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Second-order nonlinear ordinary differential equation

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Given the Second-order nonlinear ordinary differential equation

    x''(t)=1/(x(t)^2)

    Find x(t).


    2. Relevant equations

    I tried use Laplace transforms, and solving it using linear methods but that is not useful.


    3. The attempt at a solution

    I tried to find t(x) and got to dt=dx/((C-2GM/x)^0.5) or something like that.

    I guess you could find t(x) then find [inverse t(x)] = x(t)
    But I would like to know how to solve it properly really.
     
  2. jcsd
  3. Nov 20, 2009 #2

    tiny-tim

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    Hi lewis198! :smile:

    Standard trick: multiply both sides by x'(t) :wink:
     
  4. Nov 20, 2009 #3

    lanedance

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    similar outcome is to susbtitute to get a seprable equation then intgerate twice (ash means derivative w.r.t. t)
    u = x'
    then
    x" = u' = (du/dx)(dx/dt) = (du/dx)
     
  5. Nov 21, 2009 #4
    Hi lanedance, I tried that method before and got

    (1/2)*(u^2)=1/(x^2)

    dt=dx/((C-2GM/x)^0.5).

    this will therefore give me an integral t(x).

    But I need x(t). It will be quite messy doing the inverse won't it?

    I'm not sure where to go from multiplying LHS and RHS by x'(t).

    Is there a more elegant way to get x(t)? For example if I had

    ax''+bx'+cx=f(x)

    I could get y=A*e^(mt)+B*e(mt)

    But since in my characteristic equation b=0 and c=0 my m quadratic equation is void.
     
  6. Nov 25, 2009 #5
    I'll just find t(x) then t-1(x)
     
  7. Nov 25, 2009 #6

    tiny-tim

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    (try using the X2 tag just above the Reply box :wink:)

    No, the RHS is wrong and you've left out the constant of integration. :wink:
     
  8. Nov 25, 2009 #7

    lanedance

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    yeah i'm not too sure, don't know if Tim has any other ideas, but simple generic general solutions don't always apply to non-linear de's - this one is gets a little crazy near x=0, and tends to a straight line for x>>1

    So the general solution may not be able to be solved simply for t^(-1). That said if you have the right boundary conditions, this one could simplify a bit... (in particular if you could set the 1st constant of integration to zero)
     
    Last edited: Nov 25, 2009
  9. Nov 25, 2009 #8

    Mark44

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    ash?
     
  10. Nov 25, 2009 #9

    lanedance

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    there's a fire... few mistakes, so i clarified below
    --------------------------------------------------------------------------------

    similar outcome is to susbtitute to get a separable equation then integrate twice (dash means derivative w.r.t. t)
    [tex] u = x' = \frac{dx}{dt} [/tex]
    then
    [tex] x'' = u' = \frac{du}{dt} =\frac{du}{dx} \frac{dx}{dt} = \frac{du}{dx}u = \frac{1}{x^2} [/tex]
     
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