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.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{5}\mathbf{y}$

Short answer

So, the solution of the given equation is$\mathrm{y}={\mathrm{Ce}}^{5\mathrm{x}}$.

Step-by-step solution

Definition of transient term

A transient term means that you (or someone or something) will be moving on from where you are now.

Given data

Consider the differential equation,

$\frac{\mathrm{dy}}{\mathrm{dx}}=5\mathrm{y}\dots \dots \left(1\right)$

The objective is to find the general solution to the differential equation (1).

Finding Integrated factor

The standard form of linear differential equation of first order is,

$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{P}\left(\mathrm{x}\right)\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$

Rewrite the differential equation (1) in the standard form as,

$\frac{\mathrm{dy}}{\mathrm{dx}}-5\mathrm{y}=0\dots \dots \left(2\right)$

Compare the differential equation with standard form, identify it as,

$\mathrm{P}\left(\mathrm{x}\right)=-5\text{and}\mathrm{f}\left(\mathrm{x}\right)=0$

The functions and are continuous on.

The integrating factor is,

$$\begin{gathered} {}^{\int \mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}={\mathrm{e}}^{\int -5\mathrm{dx}} \\ ={\mathrm{e}}^{-5\mathrm{x}} \end{gathered}$$

Finding general solution

Multiply the differential equation (2) with the integrating factor to get,

$$\begin{gathered} \frac{dy}{dx}{e}^{-5x}-5y{e}^{-5x} \\ =0·e^{-5x}\frac{d}{dx}\left[e^{-5x}y\right] \\ =0 \end{gathered}$$

Integrate on both sides to get,

role="math" localid="1663832160615" $$\begin{gathered} \int \frac{d}{dx}\left[e^{-5x}y\right]dx \\ =\int 0dx{e}^{-5x}y \\ =Cy \\ =C{e}^{3x} \end{gathered}$$

Therefore, the general solution of the differential equation (1) is, $\mathrm{y}={\mathrm{Ce}}^{5\mathrm{x}}$.

The functions and are continuous on, so that the largest interval is,

$\mathrm{I}=-\infty <\mathrm{x}<\infty$

Suppose a term ${\mathrm{ce}}^{-\mathrm{x}}\to 0$as $x\to \infty$then the term is called transient term.

In this problem, as,$y\left(x\right)=C{e}^{5x}\to \infty$

Hence, there is no transient term in the general solution of differential equation (1).