2)}^2}$

Equation first becomes as,

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

Taking inverse Laplace transform we get

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

Hence

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