 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-2/14a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Substitution of

y = e^{- x}

gives

(x + λ + 1) e^{- x} - (x + λ) e^{- x} - e^{- x} = 0.

For a second solution

y = a x + b

aλ - b = 0

. Taking

a = 1

gives the linearly independent solution

y_{1} = x + λ .

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