 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-4/15a/iii/solution

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

Since

A^{'} / A = - i ℏ B = - \frac{i}{2} tan (ϕ - i t)

and

\frac{d}{d t} lo g cos (ϕ - i t) = i tan (ϕ - i t),

integration gives

A (t) = A_{0} [cos (ϕ - i t)]^{- 1/2},

with the branch chosen continuously from the initial value. The constant

A_{0}

fixes normalization.

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