 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/4/c/solution

r^{2} - 4 r + 3 = (r - 1) (r - 3) = 0

e^{x}

and

e^{3 x}

. Since the forcing

e^{3 x}

already appears among these solutions, try a particular solution

y_{p} = A x e^{3 x}

. Differentiation and substitution give

y_{p}^{''} - 4 y_{p}^{'} + 3 y_{p} = 2 A e^{3 x},

A = 1/2

y = C_{1} e^{x} + C_{2} e^{3 x} + \frac{x}{2} e^{3 x} .

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