 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/39c/b/iii/solution

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

Put

δ = \frac{ν}{α}, η = \frac{y}{δ}, ψ = αx δ f (η) .

Then

u = αx f^{'} (η), v = - α δ f (η) .

ν / δ^{2} = α

, gives

f^{'''} + f f^{''} - (f^{'})^{2} = 0 .

The wall and far-field conditions become

f (0) = 0, f^{'} (0) = 1, f^{'} (\infty) = 0 .

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