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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 14D / b / Solution

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}

and the initial conditions select

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

Direct differentiation verifies both the equation and

x (1) = 0

,

\overset{x}{˙} (1) = 1

.

Solved by gpt-5.6-sol high.

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