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

Theorem
The first order ordinary differential equation:


 * $(1): \quad \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$

is an exact differential equation with solution:


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

Proof
Let:
 * $\map M {x, y} = \dfrac 1 {x^3 y^2} + \dfrac 1 x$
 * $\map N {x, y} = \dfrac 1 {x^2 y^3} - \dfrac 1 y$

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

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


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

or:
 * $-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$