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 solution of the given equation is $y=cx-x\cos x,0<x<\infty$.

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

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

The objective is to find the general solution of this differential equation.

Give the largest interval I over which the general solution is defined.

And identify whether there are any transient terms in the general solution.

Evaluation

Rewrite the given differential equation as,$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{x}}\mathrm{y}=\mathrm{xsinx}$

Divide x by on both sides.

$\frac{\mathrm{dy}}{\mathrm{dx}}+\left(-\frac{1}{\mathrm{x}}\right)\mathrm{y}=\mathrm{xsinx}\dots \dots \left(1\right)$

Compare the differential equation (1) with the standard form of the first-order linear differential equation,

$$\begin{gathered} \frac{dy}{dx}+P\left(x\right)y=f\left(x\right) \\ P\left(x\right)=-\frac{1}{x}\text{and}f\left(x\right)=x\sin x \end{gathered}$$

Seek a solution of equation (1) on an interval l for which both coefficient functions P and f are continuous.

As identify $P\left(x\right)=-\frac{1}{x}andf\left(x\right)=x\sin x$, observe that the functions P and f are continuous on $(0,\infty )$.

Therefore, solution of the above differential equation is defined on $(0,\infty )$.

Solving a linear first-order equation

(i) Put the linear equation into the standard form $\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)$and from it identify p(x) and then find the integrating factor

As , integrating factor is $e^{\int P\left(x\right)dr}$

role="math" localid="1663849664146" $$\begin{gathered} \mu =e^{\int \frac{1}{x}dx} \\ =e^{-\ln x}\text{Use In}x\text{instead of}\ln \left|x\right|\text{since}x\in (0,\infty ) \\ =e^{\ln x^{-1}} \\ =x^{-1} \end{gathered}$$

(ii) Multiply both sides of the standard form equation by the integrating factor.

The left-hand side of the resulting equation is automatically the derivative of the product of the integrating factor $e^{\int P\left(x\right)dx}$and y:

Multiply (1) by integrating factor gives us

$$\begin{gathered} x^{-1}\frac{dy}{dx}+x^{-1}·\left(-\frac{1}{x}\right)y=x^{-1}·x\sin x \\ \frac{1}{x}\frac{dy}{dx}+\left(-\frac{1}{x^2}\right)y=\sin x \\ \frac{1}{x}\frac{dy}{dx}+y\frac{d}{dx}\left(\frac{1}{x}\right)=\sin x \\ \frac{d}{dx}\left[\frac{y}{x}\right]=\sin x \end{gathered}$$

(iii) Integrate both sides of the last equation and solve for y.

Integrate on both sides of the equation,

$\frac{d}{dx}\left[\frac{y}{x}\right]=\sin x$

With respect to, to get

$$\begin{gathered} \frac{y}{x}=-\cos x+c \\ y=cx-x\cos x,0<x<\infty \end{gathered}$$

Draw the graph

Graph of the solution $y=cx-x\cos x,0<x<\infty$is shown as below: -