 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-1/17b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

Let

e_{n} = y^{n} - y (t_{n})

. Subtracting the exact one-step relation from the numerical method and applying the stated Lipschitz continuity gives

∥ e_{n + 1} ∥ \leq (1 + h L) ∥ e_{n} ∥ + ∥ τ_{n + 1} ∥.

∥ τ_{n + 1} ∥ \leq C h^{p + 1}

∥ e_{n} ∥ \leq (1 + h L)^{n} ∥ e_{0} ∥ + C h^{p + 1} \sum_{j = 0}^{n - 1} (1 + h L)^{j} .

For

nh \leq t^{*}

(1 + h L)^{n} \leq e^{nh L} \leq e^{t^{*} L}

and

h^{p + 1} \sum_{j = 0}^{n - 1} (1 + h L)^{j} \leq \frac{e ^{t^{*} L} - 1}{L} h^{p} .

Consequently

0 \leq n \leq ⌊ t^{*} / h ⌋ max ∥ y^{n} - y (nh) ∥ \leq e^{t^{*} L} ∥ e_{0} ∥ + O (h^{p}) .

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