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

Theorem

The first order ordinary differential equation:

$(1): \quad \paren {x^2 - 2 y^2} \rd x + x y \rd y = 0$

is a homogeneous differential equation with solution:

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

Proof

$(1)$ can also be rendered:

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

Let:

$\map M {x, y} = x^2 - 2 y^2$

$\map N {x, y} = x y$

Put $t x, t y$ for $x, y$:

\(\ds \map M {t x, t y}\)

\(=\)

\(\ds \paren {t x}^2 - 2 \paren {t y}^2\)

\(\ds \)

\(=\)

\(\ds t^2 \paren {x^2 - 2 y^2}\)

\(\ds \)

\(=\)

\(\ds t^2 \, \map M {x, y}\)

\(\ds \map N {t x, t y}\)

\(=\)

\(\ds t x t y\)

\(\ds \)

\(=\)

\(\ds t^2 \paren {x y}\)

\(\ds \)

\(=\)

\(\ds t^2 \, \map N {x, y}\)

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:

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

Hence:

\(\ds \ln x\)

\(=\)

\(\ds \int \frac {\d z} {-\frac {1 - 2 z^2} {z} - z} + C_1\)

\(\ds \)

\(=\)

\(\ds \int \frac {z \rd z} {z^2 - 1} + C_1\)

\(\ds \)

\(=\)

\(\ds \frac 1 2 \, \map \ln {z^2 - 1} + C_1\)

Primitive of $\dfrac x {x^2 - a^2}$

\(\ds \leadsto \ \ \)

\(\ds \map \ln {z^2 - 1}\)

\(=\)

\(\ds \map \ln {x^2} + \ln C\)

\(\ds \leadsto \ \ \)

\(\ds z^2 - 1\)

\(=\)

\(\ds C x^2\)

\(\ds \leadsto \ \ \)

\(\ds \frac {y^2} {x^2} - 1\)

\(=\)

\(\ds C x^2\)

\(\ds \leadsto \ \ \)

\(\ds y^2\)

\(=\)

\(\ds x^2 + C x^4\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $1 \ \text{(a)}$