 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/15/a/ii/solution

(r + 1) (r + 2)

A e^{- x} + B e^{- 2 x}

e^{- x}

already occurs in that solution, so a constant multiple of

e^{- x}

y_{p} = C x e^{- x}

instead. Its derivatives are

C (1 - x) e^{- x}

and

C (x - 2) e^{- x}

, so substitution gives

C e^{- x}

and fixes

C = 1

. Thus

y = A e^{- x} + B e^{- 2 x} + x e^{- x} .

The extra factor of

x

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