Question: Solve the differential equation${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{,}\mathit{y}\mathbf{0}\mathbf{=}{\mathit{y}}_{\mathbf{0}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$, where$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{ll}1,& 0<t<a,\\ 0,& t>a.\end{array}$

As in Problems 6 and 7, find the Green function for the problem and use it in equation (12.4). Consider the cases$\mathit{t}\mathbf{<}\mathit{a}$ and$\mathit{t}\mathbf{>}\mathit{a}$separately.

The differential equation$y^{\text{'}\text{'}}+2y+y=f\left(t\right),y=y{0}^{\text{'}}=0$ where $f\left(t\right)=\left\{\begin{array}{cc}1,& 0<t<a,\\ 0,& t>a\end{array}\right.$.

In mathematics, a Green's function is the impulse response of an inhomogeneous linear differential operator defined on a domain with specified initial conditions or boundary conditions.

$$\begin{gathered} \mathrm{Let}\mathrm{G}\left(\mathrm{t},{\mathrm{t}}^{\text{'}}\right)\mathrm{be}\mathrm{the}\mathrm{Green}\mathrm{function}.\mathrm{Then},\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{differential}\mathrm{equation} \\ \frac{{\mathrm{d}}^2}{{\mathrm{d}}^2}\mathrm{G}\left(\mathrm{t},{\mathrm{t}}^{\mathrm{r}}\right)+2\frac{\mathrm{d}}{\mathrm{du}}\mathrm{G}\left(\mathrm{t},{\mathrm{t}}^{\text{'}}\right)+\mathrm{G}\left(\mathrm{t},{\mathrm{t}}^{\text{'}}\right)=\mathrm{\delta }\left({\mathrm{t}}^{\text{'}}-\mathrm{t}\right)\mathrm{with}\mathrm{G}=0,\frac{\mathrm{di}}{\mathrm{dt}}=0\mathrm{at}\mathrm{t}=0. \\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{this}\mathrm{can}\mathrm{be}\mathrm{given}\mathrm{as}: \\ \begin{array}{c}\mathrm{G}\left(\mathrm{t},{\mathrm{t}}^{\text{'}}\right)={\mathrm{L}}^{-1}\left\{\frac{1}{{(\mathrm{p}+1)}^2}\mathrm{L}\left\{\mathrm{\delta }\left({\mathrm{t}}^{\text{'}}-\mathrm{t}\right)\right.\right.\\ \mathrm{G}\left(\mathrm{t},{\mathrm{t}}^{\text{'}}\right)={\mathrm{L}}^{-1}\left\{\frac{1}{{(\mathrm{p}+1)}^2}{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}\mathrm{p}}\right\}\\ =\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right){\mathrm{e}}^{-\left(\mathrm{t}-{\mathrm{t}}^{\mathrm{r}}\right)}\mathrm{u}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\\ =\left\{\begin{array}{cc}0& \mathrm{t}<{\mathrm{t}}^{\text{'}}\\ \left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right){\mathrm{e}}^{-\left(\mathrm{t}-{\mathrm{l}}^{\text{'}}\right)}& \mathrm{t}>{\mathrm{t}}^{\text{'}}\end{array}\right.\end{array} \\ \mathrm{y}\left(\mathrm{t}\right)=\underset{{\mathrm{I}}_1}{\underbrace{{\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\cdot 1{\mathrm{dt}}^{\text{'}}}}=\underset{{\mathrm{I}}_2}{\underbrace{{\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)}{\mathrm{tdt}}^{\text{'}}}}-\underset{{\mathrm{I}}_1}{\underbrace{{\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\left(\mathrm{l}-{\mathrm{t}}^{\text{'}}\right)}{\mathrm{t}}^{\text{'}}{\mathrm{dt}}^{\text{'}}}} \end{gathered}$$

Evaluating, the integral I₁ we get

$$\begin{gathered} {\mathrm{I}}_1=\left({\left.{\mathrm{te}}^{-\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)}\right|}_{\mathrm{t}}^0\right. \\ \begin{array}{c}=\left({\mathrm{te}}^{-(\mathrm{t}-\mathrm{t})}-{\mathrm{te}}^{-(\mathrm{t}-0)}\right)\\ =\left(\mathrm{t}-{\mathrm{te}}^{-\mathrm{t}}\right)\end{array} \end{gathered}$$

And, evaluating the second integral I₂, we get

${\mathrm{I}}_2={\int }_0^{\mathrm{t}}{\mathrm{t}}^{\text{'}}{\mathrm{e}}^{-\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)}{\mathrm{dt}}^{\text{'}}$

Using, integration by parts where

$\begin{array}{l}\mathrm{u}={\mathrm{t}}^{\text{'}}\\ \mathrm{du}={\mathrm{dt}}^{\text{'}}\end{array}$

And,

$\begin{array}{l}dv=e^{-\left(l-t^{\text{'}}\right)}d{t}^{\text{'}}\\ v=e^{-\left(t-t^{\text{'}}\right)}\end{array}$

Thus, we have

$\begin{array}{c}{I}_2=\left({\left.uv\right|}_t^0-{\int }_0^{t}vdu\right.\\ =\left({\left.t^{\text{'}}{e}^{-\left(t-t^{\text{'}}\right)}\right|}_t^0-{\int }_0^t{e}^{-(t-t)}d{t}^{\text{'}}\right.\\ =\left(t{e}^{-(t-t)}-0\right)-\left({\left.e^{-\left(t-t^{\text{'}}\right)}\right|}_t^0\right.\\ =\left(t\right)-\left(e^{-\left(0\right)}-e^{-t}\right)\\ =t-1+e^{-t}\end{array}$

Now, that we have evaluated the integral I₁ and the integral I₂, we can now find the integral y(t)

data-custom-editor="chemistry" $$\begin{gathered} y\left(t\right)=I_1-I_2 \\ \mathrm{Thus},\mathrm{we}\mathrm{have} \\ y\left(t\right)=\left(t-t{e}^{-t}\right)-\left(t-1+e^{-t}\right) \\ \mathrm{Hence},\mathrm{we}\mathrm{get} \\ \begin{array}{l}y\left(t\right)=t-t{e}^{-t}-t+1-e^{-t}\\ y\left(t\right)=1-t{e}^{-t}-e^{-t}\end{array} \end{gathered}$$

Which is the solution to the given differential, equation provided that

t < a

Thus, the genral solution to the given differential equation, is thus

$y\mathit{\left(}t\mathit{\right)}\mathit{=}\{\begin{array}{ll}{e}^{-(t-a)}(t+1-a)-e^{-t}(t+1),& t>a\\ 1-t{e}^{-t}-e^{-t},& t<a\end{array}$

Therefore, the solution of the differential equation is:

$y\mathit{\left(}t\mathit{\right)}\mathit{=}\{\begin{array}{ll}{e}^{-(t-a)}(t+1-a)-e^{-t}(t+1),& t>a\\ 1-t{e}^{-t}-e^{-t},& t<a\end{array}$