Nonseparable differential equations

Question

How can I solve the equation $\frac{dy}{dx} =\frac{x^2-y}{x-y^2}$? I've tried few substitutions such as $y=xv$ and $y=x/v$ but all to no avail! Please, help.

Answer 1

HINT: Rewrite the equation in the form: $$(x^2-y)dx + (y^2-x)dy=0$$ This is an exact equation.

Ans: $\frac{x^3}{3}-xy+\frac{y^3}{3}=C$.

Answer 2

$$\frac { dy }{ dx } =\frac { x^{ 2 }-y }{ x-y^{ 2 } } \\ \quad \quad \quad \quad $$ $$\Downarrow \\$$$$ \left( x^{ 2 }-y \right) dx-\left( x-y^{ 2 } \right) dy=0\\ \quad \quad \quad \quad \quad $$$$\Downarrow \\ \left( { x }^{ 2 }dx+{ y }^{ 2 }dy \right) -\left( ydx+xdy \right) =0\\ \quad \quad \quad \quad \quad \quad $$$$\Downarrow \\ d\left( \frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } \right) -d\left( xy \right) =0\\ \quad \quad \quad \quad \quad \quad $$$$\Downarrow \\ d\left( \frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } -xy \right) =0\\ \quad \quad \quad \quad \quad \quad \quad $$ $$\Downarrow \\$$

$$ \frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } -xy=C$$