 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/31d/d/solution

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

Put

v = e^{S}

and

p = S^{'}

p^{'} + p^{2} = \frac{1}{4} (1 + \frac{2}{x} - \frac{1}{x ^{2}}) .

Write

p \sim b_{0} + \frac{b _{1}}{x} + \frac{b _{2}}{x ^{2}} + \dots .

Matching the constant,

x^{- 1}

, and

x^{- 2}

b_{0}^{2} = \frac{1}{4}, 2 b_{0} b_{1} = \frac{1}{2}, b_{1}^{2} + 2 b_{0} b_{2} - b_{1} = - \frac{1}{4} .

For the recessive branch,

b_{0} = - \frac{1}{2}, b_{1} = - \frac{1}{2}, b_{2} = 1,

S_{-} (x) = - \frac{x}{2} - \frac{1}{2} lo g x - \frac{1}{x} + O (x^{- 2}) .

Since

y = e^{x /2} x^{- 1/2} v

, this gives

y_{-} (x) = x^{- 1} exp (- \frac{1}{x} + O (x^{- 2})) = \frac{1}{x} - \frac{1}{x ^{2}} + O (x^{- 3}) .

This agrees with the first two terms of part (a),

U (x) \sim \frac{1}{x} - \frac{1}{x ^{2}} + \frac{2}{x ^{3}} - \dots .

Continuing the Riccati recursion reproduces the factorial asymptotic series, so

U

is proportional to this recessive branch.

For the dominant branch,

b_{0} = \frac{1}{2}, b_{1} = \frac{1}{2}, b_{2} = 0,

and in fact

S_{+} (x) = \frac{x}{2} + \frac{1}{2} lo g x + constant

solves the phase equation exactly. It follows that

v_{+} (x) ≍ e^{x /2} x^{1/2}, y_{+} (x) ≍ e^{x} .

Indeed,

e^{x}

is an exact second solution of the original differential equation.

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