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

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

Let:
 * $\map M {x, y} = y^2$
 * $\map N {x, y} = x y - x^2$

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:


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

By Solution to Homogeneous Differential Equation, its solution is:
 * $\ds \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$

where:
 * $\map f {y, x} = \dfrac {x y - x^2} {y^2}$

Hence: