 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-4/37a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Divide the two outer boundary equations respectively by

z

and leave the derivative equation as written:

A / z^{2} + B + C z + D z^{3} = 0, - A / z^{2} + B + 2 C z + 4 D z^{3} = 0.

Their difference gives

C z + 3 D z^{3} = 2 A / z^{2}

z \to \infty

, the inner equations require

A, B \to 1/4

. The two outer equations at leading order thus give

C z + D z^{3} = - 1/4

and

C z + 3 D z^{3} = 0

, so

C = - \frac{3}{8 z} + O (z^{- 2}), D = \frac{1}{8 z ^{3}} + O (z^{- 4}) .

The inner equations also give

A + B = 1/2 - C - D

and

B - A = - 2 C - 4 D

. Solving and substituting yields

A = \frac{1}{4} - \frac{3}{16 z} + O (z^{- 2}), B = \frac{1}{4} + \frac{9}{16 z} + O (z^{- 2}), C \sim - \frac{3}{8 z}, D \sim \frac{1}{8 z ^{3}} .

Replacing

z

b / a

gives all four requested approximations.

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