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  1. Apr 1, 2010 #1
    1. The problem statement, all variables and given/known data

    yy'' = (y')2 - (y')3

    2. Relevant equations

    3. The attempt at a solution

    y' = p(y)
    y'' - p'p

    yp' = p - p2

    dp/p + dp/(1-p) = dy/y

    ln|p|+ln|1-p| = ln|y|+c

    p-p2 = cy

    y' - y'2 = cy

    now what?
    how do I solve for y?
    I think I'm missing some stupid algebra thingy here, but can't figure it out.
  2. jcsd
  3. Apr 1, 2010 #2


    Staff: Mentor

    What does p(y) mean? I would normally take this to mean "p of y". Do you mean y' = py; i.e. p times y?
    What is y'' - p'p? That's not an equation. How does it relate to the equation above it?
  4. Apr 2, 2010 #3
    y' = p(y)
    p is a function of y
    should be y = p'p, not y - p'p.
    y'' = p'(y)y' (chain rule)
    y'' = p'(y)p(y) or p'p. :)
  5. Apr 2, 2010 #4


    User Avatar
    Homework Helper
    Gold Member

    Errm... [itex]\frac{d}{dp}\ln|1-p|=-\frac{1}{1-p}\neq\frac{1}{1-p}[/itex] :wink:
  6. Apr 2, 2010 #5
    oh. of course. stupid me. Now it all comes together.
    Thanks for pointing this out.

    It's always the little stuff that makes it problematic.
    Have a nice day! :)
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