 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/4/b/solution

For a Frobenius exponent

r

, put

p_{0} = lim_{x \to 0} x P (x) = 1

and

q_{0} = lim_{x \to 0} x^{2} Q (x) = 0

r (r - 1) + p_{0} r + q_{0} = r^{2} = 0,

r_{1} = r_{2} = 0.

A fundamental pair has the repeated-root Frobenius form

y_{1} = \sum_{n = 0}^{\infty} u_{n} x^{n}, y_{2} = y_{1} ln x + \sum_{n = 0}^{\infty} v_{n} x^{n}, y = C_{1} y_{1} + C_{2} y_{2}

on a positive-

x

interval. In this case a real analytic solution is particularly simple:

y_{1} = 1 - x

satisfies the equation exactly. A convenient logarithmic companion, obtained below, is

y_{2} = x + (1 - x) ln [x / (1 - x)]

for

0 < x < 1

. It is independent of

y_{1}

because of its logarithmic divergence at zero.

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