 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2021-resit/10/b/solution

f_{p} = a t^{3} + b t^{2} + c t + d

. Substitution into

f_{p}^{''} + 3 f_{p}^{'} + 2 f_{p}

gives

2 a t^{3} + (9 a + 2 b) t^{2} + (6 a + 6 b + 2 c) t + (2 b + 3 c + 2 d) .

Matching successive coefficients with the forcing gives

a = 1

b = - 3

c = 4

and

d = 1

f_{p} (t) = t^{3} - 3 t^{2} + 4 t + 1.

Indeed

f_{p}^{'} = 3 t^{2} - 6 t + 4

and

f_{p}^{''} = 6 t - 6

give

f_{p}^{''} + 3 f_{p}^{'} + 2 f_{p} = 2 t^{3} + 3 t^{2} - 4 t + 8

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

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