 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2/5c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The identity

(n + 1)^{3} - 2 n^{3} + (n - 1)^{3} = 6 n

shows that

y_{n}^{(p)} = h^{3} n^{3} /6

is a particular solution. Combining it with the homogeneous solution from part (b) gives

y_{n} = A + B n + \frac{h ^{3} n ^{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}, y_{N} - y (1) = - \frac{h ^{2}}{6} .

Thus the absolute endpoint error is $h^{2} /6$ , consistent with the second-order truncation error of the centred scheme.

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