 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/8/a/solution

Away from

y = \pm 1

, divide by the coefficient of

x^{''}

to obtain the normal form

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

Thus

y = 1 and y = - 1

are the only finite singular points. For a second-order linear differential equation,

y_{0}

(y - y_{0}) P (y)

and

(y - y_{0})^{2} Q (y)

extend analytically to

y_{0}

. At

y_{0} = 1

these functions are

\frac{1}{y + 1}, \frac{y - 1}{y + 1},

and at

y_{0} = - 1

they are

\frac{1}{y - 1}, \frac{y + 1}{y - 1} .

Each pair is analytic at its respective point. Therefore

both finite singular points are regular

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