= Solution
For example, choose $h=1/4$ and $k=1/2$ around $(a,b)=(0,0)$. On the rectangle, $1+y\geq1/2$, so $f=-\sqrt{1+y}$ is continuous, $|f|\leq\sqrt{3/2}$ and $|\partial f/\partial y|\leq1/\sqrt2$. The <mean value theorem> therefore supplies a <Lipschitz constant> $L=1/\sqrt2$. These choices even satisfy $Mh<k$ and $Lh<1$.
While $y>-1$, <separation of variables> gives $2\sqrt{1+y}=2-x$. Therefore the uniquely determined branch up to its first arrival at $-1$ is
$$
\boxed{y(x)=\left(1-\frac x2\right)^2-1,\qquad 0\leq x\leq2.}
$$
It reaches $y=-1$ at $x=2$, with derivative zero. The restriction $x\leq2$ is essential: extending the same quadratic beyond two would make its derivative positive and violate the original differential equation. Beyond that time the <waiting-time continuation at a non-Lipschitz equilibrium> is nonunique.
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