 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/1a/solution

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

y_{n + 2} - 4 y_{n + 1} + 4 y_{n} = 0

(r - 2)^{2}

, so

y_{n}^{(h)} = (C + D n) 2^{n} .

y_{n} = an + b

. The left-hand side becomes

an + (b - 2 a),

so matching

n

gives

a = 1

and

b = 2

. Hence

y_{n} = (C + D n) 2^{n} + n + 2.

The conditions

y_{0} = 1

and

y_{1} = 0

give

C = - 1

and

D = - 1/2

. Therefore

y_{n} = n + 2 - (1 + \frac{n}{2}) 2^{n} .

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