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

Theorem
The first order ODE:
 * $(1): \quad \left({3 x^2 - y^2}\right) \mathrm d y - 2 x y \, \mathrm d x = 0$

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

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

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

but that $(1)$ is not exact.

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

Let:
 * $P \left({x, y}\right) = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$

Thus:

It can be observed that:

Thus $\dfrac {P \left({x, y}\right)} {M \left({x, y}\right)}$ is a function of $y$ only.

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


 * $\mu \left({y}\right) = e^{\int -h \left({y}\right) \mathrm d y}$

Hence:

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

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

which is now exact.

Let $M$ and $N$ be redefined as:


 * $M \left({x, y}\right) = - \dfrac {2 x} {y^3}$
 * $N \left({x, y}\right) = \dfrac {3 x^2} {y^4} - \dfrac 1 {y^2}$

By Solution to Exact Differential Equation, the solution to $(1)$ is:
 * $f \left({x, y}\right) = C$

where:

Hence:

and:

Thus:
 * $f \left({x, y}\right) = \dfrac 1 y - \dfrac {x^2} {y^3}$

and by Solution to Exact Differential Equation, the solution to $(1)$ is:


 * $\dfrac 1 y - \dfrac {x^2} {y^3} = C$