Coupled 2nd-Order Non-linear ODEs

[cepheid]

Homework Statement

I'm trying to solve the equations:

$$\ddot{\phi} + 2\left(\frac{\cos \theta}{\sin \theta}\right) \dot{\theta}\dot{\phi} =0$$​

and

$$\ddot{\theta} - \sin \theta \cos \theta \dot{\phi^2} =0$$​

for $\theta(\lambda), \phi(\lambda)$ where the dots represent differentiation w.r.t the parameter $\lambda$.

Homework Equations

Chain rule, other differentiation tricks

The Attempt at a Solution

I've tried to tackle the first equation by writing:

$$\frac{\cos \theta}{\sin \theta} \dot{\theta} = \frac{1}{\sin \theta} \frac{d}{d\theta}(\sin \theta) \dot{\theta} = \frac{d}{d\lambda}[\ln (\sin \theta) ]$$​

I'm wondering if it's correct to do that. It seems to follow from the chain rule, but I'm not sure. If so, then the equation becomes:

$$\frac{\ddot{\phi}}{\dot{\phi}} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ]$$

$$\frac{1}{\dot{\phi}}\frac{d \dot{\phi}}{d\lambda} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ]$$

$$\frac{d \ln ( \dot{\phi})}{d\lambda} = -2\frac{d}{d\lambda}[\ln (\sin \theta) ]$$​

Then I integrated both sides w.r.t. lambda:

$$\ln ( \dot{\phi}) = - 2\ln (\sin \theta) + \mathrm{~const.}$$

$$\dot{\phi} \propto \frac{1}{\sin^2 \theta}$$​

Then I tried substituting this result into the second ODE, but it gave me:

$$\ddot{\theta} \propto \frac{\cos \theta}{\sin^3 \theta}$$​

which I don't know what to do with.

 

[cepheid]

Just bumping the thread. Any insight you might have, esp. regarding the last line, would be most helpful.