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

Theorem
The first order ordinary differential equation:


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

is an exact differential equation with solution:


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

This can also be presented as:
 * $\dfrac {\d y} {\d x} = -\dfrac {\dfrac y {1 - x^2 y^2} - 1} {\dfrac x {1 - x^2 y^2} }$

Proof
First express $(1)$ in the form:
 * $(2): \quad \paren {\dfrac y {1 - x^2 y^2} - 1} + \paren {\dfrac x {1 - x^2 y^2} } \dfrac {\d y} {\d x}$

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

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:

Substitute $z = x y$ to obtain:
 * $\dfrac {\d z} {\d x} = y$

This gives:

and:

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

and by Solution to Exact Differential Equation, the solution to $(1)$, after simplification, is:
 * $\map \ln {\dfrac {1 + x y} {1 - x y} } - 2 x = C$