Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ia/paper-2/7b/a/solution

Substitute and use to cancel the terms proportional to . The remaining ordinary differential equation is
On an interval where , divide through and put :
Thus reduction of order gives
A nonzero multiplicative constant gives an independent solution; adding a constant inside the integral adds a multiple of . The Wronskian is proportional to and is nonzero on the interval. The undivided equation remains meaningful at a zero of , but this integral representation then needs continuation rather than direct division.

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