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

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathit{y}\mathbf{=}\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{;}\mathbf{} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{} \\ \mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}10,0⩽t⩽10,\\ 20,10<t<20,\\ 0,20<t\end{array} \end{gathered}$$

The use of Laplace transformation is to convert differential equation into differential equations into 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

$y\text{'}\text{'}+2y\text{'}+10y=g\left(t\right)$

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

$$\begin{gathered} g\left(t\right)=10,0⩽t⩽10 \\ =20,10⩽t⩽20 \end{gathered}$$

Using rectangular and unit function we can write

$$\begin{gathered} g\left(t\right)=10-10u(t-10)+20u(t-10)-20u(t-20) \\ =10+10u(t-10)-20u(t-20) \end{gathered}$$

Taking Laplace transform of initial value problem is

$\mathcal{L}y\text{'}\text{'}\left(s\right)+2\mathcal{L}y\text{'}\left(s\right)+10\mathcal{L}y\left(s\right)=\mathcal{L}\left[g\right(l\left)\right]$

$$\begin{gathered} s^2\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+2s\mathcal{L}y\left(s\right)-2y\left(0\right)+10y\left(s\right)=\mathcal{L}\mid 10+10u(t-10)-20u(t-20)] \\ {s}^2\mathcal{L}y(s)+s-0+2s\mathcal{L}y(s)+2+10\mathcal{L}y(s)=\frac{10}{s}+\frac{10c}{s}-\frac{20e^{10s}}{s} \\ \left(s^2+2s+10\right)\mathcal{L}y(s)+(s+2)=\frac{10}{s}+\frac{10t^{-11)}}{s}-\frac{20e^{-10s}}{s} \end{gathered}$$

$$\begin{gathered} \mathcal{L}y\left(s\right)=-\frac{s+2}{\left(s^2+2s+10\right)}+\frac{10}{s\left(s^2+2s+10\right)}+\frac{10e^{-10s}}{s\left(s^2+2s+10\right)}-\frac{20e^{-10s}}{s\left(s^2+2s+10\right)} \\ =-\frac{s+1}{{(s+1)}^2+9}-\frac{1}{{(s+1)}^2+9}+\frac{10}{s\left(s^2+2s+10\right)}+\frac{10e^{-10s}}{s\left(s^2+2s+10\right)}-\frac{20e^{-10s}}{s\left(s^2+2s+10\right)} \end{gathered}$$

Using partial fraction

$\frac{10}{s\left(s^2+2s+10\right)}=\frac{-s-2}{s^2+2s+10}+\frac{1}{s}$

Equation first becomes as

$\mathcal{L}y\left(s\right)=-\frac{s+1}{{(s+1)}^2+9}-\frac{1}{{(s+1)}^2+9}+\frac{-s-2}{s^2+2s+10}+\frac{1}{s}+\left[\frac{-s-2}{s^2+2s+10}+\frac{1}{s}\right]e^{-10s}-\left[\frac{-s-2}{s^2+2s+10}+\frac{1}{s}\right]2e^{-20s}$

$$\begin{gathered} y\left(t\right)=-e^{-t}\cos 3t-\frac{1}{3}{e}^{-t}\sin 3t-e^{-t}\cos 3t-\frac{1}{3}{e}^{-t}\sin 3t+1-e^{-(t-10)}\cos 3(t-10)-\frac{1}{3}{e}^{-(t-10)}\sin 3(t-10)+u(t-10) \\ +2e^{-(t-20)}\cos 3(t-20)+\frac{2}{3}{e}^{-(t-20)}\sin 3(t-20)-2u(t-20) \\ =-2e^{-t}\cos 3t-\frac{2}{3}{e}^{-t}\sin 3t+1-e^{-(t-10)}\cos 3(t-10)-\frac{1}{3}{e}^{-(t-10)}\sin 3(t-10)+u(t-10) \\ +2e^{-(t-20)}\cos 3(t-20)+\frac{2}{3}{e}^{-(t-20)}\sin 3(t-20)-2u(t-20) \end{gathered}$$

hence

$$\begin{gathered} y\left(t\right)=-2e^{-t}\cos 3t-\frac{2}{3}{e}^{-t}\sin 3t+1-e^{-(t-10)}\cos 3(t-10)-\frac{1}{3}{e}^{-(t-10)}\sin 3(t-10)+u(t-10) \\ +2e^{-(t-20)}\cos 3(t-20)+\frac{2}{3}{e}^{-(t-20)}\sin 3(t-20)-2u(t-20) \end{gathered}$$