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Can someone tell me how to solve the ODE,

d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?

d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?

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- Thread starter wofsy
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Can someone tell me how to solve the ODE,

d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?

d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?

- #2

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

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thanks

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HallsofIvy

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Mute

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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.

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HallsofIvy

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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.

Let v= dx/dt so that d

[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.

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Mute

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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.

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

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- #8

HallsofIvy

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

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