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Math Help - Help with this DE problem

  1. #16
    Member
    Joined
    Sep 2009
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    82

    ok:

    Quote Originally Posted by whitepenguin View Post
    y' = \frac{y}{3x - y^2}
    Thank you
    \dfrac{dy}{dx}=\dfrac{y}{3x-y^2}
    \dfrac{dy}{dx}(3x-y^2) = y
    (3x-y^2)dy = ydx
    (3x-y^2)dy-ydx=0

    now this is in the form of what my professor calls "exact" except we still need an integrating factor. They are quite difficult to find, for me, and honestly i cannot get this one to work out so im sorry but, i dont think i can help.
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  2. #17
    MHF Contributor Calculus26's Avatar
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    there is no one method of finding int factor but there are special cases

    Two things

    1)If 1/M (dN/dx - dM/dy) = f(y) then int factor is e^(integralf(y))


    2) If 1/N (dM/dy- dN/dx) = f(x) tehn int factor is e^(integralf(x))


    In your case 1) applies and 1/y^4 is the int factor
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