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

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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:

\(\ds \frac {\partial M} {\partial y}\) \(=\) \(\ds \frac 1 {1 - x^2 y^2} + \frac {2 x^2 y^2} {\paren {1 - x^2 y^2}^2}\)
\(\ds \) \(=\) \(\ds \frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2}\)
\(\ds \frac {\partial N} {\partial x}\) \(=\) \(\ds \frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2}\) similarly


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:

\(\ds f\) \(=\) \(\ds \map M {x, y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \int \paren {\dfrac y {1 - x^2 y^2} - 1} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \int \dfrac {y \rd x} {1 - x^2 y^2} - \int \rd x + \map g y\)

Substitute $z = x y$ to obtain:

$\dfrac {\d z} {\d x} = y$

This gives:

\(\ds f\) \(=\) \(\ds \map M {x, y} \rd x + \map g y\)
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {\frac {1 + z} {1 - z} } - \int \rd x + \map g y\)
\(\ds \leadsto \ \ \) \(\ds \map M {x, y} \rd x + \map g y\) \(=\) \(\ds \frac 1 2 \map \ln {\frac {1 + x y} {1 - x y} } - x + \map g y\)

and:

\(\ds f\) \(=\) \(\ds \int \map N {x, y} \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \int \paren {\frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2} } \rd y + \map h x\)
\(\ds \) \(=\) \(\ds \frac 1 2 \map \ln {\frac {1 + x y} {1 - x y} } + \map h x\)


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:

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

$\blacksquare$


Sources

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