First Order ODE/(y^2 - 3 x y - 2 x^2) dx = (x^2 - x y) dy

Theorem
The first order ordinary differential equation:


 * $(1): \quad \paren {y^2 - 3 x y - 2 x^2} \rd x = \paren {x^2 - x y} \rd y$

is a homogeneous differential equation with solution:


 * $y^2 x^2 - 2 y x^3 + x^4 = C$

Proof
$(1)$ can also be rendered:
 * $\dfrac {\rd y} {\rd x} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$

Let:
 * $\map M {x, y} = y^2 - 3 x y - 2 x^2$
 * $\map N {x, y} = x^2 - x y$

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.

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

where:
 * $\map f {x, y} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$

Hence:

Substituting back for $z$ and tidying up, the result is obtained:
 * $y^2 x^2 - 2 y x^3 + x^4 = C$