Difference equation by Z-transform

Short answer

Z-transform yields X(z)=z/(( z-5)(z+1)). Partial fractions: X(z)=(1/6)[z/(z-5)-z/(z+1)]. Inverse gives x(k)=[5^k - (-1)^k]/6.

Apply Z-transform to solve the difference equation x(k + 2) − 4x(k + 1) − 5x(k) = 0, given that x(0) = 0, x(1) = 1.

x(k + 2) − 4x(k + 1) − 5x(k) = 0, x(0) = 0, x(1) = 1.

Step 1: Z-transform.

[z²X − z·1] − 4[zX] − 5X = 0
(z² − 4z − 5) X = z
X(z) = z/((z − 5)(z + 1))

Step 2: Partial fractions on X/z.

X/z = 1/((z−5)(z+1)) = (1/6) [ 1/(z − 5) − 1/(z + 1) ]
X(z) = (1/6) [ z/(z − 5) − z/(z + 1) ]

Step 3: Invert.

Answer: x(k) = [5ᵏ − (−1)ᵏ]/6

Check: x(0) = 0, x(1) = (5 + 1)/6 = 1, x(2) = 4x(1) + 5x(0) = 4 and (25 − 1)/6 = 4 ✓; x(3) = 4·4 + 5·1 = 21 and (125 + 1)/6 = 21 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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