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Non-linear second order from calculus of variation I can't solve

  1. Jul 23, 2013 #1
    Hi,

    I derived the equation:

    [tex]1+(y')^2-y y''-2y\left(1+(y')^2\right)^{3/2}=0[/tex]

    Letting [itex]y'=p[/itex] and [itex]y''=p\frac{dp}{dy}[/itex], I obtain:

    [tex]\frac{dp}{dy}=\frac{1+p^2-2y(1+p^2)^{3/2}}{yp}[/tex]

    I believe it's tractable in p because Mathematica gives a relatively simple answer:

    [tex]p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}\\
    \\
    -\frac{i \sqrt{-y^2+y^4-4 y^2 C[1]+4 C[1]^2}}{y^2-2 C[1]}
    \end{cases}[/tex]

    I don't see how to get to that answer and I was wondering if someone here could help me with this.

    Thanks,
    Jack
     
  2. jcsd
  3. Jul 23, 2013 #2

    SteamKing

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    What does 'C[1]' represent?
     
  4. Jul 23, 2013 #3
    I should have changed that to just c. It's a constant:

    [tex]p=\begin{cases}\frac{i \sqrt{-y^2+y^4-4 y^2 c+4 c^2}}{y^2-2 c}\\
    \\
    -\frac{i \sqrt{-y^2+y^4-4 y^2 c+4 c^2}}{y^2-2 c}
    \end{cases}[/tex]
     
  5. Jul 24, 2013 #4

    fzero

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    Try the substitution

    $$ w = \frac{\sqrt{1+p^2}}{y}.$$

    You should find that

    $$ \frac{dw}{dy} = -2y w^2,$$

    which is separable.
     
  6. Jul 24, 2013 #5
    Hi !

    A laborious method below :
     

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  7. Jul 24, 2013 #6
    You guys are just shear art in here. [itex]\frac{dw}{dy}=-2w^2y[/itex] was tough for me to go through but I got it. Still working through Jacquelin's. I'm sure it's correct though. :)
     
    Last edited: Jul 24, 2013
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