1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Weird ODE containing cos.

  1. Nov 11, 2008 #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. Nov 11, 2008 #2


    User Avatar
    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. Nov 11, 2008 #3


    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook