Дифф. Уравнение для прямых в полярных координатах

[Бабай]

Привет формучанам!

Я вот послал уже это сообщение в одном англоязычном форуме, но пока не получил ответа. И вспомнил про ваш форум.

Извините за то, что оно на английском. Сейчас времени нет переводить. Я его скопировал:

1. The problem statement, all variables and given/known data

Problem from Arnold's "Mathematical Methods of Classical Mechanics" on page 59.

Find the differential equation for the family of all straight lines in the plane in polar coordinates.

2. Relevant equations

$\Phi = \displaystyle\int^{t_2}_{t_1} \sqrt {{\dot{r}}^2 + r^2{}\dot{\phi}^2}\,dt$

$\frac {d}{dt}\frac {\partial{L}}{\partial{\dot{q_i}}} - \frac {\partial{L}}{\partial{q_i}} = 0$


3. The attempt at a solution

L is the integrand.

We have two equations:

$\frac {d}{dt}\frac {\dot{r}}{\sqrt {{\dot{r}}^2 + r^2{}\dot{\phi}^2}} = \frac {r\dot{\phi}^2}{\sqrt {{\dot{r}}^2 + r^2{}\dot{\phi}^2}}$

$\frac {d}{dt}\frac {r^2\dot{\phi}}{\sqrt {{\dot{r}}^2 + r^2{}\dot{\phi}^2}} = 0$

The second gives: $\frac {r^2\dot{\phi}}{\sqrt {{\dot{r}}^2 + r^2{}\dot{\phi}^2}} = c$

If c=0 we have the derivative of phi zero, that is phi would be constant and we are (essentially) done. If phi is constant, we have a bundle of lines passing through the origin. But what about r? Do we get problems in the origin? Polar coordinates are not defined there, are they?
If c is not zero, then the first equation can be rewritten as:
$\frac {d}{dt}\frac {\dot{r}}{r^2\dot{\phi}} = \frac {\dot{\phi}}{r}$

I cannot get an idea of how to solve it for r-dot AND phi-dot. Tried to differentiate the left side and see what happens, but somehow nothing attractive comes out.
So how to proceed? Is it the right way? Or is there anything I can't see at a glance that helps?

[V.V.]

А что мешает сделать естественные замены функций, решить в терминах $x$, $y$ и перейти обратно к $r$, $\varphi$?