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

Theorem
The first order ordinary differential equation:


 * $(1): \quad \left({y^2 - 3 x y - 2 x^2}\right) \mathrm d x = \left({x^2 - x y}\right) \mathrm d y$

is a homogeneous differential equation with solution:


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

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

Let:
 * $M \left({x, y}\right) = y^2 - 3 x y - 2 x^2$
 * $N \left({x, y}\right) = x^2 - x y$

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:
 * $\displaystyle \ln x = \int \frac {\mathrm d z} {f \left({1, z}\right) - z} + C$

where:
 * $f \left({x, y}\right) = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$

Hence:

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