Weird ODE containing cos.

  1. 1. The problem statement, all variables and given/known data

    I can not determine the solution to the diff. eq.

    2. Relevant equations

    [tex]\ddot{x}+k \cos{x}=0[/tex].

    The constant k is postive.

    3. The attempt at a solution

    I tried solving it with the methods I know and it all ended up in a big mess. I am just not used to the second term, \cos. Doe's anyone know what to do? :confused:
  2. jcsd
  3. Office_Shredder

    Office_Shredder 4,487
    Staff Emeritus
    Science Advisor
    Gold Member

    It's GOT to be something of the form

    [tex]x=arccos(y)[/tex] for some y a function of t, based on an argument of 'this is really fricking hard otherwise'. Try the substitution and see what you get
  4. HallsofIvy

    HallsofIvy 41,265
    Staff Emeritus
    Science Advisor

    Since the independent variable, which I will call "t", does not appear explicitely in the equation, this is a candidate for "quadrature". Let [itex]y= dx/dt[/itex]. Then [itex]d^2x/dt^2= (dy/dx)(dx/dt)[/itex] by the chain rule. But since [itex]y= dx/dt[/itex], that is [itex]d^2x/dt^2= y dy/dx[/itex] and the equation converts to the first order equation, for y as a function of x, [itex]y dy/dx= -cos(x)[/itex] which is [itex]y dy= -cos(x)dx[/itex]. Integrating, [itex](1/2)y^2= sin(x)+ C[/itex]. Now we have [itex]y^2= -2 sin(x)+ 2C[/itex] or [itex]y= dx/dt= \sqrt{2C- 2sin(x)}[/itex] so [itex]dx= \sqrt{2C- 2sin(x)}dx[/itex].

    I am not sure that is going to be easy to integrate, but that gives the solution.
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