 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/8/a/solution

First simplify the coefficient of

y

using

\frac{m ^{3}}{x - m} + x^{2} + m x + m^{2} = \frac{x ^{3}}{x - m} .

After dividing by

x (x - m)

y^{''} + P (x) y^{'} + Q (x) y = 0, P (x) = \frac{1}{x} + \frac{1}{x - m}, Q (x) = \frac{ω ^{2} x ^{2}}{( x - m ) ^{2}} .

At zero,

x P \to 1

and

x^{2} Q \to 0

, and both products are analytic there. At

m

(x - m) P \to 1

and

(x - m)^{2} Q \to m^{2} ω^{2}

, again with analytic extensions. Thus

x = 0 and x = m are regular singular points

, and there are no other finite singular points.

Infinity must also be considered because the question asks for all singular points. Put

t = 1/ x

and

Y (t) = y (1/ t)

Y

in its normal-form equation is

Q (1/ t) / t^{4} \sim ω^{2} / t^{4}

t^{2}

does not make it analytic at zero. Hence

x = \infty is an irregular singular point

 New to topics? Read the docs here!