
#1
Jun308, 11:21 AM

P: 707

Can someone tell me how to solve the ODE,
d2x/dt2 = cosx d2y/dt2 = cosy in the plane? 



#2
Jun308, 02:34 PM

P: 664

You can use seperation of variables and then integrate twice (with an integration constant!)




#3
Jun308, 03:53 PM

P: 707

thanks




#4
Jun408, 07:38 AM

Math
Emeritus
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Thanks
PF Gold
P: 38,879

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.




#5
Jun408, 12:10 PM

HW Helper
P: 1,391

[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. 



#6
Jun408, 05:29 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

If d^{2}x/dt^{2}= 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 d^{2}x/dt^{2}= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v^{2}= 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. 



#7
Jun408, 11:36 PM

HW Helper
P: 1,391

[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 



#8
Jun508, 06:59 AM

Math
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PF Gold
P: 38,879

And how do you know that was how it was "intended to be read"?




#9
Jun508, 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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