The breakdown of the solution to the ODE $y' = \sqrt{4-2y}$. Suppose $y=y(x)$ be a real valued function such that 
$$
y(x) = \int_0^x \sqrt{4-2y(t)} dt.
$$
By differentiating both side with respect to $x$, we get the ODE
$$
y' = \sqrt{4-2y}, \quad y(0)=0.
$$
This is a separable equation which has 
$$
y_1(x) = -\frac 12 (2-x)^2 + 2
$$
as the solution. By inspection,
$$\begin{align}
y_1'(x) &=2-x, \\
\sqrt{4-2y_1(x)} &= |2-x|
\end{align}$$
so this solution works for $x\le 2$.

Is it possible to continue to solution pass $x=2$? I know one trivial answer which is
  $$
y(x) = \begin{cases} -\frac 12 (2-x)^2 + 2 &; x\le 2 \\
2 &; x>2.
\end{cases}
$$

Using the notation from the Picard-Lindelof theorem, we have 
$$
y' = f(y)
$$
where $f(y) = \sqrt{4-2y}$ so $f$ is not Lipschitz at $y=2$ so the breakdown at $x=2$ (which causes $y(2)=2$) is to be expected. However, I don't know much about ODE apart from the general local existence result so I'd be very glad if someone could tell me if there is any other possible extension of the solution pass $x=2$.

$$
\begin{align}
y'(x) &= \sqrt{4-2y(x)}\\
y'(x) &= \sqrt{2}\sqrt{2-y(x)}\\
\dfrac{y'(x)}{\sqrt{2-y(x)}} &= \sqrt{2}\\
\int\dfrac{y'(x)}{\sqrt{2-y(x)}} dx&= \int\sqrt{2}dx\\
-2\sqrt{-y(x)+2}&=\sqrt{2}x+K_1\\
y(x)&=-\dfrac{1}{4}(\sqrt{2}x+K_1)^2-2\\
\end{align}
$$
For, $y(0)=0$: 
$$K_1=\{-2;2\}.$$
Therefore, 
$$
y(x) = \begin{cases}
-\dfrac{1}{2}x(x-4), & \mbox{if } K_1 = -2 \mbox{, } x\geq 2 \\ 
-\dfrac{1}{2}x(x+4), & \mbox{if } K_1 = 2 
\end{cases}
$$
