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:

\(\ds \frac {\partial M} {\partial y}\)

\(=\)

\(\ds \frac {6 x} {2 y^4}\)

\(\ds \dfrac {\partial N} {\partial x}\)

\(=\)

\(\ds \dfrac {6 x} {y^4}\)

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:

\(\ds \dfrac {\partial f} {\partial x}\)

\(=\)

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

\(\ds \dfrac {\partial f} {\partial y}\)

\(=\)

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

Hence:

\(\ds f\)

\(=\)

\(\ds \map M {x, y} \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds \int - \dfrac {2 x} {y^3} \rd x + \map g y\)

\(\ds \)

\(=\)

\(\ds -\dfrac {x^2} {y^3} + \map g y\)

and:

\(\ds f\)

\(=\)

\(\ds \int \map N {x, y} \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds \int \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y + \map h x\)

\(\ds \)

\(=\)

\(\ds -\dfrac {x^2} {y^3} + \dfrac 1 y + \map h x\)

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:

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

$\blacksquare$