Unique solution for ode $y' = {\sqrt{1-y^2}} $ I was given to solve the next ode:
$y' = {\sqrt{1-y^2}} $
I found its solution: $y=sin(x+c)$
Now, I'm given that $y(0)=0$ and asked to show the only solution is $y=sin(x)$ in the region $(-\infty,\infty)$.
I get that c=$\pi k$, and therefore infinite number of solutions. 
Can I get any help?
 A: As written, your initial-value problem has a unique continuously-differentiable solution on $\mathbf{R}$:
$$
y(x) = \begin{cases}
  -1 & x < -\pi/2, \\
  \sin x & -\pi/2 \leq x \leq \pi/2, \\
  1 & \pi/2 < x.
\end{cases}
$$
The ODE $y' = \sqrt{1 - y^{2}} \geq 0$ implies $y$ is globally non-decreasing. Particularly, the sine function itself is not a solution.
As you've found, separating variables and integrating gives $\arcsin y = x + c$; the initial condition $y(0) = 0$ gives $c = 0$, so $y = \sin x$ in some neighborhood of $0$. (The choices $c = 2k\pi$ give precisely the same function, since $\sin(x + 2k\pi) = \sin x$ for all $x$, while $c = (2k - 1)\pi$ isn't a solution, since if $y = -\sin x$, then $y' = -\cos x = -\sqrt{1 - y^{2}} \neq \sqrt{1 - y^{2}}$.)
Patching the sine function at $x = \pm \pi/2$ with the signum function
$$
\operatorname{sgn} x = \frac{x}{|x|},\quad x \neq 0,
$$
gives a unique, continuously-differentiable solution, as is straightforward to show analytically and visually apparent in the slope field:

A: Let $y=\sin (x+c)$ obviously, $y'=\cos (x+c)$.
Now, the initial condition says that $\sin(c)=0$. Here we have two choices (adding multiples $2\pi$ yield the same solutions): $c=0$ or $c=\pi$.
What happens if we choose $c=0$? Well, than we have that, for example, $y'(0)=\cos(c)=1$. Nothing wrong there. But if we choose $c=\pi$? 
Then we'll have $y'(0)=\cos(c)=-1$. This says that our $y$, with a choice of $c=\pi$ doesn't solve the differential equation (There's a square root!).
So we have to IF $y=\sin (x+c)$ is a solution, AND $y(0)=0$, we have that $c=0$.
This however, doesn't show that it is the only solution. For that, you need the Picard–Lindelöf theorem.
A: But the answer they want you to prove is wrong! The general solution to that differential equation is $y=\sin(x+C)$, or $y=-1$, or $y=1$. Note the presence of two singular solutions where $y^{\prime}$ starts out zero and stays zero. But this doesn't apply in this case, one might think, because the initial condition is $y(0)=0$. So the only solution that satisfies the initial condition and the constraint that $y^{\prime}\ge0$ is $y(x)=\sin(x)$.  
But look at what happens when $x=\frac{\pi}2$. Then $y(x)=1$, a singular solution. And $y(x)$ can't be $\sin(x)$ for $x\in(\frac{\pi}2,\frac{3\pi}2)$ because then $y^{\prime}(x)=\cos(x)$ would be negative, in contradiction to the problem statement. Thus the solution turns out to be
$$y(x)=\left\{\begin{array}{rr}\sin(x),&0\le x\le\frac{\pi}2\\1,&x\ge\frac{\pi}2\end{array}\right.$$
Of the two singular solutions, $y=-1$ is unstable and $y=+1$ is stable and if the initial conditions start out at the first solution, at any time $x_0$ the system could switch to $y=\sin(x-x_0-\pi/2)$ and $\pi$ later arrive at $y(x)=1$ and stay there.
