First Order ODE/x^2 y' - 3 x y - 2 y^2 = 0

Theorem
The first order ordinary differential equation:


 * $(1): \quad x^2 \dfrac {\d y} {\d x} - 3 x y - 2 y^2 = 0$

is a homogeneous differential equation with solution:


 * $y = C x^2 \paren {x + y}$

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

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.

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

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