Find the general integral of (2x^2yln y-x)y'=y

Find the general integral of $(2x^2y\ln y-x)y'=y$.

Step 1: Invert and rearrange

$\dfrac{dx}{dy}=\dfrac{2x^2y\ln y-x}{y}=2x^2\ln y-\dfrac{x}{y}$, i.e. $\dfrac{dx}{dy}+\dfrac{x}{y}=2x^2\ln y$.

Step 2: Substitute $v=1/x$

Dividing by $x^2$ and using $v'=-x^{-2}dx/dy$: $-v'+\dfrac{v}{y}=2\ln y$, i.e. $v'-\dfrac{v}{y}=-2\ln y$.

Step 3: Solve the linear equation

With $\mu(y)=1/y$: $\dfrac{d}{dy}\!\left(\dfrac{v}{y}\right)=-\dfrac{2\ln y}{y}$. With $u=\ln y$, $\displaystyle\int-\frac{2\ln y}{y}dy=-(\ln y)^2+C$, so $v=Cy-y(\ln y)^2$.