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Consider the differential equation$\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{鈥�}\mathbf{af}\left(t\right)\mathbf{=}\mathbf{g}\left(t\right)$ where g(t) is a smooth function and 'a' is a constant. Then the general solution will be role="math" localid="1660806053060" $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{at}}\mathbf{鈭�}{\mathbf{e}}^{\mathbf{-}\mathbf{at}}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$.

Consider the differential equation as follows.

$\mathrm{f}\text{'}\left(\mathrm{t}\right)鈥�2\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}$

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

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

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

Substitute the value${\mathrm{e}}^{2\mathrm{t}}$ for g(t) and 2 for f (t) in $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}鈭�{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$as follows.

role="math" localid="1660806345559" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}鈭�{\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2t}鈭�{\mathrm{e}}^{2t}脳0脳\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}鈭�0\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left(\mathrm{C}\right) \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C}+{\mathrm{e}}^2\mathrm{C} \end{gathered}$$

Hence, the solution for the linear differential equation $\mathrm{f}\text{'}\left(\mathrm{t}\right)+2\mathrm{f}\left(\mathrm{t}\right)=0$is.

role="math" localid="1660806669972" $$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\mathrm{C}+{\mathrm{e}}^2\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}+2}\mathrm{C} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}+2}\left(鈭�\mathrm{C}=1\right) \end{gathered}$$

Thus, the solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}+2}$.