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.
Differential Equation: (1-xy)^-2 dx + [y^2 + x^2 (1-xy)^-2] dy = 0
New forum topics
- Please help me solve this problem using WSD (Working Stress Design) Method
- Physics: Uniform Motion
- Rescue at Sea
- Using two pumps [ Temporarily locked ]
- Emptying a Tank [ Temporarily locked ]
- Speed of a Plane [ Temporarily locked ]
- Range of an Airplane [ Temporarily locked ]
- Chemistry: Sugar Molecules
- Moving Walkways [ Temporarily locked ]
- Physics: Uniform Motion [ Temporarily locked ]
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