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Thread: Exact? Differential equation

  1. #1
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    Exact? Differential equation

    solve and Test for exactness... if not use another method
    $\displaystyle (1-xy)^{-2}dx + (y^2+x^2(1-xy)^{-2})dy = 0$



    this is not exact ... but i dont know what other method to use
    need suggestion
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  2. #2
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    Krizalid's Avatar
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    The following differential equation

    $\displaystyle M(x,y)+N(x,y)\frac{dy}{dx}=0,$

    is exact iff $\displaystyle \frac{{\partial N}}
    {{\partial x}} = \frac{{\partial M}}
    {{\partial y}}.$

    Did you check if this one is exact or not? Otherwise, there're 4 methods to transform an inexact ODE into a exact one.
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  3. #3
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    yep The equation is exact when i simplify....

    i need help on something else


    its on solving the differential equation
    since
    $\displaystyle \frac{\partial{F}}{\partial{y}} = y^2 + x^2(1-xy)^{-2}$

    $\displaystyle \frac{\partial{F}}{\partial{x}} = (1-xy)^{-2}$

    now im integrating the dx component since it is the easiest to integrate:
    $\displaystyle \int \frac{\partial{F}}{\partial{x}} = \int {(1-xy)}^{-2} = \frac{1}{(y(1-xy))} + Q(y) $ - since y is a constant

    ^
    now the previous part is differentated w/ respect to y and equate with $\displaystyle \frac{\partial{F}}{\partial{y}}$:

    so
    $\displaystyle \frac{x}{y(1-xy)^2} - \frac{1}{y^2(1-xy)} + Q'(y) = y^2 + x^2(1-xy)^{-2} $
    ^
    this part is my problem... i need to make it into a function of a y variable but i cannot do it correctly
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