 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/15b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Differentiating the ansatz gives

\dot{ϕ} = \frac{2 λ ϕ _{0}}{s i n h ( 2 λ t )} .

The Raychaudhuri equation therefore becomes

\dot{H} = - 2 λ^{2} \frac{ϕ _{0}^{2}}{M _{Pl}^{2}} csch^{2} (2 λ t) .

Since

d [coth (2 λ t)] / d t = - 2 λ csch^{2} (2 λ t)

, the requested specific solution takes the integration constant to be zero:

H (t) = λ \frac{ϕ _{0}^{2}}{M _{Pl}^{2}} coth (2 λ t) .

Now

\overset{a}{˙} / a = H

, so

lo g a = \frac{ϕ _{0}^{2}}{2 M _{Pl}^{2}} lo g sinh (2 λ t) + lo g a_{0},

and

a (t) = a_{0} [sinh (2 λ t)]^{ϕ_{0}^{2} / (2 M_{Pl}^{2})} .

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