Difference eq y(n+2)+2y(n+1)+y(n)=n

Equation: y(n + 2) + 2y(n + 1) + y(n) = n, y(0) = y(1) = 0 (n plays the role of k).

Step 1: z-transform (Z[n] = z/(z − 1)²). With zero initial values:

(z² + 2z + 1) Y = z/(z − 1)²
Y(z) = z / [(z − 1)²(z + 1)²]

Step 2: partial fractions.

Y/z = 1/[(z − 1)²(z + 1)²] = A/(z − 1) + B/(z − 1)² + C/(z + 1) + D/(z + 1)²

B = 1/(1 + 1)²  = 1/4
D = 1/(−1 − 1)² = 1/4
A = d/dz (z + 1)⁻² at z = 1 = −2/(z+1)³ = −1/4
C = d/dz (z − 1)⁻² at z = −1 = −2/(z−1)³ = +1/4

Y(z) = −¼ z/(z − 1) + ¼ z/(z − 1)² + ¼ z/(z + 1) + ¼ z/(z + 1)²

Step 3: inverse using Z⁻¹[z/(z−1)] = 1, Z⁻¹[z/(z−1)²] = n, Z⁻¹[z/(z+1)] = (−1)ⁿ, Z⁻¹[z/(z+1)²] = −n(−1)ⁿ:

y(n) = −¼ + ¼ n + ¼ (−1)ⁿ − ¼ n (−1)ⁿ

Answer: y(n) = ¼ [ n − 1 + (1 − n)(−1)ⁿ ].

Check: the formula gives y(0..4) = 0, 0, 0, 1, 0, and the recurrence y(n+2) = n − 2y(n+1) − y(n) gives y(2) = 0, y(3) = 1, y(4) = 2 − 2·1 − 0 = 0 ✓.