 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/37b/a/iv/solution

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

Δ y_{h} = A cos ϕ + B sin ϕ .

For the constant and

cos 2 ϕ

Δ y_{p} = \frac{3 M}{2 b ^{2}} + \frac{M}{2 b ^{2}} cos 2 ϕ .

Indeed, applying

d^{2} / d ϕ^{2} + 1

multiplies the

cos 2 ϕ

- 3

. Thus

Δ y = A cos ϕ + B sin ϕ + \frac{M}{2 b ^{2}} (3 + cos 2 ϕ) .

The

B sin ϕ

term only changes the definition of

b

at this order, so one may set

B = 0

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