Question

Solve the differential equation$\mathbf{f}\mathbf{\text{'}}\left(t\right)\mathbf{-}\mathbf{2}\mathbf{f}\left(t\right)\mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$and find the solution of the differential equation.

Short answer

The solution is $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{te}}^{2\mathrm{t}}+{\mathrm{Ce}}^{2\mathrm{t}}$.

Step-by-step solution

Definition of first order linear differential equation

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

Determination of the solution

Consider the differential equation as follows.

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

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

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

role="math" localid="1660799880430" $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt}$

Compute the calculation of the solution.

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

$$\begin{gathered} \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\int {\mathrm{e}}^{-\mathrm{at}}\mathrm{g}\left(\mathrm{t}\right)\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int {\mathrm{e}}^{-2\mathrm{t}}\mathrm{x}{\mathrm{e}}^{2\mathrm{t}}\mathrm{x}\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int {\mathrm{e}}^{0\mathrm{t}}\mathrm{dt} \\ \mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\int \mathrm{tdt} \end{gathered}$$

Simplify further as follows.

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

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