Existence and Uniqueness Theorem of $y'=1+2\sqrt{x-y}$ I solved the following differential equation:
$$y'=1+2\sqrt{x-y}$$ 
the solution I found is $\sqrt{x-y}=-x+C$ and $x=y(y\ne0)$
after that I was asked the following questions:
a) Find a solution that that maintain $y(1)=3$ and find it's domain.
when I input $y(1)=3$ into the equation I get $(c-1)^2=-2$ and that's impossible. I don't understand why it is considered a solution and what is the domain.
b) Find two different solutions that maintain $y(1)=1$ and explain why it does not contradict the Existence and Uniqueness Theorem.
When I input $y(1)=1$ I get $C=1$ I think that the equation is continuous for $y\gt0$ and $y\lt 0$ so for each we have one solution. Is this true.
Also, I don't fully understand what does it mean that a differential equation has a "solution". I would like a small explanation.
Answer for any of the questions above would be highly appreciated.
 A: First of all, when a differential equation has a solution, it means there there is some differentiable function over some domain, such that, if you plug it into the differential equation, you get a true statement.
For example, if we take the function $y=x$, we have $y'=1$ and $y-x=0$ for the domain $x\in\mathbb{R}$, and the differential equation is satisfied. Great.
If $x<y$, then we immediately have a problem, because the differential equation is undefined at such points. Thus, $y(1)=3$ is not compatible with this differential equation. (I'm assuming we're not messing around with complex numbers!)
Now, if you solve your equation $\sqrt{x-y}=-x+C$ for $y$, you get:
$$y=x-(C-x)^2$$
Plugging $(1,1)$ into that, we get that $C=1$, so we have the function $y=x-(1-x)^2$, or
$$y=-(x^2-3x+1)$$
In this case, $y'=3-2x$, while the RHS of the differential equation comes out to:
$$1+2\sqrt{x+(x^2-3x+1)} = 1+2\sqrt{x^2-2x+1}$$
Now, for this to match the LHS, we need the square root to evaluate to $(1-x)$ and not $(x-1)$. This happens if $x<1$, so that solution works on the domain $(-\infty,1]$. To the right of $x=1$, that function no longer solves the equation, and we'd have to go with $y=x$ instead.
So, the point $(1,1)$ is kind of funny: it is on two different solutions coming in from the left, but only one going out to the right. This isn't the behavior we usually see from differential equations, which is why you are asked to reconcile it with the uniqueness theorem. The answer to that question is that the uniqueness theorem applies in the interior of the DiffEQ's domain of continuity, but the whole line $y=x$ is on the boundary of that domain, so we expect squirrely stuff to happen.
To be explicit, there are two solutions passing through $(1,1)$, both with full domain:
$$y_1(x)=x$$
and
$$y_2(x) = \begin{cases}-(x^2-3x+1) & x\leq 1\\ x & x>1 \end{cases}$$
