 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/4/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

(m + 1)^{2} = 0

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

. For the polynomial forcing, substitute

a x^{2} + b x + c

to obtain

a x^{2} + (4 a + b) x + (2 a + 2 b + c) = x^{2} .

Thus

a = 1

b = - 4

c = 6

. For the forcing

2 e^{- x}

, both

e^{- x}

and

x e^{- x}

K x^{2} e^{- x}

. The identity

(\frac{d ^{2}}{d x ^{2}} + 2 \frac{d}{d x} + 1) (e^{- x} v) = e^{- x} v^{''}

gives

K = 1

. The resulting general solution is

y = (C_{1} + C_{2} x + x^{2}) e^{- x} + x^{2} - 4 x + 6.

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