 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-317/3/ii/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

\frac{d P}{d r} = - \frac{G m ρ}{r ^{2}}, \frac{d m}{d r} = 4 π r^{2} ρ .

Eliminating

m

yields

\frac{1}{r ^{2}} \frac{d}{d r} (\frac{r ^{2}}{ρ} \frac{d P}{d r}) = - 4 π Gρ, P = K_{NR} ρ^{5/3} .

ρ = ρ_{c} θ^{3/2}, r = a ξ, a^{2} = \frac{5 K _{NR}}{8 π G} ρ_{c}^{- 1/3} .

\frac{1}{ξ ^{2}} \frac{d}{d ξ} (ξ^{2} θ^{'}) = - θ^{3/2}, θ (0) = 1, θ^{'} (0) = 0.

ξ_{1}

θ

, with

R = a ξ_{1}

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