 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2/20/a/solution

v = (x^{2} - 1)^{l}

gives

(x^{2} - 1) v^{'} = 2 l xv

; the identity also holds for

l = 0

. Differentiate it

l + 1

times using the Leibniz rule. The quadratic factor has only its first two derivatives nonzero, so

(x^{2} - 1) v^{(l + 2)} + 2 (l + 1) x v^{(l + 1)} + l (l + 1) v^{(l)} = 2 l [x v^{(l + 1)} + (l + 1) v^{(l)}] .

y = v^{(l)}

and collecting terms gives

(x^{2} - 1) y^{''} + 2 x y^{'} - l (l + 1) y = 0

- 1

(1 - x^{2}) y^{''} - 2 x y^{'} + l (l + 1) y = 0.

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