First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0

Theorem
The first order ODE:
 * $(1): \quad \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$

is an exact differential equation with general solution:


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

Proof
Let $M$ and $N$ be defined as:


 * $\map M {x, y} = -\dfrac {2 x} {y^3}$


 * $\map N {x, y} = \dfrac {3 x^2} {y^4} - \dfrac 1 {y^2}$

Then:

Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be exact.

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

where:

Hence:

and:

Thus:
 * $\map f {x, y} = \dfrac 1 y - \dfrac {x^2} {y^3}$

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


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