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Difference eq x(k+2)+3x(k+1)+2x(k)=0

Short answer

Take z-transform with initial conditions to get X(z)=z/[(z+1)(z+2)], partial fractions give z/(z+1)-z/(z+2). Inverse gives x(k)=(-1)^k-(-2)^k for k≥0.

Solve the difference equation by the application of z-transform: x(k + 2) + 3x(k + 1) + 2x(k) = 0 with conditions x(0) = 0, x(1) = 1.

Equation: x(k + 2) + 3x(k + 1) + 2x(k) = 0, x(0) = 0, x(1) = 1.

Step 1: take the z-transform using

  • Z[x(k+1)] = zX(z) − z x(0)
  • Z[x(k+2)] = z²X(z) − z² x(0) − z x(1)
[z²X − z²·0 − z·1] + 3[zX − z·0] + 2X = 0
(z² + 3z + 2) X = z
X(z) = z / [(z + 1)(z + 2)]

Step 2: partial fractions.

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

X(z) = z/(z + 1) − z/(z + 2)

Step 3: inverse transform.

Answer: x(k) = (−1)ᵏ − (−2)ᵏ, k ≥ 0.

Check with the recurrence x(k+2) = −3x(k+1) − 2x(k): x(2) = −3, x(3) = 7, x(4) = −15. The formula gives −3, 7, −15 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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