 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2017/6/a/ii/solution

y

and integrate. The endpoint values and integration by parts give

0 = \int_{x_{0}}^{x_{1}} [y (p y^{'})^{'} + q y^{2} + λ w y^{2}] d x = - \int_{x_{0}}^{x_{1}} (p y^{'2} - q y^{2}) d x + λ I [y] .

Since

I [y] = 1

J [y] = λ .

Thus the multiplier equals the stationary value of

J

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