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

Table of contents
- Solution 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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