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Math Help - finding a singular solution

  1. #1
    Newbie
    Joined
    Oct 2009
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    finding a singular solution

    I am having trouble with this problem. I believe that there is a singular solution involved in it yet I do not know how to find it.

    (z'')(z^2)=(z')^3

    so far i have rewritten the problem as
    (z'')(z^2)-(z')^3=0 , made the substitution u=z' and continued to solve that using separation of variables. I still do not understand how to find the singular solution though

    thank you for any help
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  2. #2
    Super Member
    Joined
    Aug 2008
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    How about we write it as just y'' y^2=(y')^3

    By inspection you can see y=k is a solution including the solution y=0 right? So letting y'=p and making that substitution, I get pp'y^2=p^3 and I can do the division to obtain \frac{dp}{dy}p^{-2}=y^{-2} only if p\neq 0 and y\neq 0. Doing that, and integrating twice, I obtain the solution:

    c_1 y+\ln(y)=x+c_2. Now note that the solutions y(x)=k are not particular cases of this solution and it looks like the general solutions are asymptotically tangent to the solution y=0 but not the other y=k solutions. I'm a little unsure about this part as to whether all the y=k solutions are singular or just the y=0 one.
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