 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/4/solution

(r - 1) (r - 3) = 0

y_{c} = A e^{x} + B e^{3 x} .

A linear trial

y_{p} = a x + b

gives

3 a x + (3 b - 4 a) = 3 x

, so

a = 1

b = 4/3

. One particular integral is therefore

y_{p} = x + 4/3

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