Answer
\[y(x)=\frac{1}{2}x^3(2\ln x-1)+Cx\]
Work Step by Step
$y'-x^{-1}y=2x^2\ln x$ ___(1)
$\frac{dy}{dx}-\frac{1}{x}y=2x^2\ln x$ ____(2)
This is linear differential equation
Integrating Factor:-
\[I(x)=e^{\int -\frac{1}{x}dx}=e^{-\ln|x|}=\frac{1}{x}\]
Multiply (2) by $I(x)$
$\frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}y=2x\ln x$
$\frac{d}{dx}\left(\frac{y}{x}\right)=2x\ln x$
Integrating,
$\frac{y}{x}=2\int x\ln xdx+C$
$C$ is constant of integration
$\frac{y}{x}=2\left[\ln x.\frac{x^2}{2}-\int\frac{1}{x}.\frac{x^2}{2}dx\right]+C$
$\frac{y}{x}=2\left[\frac{x^2}{2}\ln x-\frac{x^2}{4}\right]+C$
$y=x^3\left[\ln x-\frac{1}{2}\right]+Cx$
$y=\frac{1}{2}x^3(2\ln x-1)+Cx$
Hence general solution of (1) is $y(x)=\frac{1}{2}x^3(2\ln x-1)+Cx$.
