 Solution

ID: past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2017/2/b/ii/solution

y^{'}

with the derivative of the coefficient of

y^{''}

(r x^{2})^{'} = - r x, x^{2} r^{'} + 3 x r = 0.

A positive integrating factor is therefore

r (x) = x^{- 3}

, with any positive constant multiple equally valid. Direct calculation gives

r = x^{- 3}, L y = (x^{- 1} y^{'})^{'} + x^{- 3} y .

This uses the plus-divergence sign convention of the question. The usual minus-divergence Sturm-Liouville operator is its negative; no extra sign is inserted into the requested eigenvalue equation.

Because

r

never vanishes on the interval, multiplying the original eigenvalue equation by

r

leaves the requested weighted eigenvalue problem unchanged:

λ_{k} = - k^{2} π^{2}, y_{k} = x sin (kπ lo g x) .

The weight in the inner product is essential. The substitution

t = lo g x

gives

\int_{1}^{e} r y_{j} y_{k} d x = \int_{0}^{1} sin (jπ t) sin (kπ t) d t = {\frac{1}{2}, 0, j = k, j \neq = k .

x^{- 3}

, and the squared norm is

N_{k} = 1/2

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