 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-2/8c/b/solution

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

For

m = 1

Y = (1 - η^{2}) /2

. Expanding both sides of the perturbation equation gives

(1 - η^{2}) y^{''} - 2 η y^{'} + 6 λ y = 0.

For

y = \sum_{r \geq 0} a_{r} η^{r}

, coefficient comparison gives

a_{r + 2} = \frac{r ( r + 1 ) - 6 λ}{( r + 2 ) ( r + 1 )} a_{r} .

The even and odd series terminate exactly when

6 λ = k (k + 1)

with the corresponding parity. Thus every positive integer

k

gives the regular polynomial solution, a Legendre polynomial up to scale.

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