# ODE on the plane

1. Jun 3, 2008

### wofsy

Can someone tell me how to solve the ODE,

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

2. Jun 3, 2008

### dirk_mec1

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

3. Jun 3, 2008

thanks

4. Jun 4, 2008

### HallsofIvy

Staff Emeritus
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. Jun 4, 2008

### Mute

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.

6. Jun 4, 2008

### HallsofIvy

Staff Emeritus
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.

Last edited: Jun 4, 2008
7. Jun 4, 2008

### Mute

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

Last edited: Jun 4, 2008
8. Jun 5, 2008

### HallsofIvy

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

9. Jun 5, 2008

### wofsy

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?