First Order ODE/x y dy = x^2 dy + y^2 dx

Theorem
The first order ODE:
 * $(1): \quad x y \, \mathrm d y = x^2 \, \mathrm d y + y^2 \, \mathrm d x$

has the solution:
 * $y = x \ln y + C x$

Proof
Let $(1)$ be rearranged as:
 * $(2): \quad \dfrac {\mathrm d x} {\mathrm d y} - \dfrac x y = -\dfrac {x^2} {y^2}$

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

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 $(3)$: