Difference equation with step input

Given $x(k+2) - x(k+1) + 0.25x(k) = u(k+2)$, $x(0)=1$, $x(1)=2$, $u(k) = 1$ for $k \ge 0$.

Step 1: z-transform of both sides

Shifting theorems: $\mathcal{Z}[x(k+2)] = z^2X(z) - z^2x(0) - zx(1)$, $\mathcal{Z}[x(k+1)] = zX(z) - zx(0)$.

Right side: $u(k+2) = 1$ for all $k \ge 0$, so $\mathcal{Z}[u(k+2)] = z^2U(z) - z^2u(0) - zu(1) = \frac{z^3}{z-1} - z^2 - z = \frac{z}{z-1}$.

Left side:

$$\begin{aligned} &[z^2X - z^2(1) - z(2)] - [zX - z(1)] + 0.25X \\ &= (z^2 - z + 0.25)X(z) - z^2 - z \end{aligned}$$

Step 2: solve for $X(z)$

$$\begin{aligned} (z-0.5)^2X(z) &= z^2 + z + \frac{z}{z-1} = \frac{(z^2+z)(z-1) + z}{z-1} = \frac{z^3}{z-1} \\ X(z) &= \frac{z^3}{(z-1)(z-0.5)^2} \end{aligned}$$

Step 3: inverse by residues of $X(z)z^{k-1} = \frac{z^{k+2}}{(z-1)(z-0.5)^2}$

At $z = 1$:

$$K_1 = \left.\frac{z^{k+2}}{(z-0.5)^2}\right|_{z=1} = \frac{1}{0.25} = 4$$

At $z = 0.5$ (double pole):

$$\begin{aligned} K_2 &= \left.\frac{d}{dz}\left[\frac{z^{k+2}}{z-1}\right]\right|_{z=0.5} = \left.\frac{z^{k+1}[(k+2)(z-1) - z]}{(z-1)^2}\right|_{z=0.5} \\ &= \frac{0.5^{k+1}[-0.5(k+2) - 0.5]}{0.25} = -2(0.5)^{k+1}(k+3) \\ &= -(k+3)(0.5)^k \end{aligned}$$

Result

$$x(k) = 4 - (k+3)(0.5)^k, \quad k = 0, 1, 2, \dots$$

Check with the recursion $x(k+2) = x(k+1) - 0.25x(k) + 1$

Final value: $\lim_{z\to1}(z-1)X(z) = 1/0.25 = 4$, which matches $x(\infty) = 4$.

Answer: $x(k) = 4 - (k+3)(0.5)^k$; the output rises from 1 and settles at 4.