How to solve a differential equation with a cosine term?

[Reid]

Homework Statement

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

Homework Equations

\ddot{x}+k \cos{x}=0.

The constant k is postive.

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?

 

[Office_Shredder]

It's GOT to be something of the form

x=arccos(y) 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

 

[HallsofIvy]

Homework Statement

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

Homework Equations

\ddot{x}+k \cos{x}=0.

The constant k is postive.

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?

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

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