Find the general integral of x dx=((x^2)/(y)-y^3)dy

Find the general integral of $x\,dx=\left(\dfrac{x^2}{y}-y^3\right)dy$.

Explanation of Method

Treating $x$ as a function of $y$ gives $\dfrac{dx}{dy}-\dfrac{x}{y}=-y^3x^{-1}$, a Bernoulli equation ($n=-1$); substitute $v=x^2$.

Multiplying $\dfrac{dx}{dy}-\dfrac{x}{y}=-\dfrac{y^3}{x}$ by $2x$: $\dfrac{d(x^2)}{dy}-\dfrac{2x^2}{y}=-2y^3$, i.e. $v'-\dfrac{2v}{y}=-2y^3$.

With $\mu(y)=1/y^2$: $\dfrac{d}{dy}\!\left(\dfrac{v}{y^2}\right)=-2y$, so $\dfrac{v}{y^2}=-y^2+C$, i.e. $v=Cy^2-y^4$.

Final answer:

\[\boxed{\,x^2=Cy^2-y^4,\qquad C\in\mathbb{R}}\]