In Problems 25 - 32, solve the given initial value problem using the method of Laplace transforms.

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{;} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \\ \mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}{e}^{-t},0⩽t<3,\\ 1,3<t\end{array} \end{gathered}$$

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

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

Where, F(s) = Laplace Transform

S= Complex number

t= real number >= 0

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

Given initial value problem

$y\text{'}\text{'}+3y\text{'}+2y=g\left(t\right)$

Where.$y\left(0\right)=2andy\text{'}\left(0\right)=-1$Also

$$\begin{gathered} g\left(t\right)=e^{-t},0⩽t⩽3 \\ =1,3<t \end{gathered}$$

Using rectangular and unit function we can write

$$\begin{gathered} g\left(t\right)=e^{-t}\left[u\right(t)-u(t-3\left)\right]+1u(t-3) \\ =e^{-t}-e^{-t}u(t-3)+u(t-3) \end{gathered}$$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+3\mathcal{L}y\text{'}\left(s\right)+2\mathcal{L}y\left(s\right)=\mathcal{L}\left[g\right(t\left)\right]$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+3s\mathcal{L}y\left(s\right)-3y\left(0\right)+2y\left(s\right)=\mathcal{L}\left[e^{-t}-e^{-t}u(t-3)+u(t-3)\right] \\ {s}^2\mathcal{L}y\left(s\right)-2s+1+3s\mathcal{L}y\left(s\right)-6+2\mathcal{L}y\left(s\right)=\frac{1}{s+1}-e^{-3s}\frac{e^{-3}}{s+1}+e^{-3s}\frac{1}{s} \\ \left(s^2+3s+2\right)\mathcal{L}y\left(s\right)-2s-5=\frac{1}{s+1}-e^{-3s}\frac{e^{-3}}{s+1}+e^{-3s}\frac{1}{s} \end{gathered}$$

$\mathcal{L}y\left(s\right)=\frac{2s+5}{\left(s^2+3s+2\right)}+\frac{1}{(s+1)\left(s^2+3s+2\right)}-e^{-3s}\frac{e^{-3}}{(s+1)\left(s^2+3s+2\right)}+e^{-3s}\frac{1}{s\left(s^2+3s+2\right)}\cdots \left(1\right)$

Using partial fraction

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

$$\begin{gathered} y\left(t\right)=3e^{-t}-e^{-2t}+e^{-2t}-e^{-t}+t{e}^{-2t} \\ -e^{-3}\left[e^{-2(t-3)}-e^{-(t-3)}+(t-3)e^{-(t-3)}\right]u(t-3)+\left[\frac{1}{2}-e^{-(t-3)}+\frac{1}{2}{e}^{-2(t-3)}\right]u(t-3) \\ =2e^{-t}+t{e}^{-2t}-\left[e^{-2t+3}-e^{-t}+(t-3)e^{-t}-\frac{1}{2}+e^{-(t-3)}-\frac{1}{2}{e}^{-2(t-3)}\right]u(t-3) \end{gathered}$$

hence

$\left.y\left(t\right)=2e^{-t}+t{e}^{-2t}-\left[e^{-2t+3}-e^{-t}+(t-3)e^{-t}-\frac{1}{2}+e^{-(t-3)}-\frac{1}{2}{e}^{-2(t-3)}\right]u(t-3)\right]$