= Solution
The <vector field> is continuous on the cube. Part (b) gives $M=1$, and part (c) supplies a uniform <Lipschitz constant> $L=1$. In the quantitative <Picard-Lindelöf theorem>, the independent-variable half-width and the dependent-variable radius are both one, so
$$
h=\min\{1,1/1\}=1.
$$
Therefore
$$
\boxed{\text{There is a unique solution for every }x\in[-1,1].}
$$
The global derivative bound also makes the cube condition explicit. Along any solution segment starting at the initial point,
$$
\|Y(x)-B\|_\infty
=\left\|\int_0^xF(s,Y(s))\,ds\right\|_\infty\leq|x|.
$$
Thus it cannot leave the dependent-variable ball before $|x|=1$. Bounded derivatives give limiting endpoint values, and local existence extends to them because $F$ is defined continuously on all space. Uniform Lipschitz continuity ensures that all such local pieces agree on overlaps, proving uniqueness throughout the closed interval.
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