 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/9e/a/solution

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

Since

a (τ) = (τ / τ_{0})^{2}

, the conformal Hubble parameter is

H = 2/ τ

. With

Ω_{M} = 1

, the mode equation becomes

δ^{''} + \frac{2}{τ} δ^{'} - \frac{6}{τ ^{2}} δ = 0.

Substitution of

δ = τ^{β}

gives

β (β - 1) + 2 β - 6 = (β - 2) (β + 3) = 0.

The two independent solutions are therefore

τ^{2}

and

τ^{- 3}

, so

δ (k, τ) = A (k) τ^{2} + B (k) τ^{- 3} .

 New to topics? Read the docs here!