 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/6/b/solution

8 e^{- 2 x}

2 e^{- 2 x}

. The term

e^{- x}

is resonant with a homogeneous solution, so use

a x e^{- x}

for its forcing; substitution gives

a = 1

Thus

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

. The two initial conditions reduce to

C_{1} + C_{2} = 3

and

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

, so

C_{1} = 1

C_{2} = 2

. The solution of the initial value problem is

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

The extra factor

x

in the resonant particular solution is essential.

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