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Nonlinear ODE

  1. Jun 4, 2016 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    $$y''+6y^{2/3}=0$$

    2. Relevant equations
    Nothing comes to mind

    3. The attempt at a solution
    I don't really know where to start. Any tricks or tips are appreciated. This isn't a homework question, but I posted here since I didn't know where else to post.

    Thanks for your time
     
  2. jcsd
  3. Jun 4, 2016 #2

    SteamKing

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    If it's not a HW problem, you can post it in one of the Technical Math Forums, like Calculus or Differential Equations. That's what those Fora are for.

    As to the solution to this particular ODE, probably a substitution of some kind to reduce the second order equation to a first order equation is the best approach.

    This article shows some general techniques for such transformations:

    http://users.math.msu.edu/users/gnagy/teaching/13-fall/mth340/L09-340.pdf
     
  4. Jun 4, 2016 #3

    joshmccraney

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    Thanks for the response. Should I move this thread then?

    Also, I found that link before posting, but it doesn't apply to this particular problem since the nonlinear ##2/3## exponent. Any other ideas or sources?
     
  5. Jun 4, 2016 #4

    Ray Vickson

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    Maple gets an "implicit" solution, defined via an equation of the form
    [tex] a - x = \int_0^{y(x)} \frac{dt}{\sqrt{b-t^{5/3}}} [/tex]
     
  6. Jun 4, 2016 #5

    joshmccraney

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    I didn't even think about letting a computer try it. Is that function at all related to the hypergeometric function? I ask because when solving in mathematica I also receive an implicit solution involving the hypergeometric function. Your solution looks much simpler, though.

    Any ideas how Maple arrived at such a solution?

    Also, if a boundary condition was ##y'(0)=0## then the maple solution would be $$-1=\frac{1}{\sqrt{b-y^{5/3}}}y'(x) \implies -1=0$$ Then doesn't this boundary condition precludes this solution?
     
    Last edited: Jun 5, 2016
  7. Jun 5, 2016 #6

    pasmith

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    Equations of the form [itex]y'' = f(y)[/itex] can always be reduced to [itex]\frac12 y'^2 = \int f(y)\,dy[/itex] by multiplying both sides by [itex]y'[/itex] and integrating.
     
  8. Jun 5, 2016 #7

    epenguin

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    Always useful to ask have I seen anything like this before? (Polya principle). You probably have, and you will again.

    LHS acceleration, RHS a force as function of distance y.

    When made definite integral, RHS minus potential energy difference LHS kinetic energy difference

    Discovery of conservation of energy., kinetic plus potential.

    Owed to
    https://www.aps.org/publications/apsnews/200812/physicshistory.cfm
     
    Last edited: Jun 7, 2016
  9. Jun 9, 2016 #8

    joshmccraney

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    Thanks pastime and epenguin!

    It does look like ##y=-x^6/125## comes close to analytically solving (everything except the boundary condition ##y(1)=0##). I forgot to mention the two associated boundary conditions: ##y'(0)=0## and ##y(1)=0##.
     
    Last edited: Jun 9, 2016
  10. Jun 10, 2016 #9

    pasmith

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    y = 0 is a solution...
     
  11. Jun 12, 2016 #10

    joshmccraney

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    It totally is! But it's trivial for the context of the equation. :(
     
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