= Solution
{c}
Here $y'=x/(1+y^2)$ and $y(0)=1$. Since $x\geq0$, the solution is nondecreasing, so $y(x)\geq1$. In fact $0\leq y'\leq1/2$ on $[0,1]$, which also bounds the solution and ensures it extends across this entire interval. The smooth right-hand side makes $y$ twice continuously differentiable, with <second derivative>
$$
y''=\frac1{1+y^2}-\frac{2x^2y}{(1+y^2)^3}
=\frac{(1+y^2)^2-2x^2y}{(1+y^2)^3}.
$$
For $y\geq1$ and $0\leq x\leq1$,
$$
(1+y^2)^2\geq4y^2\geq2y\geq2x^2y,
$$
so $y''\geq0$. The subtracted term in the first expression is nonnegative, giving the useful bound
$$
\boxed{0\leq y''(x)\leq\frac1{1+y(x)^2}\leq\frac12\qquad(0\leq x\leq1).}
$$
The integral form of the <Taylor remainder> now gives
$$
\begin{aligned}
\tau_n
&=\frac{y(x_{n+1})-y(x_n)}h-f(x_n,y(x_n))\\
&=\frac1h\int_{x_n}^{x_{n+1}}(x_{n+1}-t)y''(t)\,dt.
\end{aligned}
$$
Therefore the <normalized local truncation error> satisfies the stronger estimate
$$
\boxed{0\leq\tau_n\leq\frac h4\leq\frac h2.}
$$
This proves the requested bound for $n=1,\ldots,N-1$, and also for $n=0$, which will be needed to start the <global error> estimate. For $N=1$ the stated range is empty, but the estimate at $n=0$ still applies.
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