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Easy nonlinear ODE

  1. Jan 5, 2009 #1
    Hello, i am trying to solve this nonlinear ODE

    y'y''=-1

    can someone help me?

    p.s maybe 2y'y''=-2 => (y'y')'=-2...
     
  2. jcsd
  3. Jan 5, 2009 #2

    Dick

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    That's a good start. Now integrate ((y')^2)'=(-2).
     
  4. Jan 5, 2009 #3
    οκ. [tex](y')^2=-2x+c[/tex] and

    [tex]y(x)=\int\sqrt{-2x+c}\hspace{3}dx+c_2=[/tex][tex]-\frac{2\sqrt{2}}{3}\sqrt{(c_1 - x)^3}+c_2[/tex]

    but i see that [tex]y(x)=\frac{2\sqrt{2}}{3}\sqrt{(c_1 - x)^3}+c_2[/tex] is sol'n too. How to show that?
     
    Last edited: Jan 5, 2009
  5. Jan 5, 2009 #4

    Dick

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    If (y')^2=c-2x then y' is either +sqrt(c-2x) or -sqrt(c-2x). There are two solutions.
     
  6. Jan 5, 2009 #5
    Thanx! o:)
     
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