First Order ODE/(3 x^2 - y^2) dy - 2 x y dx = 0

Theorem
The first order ODE:
 * $(1): \quad \paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$

has the general solution:
 * $\dfrac 1 y - \dfrac {x^2} {y^3} = C$

This can also be presented in the form:
 * $\dfrac {\d y} {\d x} = \dfrac {2 x y} {3 x^2 - y^2}$

Proof
We note that $(1)$ is in the form:
 * $\map M {x, y} \rd x + \map N {x, y} \rd y = 0$

but that $(1)$ is not exact.

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

Let:
 * $\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

It can be observed that:

Thus $\dfrac {\map P {x, y}} {\map M {x, y} }$ is a function of $y$ only.

So Integrating Factor for First Order ODE: Function of One Variable can be used:


 * $\map \mu y = e^{\int -\map h y \rd y}$

Hence:

Thus an integrating factor for $(1)$ has been found:
 * $\mu = \dfrac 1 {y^4}$

which yields, when multiplying it throughout $(1)$:
 * $\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$

which is now exact.

By First Order ODE: $\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$, its solution is:
 * $\dfrac 1 y - \dfrac {x^2} {y^3} = C$