 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1/13a/b/solution

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

Because the parameter enters through

λ w y

L (y; λ_{0}) = L (y; λ) - (λ - λ_{0}) w y = y^{m + 1} - ε^{m} μ w y .

Use this as

f

in part (a). The necessary orthogonality condition is

0 = \int_{0}^{1} y_{0} (y^{m + 1} - ε^{m} μ w y) d x .

With

y = ε y_{0} + ε^{2} y_{1}

and

μ = O (1)

y^{m + 1} = ε^{m + 1} y_{0}^{m + 1} + O (ε^{m + 2}), w y = εw y_{0} + O (ε^{2}) .

Consequently

0 = ε^{m + 1} [\int_{0}^{1} y_{0}^{m + 2} d x - μ \int_{0}^{1} w y_{0}^{2} d x] + O (ε^{m + 2}) .

The normalization makes the second integral one. Divide by

ε^{m + 1}

to obtain

μ = \int_{0}^{1} y_{0}^{m + 2} d x + O (ε) .

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