First Order ODE/(x^2 - 2 y^2) dx + x y dy = 0

Theorem
The first order ordinary differential equation:


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

is a homogeneous differential equation with solution:


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

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

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

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

Hence: