 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/17/b/ii/solution

A quadratic polynomial is sufficient: set

y_{p} = A x^{2}

. Its fifth derivative vanishes, so

y_{p}^{(5)} - y_{p}^{'} = - 2 A x

. Matching the forcing gives

A = - 1/2

y_{p} = - \frac{x ^{2}}{2} .

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