Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 2B (Differential Equations)

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- Solution 2B

Solution

The characteristic polynomial of the linear recurrence relation is

r^{3} - 6 r^{2} + 12 r - 8 = (r - 2)^{3} .

The repeated-root rule therefore gives

x_{n} = (A + B n + C n^{2}) 2^{n} .

The condition

x_{0} = 0

gives

A = 0

. The other two conditions give

2 (B + C) = 4, 4 (2 B + 4 C) = 24,

so

B + C = 2

and

B + 2 C = 3

. Thus

B = C = 1

, and

x_{n} = n (n + 1) 2^{n} .

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