 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability/4/b/solution

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

C_{1} e^{x} + C_{2} e^{2 x}

. Because the forcing

e^{x}

is already a homogeneous mode, the method of undetermined coefficients requires

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

rather than

A e^{x}

Here

y_{p}^{'} = A (x + 1) e^{x}

and

y_{p}^{''} = A (x + 2) e^{x}

, so

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

. Thus

A = - 1

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

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