= Solution
The identity
$$
(n+1)^3-2n^3+(n-1)^3=6n
$$
shows that $y_n^{(p)}=h^3n^3/6$ is a <particular solution>. Combining it with the homogeneous solution from part (b) gives
$$
y_n=A+Bn+\frac{h^3n^3}{6}.
$$
The condition $y_0=0$ gives $A=0$, and $y_{-1}=y_1$ gives $B=-h^3/6$. Hence
$$
y_n=\frac{h^3}{6}(n^3-n).
$$
The exact initial-value solution is $y(t)=t^3/6$. Since $N=1/h$,
$$
y_N=\frac{1-h^2}{6},
\qquad
y_N-y(1)=\boxed{-\frac{h^2}{6}}.
$$
\b[Thus the absolute endpoint error is $h^2/6$, consistent with the second-order <truncation error> of the centred scheme.]
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