Answer
$y(x)=x^2(\ln x)^2+Cx^2$
Work Step by Step
We are given:
\[x\frac{dy}{dx}-2y=2x^2\ln x\]
Dividing both sides by $x$,
\[\frac{dy}{dx}-\frac{2y}{x}=2x\ln x\]
Integrating factor,
\[I=e^{-2\int \frac{x1}{x}dx}=e^{-2\ln x}=x^{-2}\]
The equation becomes:
$\frac{d}{dx}(x^{-2}y)=2x^{-1}\ln x$
Integrating both sides:
$x^{-2}y=(\ln x)^2+C$
$C_{1}$ is constant of integration
Hence general solution is $y(x)=x^2(\ln x)^2+Cx^2$.
