- 2072 Asoj · 5 marks
Difference eq x(k+2)-3x(k+1)+2x(k)=4^k
Short answer
Take z-transform, obtain X(z)=z(z-3)/[(z-1)(z-2)(z-4)]. Partial fractions give x(k)=2^(k-1)+4^k/6-2/3.
Solve difference equation x(k + 2) - 3x(k + 1) + 2x(k) = 4ᵏ for x(0) = 0 and x(1) = 1.
Equation: x(k + 2) − 3x(k + 1) + 2x(k) = 4ᵏ, x(0) = 0, x(1) = 1.
Step 1: z-transform (Z[4ᵏ] = z/(z − 4)):
[z²X − z] − 3[zX] + 2X = z/(z − 4)
(z² − 3z + 2) X = z + z/(z − 4)
(z − 1)(z − 2) X = z (z − 3)/(z − 4)
X(z) = z (z − 3) / [(z − 1)(z − 2)(z − 4)]
Step 2: partial fractions.
X(z)/z = (z − 3)/[(z − 1)(z − 2)(z − 4)] = A/(z − 1) + B/(z − 2) + C/(z − 4)
A = (1 − 3)/[(1 − 2)(1 − 4)] = (−2)/(3) = −2/3
B = (2 − 3)/[(2 − 1)(2 − 4)] = (−1)/(−2) = 1/2
C = (4 − 3)/[(4 − 1)(4 − 2)] = 1/6
X(z) = −(2/3) z/(z − 1) + ½ z/(z − 2) + (1/6) z/(z − 4)
Step 3: inverse.
Answer: x(k) = ½·2ᵏ + (1/6)·4ᵏ − 2/3 = 2^{k−1} + 4ᵏ/6 − 2/3.
Check: x(0) = 0 ✓, x(1) = 1 ✓, x(2) = 4 (recurrence: 3·1 − 0 + 1 = 4 ✓), x(3) = 3·4 − 2·1 + 4 = 14 ✓ (formula: 4 + 64/6 − 2/3 = 14).
✓ Answer checked by Claude Opus 5.5 (high effort)
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