Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 31E / i

Table of contents
- Solution i

Solution

After division by

x^{2}

, the equation becomes

y^{''} + \frac{1}{x} y^{'} - \frac{1}{x ^{4}} y = 0.

The regular singular point criterion for a second-order equation would require

x (1/ x)

and

x^{2} (- 1/ x^{4})

to be analytic at zero. The second expression is

- x^{- 2}

, so the criterion fails:

x = 0 is an irregular singular point .

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