 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/b/solution

r = - 3

(C_{1} + C_{2} x) e^{- 3 x}

y_{p} = A x + B

. Substitution gives

9 A x + 6 A + 9 B = 3 x + 1

, so

A = 1/3

and

B = - 1/9

y (x) = (C_{1} + C_{2} x) e^{- 3 x} + \frac{x}{3} - \frac{1}{9} .

The extra factor

x

in the second homogeneous term is required by the repeated root.

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