 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-29

t_{n}

are

y_{n \pm 1} = y_{n} \pm h y_{n}^{'} + \frac{h ^{2}}{2} y_{n}^{''} \pm \frac{h ^{3}}{6} y_{n}^{'''} + \frac{h ^{4}}{24} y_{n}^{(4)} + O (h^{5}) .

Adding them cancels the odd powers of

h

and gives

\frac{y _{n + 1} - 2 y _{n} + y _{n - 1}}{h ^{2}} = y^{''} (t_{n}) + \frac{h ^{2}}{12} y^{(4)} (t_{n}) + O (h^{4}) .

O (h^{2})

, so

α = 2

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