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.

$\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{sinx}$

Short answer

So, the general solution of the given differential equation is$y\left(x\right)=\frac{3}{2}+c{x}^{-2}$

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 following differential equation,

$x\frac{dy}{dx}+2y=3$

Dividing both sides by x obtain,

$\frac{dy}{dx}+\frac{2}{x}y=\frac{3}{x}\dots \dots \left(1\right)$

Evaluation

This is the linear differential equation of first order is of the form $\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$.

In this problem, $P\left(x\right)=\frac{2}{x}andf\left(x\right)=\frac{3}{x},P$and f are continuous on $(0,\infty )$. The integrating factor is,

$$\begin{gathered} =e^{2\ln x}\text{Since}a\ln x=\ln x^a \\ =e^{\ln x^2}\text{Since}{e}^{\ln a}=a \\ =x^2 \end{gathered}$$

Multiply equation 91) with this integrating factor,

$$\begin{gathered} x^2\frac{dy}{dx}+x^2·\frac{2y}{x}=\frac{3}{x}·x^2 \\ {x}^2\frac{dy}{dx}+2xy=3x \end{gathered}$$

 Finding general solution

This can be written in total differential as,

$\frac{d}{dx}\left[x^{2}y\right]=3x$

Integrate both sides with respect to x as,

$$\begin{gathered} y=\frac{3}{2}+\frac{c}{x^2} \\ y=\frac{3}{2}+c{x}^{-2} \end{gathered}$$

Hence, the solution of the given differential equation is $y\left(x\right)=\frac{3}{2}+c{x}^{-2}$.

The solution exists for $-\infty <x<\infty$.

In the solution $y_c\to 0as x\to \infty$then the term is called transient term.

The solution can be divided into two terms and .

It is clear that,$y_c=c{x}^{-2}\to 0asx\to \infty$ ,

Therefore, $y_c=c{x}^{-2}$is a transient term.