Question

Find the general solution of the given differential equation. Give the largest interval / over which the general solution is defined. Determine whether there are any transient terms in the general solution.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{3}\mathbf{x}}$

Short answer

So, the solution of the given equation is $\mathrm{y}=\frac{1}{4}{\mathrm{e}}^{3\mathrm{x}}+{\mathrm{Ce}}^{-\mathrm{x}}$.

Step-by-step solution

Given data

Consider the following differential equation:

$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}={\mathrm{e}}^{3\mathrm{x}}\dots \dots \left(1\right)$

The objective is to find the general solution for the given differential equation.

Evaluation

The functions P(x) =1 and $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{3\mathrm{x}}$are continuous on $(-\infty ,\infty )$.

The integrating factor is,

$$\begin{gathered} e^{\int /\left(x\right)dx}=e^{\int 1dx} \\ =e^x \end{gathered}$$

Thus, the integrating factor is $e^x$.

Finding the largest interval by integrating the function

Multiply the equation (1) with the integrating factor e^x to get,

$$\begin{gathered} \frac{d}{dx}\left[e^{x}y\right]=e^{4x} \\ d\left[e^{x}y\right]=e^{4x}dx \end{gathered}$$

Integrate on both sides to get,

$$\begin{gathered} {\mathrm{e}}^{\mathrm{x}}\mathrm{y}=\frac{{\mathrm{e}}^{4\mathrm{x}}}{4}+\mathrm{C} \\ \mathrm{y}=\frac{1}{4}{\mathrm{e}}^{3\mathrm{x}}+{\mathrm{Ce}}^{-\mathrm{x}} \end{gathered}$$

Thus, the general solution of the given differential equation is,$y=\frac{1}{4}{e}^{3x}+C{e}^{-x}$ .

And, the solution is defined on, so that the largest interval is, $\mathrm{I}=(-\infty ,\infty )$.

Finding transient term

In general solution, ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{1}{4}{\mathrm{e}}^{3\mathrm{x}}$and $Y_{\epsilon }=C{e}^{-x}$

For large values of :

$$\begin{gathered} y_p\left(x\right)=\frac{1}{4}{e}^{3x}\to \infty \text{as}x\to \infty \\ {y}_e\left(x\right)=C{e}^{-x}\to 0\text{as}x\to \infty \end{gathered}$$

Transient term of the general solution of ordinary differential equation is term that approaches zero as $x\to \infty$.

The term ${\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{Ce}}^{-\mathrm{x}}$is a transient term in the general solution since $y_c\to 0$as $x\to \infty$, Therefore, the term $y_c\left(x\right)=C{e}^{-x}$is a transient term in the general solution.