 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/13/b/solution

r^{2} + 7 r + 12 = (r + 3) (r + 4)

C_{1} e^{- 3 x} + C_{2} e^{- 4 x}

y_{p} = K x e^{- 3 x}

. Substitution gives

y_{p}^{''} + 7 y_{p}^{'} + 12 y_{p} = K e^{- 3 x}

, hence

K = 2

The full solution is

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

. At zero, the two initial conditions give

C_{1} + C_{2} = 1

and

2 - 3 C_{1} - 4 C_{2} = 0

. Solving yields

C_{1} = 2

C_{2} = - 1

, so

y (x) = (2 + 2 x) e^{- 3 x} - e^{- 4 x} .

 New to topics? Read the docs here!