 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/1/c/solution

Put

Y (θ) = y (cos θ)

, with

0 < θ < π

. The chain rule gives

y^{'} = - \frac{Y ^{'}}{s i n θ}, y^{''} = \frac{Y ^{''}}{s i n ^{2} θ} - \frac{c o s θ}{s i n ^{3} θ} Y^{'} .

Consequently the original differential expression becomes

- Y^{''} - 2 cot θ Y^{'}

. If

Y = F / sin θ

Y^{''} + 2 cot θ Y^{'} = \frac{F ^{''} + F}{s i n θ} .

Taking

F = sin n θ

therefore yields

- Y^{''} - 2 cot θ Y^{'} = (n^{2} - 1) Y, y_{n} (x) = \frac{sin ( n arccos x )}{1 - x ^{2}} .

The apparent endpoint singularities are removable. Expansion at

θ = 0, π

gives

y_{n} (1) y_{n}^{'} (1) = n, = \frac{n ( n ^{2} - 1 )}{3}, y_{n} (- 1) y_{n}^{'} (- 1) = (- 1)^{n - 1} n, = (- 1)^{n} \frac{n ( n ^{2} - 1 )}{3} .

Thus both endpoint regularity conditions hold. This is the Chebyshev polynomial of the second kind

U_{n - 1} (x)

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