91Ó°ÊÓ

Consider the differential equation$\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{-}\mathbf{af}\left(t\right)\mathbf{=}\mathbf{g}\left(t\right)\mathbf{}\mathbf{where}\mathbf{}\mathbf{g}\left(t\right)$ where is a smooth function and 'a' is a constant. Then the general solution will be $\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{∫}{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{g}\left(t\right)\mathbf{dt}$.

Consider the differential equation as follows.

$\mathrm{f}\text{'}\left(\mathrm{t}\right)-\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}$

Now, the differential equation is in the form as follows.

$\mathrm{f}\text{'}\left(\mathrm{t}\right)-\mathrm{af}\left(\mathrm{t}\right)=\mathrm{g}\left(\mathrm{t}\right)$, where g (t) is a smooth function, then the general solution will be as follows.

$\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}∫{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$

Substitute the value$\mathrm{t}\mathrm{for}\mathrm{g}\left(\mathrm{t}\right)\mathrm{and}1\mathrm{for}\mathrm{a}\mathrm{in}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}∫{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}∫{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{1\mathrm{t}}∫{\mathrm{e}}^{-1\mathrm{t}}\mathrm{x}\mathrm{t}\mathrm{x}\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}∫{\mathrm{e}}^{-1}\mathrm{tdt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}} \end{gathered}$$

Using substitution method, solve the integral $\mathrm{l}=∫{\mathrm{e}}^{-\mathrm{t}}\mathrm{tdt}$by integration by parts as follows.

$$\begin{gathered} \mathrm{u}=\mathrm{t} \\ \mathrm{du}=\mathrm{dt} \\ ∫\mathrm{dv}=∫{\mathrm{e}}^{-1}\mathrm{dt} \\ \mathrm{v}={\mathrm{e}}^{-\mathrm{t}} \end{gathered}$$

Substitute the value $\mathrm{t}\mathrm{for}\mathrm{du}\mathrm{and}-{\mathrm{e}}^{-\mathrm{t}}\mathrm{for}\mathrm{v}\mathrm{in}∫\mathrm{udv}=\mathrm{uv}-∫\mathrm{vdu}$follows.

$$\begin{gathered} ∫\mathrm{udv}=\mathrm{v}-∫\mathrm{vdu} \\ =-{\mathrm{te}}^{-\mathrm{t}}-∫-{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt} \\ =-{\mathrm{te}}^{-\mathrm{t}}+∫{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt} \\ \mathrm{l}=-{\mathrm{te}}^{-\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}+\mathrm{C} \end{gathered}$$

Substitute the value $-{\mathrm{te}}^{-\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}+\mathrm{C}\mathrm{for}\mathrm{l}\mathrm{in}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\mathrm{l}$as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\mathrm{l} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left(-{\mathrm{te}}^{-\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}+\mathrm{C}\right) \\ \mathrm{f}\left(\mathrm{t}\right)=-{\mathrm{te}}^{-\mathrm{t}}{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{t}}+\mathrm{Ce} \\ \mathrm{f}\left(\mathrm{t}\right)=-\mathrm{t}-1+{\mathrm{Ce}}^{\mathrm{t}} \end{gathered}$$

Hence, the solution for the linear differential equation $\mathrm{f}\text{'}\left(\mathrm{t}\right)+\mathrm{f}\left(\mathrm{t}\right)=\mathrm{t}\mathrm{is}\mathrm{f}\left(\mathrm{t}\right)=-\mathrm{t}-1+{\mathrm{Ce}}^{\mathrm{t}}$