Answer
$y=cx + x^2\ln x-x^2$
Work Step by Step
We are given:
$xy'-y=x^2 \ln x$
$\frac{xy'}{x}-\frac{y}{x}=\frac{x^2 \ln x}{x}$
$y'-x^{-1}y=x \ln x$
The general solution is given by:
$y=e^{\int \frac{1}{x}}dx(c+\int e^{-\int \frac{1}{x}dx}x \ln x dx)$
Simplify:
$y=x(c+\int \ln x dx)=cx + x^2\ln x-x^2$
