# System of Implicit Non-Linear First Order ODEs

I have an extremely messy system of differential equations. Can anyone offer any ideas for a general solution?  p(t) is a function of t, and A is a constant.

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I have an extremely messy system of differential equations. Can anyone offer any ideas for a general solution?  p(t) is a function of t, and A is a constant.
An idea would be to rewrite ##\cos \arctan \frac{\dot{y}}{\dot{x}}## as ##\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}}##, and similar for the other equation. Then try to simplify it a bit.

Didn't think of that, thanks! I'll keep working on it.

pasmith
Homework Helper
I have an extremely messy system of differential equations. Can anyone offer any ideas for a general solution?  p(t) is a function of t, and A is a constant.
You may be best served by switching to intrinsic coordinates $(s, \psi)$ where $$\dot x = \dot s \cos\psi, \\ \dot y = \dot s \sin \psi.$$ Your system is then $$(\dot s - p) \cos \psi = 0 \\ (\dot s - p) \sin \psi = -At.$$ Now either $\dot s = p$ or $\cos \psi = 0$. The first is impossible unless $A = 0$. The second requires that $\dot x = 0$ and $\dot y = \pm p - At$.

If $A = 0$ then $\dot s = p$ and $\psi$ can be an arbitrary function of $t$.

You may be best served by switching to intrinsic coordinates $(s, \psi)$ where $$\dot x = \dot s \cos\psi, \\ \dot y = \dot s \sin \psi.$$ Your system is then $$(\dot s - p) \cos \psi = 0 \\ (\dot s - p) \sin \psi = -At.$$ Now either $\dot s = p$ or $\cos \psi = 0$. The first is impossible unless $A = 0$. The second requires that $\dot x = 0$ and $\dot y = \pm p - At$.

If $A = 0$ then $\dot s = p$ and $\psi$ can be an arbitrary function of $t$.
Sorry, but it looks like something went wacky with your latex code. Could you retype it? It's hard to understand what your saying.

Sorry, but it looks like something went wacky with your latex code. Could you retype it? It's hard to understand what your saying.
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Basically he says to substitute ##\dot x = \dot s \cos\psi##, ##\dot y = \dot s \sin \psi##. Then it's very simple to arrive at a solution.

BTW, perhaps the first solution should have been ##\dot x=0##, ##\dot y=p\pm At##.