 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/b/solution

y_{n}

y_{m}

and subtract the equation for

y_{m}

multiplied by

y_{n}

. Integration gives

(n^{2} - m^{2}) \int_{- 1}^{1} w y_{n} y_{m} d x = [p (y_{n} y_{m}^{'} - y_{m} y_{n}^{'})]_{- 1}^{1} = 0.

The boundary expression is zero because

p \to 0

and the functions and first derivatives remain finite. Since

n, m

are distinct positive integers, this proves

\int_{- 1}^{1} y_{n} y_{m} 1 - x^{2} d x = 0.

\int_{- 1}^{1} p y_{n}^{'} y_{m}^{'} d x = [p y_{n}^{'} y_{m}]_{- 1}^{1} - \int_{- 1}^{1} (p y_{n}^{'})^{'} y_{m} d x = (n^{2} - 1) \int_{- 1}^{1} w y_{n} y_{m} d x = 0.

Thus the new weight is

w_{1} (x) = (1 - x^{2})^{3/2} .

n = 1

the derivative is identically zero, which is consistent with the identity.

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