The general solution of the differential equation $(2x-10y^3)dy+ydx=0,y\ne 0$ is 

$\begin{array}{l}\left.x-10y^3\right)\cdot \frac{dy}{dx}=-y\\ \left(2x-10y^3\right)=-y\cdot \frac{dx}{dy}\end{array}$

$\frac{dx}{dy}=-\frac{2x}{y}+10y^2$

$\frac{dx}{dy}+\frac{2x}{y}=+10y^2$

Linear differential equation in x

$p\left(y\right)=\frac{2}{y}$ and $Q\left(y\right)=10y^2$

General solution is

$x{y}^2-2y^5=c$