 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/38d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

u = - ψ_{y}, v = ψ_{x},

which satisfies continuity identically. Introduce

D = (\frac{μ ^{2}}{ρB})^{1/3}, U_{0} = (\frac{B ^{2}}{ρ μ})^{1/3},

so that

Δ = D x^{2/5}, U = U_{0} x^{1/5}, η = \frac{y}{D x ^{2/5}} .

ψ = U Δ f (η) = U_{0} D x^{3/5} f (η) .

Then

u = - U_{0} x^{1/5} f^{'}

and

v = U_{0} D x^{- 2/5} (\frac{3}{5} f - \frac{2}{5} η f^{'}) .

A direct differentiation shows that the terms containing

η f^{'} f^{''}

cancel:

u u_{x} + v u_{y} = U_{0}^{2} x^{- 3/5} [\frac{1}{5} (f^{'})^{2} - \frac{3}{5} f f^{''}] .

Also,

u_{yy} = - \frac{U _{0}}{D ^{2}} x^{- 3/5} f^{'''},

while the delta scaling gives

b (x) δ (y) = \frac{B}{D} x^{- 3/5} δ (η) .

D

and

U_{0}

imply

\frac{ρ U _{0} D ^{2}}{μ} = 1, \frac{B D}{μ U _{0}} = 1.

f^{'''} + \frac{1}{5} (f^{'})^{2} - \frac{3}{5} f f^{''} = δ (η) .

Thus

c_{1} = \frac{1}{5}, c_{2} = - \frac{3}{5} .

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