Question

Find the general solution of the differential equation $x\frac{dy}{dx}+2y=x^2\text{log}x$

Answer 1

Given differential equation :

$x\frac{dy}{dx}+2y=x^2\text{log}x$

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

Given differential equation is of the form
$\frac{dy}{dx}+Py=Q$

By comparing both the equations, we get

$P=\frac{2}{x}$ and $Q=x\text{log}x$

The general solution of the given differential equation is

$y(I.F)=\int (Q\times I.F.)dx+c$ ....(i)

Firstly, we need to find I.F.

$I.F.=e^{\int pdx}$

$I.F.=e^{\int \frac{2}{x}dx}$

$I.F.=e^{2\text{log}x}$

$I.F.=x^2(\because e^{\text{log}x}=x)$

Substituting the value of I.F in (i), we get

$y\times x^2=\int x\text{log}x.x^{2}dx+c$

$y{x}^2=\int x^3.\text{log}xdx+c$ ........(ii)

Using Integration by parts :

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left[f^{\prime }\right(x)\int g(x\left)dx\right]dx$

Taking $f\left(x\right)=\text{log}x$ and $g\left(x\right)=x^3$

$y{x}^2=\text{log}x\int x^{3}dx-\int [\frac{d}{dx}\text{log}x\int x^3]dx$

$y{x}^2=\text{log}x\left(\frac{x^4}{4}\right)-\int \frac{1}{x}\left(\frac{x^4}{4}\right)dx+c$

$y{x}^2=\text{log}x\left(\frac{x^4}{4}\right)-\int \left(\frac{x^3}{4}\right)dx+c$

$y{x}^2=\frac{x^4\text{log}x}{4}-\frac{x^4}{4\times 4}+c$

$y{x}^2=\frac{x^4\text{log}x}{4}-\frac{x^4}{16}+c$

$y=\frac{4x^2\text{log}x}{16}-\frac{x^2}{16}+c{x}^{-2}$

$\therefore y=\frac{x^2}{16}(4\text{log}x-1)+c{x}^{-2}$