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How to solve a second order diff eq?

  1. Feb 13, 2013 #1
    If have this equation:

    [itex] \frac {d^2x}{dt^2}=-\frac{x}{1+x^2} [/itex]

    How do I solve it?
     
  2. jcsd
  3. Feb 13, 2013 #2

    mfb

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    Did you try the usual methods, like separation of variables or a direct integration?
     
  4. Feb 13, 2013 #3

    I like Serena

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    How about numerical methods, like they are applied here?
    (I do not believe there is a nice analytic solution.)
     
  5. Feb 13, 2013 #4

    lurflurf

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    $$\int \! \dfrac{d^2x}{dt^2}\dfrac{dx}{dt} \, \mathrm{d}t=-
    \int \! \frac{x}{1+x^2}\dfrac{dx}{dt} \, \mathrm{d}t$$
     
  6. Feb 14, 2013 #5
    The solution to this equation should be:

    [itex] x(t) = sin(t) [/itex]

    I simply don't know how to get there ...

    I have integrated but I get something like integ of 1/sqr(-ln(1+x^2)) dx?! :(
     
  7. Feb 14, 2013 #6
  8. Feb 14, 2013 #7

    mfb

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    This would give ##x''=-x##, and ##1+x^2 \neq 1## (in general), so it is not a solution.

    The first integral should give something like ln(1+x^2).
     
  9. Feb 14, 2013 #8
    @ mfb I am out of ideas. I have a problem that I have been struggling with for half a year and still haven't solved it. I posted it once but didn't get an exact complete solution. https://www.physicsforums.com/showthread.php?t=668751&goto=newpost It has to be done without conservation of energy. (Just using forces) Thanks in advance.
     
  10. Feb 14, 2013 #9
    introduce a new variable
    [itex]p=\frac{dx}{dt}[/itex]

    by the chain rule we have:
    [itex]\frac{d^2x}{dt^2}=\frac{dp(x(t))}{dx}=\frac{dp}{dx}\frac{dx}{dt}=\frac{dp}{dx}p[/itex]
    This means that
    [itex]pdp = \frac{-1}{1+x^2}dx[/itex]

    integrate!

    [itex]\int pdp = \int \frac{-1}{1+x^2}dx[/itex]
    [itex]\frac{1}{2}p^2 = -\frac{1}{2}\ln(1+x^2)+C_1[/itex]

    and because p is the derivative dx/dt, we get

    [itex]\frac{dx}{\sqrt{-\ln(1+x^2)+C_1}} = 1[/itex]

    integrate!

    [itex]\int\frac{1}{\sqrt{\ln(1+x^2)+C_1}}dx = t + C_2[/itex]


    Well, not a very nice solution because it's implicit.
     
  11. Feb 14, 2013 #10

    I like Serena

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    Let us denote ##\dot x## to be the same as ##dx \over dt##.

    Then:
    $$\ddot x = - \frac{x}{1+x^2}$$
    $$2 \ddot x \dot x = -\frac{1}{1+x^2} \cdot 2x \cdot \dot x$$
    $$\dot x^2 = -\ln(1+x^2) + C$$

    And there it ends! :cry:
    No nice integration after this!
     
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