Difference eq x(k+2)+6x(k+1)+9x(k)=2^k

Equation: x(k + 2) + 6x(k + 1) + 9x(k) = 2ᵏ, x(0) = x(1) = 0.

Step 1: z-transform. With zero initial values the shift terms vanish:

z²X + 6zX + 9X = z/(z − 2)
(z + 3)² X = z/(z − 2)
X(z) = z / [(z − 2)(z + 3)²]

Step 2: partial fractions.

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

A = 1/(2 + 3)²   = 1/25
C = 1/(−3 − 2)   = −1/5
A + B = 0        → B = −1/25

X(z) = (1/25) z/(z − 2) − (1/25) z/(z + 3) − (1/5) z/(z + 3)²

Step 3: inverse. Z⁻¹[z/(z + 3)²] = k(−3)^{k−1}, so −(1/5)·k(−3)^{k−1} = (1/15) k (−3)ᵏ.

Answer: x(k) = [2ᵏ − (−3)ᵏ]/25 + (k/15)(−3)ᵏ, k ≥ 0.

Check: x(0) = 0, x(1) = (2 + 3)/25 − 3/15 = 0 ✓, x(2) = −5/25 + 18/15 = 1 (recurrence: 1 − 0 − 0 = 1 ✓), x(3) = −4 ✓.