 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/15b/c/solution

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

d τ = \frac{d t}{a ( t )} .

For

a = a_{0} e^{H_{inf} t}

, choose

τ \to 0

t \to + \infty

. Integration gives

τ = - \frac{e ^{- H_{inf} t}}{a _{0} H _{inf}} = - \frac{1}{a H _{inf}}, - \infty < τ < 0,

Writing a prime for

d / d τ

, the relations

\frac{d}{d t} = \frac{1}{a} \frac{d}{d τ}, \frac{a ^{'}}{a} = - \frac{1}{τ}

ϕ_{k}^{''} + 2 \frac{a ^{'}}{a} ϕ_{k}^{'} + c^{2} k^{2} ϕ_{k} = 0.

ϕ_{k} = a ϕ_{k} = - \frac{1}{H _{inf} τ} ϕ_{k} .

A direct substitution gives

ϕ_{k}^{''} + (c^{2} k^{2} - \frac{a ^{''}}{a}) ϕ_{k} = 0.

Here

a^{''} / a = 2/ τ^{2}

\frac{d ^{2} ϕ _{k}}{d τ ^{2}} + (c^{2} k^{2} - \frac{2}{τ ^{2}}) ϕ_{k} = 0 .

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