 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/8/b/solution

t = y - 1

. Then

y^{2} - 1 = t (t + 2)

x = t^{σ} \sum_{j = 0}^{\infty} a_{j} t^{j}, a_{0} \neq = 0.

t^{σ - 1}

[2 σ (σ - 1) + σ] a_{0} = σ (2 σ - 1) a_{0} .

σ (2 σ - 1) = 0, σ_{1} = 0, σ_{2} = \frac{1}{2} .

Their difference is not an integer, so the Frobenius method produces two independent series without a logarithmic term:

x_{1} (y) = j = 0 \sum \infty a_{j} (y - 1)^{j}, x_{2} (y) = (y - 1)^{1/2} j = 0 \sum \infty b_{j} (y - 1)^{j}, a_{0} b_{0} \neq = 0.

Choose a branch of

(y - 1)^{1/2}

, or take

y > 1

for real solutions. A general local solution is

C_{1} x_{1} + C_{2} x_{2}

by linear independence and the superposition principle. The analytic factors converge at least for

∣ y - 1∣ < 2

, up to the next singular point at

y = - 1

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