 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/5c/a/solution

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

For a normalized equation

y^{''} + P (x) y^{'} + Q (x) y = 0,

x_{0}

is an ordinary point when

P, Q

are analytic there. It is a regular singular point when

(x - x_{0}) P (x)

and

(x - x_{0})^{2} Q (x)

Here

P (x) = \frac{1}{x} - 1, Q (x) = \frac{λ}{x},

x = 0

y = \sum_{n = 0}^{\infty} a_{n} x^{n} .

Equating the coefficient of

x^{n}

gives

(n + 1)^{2} a_{n + 1} + (λ - n) a_{n} = 0,

and hence

a_{n + 1} = \frac{n - λ}{( n + 1 ) ^{2}} a_{n} .

Therefore

a_{n} = a_{0} \frac{\prod _{j = 0}^{n - 1} ( j - λ )}{( n ! ) ^{2}}, y = a_{0} \sum_{n = 0}^{\infty} \frac{\prod _{j = 0}^{n - 1} ( j - λ )}{( n ! ) ^{2}} x^{n} .

The series terminates exactly when

λ = N

is a nonnegative integer: the factor with

j = N

then makes

a_{N + 1} = 0

. Thus the polynomial solutions occur for

λ = 0, 1, 2, \dots .

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