 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2018/8/a/solution

r^{2} + 5 r + 4 = (r + 1) (r + 4) = 0

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

. Write the forcing as

30 e^{x} + 30 e^{- x}

3 e^{x}

for the first term. Because

e^{- x}

K x e^{- x}

for the second; substitution gives

3 K = 30

. Therefore

y = C_{1} e^{- x} + C_{2} e^{- 4 x} + 3 e^{x} + 10 x e^{- x} .

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