Solution
= Solution
Two solutions for $x\geq2$ are
$$
\boxed{y_1(x)=-1,\qquad y_2(x)=-1-\frac{(x-2)^2}{4}.}
$$
For the second, $y_2'=-(x-2)/2=-\sqrt{|1+y_2|}$ because $x\geq2$; both have value $-1$ and derivative zero at two. They are distinct for every $x>2$. Local <Lipschitz continuity> fails at the equilibrium, so the existence of these two continuations is compatible with the <Picard-Lindelöf theorem>.