 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2/13/c/solution

(r + 1) (r + 2) = 0

y_{h} = C_{1} e^{- x} + C_{2} e^{- 2 x}

A x + B

gives

2 A = 1

and

3 A + 2 B = 0

, so it is

x /2 - 3/4

. A trial

C e^{3 x}

gives

(9 + 9 + 2) C = 1

, so

C = 1/20

. Therefore

y (x) = C_{1} e^{- x} + C_{2} e^{- 2 x} + \frac{x}{2} - \frac{3}{4} + \frac{1}{20} e^{3 x} .

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