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Solving Equation with Second Derivative

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve the following equation for x(y). (use no differential functions)


    x(0)' and x(0) are known.

    2. Relevant equations


    3. The attempt at a solution
    I'm a bit unsure as to what to do next but I can easily make a messy formula up to approximate the result.


    k1 and k2 are a constant value.
    The bigger E is, the bigger the error. For an infinite recursive use of this formula with an infinitesimal E the right result would be achieved. I don't know what it converges to though.

    Any help for this problem or/and similar problems would be great. Thank you.
    Last edited: Jan 4, 2012
  2. jcsd
  3. Jan 4, 2012 #2


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    What does "use no differential functions" mean?

    [strike]Why don't you just integrate the whole thing twice, or separate variables?[/strike]

    [edit]It was a bit more complex than I thought - never mind the second line there[/edit]
    Last edited: Jan 4, 2012
  4. Jan 4, 2012 #3
    the equation does not contain explicit dependency on y, so try multiplying both sides on dx/dy. You'll get a sort of "conservation of energy" for the problem:
    (dx/dy)^2/2 + 1/x = const
    const depends on the initial conditions. This way you are left with solving 1st order diff equation
  5. Jan 4, 2012 #4
    It means that I want something with which I can calculate values directly. I do not want Integrals or Derivatives in my final formula and I think this is very much possible.

    I'm not sure what you mean by explicit but for a given k1, k2 and y there's a single well defined value of x.
    Also I'm sorry I'm not very good at handling differentials, I don't even have much practice with calculus. I don't understand how you got that from multiplying the first equation by dx/dy. (assuming that was what you did?)
    Last edited: Jan 4, 2012
  6. Jan 4, 2012 #5
    so you just need an answer, right?
  7. Jan 4, 2012 #6
    I've had my time for solving it alone for fun. But it doesn't look so simple and I am extremely curious as for the answer.

    So, yes please.
    I'm trying to find an answer but with the proof too. Otherwise I'll just forget it because I won't understand it.
    If you could do that, it'd be really nice. Thanks.
  8. Jan 4, 2012 #7
    try wolframalpha.com...
  9. Jan 4, 2012 #8


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    Your equation is
    [tex]\frac{d^2x}{dy^2}= x^{-2}[/tex]
    As quZz said, y does not appear explictly so you can use "quadrature". let v= dx/dy. Then
    [tex]\frac{d^2x}{dy^2}= \frac{dv}{dy}= \frac{dv}{dx}\frac{dx}{dy}= v\frac{dv}{dx}= x^{-2}[/tex]

    That is now a separable first order equation:
    [tex]v dv= x^{-2}dx[/tex]
    [tex]\frac{1}{2}v^2= -x^{-1}+ C[/tex]

    [tex]v^2= 2(C- x^{-1})[/tex]
    [tex]v= \frac{dx}{dy}= 2\sqrt{C- x^{-1}}[/tex]
    which is also a separable first order equation:
    [tex]\frac{dx}{\sqrt{C- x^{-1}}}= 2dy[/tex]
  10. Jan 4, 2012 #9
    I didn't know how to write second derivatives in Wolfram Alpha and I thought it wouldn't give a proof for such a complex problem.

    Anyway thanks a lot. Problem solved.

    Edit: There's a slight mistake on the your result by the way, v^2=2*(C-x^-1). It's times 2, not times "2 squared". It was much easier to follow then wolfram though. Thanks.
    Last edited: Jan 4, 2012
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