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

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

Step 1: take the z-transform using Z[x(k + 2)] = z²X − z²x(0) − z x(1) and Z[x(k + 1)] = zX − z x(0):

[z²X − z²(0) − z(2)] + 2[zX − z(0)] + 3X = 0

(z² + 2z + 3) X(z) = 2z

Step 2: solve for X(z):

X(z) = 2z/(z² + 2z + 3)

Step 3: invert. Roots of z² + 2z + 3: z = −1 ± i√2, modulus r = √3, and r cos θ = −1, r sin θ = √2, so cos θ = −1/√3, θ = cos⁻¹(−1/√3) ≈ 125.26° (2.186 rad).

Write the denominator as z² − 2r cos θ z + r². Using Z[rᵏ sin kθ] = r sin θ · z/(z² − 2r cos θ z + r²), with r sin θ = √2:

X(z) = (2/√2) · √2 z/(z² + 2z + 3) = √2 · Z[(√3)ᵏ sin kθ]

Answer:

x(k) = √2 (√3)ᵏ sin(kθ), θ = cos⁻¹(−1/√3), k = 0, 1, 2, …

Check with the recurrence x(k + 2) = −2x(k + 1) − 3x(k):

k : 0   1   2   3   4    5
x : 0   2  -4   2   8  -22

The formula gives x(1) = √2·√3·√(2/3) = 2, x(2) = −4, x(3) = 2, matching the recurrence.