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Weird ODE containing cos.

by Reid
Tags: weird
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Reid
#1
Nov11-08, 10:31 AM
P: 36
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?
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Office_Shredder
#2
Nov11-08, 10:37 AM
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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
HallsofIvy
#3
Nov11-08, 12:51 PM
Math
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Thanks
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P: 39,553
Quote Quote by Reid View Post
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?
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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