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Math Help - difficult(for me) DE

  1. #1
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    difficult(for me) DE

    can someone here help me out with an idea how to solve this DE :
    2xyy' - x^2y'^2 - y^2 = 2S
    S is real constant. i know that this is DE that cannot be solved according to derivative, or it is but if i solved it, i mean if i write it down this way :
    y'=F(x,y)
    i think it would be impossible to solve it, but this DE can't be written down in form y=F(x,y') or x=F(y,y') easily either. of course i can try to solve it some of thease ways, but maybe there is some trick hiding here?
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  2. #2
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    How about solving for y'. Don't worry about the \pm part, just use + for now. Tell you what, I went through it quick so you can check for errors if you want to. Doing that I get:

    \left(y+\sqrt{y^2+k}\right)dx-xdy=0

    and I think an integrating factor in terms of y can be found to make it exact.
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  3. #3
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    Quote Originally Posted by noteiler View Post
    can someone here help me out with an idea how to solve this DE :
    2xyy' - x^2y'^2 - y^2 = 2S
    S is real constant. i know that this is DE that cannot be solved according to derivative, or it is but if i solved it, i mean if i write it down this way :
    y'=F(x,y)
    i think it would be impossible to solve it, but this DE can't be written down in form y=F(x,y') or x=F(y,y') easily either. of course i can try to solve it some of thease ways, but maybe there is some trick hiding here?
    Note that your ODE can be written as

    \left(xy'-y\right)^2 = -2 S

    or

     <br />
x y' - y = \pm \sqrt{-2S}.<br />

    It's now linear!
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  4. #4
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    thanks alot, i knew that i didn't see something
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