## Weird ODE containing cos.

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

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

2. Relevant equations

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

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?

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 Blog Entries: 1 Recognitions: Homework Help 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

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 Quote by Reid 1. The problem statement, all variables and given/known data I can not determine the solution to the diff. eq. 2. Relevant equations $$\ddot{x}+k \cos{x}=0$$. 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?
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 $dx= \sqrt{2C- 2sin(x)}dx$.