 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/8/a/solution

Divide by

2 x^{2}

to put the ordinary differential equation into normalized form

y^{''} + P (x) y^{'} + Q (x) y = 0

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

The coefficients are singular at zero, but

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

are both analytic there. This is precisely the regular singular point criterion for a second-order equation, so $x = 0$ is a regular singular point.

 New to topics? Read the docs here!