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.

$\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{x}$

Short answer

So, the general solution of the given differential equation is $y=\frac{1}{7}{x}^3-\frac{1}{5}x+c{x}^{-4};\text{for}0<x<\infty$.

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,

$\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+4\mathrm{y}={\mathrm{x}}^3-\mathrm{x}$

The objective is to find the general solution of the differential equation. Also give the largest interval I over which the general solution defined. Also determine whether there are any transient terms in the general solution.

Write the given differential equation in standard form as shown:

$$\begin{gathered} x\frac{dy}{dx}+4y=x^3-x\text{(Given equation)} \\ \frac{dy}{dx}+\frac{4}{x}y=x^2-1\dots \dots \text{(i)} \end{gathered}$$

Evaluation

The standard form of the linear differential equation is,

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

Clearly equation (i) is in the standard form.

Compare the differential equation (i) with standard form and identify,

$P\left(x\right)=\frac{4}{x}\text{and}f\left(x\right)=x^2-1$

Note that $P\left(x\right)=\frac{4}{x}$is continuous on the interval $(0,\infty )$.

And $f\left(x\right)=x^2-1$is continuous on the interval $(0,\infty )$.

Integrating of the equation

The integrating factor is,

$$\begin{gathered} e^{\int f\left(x\right)dx}=e^{\int \frac{4}{x}dx} \\ =e^{4\int \frac{1}{x}dx} \\ =e^{4\ln x}\left[\text{use}\ln x\text{instead of}\ln \right|x|\text{since}x>0 \\ =e^{\ln x^4} \\ =x^4 \end{gathered}$$

Multiply both sides of differential equation (i) with integrating factor.

$$\begin{gathered} x^4\left(\frac{dy}{dx}+\frac{4}{x}y\right)=\left(x^2-1\right)x^4 \\ {x}^4\frac{dy}{dx}+4x^{3}y=x^6-x \\ \frac{d}{dx}\left(x^{4}y\right)=x^6-x^4 \end{gathered}$$

Finding general solution by integrating

Take integration on both sides,

$$\begin{gathered} \int \frac{d}{dx}\left(x^{4}y\right)=\int \left(x^6-x^4\right)dx \\ y=\frac{1}{7}{x}^7{x}^{-4}-\frac{1}{5}{x}^5{x}^{-4}+c{x}^{-4} \\ y=\frac{1}{7}{x}^3-\frac{1}{5}x+c{x}^{-4} \end{gathered}$$

Finding transient term

In the general solution $y\left(x\right)=\frac{1}{7}{x}^3-\frac{1}{5}x+c{x}^{-4}$, complementary function is $y_e\left(x\right)=c{x}^{-4}$and particular solution is $y_p\left(x\right)=\frac{1}{7}{x}^3-\frac{1}{5}x$.

For large value of x, value of y_ex is negligible. (Because, as ${\mathrm{x}}^{-4}\to 0\mathrm{as}{\mathrm{x}}^{-4}\to 0$).

Therefore $Y_{\epsilon }\left(x\right)=c{x}^{-4}$a transient term in the general solution.