---
title: "Difference eq x(k+2)+6x(k+1)+9x(k)=2^k"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:difference-eq-forced-2k-9"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/difference-eq-forced-2k-9
site: BCT Engineering
---

# Difference eq x(k+2)+6x(k+1)+9x(k)=2^k

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2071 Bhadra
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/difference-eq-forced-2k-9
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Using z-transform solve the difference equation: x(k + 2) + 6x(k + 1) + 9x(k) = 2ᵏ; x₀ = x₁ = 0.

## Short answer

Z-transform gives X(z)=z/[(z-2)(z+3)^2]. Partial fractions and inverse yield x(k)=[2^k-(-3)^k]/25 + (k/15)(-3)^k.

## Answer

**Equation:** x(k + 2) + 6x(k + 1) + 9x(k) = 2ᵏ, x(0) = x(1) = 0.

**Step 1: z-transform.** With zero initial values the shift terms vanish:

```text
z²X + 6zX + 9X = z/(z − 2)
(z + 3)² X = z/(z − 2)
X(z) = z / [(z − 2)(z + 3)²]
```

**Step 2: partial fractions.**

X/z = 1/[(z − 2)(z + 3)²] = A/(z − 2) + B/(z + 3) + C/(z + 3)²

```text
A = 1/(2 + 3)²   = 1/25
C = 1/(−3 − 2)   = −1/5
A + B = 0        → B = −1/25
```

X(z) = (1/25) z/(z − 2) − (1/25) z/(z + 3) − (1/5) z/(z + 3)²

**Step 3: inverse.** Z⁻¹[z/(z + 3)²] = k(−3)^{k−1}, so −(1/5)·k(−3)^{k−1} = (1/15) k (−3)ᵏ.

**Answer:** x(k) = [2ᵏ − (−3)ᵏ]/25 + (k/15)(−3)ᵏ, k ≥ 0.

Check: x(0) = 0, x(1) = (2 + 3)/25 − 3/15 = 0 ✓, x(2) = −5/25 + 18/15 = 1 (recurrence: 1 − 0 − 0 = 1 ✓), x(3) = −4 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Zero initial values mean
   A. the shift terms vanish: (z+3)^2 X = z/(z-2)
   B. X = 0
   C. z^2 X = z
   D. extra z terms remain
   Answer: A. Only forcing remains. (Key point: Zero initial)

2. X(z) =
   A. z/[(z-2)(z-3)^2]
   B. z/[(z-2)(z+3)^2]
   C. 1/[(z-2)(z+3)^2]
   D. z/(z+3)^2
   Answer: B. From the equation. (Key point: Transform)

3. Partial fractions of X/z have
   A. only two terms
   B. A/(z-2) + C/(z+3)^2
   C. A/(z-2) + B/(z+3) + C/(z+3)^2
   D. terms with z^3
   Answer: C. Repeated factor. (Key point: Repeated root)

4. x(k) =
   A. [2^k + (-3)^k]/25
   B. [2^k - 3^k]/25 + k 3^k/15
   C. [2^k - (-3)^k]/25 - (k/15)(-3)^k
   D. [2^k - (-3)^k]/25 + (k/15)(-3)^k
   Answer: D. x(0)=x(1)=0 check. (Key point: Result)

## Close variants set in other exams

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