Hello, can anyone solve this equation?

I can't figure it out,

$(1-xy)^{-2} dx + \left[ y^2 + x^2 (1-xy)^{-2} \right] dy = 0$

Thanks.

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

Check for exactness:

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

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

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

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

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

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

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

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

To solve this type of equation, see this page: https://mathalino.com/node/494

More information about text formats

Follow @iMATHalino

MATHalino

This is an exact equation

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

Check for exactness:

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

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

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

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

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

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

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

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

To solve this type of equation, see this page: https://mathalino.com/node/494

## Add new comment