 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/15/b/solution

The repeated factor in the differential operator suggests

y = e^{- x} u (x)

. Direct differentiation gives

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

, so

u^{''} = 4

and

u = 2 x^{2} + B x + C .

The first value gives

C = 1

. The second gives

u (1) = ey (1) = 1

, hence

2 + B + 1 = 1

and

B = - 2

. Therefore

y (x) = e^{- x} (2 x^{2} - 2 x + 1), y (2) = 5 e^{- 2} .

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