# ODE on the plane

wofsy
Can someone tell me how to solve the ODE,

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

dirk_mec1
You can use seperation of variables and then integrate twice (with an integration constant!)

wofsy
thanks

Homework Helper
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.

Homework Helper
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,

$$\dot{x(t)} - \dot{x(t_0)} = \int_{t_0}^{t}d\tau~\cos y(\tau)$$

which isn't so useful if you can't solve for what y(t) is.

Homework Helper
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= $\sqrt{2(sin(x)+ C)}$ or
$$\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt$$
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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Homework Helper
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= $\sqrt{2(sin(x)+ C)}$ or
$$\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt$$
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

$$\frac{d^2x}{dt^2} = -\cos x \frac{d^2y}{dt^2} = -\cos y$$

i.e.,

$$\frac{d^2x}{dt^2} = -\cos y$$
and
$$\cos x \frac{d^2y}{dt^2} = \cos y$$

This is why I always use some sort of punctuation in between separate equations written on the same line. =P

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