 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-2/14d/b/solution

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

Substituting

x = t^{1/2} y (lo g t)

gives

y^{''} + (ϵ^{- 2} - \frac{1}{4}) y = 0.

Thus

k = ϵ^{- 2} - \frac{1}{4}

x (t) = \frac{t}{k} sin (k lo g t) .

Direct differentiation verifies both the equation and

x (1) = 0

\overset{x}{˙} (1) = 1

 New to topics? Read the docs here!