 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/13/a/solution

r^{2} + 2 r = r (r + 2) = 0

C_{1} + C_{2} e^{- 2 x}

A x^{2} + B x

. Substitution gives

2 A + 4 A x + 2 B = 4 x

, so

A = 1

and

B = - 1

. Therefore the general solution is

y = x^{2} - x + C_{1} + C_{2} e^{- 2 x} .

As a direct check, the polynomial part contributes

2 + 2 (2 x - 1) = 4 x

, while both complementary terms contribute zero.

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