 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/8/iii/solution

Substitute

y = v P_{n}

into the Legendre differential equation and expand using the product rule. The terms proportional to

v

cancel because

P_{n}

is itself a solution. What remains is

(1 - x^{2}) P_{n} v^{''} + [2 (1 - x^{2}) P_{n}^{'} - 2 x P_{n}] v^{'} = 0.

Multiplying by

P_{n}

identifies an exact derivative:

\frac{d}{d x} [(1 - x^{2}) P_{n}^{2} v^{'}] = 0.

The latter form is convenient for reduction of order on intervals where

P_{n} \neq = 0

and

x \neq = \pm 1

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