 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/17c/b/solution

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

Expand the exact solution about

t_{n + 1} = t

. The left side is

y (t + h) - y (t) = h y^{'} + \frac{h ^{2}}{2} y^{''} + \frac{h ^{3}}{6} y^{'''} + \frac{h ^{4}}{24} y^{(4)} + O (h^{5}) .

Since

f (t, y (t)) = y^{'} (t)

, the right side expands as

h y^{'} + h^{2} (1 + 2 α + β) y^{''} + \frac{h ^{3}}{2} (1 - β) y^{'''} + \frac{h ^{4}}{6} (1 + 2 α + β) y^{(4)} + O (h^{5}) .

The

h y^{'}

terms agree for every

α, β

, so the method is always consistent and is generically of order one.

The method is explicit exactly when

1 + α = 0

. With

α = - 1

, its

h^{2}

defect vanishes only for

β = 3/2

, so a generic explicit member has order one; that exceptional explicit member has order two and cannot satisfy the next order condition.

For the unrestricted method to have order at least three, the

h^{2}

and

h^{3}

coefficients must agree:

1 + 2 α + β = \frac{1}{2}, \frac{1 - β}{2} = \frac{1}{6} .

Solving gives

α = - \frac{7}{12}, β = \frac{2}{3} .

For these values,

1 + 2 α + β = 1/2

, so the right-side

h^{4}

1/12

, different from the left-side coefficient

1/24

. Hence the order is exactly three.

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