Difference equation by Z-transform

Short answer

After Z-transform, X(z)=z/(( z-2)(z+1)). Partial fractions give X(z)=(1/3)[z/(z-2)-z/(z+1)]. Inverse yields x(k)=[2^k - (-1)^k]/3.

Apply Z-transform to solve the difference equation x(k + 2) − x(k + 1) − 2x(k) = 0, given that x(0) = 0, x(1) = 1.

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

Step 1: Z-transform.

[z²X − z²·0 − z·1] − [zX − z·0] − 2X = 0
(z² − z − 2) X = z
X(z) = z/((z − 2)(z + 1))

Step 2: Partial fractions on X/z.

X/z = 1/((z−2)(z+1)) = (1/3) [ 1/(z − 2) − 1/(z + 1) ]
X(z) = (1/3) [ z/(z − 2) − z/(z + 1) ]

Step 3: Invert using Z⁻¹[z/(z − a)] = aᵏ.

Answer: x(k) = [2ᵏ − (−1)ᵏ]/3

Check: x(0) = 0, x(1) = (2 + 1)/3 = 1, x(2) = x(1) + 2x(0) = 1 and (4 − 1)/3 = 1 ✓; x(3) = x(2) + 2x(1) = 3 and (8 + 1)/3 = 3 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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