exact differential equation

(x2-4xy-2y2)dx+(y2-4xy-2x2)dy=0

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  1. I assume you are looking for a solution?
    We want a function F(x,y) such that

    Fx(x,y) = x^2 - 4xy - 2y^2
    and
    Fy(x,y) = y^2 - 4xy - 2x^2

    To make sure there is such an F, let
    M = x^2 - 4xy - 2y^2
    N = y^2 - 4xy - 2x^2

    My = -4x - 4y
    Nx = -4y - 4x

    So, things look good.

    Our solution is
    F(x,y) = 1/2 x^3 - 2x^2y - 2xy^2 + 1/3 y^3 = C

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  2. Your differential equation is of the form
    M(x,y) dx + N(x,y) dy = 0
    with dN/dx = dM/dy (those being partial derivatives)
    The last equation is a necessary and sufficient condition for a solution of the form
    f(x,y) = 0
    In your case
    df/dx (partial) = x^2 -4xy -2y^2
    and
    df/dy (partial) = y^2 -4xy -2x^2

    f(x,y) = x^3/3 -2yx^2 -2y^2x + g(y)
    y^2 -4xy -2x^2 = -2x^2 -4yx -dg/dy
    -dg/dy = y^2
    g(y) = -y^3/3 + C

    Thus the solution is
    x^3/3 -2yx^2 -2y^2x -y^3/3 = C

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  3. Steve and I disagree on a couple of terms. He could be right. My math is rusty. Check the math yourself. There are some good tutorials on the "exact differential" method online.

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  4. Steve's first term should have read (1/3)x^3 while the last term should be positive (+(1/3)y^3).

    It was probably a typo in each case, so there was not really a disagreement, especially everyone uses the same method of solution.

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  5. Finally answer will be x^3+y^3-6xy(x+y)=c

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