 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/15b/b/solution

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

Separate continuity gives

ρ_{r} = ρ_{r 0} a^{- 4} = β \frac{3 c ^{2} H _{0}^{2}}{8 π G} a^{- 4}, ρ_{Λ} = ρ_{Λ0} = \frac{3 c ^{2} H _{0}^{2}}{8 π G},

t = t_{0}

, where

a = 1

H_{0}^{2} + K c^{2} = (β + 1) H_{0}^{2},

K c^{2} = β H_{0}^{2} .

At general

a

H^{2} + \frac{β H _{0}^{2}}{a ^{2}} = H_{0}^{2} (\frac{β}{a ^{4}} + 1),

which rearranges to

H^{2} = \frac{H _{0}^{2}}{a ^{4}} (β - β a^{2} + a^{4}) .

Part (a), applied to

ρ = ρ_{r} + ρ_{Λ}

and

P = ρ_{r} /3 - ρ_{Λ}

, gives

\dot{H} + H^{2} = - H_{0}^{2} (\frac{β}{a ^{4}} - 1) .

Subtracting the displayed formula for

H^{2}

yields

\dot{H} = - β \frac{H _{0}^{2}}{a ^{4}} (2 - a^{2}) .

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