 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-25

Here

ϕ (u, h) = \frac{1}{4} [f (u) + 3 f (u + \frac{2 h}{3} f (u))] .

For

0 < h \leq 1

f

gives

∣ ϕ (u, h) - ϕ (v, h) ∣ \leq (K + \frac{K ^{2}}{2}) ∣ u - v ∣,

so part (ii) applies with a constant independent of sufficiently small

h

The exact solution has the Taylor expansion>

y (t + h) = y (t) + h f (y (t)) + \frac{h ^{2}}{2} f^{'} (y (t)) f (y (t)) + O (h^{3}) .

Meanwhile,

f (y + \frac{2 h}{3} f (y)) = f (y) + \frac{2 h}{3} f^{'} (y) f (y) + O (h^{2}),

so one numerical step from

y

y + h ϕ (y, h) = y + h f (y) + \frac{h ^{2}}{2} f^{'} (y) f (y) + O (h^{3}) .

The local error is therefore

O (h^{3})

. Taking

p = 2

in part (ii) proves the required second-order global-error bound. This method is a two-stage Runge-Kutta method.

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