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Math Help - about differential equation

  1. #1
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    about differential equation

    it seems a pretty easy question,but really confuses me.
    anyway,it is a differential equation:
    (2xy-y^3)y`+1=0

    many thanks
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  2. #2
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    Integrable Impliciation

    f(x,y)'=f'(x,y)y'... Abuse of notation perhaps.

    \int (2xy+y^3)y' +1 dy = \int 0 dy

    xy^2+\frac{y^4}4+y=C

    Now the hard part: isolate y.
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  3. #3
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    i'm afraid that you got a wrong answer.. Still,i appreciate your help
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  4. #4
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    Quote Originally Posted by Media_Man View Post
    f(x,y)'=f'(x,y)y'... Abuse of notation perhaps.

    \int (2xy+y^3)y' +1 dy = \int 0 dy

    xy^2+\frac{y^4}4+y=C

    Now the hard part: isolate y.
    No, that's incorrect. x is not a constant and cannot be treated as one in the first integral.

    farlist, that is NOT "a pretty easy question" but here is what I came up with: (2xy+ y^3)\frac{dy}{dx}+ 1= y(2x+ y^2)\frac{dy}{dx}+ 1= 0. Just because it looked like it might simplify things, I let u= 2x+y^2. Then \frac{du}{dx}= 2+ 2y\frac{dy}{dx} so that y\frac{dy}{dx}= \frac{1}{2}\frac{du}{dx}- 1 and y(2x+ y^2)\frac{dy}{dx}= uy\frac{dy}{dx}= \frac{1}{2}u\frac{du}{dx}- u. Putting that into the equation: (2x+ y^2)\left(y\frac{dy}{dx}\right)+ 1= 0 becomes \frac{1}{2}u\frac{du}{dx}- u+ 1= 0. That's a separable equation and should be easy to solve.
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  5. #5
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    We can make dx+(2xy-y^3)dy=0 exact via the integrating factor e^{y^2}. Check the book on the \frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right) form. This yields:

    e^{y^2}dy+e^{y^2}(2xy-y3)dy=0

    Then the solution is F(x,y)=xe^{y^2}+T'(y)

    Solving for T'(y) we get T=-1/2 e^{y^2}(y^2-1)

    The solution is then:

    xe^{y^2}-1/2e^{y^2}(y^2-1)=c

    in which  y(x) can be solved for explicitly (how?).

    And this is how I checked the solution in Mathematica:

    Code:
    n[7]:=
    ysol = First[Derivative[1][y][x] /. 
        Solve[D[x*Exp[y[x]^2] - 
            (1/2)*Exp[y[x]^2]*(y[x]^2 - 1), 
           x] == 0, Derivative[1][y][x]]]
    Simplify[(2*x*y[x] - y[x]^3)*ysol + 1]
    
    Out[7]=
    1/(y[x]*(-2*x + y[x]^2))
    
    Out[8]=
    0
    Last edited by shawsend; May 27th 2009 at 10:08 AM.
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  6. #6
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    ok.maybe it is not that EASY.Actually what i figure out is exactly the same as Shawsend's anwser,but i didn't check it in Matlab.
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