# Differential Equation: (1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2] dy = 0

Hello, can anyone solve this equation?

I can't figure it out,

(1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2] dy = 0

Thanks.

Rate:

### This is an exact equation

This is an exact equation
$(1 - xy)^{-2} \, dx + \left[ y^2 + x^2 (1 - xy)^{-2} \right] \, dy = 0$

Check for exactness:

$M = (1 - xy)^{-2}$

$\dfrac{\partial M}{\partial y} = -2(1 - xy)^{-3}(-x)$

$\dfrac{\partial M}{\partial y} = 2x(1 - xy)^{-3}$

$N = y^2 + x^2 (1 - xy)^{-2}$

$\dfrac{\partial N}{\partial x} = -2x^2(1 - xy)^{-3}(-y) + 2x(1 - xy)^{-2}$

$\dfrac{\partial N}{\partial x} = 2x^2y(1 - xy)^{-3} + 2x(1 - xy)^{-2}$

$\dfrac{\partial N}{\partial x} = 2x(1 - xy)^{-3} \left[ xy + (1 - xy) \right]$

$\dfrac{\partial N}{\partial x} = 2x(1 - xy)^{-3}$

$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$, hence, exact!