 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-310/2/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

δ_{m}^{''} + H δ_{m}^{'} ≃ 4 π G a^{2} \overset{ρ}{ˉ}_{m} δ_{m} = \frac{3}{2} H_{0}^{2} Ω_{m, 0} a^{- 1} δ_{m} = \frac{3 C}{2 y} δ_{m} .

From

y^{'} = y H

and part (b),

y^{'2} = C (1 + y), y^{''} = \frac{C}{2}, H y^{'} = \frac{C ( 1 + y )}{y} .

The expanding branch has

y^{'} > 0

. Applying the chain rule yields

δ_{m}^{''} = C (1 + y) \frac{d ^{2} δ _{m}}{d y ^{2}} + \frac{C}{2} \frac{d δ _{m}}{d y}, H δ_{m}^{'} = \frac{C ( 1 + y )}{y} \frac{d δ _{m}}{d y} .

Substitute these expressions and divide by

C (1 + y) > 0

\frac{d ^{2} δ _{m}}{d y ^{2}} + \frac{2 + 3 y}{2 y ( 1 + y )} \frac{d δ _{m}}{d y} - \frac{3}{2 y ( 1 + y )} δ_{m} = 0.

y > 0

; it keeps radiation in the expansion rate and matter in the slowly varying self-gravity source.

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