 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/15/a/ii/solution

- 1

, suggesting the substitution

y = e^{- x} v

y^{'} = e^{- x} (v^{'} - v), y^{''} = e^{- x} (v^{''} - 2 v^{'} + v),

y^{''} + 2 y^{'} + y = e^{- x} v^{''}

v^{''} = 2 x

. Two integrations give

v = \frac{x ^{3}}{3} + C_{1} x + C_{2} .

Therefore

y (x) = e^{- x} (\frac{x ^{3}}{3} + C_{1} x + C_{2}) .

The two arbitrary constants account for the repeated-root homogeneous solution.

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