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

Theorem
The first order ODE:
 * $(1): \quad y^2 \, \mathrm d x = \left({x^3 - x y}\right) \mathrm d y$

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

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

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

where:
 * $P \left({y}\right) = \dfrac 1 y$
 * $Q \left({y}\right) = \dfrac 1 {y^2}$
 * $n = 3$

and so is an example of Bernoulli's equation.

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

where:
 * $\mu \left({y}\right) = e^{\left({1 - n}\right) \int P \left({y}\right) \, \mathrm d y}$

Thus $\mu \left({y}\right)$ is evaluated:

and so substituting into $(3)$: