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

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

Let:
 * $M \left({x, y}\right) = y^2$
 * $N \left({x, y}\right) = x y - x^2$

Put $t x, t y$ for $x, y$:

Thus both $M$ and $N$ are homogeneous functions of degree $2$.

Thus, by definition, $(1)$ is a homogeneous differential equation:


 * $\dfrac {\mathrm d x} {\mathrm d y} = \dfrac {x^2 - x y} {y^2}$

By Solution to Homogeneous Differential Equation, its solution is:
 * $\displaystyle \ln x = \int \frac {\mathrm d z} {f \left({1, z}\right) - z} + C$

where:
 * $f \left({y, x}\right) = \dfrac {x y - x^2} {y^2}$

Hence: