Weird ODE containing cos.


by Reid
Tags: weird
Reid
Reid is offline
#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?
Phys.Org News Partner Science news on Phys.org
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)
Office_Shredder
Office_Shredder is offline
#2
Nov11-08, 10:37 AM
Mentor
P: 4,499
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
HallsofIvy is offline
#3
Nov11-08, 12:51 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881
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.


Register to reply

Related Discussions
x^x. this is a weird one General Math 17
Fraction of a fraction General Math 16
Why would a circuit have a zero ohm resistor (3 black bands) in it? Electrical Engineering 13
something weird... General Physics 7