Bernoulli's Equation/y^2 dx = (x^3 - x y) dy

Theorem
The first order ODE:
 * $(1): \quad y^2 \rd x = \paren {x^3 - x y} \rd y$

has the general solution:
 * $3 y = 2 x^2 + C x^2 y^2$

Proof
Dividing $(1)$ by $y^2$ and rearranging:
 * $(2): \quad \dfrac {\d x} {\d y} + \dfrac x y = \dfrac {x^3} {y^2}$

It can be seen that $(2)$ is in the form:
 * $\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$

where:
 * $\map P y = \dfrac 1 y$
 * $\map Q y = \dfrac 1 {y^2}$
 * $n = 3$

and so is an example of Bernoulli's equation.

By Solution to Bernoulli's Equation it has the general solution:
 * $(3): \quad \displaystyle \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$

where:
 * $\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$

Thus $\map \mu y$ is evaluated:

and so substituting into $(3)$: