Question

Solve the differential equation$\mathit{f}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{2}\mathit{f}\left(t\right)\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{t}}$and find the solution of the differential equation.

Short answer

The solution is$f\left(t\right)=\frac{1}{5}{e}^{3t}+C{e}^{-2t}$

Step-by-step solution

Definition of first order linear differential equation

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

Determination of the solution

Consider the differential equation as follows.

$f\text{'}\left(t\right)+2f\left(t\right)=e^{3t}$

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

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

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

Compute the calculation of the solution.

Substitute the value $e^{3t}$for $g\left(t\right)$and -2 for a in $\text{f}\left(t\right)=e^{at}\int e^{-at}g\left(t\right)dt$as follows.

role="math" localid="1665039959672" $$\begin{gathered} \text{f}\left(t\right)=e^{at}\int e^{-at}g\left(t\right)dt \\ f\left(t\right)=e^{-2t}\int e^{2t}\times e^{3t}\times dt \\ f\left(t\right)=e^{-2t}\int e^{\left(2+3\right)t}dt \\ f\left(t\right)=e^{-2t}\int e^{5t}dt \end{gathered}$$

Using substitution method in the integral as follows.

$\begin{array}{rcl}y& =& 5t\\ dy& =& 5dt\\ \frac{dy}{5}& =& dt\end{array}$

Substitute the value, 5t for y and $\begin{array}{rcl}& & \frac{dy}{5}\end{array}$for dt in $\begin{array}{rcl}f\left(t\right)& =& e^{-2t}\int e^{5t}dt\end{array}$as follows.

$\begin{array}{rcl}f\left(t\right)& =& e^{-2t}\int e^{5t}dt\\ f\left(t\right)& =& e^{-2t}\int e^y\frac{dy}{5}\\ f\left(t\right)& =& \frac{1}{5}{e}^{-2t}\left(e^y+C\right)\end{array}$

Now, undo the substitution as follows.

$\begin{array}{rcl}f\left(t\right)& =& \frac{1}{5}{e}^{-2t}\left(e^y+C\right)\\ f\left(t\right)& =& \frac{1}{5}{e}^{-2t}\left(e^{5t}+C\right)\\ & =& \frac{1}{5}{e}^{-2t}{e}^{5t}+e^{-2t}C\\ f\left(t\right)& =& \frac{1}{5}{e}^{3t}+e^{-2t}C\end{array}$

Hence, the solution for the linear differential equation$\begin{array}{rcl}f\text{'}\left(t\right)+2f\left(t\right)& =& e^{3t}\end{array}$ is$\begin{array}{rcl}f\left(t\right)& =& \frac{1}{5}{e}^{3t}+e^{-2t}C\end{array}$ .