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Difference eq x(k+2)-4x(k+1)+4x(k)=0, x(0)=0

Short answer

Transform with x(0)=0, x(1)=1 yields X(z)=z/(z−2)². Invert using Z⁻¹[z/(z−a)²]=k a^(k−1): x(k)=k·2^(k−1), k≥0.

Apply method of z-transform to solve the difference equation x(k + 2) - 4x(k + 1) + 4x(k) = 0; x(0) = 0, x(1) = 1.

Equation: x(k + 2) − 4x(k + 1) + 4x(k) = 0, x(0) = 0, x(1) = 1.

Step 1: transform. Using Z[x(k + 2)] = z²X − z²x(0) − z x(1) and Z[x(k + 1)] = zX − z x(0):

[z²X − 0 − z(1)] − 4[zX − 0] + 4X = 0

(z² − 4z + 4) X = z

Step 2: solve.

X(z) = z/(z − 2)²

Step 3: invert. Z⁻¹[z/(z − a)²] = k aᵏ⁻¹ with a = 2:

Answer: x(k) = k · 2ᵏ⁻¹, k ≥ 0

Check with x(k + 2) = 4x(k + 1) − 4x(k):

k    : 0  1  2   3   4
x(k) : 0  1  4  12  32

The formula gives x(2) = 2 · 2 = 4, x(3) = 3 · 4 = 12, x(4) = 4 · 8 = 32. ✓

✓ Answer checked by Claude Opus 5.5 (high effort)

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