Find all solutions of ${\dot x}^2=1-x^2$ and study their regularity. Can anyone help me with this exercise?
"Find all solutions defined on $[0,+\infty)$ to the differential equation
$$
\begin{cases}
{\dot x}^2=1-x^2,\\
x(0)=0,\\
\dot x(0)=1.
\end{cases}
$$
and study their regularity."
Why two initial condition ? Should I use the following identity? 
$$ \dot{(\dot x x)}=\ddot x x + \dot x \dot x.$$
Thanks in advance.

UPDATE. My ideas:
First of all we observe the presence of two equilibria ($\pm 1$). They are not solutions because they do not respect the initial conditions. Suppose $1-x^2\neq 0$. Since $\dot x (0) = 1$, for sign permanence we will have $\dot x> 0$ on $[0, \delta]$ (for a suitable $\delta>0$). We can make the square root to both members and remove the absolute value at $\dot x$, that is$$
\begin{cases}
{\dot x}=\sqrt{1-x^2},\\
x(0)=0,\\
\dot x(0)=1.
\end{cases}
$$
At this point, since $1-x^2$ is non-zero, we can divide and by separation of the variables I get that the solution on $[0, \delta]$ is $x(t)=sin (t)$. I note that I can extend this solution with uniqueness on $[0,\, \pi / 2]$. At this point I can:


*

*continue with the sin function,

*glue the $\sin$ and the constant function $1$,

*glue the $\sin$ the constant functions $\pm 1$ and $\cos$ (the gluing must be done in the points where $\sin$ and $\cos$ are equal to $1$ or $-1$).
I claim that the possible solutions are only those listed above.
Question: is it sufficiently formal? Do you agree with my solution and with my final claim?
 A: Here is some intuition that might be useful. The evolution of this ODE can be interpreted as a simple physical model. Consider a particle rolling in a potential $V(x) = x^2$ which has a tiny flat part close to $x=\pm 1$ (the total energy of this particle is $\dot{x}^2 + x^2 = 1$ which is our ODE). We start off at $x=0$ with velocity $v=1$. The particle has just enough energy to climb up and manage to arrive at the very edge of the flat part at $x=1$ with velocity $v=0$. 


Since the particle is now placed at the very edge the further evolution is ambiguous. The first option is that it just sits still forever at the edge since it's technically on a flat part so there is no force forcing it down. The second option is that it immediately starts to roll back down. The third option is that it sits still for a fixed amount of time and then starts to roll back down. If it rolls back down it will have just enough energy to climb up at $x=-1$ and the same story repeats.
The case where the particle always immediately rolls back down corresponds to $x(t) = \sin(t)$. The case where it sits still forever on the edge corresponds to
$$x(t) = \left\{\matrix{\sin(t) & 0 \leq t < \frac{\pi}{2}\\1 & \frac{\pi}{2}\leq t}\right.$$
The case where it sits still for a time $\Delta t$ corresponds to
$$x(t) = \left\{\matrix{\sin(t) & 0 \leq t < \frac{\pi}{2}\\1 & \frac{\pi}{2}\leq t < \frac{\pi}{2} + \Delta t\\\sin(t-\Delta t) & \frac{\pi}{2} +\Delta t < t < \frac{\pi}{2}+\Delta t+ \pi\\\cdots & \cdots}\right.$$
where $\cdots$ depends on what happens when it arrives at the edge at $x=-1$. Writing down an expression for the solution gets quite hairy, but based on the description above it's not hard to understand how the solutions we can get will look like. Below is a random example for $x(t)$:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
A: hint
Differentiating the first equation gives
$$2\dot x \ddot x=-2 x \dot x$$
thus
$$\ddot x+x=0$$
and
$$x=A\cos(t)+B\sin(t)$$
the initial conditions yields to
$$x=\sin(t)$$
