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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 15B / d / Solution Created 2026-09-24 Updated 2026-09-25

For

β = 4

,

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

The branch expanding from

a = 0

has

H = \frac{H _{0} ( 2 - a ^{2} )}{a ^{2}}

until it approaches

a^{2} = 2

. With

x = a^{2}

,

\overset{x}{˙} = 2 a \overset{a}{˙} = 2 a^{2} H = 2 H_{0} (2 - x), x (0) = 0.

Solving this linear equation gives the critical closed radiation--cosmological-constant solution

a (t) = 2 (1 - e^{- 2 H_{0} t}) .

For

H_{0} t ≪ 1

,

1 - e^{- 2 H_{0} t} \sim 2 H_{0} t, a (t) \sim 2 H_{0} t .

For

H_{0} t ≫ 1

,

a (t) = 2 (1 - \frac{1}{2} e^{- 2 H_{0} t} + O (e^{- 4 H_{0} t})),

so

a (t) ⟶ 2

exponentially from below.

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