Find all solutions for ODE: $x'(t) = \sqrt{1 - x(t)^2}$, where $x(0)=0$ It's given that $x(t) = \sin(t)$, satisfies this when $\cos(t) \ge 0$ (and I understand why), but how can I find the other ones?
This is in the chapter that discusses the Global Uniqueness Theorem, so I guess I either have to prove that this is the only solution, or somehow create a set of solutions from $\sin(t)$ and show that they're unique.
So far I've only managed to show that $f(t, x) = \sqrt{1 - x(t)^2}$ is Lipschitz-continuous with respect to $x$, in any point $(t_0, x_0 \neq 1)$, but I'm not sure where that gets me.
Edit: there are many great comments here (admittedly, most went over my head), but given the context in which this was given, I'm looking for a solution which uses the Uniqueness Theorem. I think it's enough to prove that $x(t)=\sin(t)$ is unique when $\cos(t) \ge 0$ and that $x(t)=-\sin(t)$ is unique when $\cos(t) \le 0$, and the way of doing this probably goes through Lipschitz-continuity (of course, we can add any $2\pi k$, e.g $x(t) = \sin(x + 2\pi k)$, but those are essentially the same exact solutions).
 A: Notice that $a^2+b^2=1\iff \exists \theta\in[0,2\pi) \mid \begin{cases} a=\cos \theta\\ b=\sin \theta\end{cases}$
The definition of $x'(t)$ leads to $x(t)^2+x'(t)^2=1$ so there exists $\theta(t)$ so that $x,x'$ are trigonometric lines of this angle.
Now since $\cos$ and $\sin$ are continuous functions, and since the ODE imposes some regularity on $x(t),x'(t)$ then the function $\theta(t)$ itself needs to be continuous.
I have discussed further about the maximality in this other post: Valid operation: differential equations
Basically solutions are piecewise $\sin$ branching from time to time with $y=\pm1$ at tangency points.

Addendum: In your case the formula for $x'$ implies $x'(t)\ge 0$ so the solution is increasing (not necessarily strictly).
In the interval $[-\frac{\pi}2,\frac{\pi}2]$ the solution is $\sin(t)$ because of the initial condition.
Since this solution reaches $\pm 1$ at the bounds of the interval, the solution is then extended by constant branches outside this interval, this is the only way to keep it increasing (in loose sense). 
So the equation $x'^2+x^2=1$ has many maximal solutions ($\sin$ on an interval containing $0$ then branching to $\pm 1$ or $\sin$ again outside), but the equation $x'=\sqrt{1-x^2}$ has an unique continuous solution due to the monotonicity constraint.
$\begin{cases}
x(t)=-1 & t<-\frac{\pi}2\\
x(t)=\sin(t) & t\in[-\frac{\pi}2,\frac{\pi}2]\\
x(t)=1 & t>\frac{\pi}2
\end{cases}$
A: You may directly solve the ode as follows,
$$\frac{dx}{\sqrt{1 - x^2}} = dt$$
Integrate, with $x(0)=0$ for the lower bounds,
$$\int_0^x \frac{du}{\sqrt{1 - u^2}} = \int_0^t ds$$
to get
$$\sin^{-1}(x)-\sin^{-1}(0)=t$$
or,
$$\sin^{-1}(x)=t$$
Therefore, the only solutions are,
$$x=\sin(t+2\pi k)$$
where $-\pi/2 \le t \le \pi/2$ and $k$ is any integer. In other words, the solution is $\sin(t)$ with domain in the 1st and 4th quadrants.
A: The following argument may help. The ODE in your problem can be written as
$$
\frac{\dot{x} x}{\sqrt{1-x^2}} = x, \qquad x(0)=0
$$
Integrating you get
$$\int^t_0x(s)\,ds= 1-\sqrt{1-x^2} = 1-\dot{x}$$
Differentiation gives
$$ \ddot{x}+x=0$$
Hence
$x(t)=a\cos(t) + b\sin(t)$. The condition $x(0)=0$ implies $x(t)=b\sin(t)$. Still, you need
to verify the original equation
$$b\cos(t)=\sqrt{1-b^2\sin^2(t)}$$
 Which holds only of $b=\pm1$. To comfort with positive sign on the right-hand side of the original equation,  we take $b=1$. Also, this is all valid as long as $|t|\leq \pi/2$.
A: We observe the equation $x'(t)=\sqrt{1-(x(t))^2}$.
$x(t)=\sin (t)$ is a solution when $\cos(t) \geq 0$, which is true for $t\in (2k\pi - \frac{\pi}{2} , 2k\pi + \frac{\pi}{2} )$ and $x(t)=-\sin (t)$ is a solution when $\cos(t) \leq 0$, which is true for $t\in (2k\pi + \frac{\pi}{2} , 2k\pi + \frac{3\pi}{2} )$
With a given condition $x(0)=0$ solution is unique on any interval that contains $0$. That gives us that $x(t)=\sin (t)$ is unique on $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Now we want to check if there is a different way to expand this solution to other intervals of the same kind. For example, let $x$ be a solution such that $x(t)=\sin (t)$ for $t\in(-\frac{\pi}{2}, \frac{\pi}{2})$, but isn't necessarily $x(t)=\sin(t)$ for $t\in (2k\pi - \frac{\pi}{2} , 2k\pi + \frac{\pi}{2} )$ (for $k\neq 0$). Now define $y(t)=x(t+2k\pi)-x(2k\pi)$ for $t\in(-\frac{\pi}{2}, \frac{\pi}{2})$, for some $k\neq0$. $$y(0)=0$$ and $$y'(t)=x'(t+2k\pi)=\sqrt{1-(x(t+2k\pi))^2}=\sqrt{1-(y(t))^2}$$ Now we have that $y=\sin$ and therefore $x(t+2k\pi)=x(2k\pi)+sin(t)$ or, to put it differently, $x(t)-\sin(t-2k\pi)=const$.
Since $\sin(t-2k\pi)=\sin(t)$, we have $x(t)-\sin(t)=const$.
We conclude that there are infinitely many solutions on $\bigcup (2k\pi - \frac{\pi}{2} , 2k\pi + \frac{\pi}{2} )$, and on each interval of the form $(2k\pi - \frac{\pi}{2} , 2k\pi + \frac{\pi}{2} )$ solution is $\sin +const$ and additive constant can be anything, except for the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, where it has to be $0$.
Now we want to check of we can expand our solution on the intervals outside of $\bigcup (2k\pi - \frac{\pi}{2} , 2k\pi + \frac{\pi}{2} )$.
Following the same reasoning as above, we get that all solutions of the equation $x'(t)=\sqrt{1-(x(t))^2}$ on the intervals outside of $\bigcup (2k\pi - \frac{\pi}{2} , 2k\pi + \frac{\pi}{2} )$ have to be $x(t)=-\sin(t)+const$.
If we want solution on the largest possible domain, let's try to find it on $\mathbb{R}$. $x$ is continuous, so it must be true that $x(\frac{\pi}{2})=1$ and $x(-\frac{\pi}{2})=-1$ (because $x=\sin$ on $(-\frac{\pi}{2}, \frac{\pi}{2})$). If $x=-\sin+c$ on $(\frac{\pi}{2}, \frac{3\pi}{2})$), we have that $c=2$ and $x(\frac{3\pi}{2}))=3$. If $x=\sin+c$ on $(\frac{3\pi}{2}, \frac{5\pi}{2})$), we have that $c=4$ and so on. This way we can find a unique solution of the problem, defined on $\mathbb{R}$.
Edit: People have already said that this is a separable ODE (and the person who asked the question said that they don't know what separable means yet). I think what I've written can be understood by someone who didn't learn anything about "systematically" solving ODEs (including learning about separable equations). However, noticing that this is a separable equation is more universal way to solve this sort of problem.
Also note that $-\sin$ and $\cos$ can be used, in some way, interchangeably.
Edit2:
Picard's theorem gives us the existence and uniqueness of the solution on some interval around $0$ (the fact that $t_0=0$ and $x(t_0)=0$ isn't important because of the zeros, but because it is a condition of the form $x(t_0)=x_0$). There is another theorem which states that under certain conditions (conditions are much weaker than what we have here), if $I$ is maximal interval on which the solution (which exists) is defined then for every compact $K$ (in $\mathbb{R}$ in this case) there exist $t\in I$ such that $x(t)$ isn't in $K$.  In our case: you have a unique solution $x=\sin$ on $(-\delta,\delta)$ for some $\delta>0$. We know that it can be expanded at least on $(-\frac{\pi}{2},\frac{\pi}{2})$ and from the theorem I've mentioned we know it can be expanded even further.
One might ask: we know we can expand it from $(-\delta,\delta)$ to $(-\frac{\pi}{2},\frac{\pi}{2})$ as a $\sin$ function, but can we expand it in a different way?  Assume we have extended our solution to $(-\frac{\pi}{2},\frac{\pi}{2})$ and assume $\delta=sup\{t\in(-\frac{\pi}{2},\frac{\pi}{2})|x=\sin$ on $(t,\frac{\pi}{2})\}$. If $\delta=\frac{\pi}{2}$, then that's it. If $\delta<\frac{\pi}{2}$, then, because of the continuity of the solution, we have $x(\delta)=\sin(\delta)$. Let's observe interval $(\delta,\delta+2\epsilon)\subset(\delta,\frac{\pi}{2})$, for some $0<\epsilon<\delta$, and put $a=\delta+\epsilon$. Using translation (just like I've used it in the answer), we get that $x(t)=\sin(t-a)+x(a)$ for $t\in(a-\epsilon,a+\epsilon)$. From continuity we get that $x(a)=\sin(\delta)+\sin(\epsilon)$. And honestly, I hoped this would give us something nice, but as far as I see we can't prove this uniqueness and what I've written isn't entirely correct. Solution still has to be sinusoidal.
The question is: what is it that you need? Do you need to have unique solution and it's fine if it's only defined in a small neighborhood of $0$? Do you need solution on the largest possible domain? I think it's really hard to describe all solutions defined on $\mathbb{R}$ in an elegant mathematical way. I think something needs to be "sacrificed" in order to have nice representation.
However, we can't say that we don't understand what these solutions look like because they are all $\sin$ accompanied by translations of domain and codomain and we know that they must be exactly $\sin$ in small neighborhoods of $2k\pi$ points.
