 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/14b/b/solution

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

P = K ρ_{c}^{1 + 1/ n} θ^{n + 1}, ρ = ρ_{c} θ^{n}, r = a ξ,

we have

\frac{1}{ρ} \frac{d P}{d r} = (n + 1) K ρ_{c}^{1/ n} \frac{d θ}{d r} .

The pressure-support equation therefore becomes

(n + 1) K ρ_{c}^{1/ n} \frac{1}{r ^{2}} \frac{d}{d r} (r^{2} \frac{d θ}{d r}) = - 4 π G ρ_{c} θ^{n} .

Choose

a^{2} = \frac{( n + 1 ) K}{4 π G} ρ_{c}^{(1 - n) / n} .

Changing from

r

ξ

\frac{1}{ξ ^{2}} \frac{d}{d ξ} (ξ^{2} \frac{d θ}{d ξ}) = - θ^{n} .

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