Questions: Nonhomogeneous Recurrence Relations and Particular Solutions

Question 1

The recurrence a(n) = 2a(n−1) + 3ⁿ has characteristic root r = 2. To find a particular solution using undetermined coefficients, which guess is correct?

• A. A·2ⁿ, matching the form of the homogeneous solution
• B. A·3ⁿ, since 3 is not a characteristic root
• C. An + B, since 3ⁿ grows like a polynomial eventually
• D. A·n·2ⁿ, multiplying by n to handle the repeated root

Answer: (B)

When f(n) = C·αⁿ, you guess A·αⁿ — provided αⁿ is not already a homogeneous solution. Here f(n) = 3ⁿ and the characteristic root is r = 2, so 3ⁿ is not a homogeneous solution. The correct guess is A·3ⁿ. Option A fails because 2ⁿ is the homogeneous solution — substituting it into the recurrence would make both sides cancel the homogeneous part, leaving an impossible equation. Option D (A·n·2ⁿ) is the fix for *that* problem, but it's not needed here since 3 ≠ 2.

Question 2

For the recurrence a(n) = 3a(n−1) + 3ⁿ, the characteristic root is r = 3. What is the correct form for a particular solution guess?

• A. A·3ⁿ — match the form of the forcing function directly
• B. A·n·3ⁿ — multiply by n since 3ⁿ matches the homogeneous solution
• C. An + B — use a polynomial since 3ⁿ and 3ⁿ cancel
• D. A·n² — escalate to a quadratic when the standard guess fails

Answer: (B)

When the natural guess (A·αⁿ) has the same form as a homogeneous solution, substituting it in causes complete cancellation, making it impossible to match the nonzero right-hand side. The fix is to multiply by n: guess A·n·3ⁿ. In general, multiply by nⁱ where i is the smallest positive integer making the guess linearly independent from all homogeneous solutions. If 3ⁿ were a repeated homogeneous root, you'd need A·n²·3ⁿ, and so on.

Question 3

The general solution to a nonhomogeneous recurrence is found by subtracting the particular solution from the homogeneous solution.

• T. True
• F. False

Answer: False

The general solution is the *sum* of the homogeneous solution and the particular solution: a(n) = aₕ(n) + aₚ(n). The homogeneous part absorbs the initial conditions (its free constants are set by them); the particular part handles the forcing function. Subtracting would give a function that satisfies neither the recurrence nor the initial conditions correctly. This mirrors the structure of linear ODEs exactly.

Question 4

After finding the general solution a(n) = aₕ(n) + aₚ(n), the initial conditions are applied only to the homogeneous part aₕ(n) to determine its free constants.

• T. True
• F. False

Answer: True

The particular solution aₚ(n) is a *specific* function with no free constants — it was fully determined by matching coefficients to the forcing function. The homogeneous solution aₕ(n) contains the free constants (e.g., C₁r₁ⁿ + C₂r₂ⁿ). You substitute the initial conditions into the full general solution a(n) = aₕ(n) + aₚ(n) to get equations for C₁, C₂, etc. The particular solution's contribution at those initial points is a known, fixed number.

Question 5

Why must you multiply your particular-solution guess by n when the natural guess has the same form as a homogeneous solution?

Model answer: A homogeneous solution already satisfies the recurrence with right-hand side zero. If you plug a particular guess of the same form into the nonhomogeneous recurrence, both sides reduce the guess to zero (the homogeneous solution 'absorbs' it), making it impossible to match the nonzero forcing function f(n). Multiplying by n produces a function in the same family (e.g., n·αⁿ instead of αⁿ) that is linearly independent from the homogeneous solutions, so it does not cancel out when substituted, and its unknown coefficient can be determined by matching f(n).

This is the discrete analog of the identical situation in solving linear ODEs with constant coefficients — when the forcing function matches a homogeneous solution, reduction of order (or the analogous n-multiplication here) is required. The mathematical reason is always the same: the operator applied to a homogeneous solution gives zero, so you must step outside that solution space.