 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/4/a/solution

Divide by the leading coefficient for

x \neq = 0, 1

y^{''} + P (x) y^{'} + Q (x) y = 0, P (x) = \frac{1}{x}, Q (x) = \frac{1}{x ( 1 - x )} .

The pole of

P

makes

x = 0

x P (x) = 1, x^{2} Q (x) = \frac{x}{1 - x}

are real analytic at zero. Therefore

x = 0 is a regular singular point .

The ordinary-point power-series theorem does not apply there, but the Frobenius method does.

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