Solutions of a homogeneous difference equation often have the form $y_{k}=r^{k}$ for some r. Find some solutions of the equation

$y_{k+3} - 2y_{k+2} - 5y_{k+1} + 6y_{k} = 0 \text{for all} k$ (4)

Question

Solutions of a homogeneous difference equation often have the form [latex] y_{k}=r^{k} [/latex] for some r. Find some solutions of the equation

[latex] y_{k+3} - 2y_{k+2} - 5y_{k+1} + 6y_{k} = 0 ext{for all} k [/latex] (4)

Accepted Answer

Substitute [latex] r^{k} [/latex] for [latex] y_{k} [/latex] in the equation and factor the left side:

[latex] r^{k+3} - 2r^{k+2} - 5r^{k+1} + 6r^{k} = 0 [/latex] (5)

[latex] r^{k}(r^{3} - 2r^{2} - 5r + 6) = 0 [/latex]

[latex] r^{k}(r - 1)(r + 2)(r - 3) = 0 [/latex] (6)

Since (5) is equivalent to (6), [latex] r^{k} [/latex] satisfies the difference equation (4) if and only if r satisfies (6). Thus [latex]1^{k} , (-2)^{k} [/latex], and [latex] 3^{k} [/latex] are all solutions of (4). For instance, to verify that [latex] 3^{k} [/latex] is a solution of (4), compute

[latex] 3^{k+3} - 2 \cdot 3^{k+2} - 5 \cdot 3^{k+1} + 6 \cdot 3^{k} [/latex]

[latex] = 3^{k}(27 - 18 - 15 + 6) = 0 ext{for all} k [/latex]

Question 4.8.4: Solutions of a homogeneous difference equation often have th...

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