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ODE on the plane

by wofsy
Tags: plane
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wofsy
#1
Jun3-08, 11:21 AM
P: 707
Can someone tell me how to solve the ODE,

d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?
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dirk_mec1
#2
Jun3-08, 02:34 PM
P: 677
You can use seperation of variables and then integrate twice (with an integration constant!)
wofsy
#3
Jun3-08, 03:53 PM
P: 707
thanks

HallsofIvy
#4
Jun4-08, 07:38 AM
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ODE on the plane

Those two equations are completely independent- so it is not necessary to "separate variables". You are just solving two separate second order differential equations. And, in fact you just need to integrate each twice.
Mute
#5
Jun4-08, 12:10 PM
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Quote Quote by HallsofIvy View Post
Those two equations are completely independent- so it is not necessary to "separate variables". You are just solving two separate second order differential equations. And, in fact you just need to integrate each twice.
Hrm? They don't look independent to me. The derivatives are with respect to t, not x or y. Integrating the x equation would give, for instance,

[tex]\dot{x(t)} - \dot{x(t_0)} = \int_{t_0}^{t}d\tau~\cos y(\tau)[/tex]

which isn't so useful if you can't solve for what y(t) is.
HallsofIvy
#6
Jun4-08, 05:29 PM
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If d2x/dt2= cos(x) is a single equation in the dependent variable x as a function of t. There is absolutely no reason to introduce y. Since the independent variable "t" does not appear in the equation, I would use "quadrature":

Let v= dx/dt so that d2x/dt2= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v2= sin(x)+ C. dx/dt= v= [itex]\sqrt{2(sin(x)+ C)}[/itex] or
[tex]\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt[/tex]
That left side is an "elliptical integral".

Of course, y will be exactly the same, though possibly with different constants of integration.
Mute
#7
Jun4-08, 11:36 PM
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Quote Quote by HallsofIvy View Post
If d2x/dt2= cos(x) is a single equation in the dependent variable x as a function of t. There is absolutely no reason to introduce y. Since the independent variable "t" does not appear in the equation, I would use "quadrature":

Let v= dx/dt so that d2x/dt2= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v2= sin(x)+ C. dx/dt= v= [itex]\sqrt{2(sin(x)+ C)}[/itex] or
[tex]\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt[/tex]
That left side is an "elliptical integral".

Of course, y will be exactly the same, though possibly with different constants of integration.
Ah, I see, I didn't parse the problem the way it was intended to be read. I read it as

[tex]\frac{d^2x}{dt^2} = -\cos x \frac{d^2y}{dt^2} = -\cos y [/tex]

i.e.,

[tex]\frac{d^2x}{dt^2} = -\cos y[/tex]
and
[tex]\cos x \frac{d^2y}{dt^2} = \cos y [/tex]


This is why I always use some sort of punctuation in between separate equations written on the same line. =P
HallsofIvy
#8
Jun5-08, 06:59 AM
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And how do you know that was how it was "intended to be read"?
wofsy
#9
Jun5-08, 07:26 AM
P: 707
I was able to get to the elliptic integral. But I have no idea what it looks like. Further, if x and y are both the same elliptic integral then the orbits in the plane should be fairly simple. But what do they look like?


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