 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/13/b/iii/solution

A linear polynomial trial

y = ux + v

has

y^{'} = u

and

y^{''} = 0

. Substituting gives

3 ux + (3 u + 3 v) = x

, so

u = 1/3

and

v = - 3 /9 = - 1/ (33)

y_{p, 2} (x) = \frac{x}{3} - \frac{1}{3 3} .

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