Constant-Coefficient Linear Difference Equations: The Characteristic Root Method | 常系数线性差分方程的特征根解法

📚 Constant-Coefficient Linear Difference Equations: The Characteristic Root Method | 常系数线性差分方程的特征根解法

In IB Mathematics Higher Level (Analysis and Approaches), the study of difference equations forms a bridge between discrete mathematics and calculus. A constant-coefficient linear difference equation is a recurrence relation where each term is a linear combination of previous terms with fixed coefficients. The characteristic root method provides a systematic algebraic technique to find closed-form solutions without iterating step by step.

在IB数学高级水平（分析与方法）课程中，差分方程的学习架起了离散数学与微积分之间的桥梁。常系数线性差分方程是一种递推关系，其中每一项都是前若干项的线性组合，且系数固定。特征根解法提供了一种系统化的代数技巧，使我们无需逐项迭代即可求出通项公式。

1. General Form and Standard Setup | 一般形式与标准设定

A second-order constant-coefficient linear difference equation has the standard form:

二阶常系数线性差分方程的标准形式为：

uₙ₊₂ + p·uₙ₊₁ + q·uₙ = f(n)

where p and q are constants. When f(n) = 0, the equation is called homogeneous; otherwise it is non-homogeneous. The solution to a non-homogeneous equation is the sum of the homogeneous solution and a particular solution, mirroring the structure of solving linear ODEs.

其中 p 和 q 为常数。当 f(n) = 0 时，方程称为齐次的；否则称为非齐次的。非齐次方程的解等于齐次解与特解之和，这与线性常微分方程的求解结构完全对应。

To find the homogeneous solution, we assume a trial solution of the form uₙ = rⁿ, where r is a constant to be determined. Substituting into the homogeneous equation uₙ₊₂ + p·uₙ₊₁ + q·uₙ = 0 gives:

为求齐次解，我们假设试探解具有 uₙ = rⁿ 的形式，其中 r 为待定常数。将其代入齐次方程 uₙ₊₂ + p·uₙ₊₁ + q·uₙ = 0 得到：

rⁿ⁺² + p·rⁿ⁺¹ + q·rⁿ = 0

Factoring out rⁿ (which is non-zero for r ≠ 0) yields the characteristic equation:

提取公因子 rⁿ（当 r ≠ 0 时不为零）即得特征方程：

r² + p·r + q = 0

2. The Characteristic Equation: Derivation | 特征方程的推导

The characteristic equation is a quadratic obtained directly from the coefficients of the recurrence. For the general second-order equation uₙ₊₂ + p·uₙ₊₁ + q·uₙ = 0, we replace uₙ₊₂ by r², uₙ₊₁ by r, and uₙ by 1. This substitution is valid because rⁿ is never zero, allowing us to divide the entire equation by rⁿ.

特征方程是直接从递推关系的系数得到的二次方程。对于一般的二阶方程 uₙ₊₂ + p·uₙ₊₁ + q·uₙ = 0，我们将 uₙ₊₂ 替换为 r²，uₙ₊₁ 替换为 r，uₙ 替换为 1。这种替换是合法的，因为 rⁿ 永不为零，可以将整个方程除以 rⁿ。

The discriminant Δ = p² − 4q determines the nature of the roots:

判别式 Δ = p² − 4q 决定了根的性质：

Discriminant 判别式	Roots 根	General Solution 通解
Δ > 0	Two distinct real roots r₁, r₂	uₙ = A·r₁ⁿ + B·r₂ⁿ
Δ = 0	One repeated real root r	uₙ = (A + B·n)·rⁿ
Δ < 0	Complex conjugate pair r = α ± iβ	uₙ = Rⁿ(A·cos(nθ) + B·sin(nθ))

3. Case I: Distinct Real Roots | 情形一：两个相异实根

When p² − 4q > 0, the characteristic equation has two distinct real roots r₁ and r₂. The general solution is uₙ = A·r₁ⁿ + B·r₂ⁿ, where A and B are arbitrary constants determined by initial conditions.

当 p² − 4q > 0 时，特征方程有两个相异的实根 r₁ 和 r₂。通解为 uₙ = A·r₁ⁿ + B·r₂ⁿ，其中 A 和 B 是由初值条件确定的任意常数。

This form works because each root independently satisfies the recurrence. By the principle of superposition, any linear combination of independent solutions is also a solution. Since r₁ⁿ and r₂ⁿ are linearly independent when r₁ ≠ r₂, their span captures the complete solution space.

这种形式之所以有效，是因为每个根都独立地满足递推方程。根据叠加原理，独立解的任意线性组合仍然是解。由于当 r₁ ≠ r₂ 时，r₁ⁿ 与 r₂ⁿ 线性无关，它们的张成空间涵盖了完整的解空间。

Example: Solve uₙ₊₂ − 5uₙ₊₁ + 6uₙ = 0, with u₀ = 1 and u₁ = 2.

示例：求解 uₙ₊₂ − 5uₙ₊₁ + 6uₙ = 0，其中 u₀ = 1，u₁ = 2。

The characteristic equation is r² − 5r + 6 = 0, which factors as (r − 2)(r − 3) = 0. Thus r₁ = 2 and r₂ = 3. The general solution is uₙ = A·2ⁿ + B·3ⁿ. Substituting initial conditions gives A + B = 1 and 2A + 3B = 2. Solving yields A = 1 and B = 0, so uₙ = 2ⁿ.

特征方程为 r² − 5r + 6 = 0，因式分解为 (r − 2)(r − 3) = 0。因此 r₁ = 2，r₂ = 3。通解为 uₙ = A·2ⁿ + B·3ⁿ。代入初值条件得 A + B = 1 和 2A + 3B = 2。联立解得 A = 1，B = 0，故 uₙ = 2ⁿ。

4. Case II: Repeated Real Root | 情形二：重根

When p² − 4q = 0, the characteristic equation has a double root r = −p/2. The naive guess uₙ = rⁿ alone is insufficient because we only get one solution. The second linearly independent solution is n·rⁿ, so the general solution becomes uₙ = (A + B·n)·rⁿ.

当 p² − 4q = 0 时，特征方程有一个二重根 r = −p/2。仅凭试探解 uₙ = rⁿ 是不够的，因为我们只能得到一个解。第二个线性无关的解是 n·rⁿ，因此通解变为 uₙ = (A + B·n)·rⁿ。

Why n·rⁿ? This is a discrete analogue of the reduction of order technique in ODEs. Substituting uₙ = n·rⁿ into the recurrence confirms it is a valid solution whenever r is a double root. The factor n introduces the required extra degree of freedom to match two initial conditions.

为什么是 n·rⁿ？这是常微分方程中降阶法的离散类比。将 uₙ = n·rⁿ 代入递推方程可以验证，只要 r 是二重根，它就是一个有效解。因子 n 引入了满足两个初始条件所需的额外自由度。

Example: Solve uₙ₊₂ − 6uₙ₊₁ + 9uₙ = 0, with u₀ = 1 and u₁ = 3.

示例：求解 uₙ₊₂ − 6uₙ₊₁ + 9uₙ = 0，其中 u₀ = 1，u₁ = 3。

The characteristic equation is r² − 6r + 9 = (r − 3)² = 0, giving a repeated root r = 3. The general solution is uₙ = (A + B·n)·3ⁿ. From u₀ = 1, we get A = 1. From u₁ = 3, we have (A + B)·3 = 3, giving A + B = 1, so B = 0. Thus uₙ = 3ⁿ.

特征方程为 r² − 6r + 9 = (r − 3)² = 0，得到重根 r = 3。通解为 uₙ = (A + B·n)·3ⁿ。由 u₀ = 1 得 A = 1。由 u₁ = 3 得 (A + B)·3 = 3，即 A + B = 1，所以 B = 0。因此 uₙ = 3ⁿ。

5. Case III: Complex Conjugate Roots | 情形三：共轭复根

When p² − 4q < 0, the roots are a complex conjugate pair r₁, r₂ = α ± iβ, where α = −p/2 and β = √(4q − p²)/2. The general solution can be written using the polar form of a complex number: r = R·e^{iθ} where R = √(α² + β²) = √q (since q = α² + β²) and θ = arctan(β/α).

当 p² − 4q < 0 时，根为一对共轭复数 r₁, r₂ = α ± iβ，其中 α = −p/2，β = √(4q − p²)/2。通解可以用复数的极坐标形式表示：r = R·e^{iθ}，其中 R = √(α² + β²) = √q（因为 q = α² + β²），θ = arctan(β/α)。

Since r₁ⁿ = Rⁿ(cos(nθ) + i·sin(nθ)) and r₂ⁿ = Rⁿ(cos(nθ) − i·sin(nθ)), any linear combination A·r₁ⁿ + B·r₂ⁿ can be rewritten using Euler’s formula. Taking real combinations yields the compact real form:

由于 r₁ⁿ = Rⁿ(cos(nθ) + i·sin(nθ))，r₂ⁿ = Rⁿ(cos(nθ) − i·sin(nθ))，任意线性组合 A·r₁ⁿ + B·r₂ⁿ 都可以通过欧拉公式改写。取实组合后得到简洁的实数形式：

uₙ = Rⁿ(A·cos(nθ) + B·sin(nθ))

Here R is the modulus of the complex root (equal to √q) and θ is the argument. Both A and B are real constants determined by initial conditions. This sinusoidal form reveals the oscillatory nature of the sequence.

这里 R 是复根的模（等于 √q），θ 是辐角。A 和 B 是由初值条件确定的实常数。这种正弦形式揭示了序列的振荡特性。

6. Worked Example: Complex Roots | 实例演练：复根情形

Example: Solve uₙ₊₂ − 2uₙ₊₁ + 2uₙ = 0, with u₀ = 1 and u₁ = 1.

示例：求解 uₙ₊₂ − 2uₙ₊₁ + 2uₙ = 0，其中 u₀ = 1，u₁ = 1。

The characteristic equation is r² − 2r + 2 = 0. Using the quadratic formula:

特征方程为 r² − 2r + 2 = 0。使用求根公式：

r = (2 ± √(4 − 8))/2 = 1 ± i

Thus α = 1, β = 1, giving R = √(1² + 1²) = √2 and θ = arctan(1) = π/4. The general solution is uₙ = (√2)ⁿ(A·cos(nπ/4) + B·sin(nπ/4)).

因此 α = 1，β = 1，得 R = √(1² + 1²) = √2，θ = arctan(1) = π/4。通解为 uₙ = (√2)ⁿ(A·cos(nπ/4) + B·sin(nπ/4))。

Using u₀ = 1: A·cos(0) + B·sin(0) = A = 1. Using u₁ = 1: (√2)(A·cos(π/4) + B·sin(π/4)) = (√2)(1·(√2/2) + B·(√2/2)) = 1 + B = 1, giving B = 0.

由 u₀ = 1：A·cos(0) + B·sin(0) = A = 1。由 u₁ = 1：(√2)(A·cos(π/4) + B·sin(π/4)) = (√2)(1·(√2/2) + B·(√2/2)) = 1 + B = 1，解得 B = 0。

Therefore the closed-form solution is uₙ = (√2)ⁿ·cos(nπ/4). This can be verified: n=0 gives 1, n=1 gives √2·cos(π/4) = 1, n=2 gives 2·cos(π/2) = 0, and so on.

因此闭式解为 uₙ = (√2)ⁿ·cos(nπ/4)。可以验证：n=0 时为 1，n=1 时为 √2·cos(π/4) = 1，n=2 时为 2·cos(π/2) = 0，依此类推。

7. Non-Homogeneous Equations | 非齐次方程

For a non-homogeneous equation uₙ₊₂ + p·uₙ₊₁ + q·uₙ = f(n), the complete solution is uₙ = uₙʰ + uₙᵖ, where uₙʰ is the homogeneous solution and uₙᵖ is a particular solution. The method of undetermined coefficients is used to find uₙᵖ.

对于非齐次方程 uₙ₊₂ + p·uₙ₊₁ + q·uₙ = f(n)，完整解为 uₙ = uₙʰ + uₙᵖ，其中 uₙʰ 为齐次解，uₙᵖ 为特解。求特解采用待定系数法。

The form of the trial particular solution depends on f(n):

特解试探解的形式取决于 f(n)：

If f(n) is a constant C, try uₙᵖ = K (a constant).

如果 f(n) 为常数 C，尝试 uₙᵖ = K（常数）。
If f(n) is a polynomial of degree m, try a general polynomial of degree m.

如果 f(n) 是 m 次多项式，尝试一个 m 次一般多项式。
If f(n) is of the form k·aⁿ, try uₙᵖ = C·aⁿ, provided a is not a characteristic root. If a is a root, multiply by n (or n² for multiplicity 2).

如果 f(n) 形如 k·aⁿ，尝试 uₙᵖ = C·aⁿ，前提是 a 不是特征根。若 a 是特征根，则乘以 n（若是二重根则乘以 n²）。

Example: Solve uₙ₊₂ − 3uₙ₊₁ + 2uₙ = 4, with u₀ = 0 and u₁ = 0.

示例：求解 uₙ₊₂ − 3uₙ₊₁ + 2uₙ = 4，其中 u₀ = 0，u₁ = 0。

The homogeneous equation r² − 3r + 2 = (r − 1)(r − 2) = 0 gives roots 1 and 2, so uₙʰ = A·1ⁿ + B·2ⁿ = A + B·2ⁿ.

齐次方程 r² − 3r + 2 = (r − 1)(r − 2) = 0 的根为 1 和 2，因此 uₙʰ = A·1ⁿ + B·2ⁿ = A + B·2ⁿ。

For the particular solution with constant forcing f(n) = 4, try uₙᵖ = K. Substituting: K − 3K + 2K = 0 ≠ 4, so a constant trial fails. The issue is that K is absorbed because r = 1 is a root. We multiply by n: try uₙᵖ = Kn. Then K(n+2) − 3K(n+1) + 2Kn = K[(n+2) − 3(n+1) + 2n] = K(−1) = 4, giving K = −4.

对于常值强迫项 f(n) = 4，尝试特解 uₙᵖ = K。代入得 K − 3K + 2K = 0 ≠ 4，常数试探失败。问题在于 K 被吸收了，因为 r = 1 是特征根。于是乘以 n：尝试 uₙᵖ = Kn。则 K(n+2) − 3K(n+1) + 2Kn = K[(n+2) − 3(n+1) + 2n] = K(−1) = 4，解得 K = −4。

So the general solution is uₙ = A + B·2ⁿ − 4n. Using u₀ = 0 gives A + B = 0; using u₁ = 0 gives A + 2B − 4 = 0. Solving gives A = −4 and B = 4. Hence uₙ = −4 + 4·2ⁿ − 4n = 4(2ⁿ − 1 − n).

因此通解为 uₙ = A + B·2ⁿ − 4n。由 u₀ = 0 得 A + B = 0；由 u₁ = 0 得 A + 2B − 4 = 0。联立解得 A = −4，B = 4。故 uₙ = −4 + 4·2ⁿ − 4n = 4(2ⁿ − 1 − n)。

8. Connection to Fibonacci-Type Sequences | 斐波那契型数列的联系

The classic Fibonacci sequence Fₙ₊₂ = Fₙ₊₁ + Fₙ is a homogeneous constant-coefficient difference equation with p = −1 and q = −1. Its characteristic equation is r² − r − 1 = 0, with roots φ = (1 + √5)/2 and ψ = (1 − √5)/2.

经典斐波那契数列 Fₙ₊₂ = Fₙ₊₁ + Fₙ 是一个齐次常系数差分方程，其中 p = −1，q = −1。其特征方程为 r² − r − 1 = 0，根为 φ = (1 + √5)/2 和 ψ = (1 − √5)/2。

With F₀ = 0 and F₁ = 1, the general solution Fₙ = A·φⁿ + B·ψⁿ yields A = 1/√5 and B = −1/√5. This is the famous Binet formula:

取 F₀ = 0，F₁ = 1，通解 Fₙ = A·φⁿ + B·ψⁿ 得到 A = 1/√5，B = −1/√5。这就是著名的比内公式：

Fₙ = (φⁿ − ψⁿ) / √5

where φ = (1 + √5)/2 is the golden ratio. This example demonstrates how the characteristic root method transforms a purely recursive definition into an explicit formula — a direct computation of the 100th term requires only a single arithmetic evaluation, not 100 iterations.

其中 φ = (1 + √5)/2 为黄金比例。这个例子展示了特征根法如何将纯递推定义转化为显式公式——计算第 100 项只需一次算术求值，而不是迭代 100 次。

9. Method Summary and Algorithm | 方法总结与算法流程

To solve any second-order constant-coefficient linear difference equation, follow these steps:

求解任意二阶常系数线性差分方程，请遵循以下步骤：

Write the equation in standard form uₙ₊₂ + p·uₙ₊₁ + q·uₙ = f(n).

将方程写成标准形式 uₙ₊₂ + p·uₙ₊₁ + q·uₙ = f(n)。
Solve the homogeneous equation by forming r² + pr + q = 0 and finding its roots.

通过构造 r² + pr + q = 0 并求根来解齐次方程。
Write the homogeneous solution according to the root type (distinct real, repeated real, or complex conjugate).

根据根的类型（相异实根、重根或共轭复根）写出齐次解。
If f(n) ≠ 0, find a particular solution using undetermined coefficients; adjust if the trial form overlaps with the homogeneous solution.

若 f(n) ≠ 0，用待定系数法求特解；若试探形式与齐次解重叠则作相应调整。
Combine uₙ = uₙʰ + uₙᵖ and use the initial conditions u₀ and u₁ to determine the constants.

将 uₙ = uₙʰ + uₙᵖ 合并，利用初值 u₀ 和 u₁ 确定常数。

This algorithm is directly applicable to IB exam problems, which typically provide initial conditions and require either a closed-form expression or verification of a given formula.

此算法可直接应用于IB考试题目，这类题目通常给出初值条件，要求写出闭式表达式或验证给定公式。

10. Common Pitfalls and Exam Tips | 常见错误与考试建议

Students frequently make several avoidable mistakes when applying the characteristic root method:

学生在应用特征根法时常犯几个可以避免的错误：

Sign errors in the characteristic equation: The equation is r² + p·r + q = 0, not r² − p·r − q = 0. Always match the coefficients directly from the given recurrence.

特征方程符号错误：方程是 r² + p·r + q = 0，而非 r² − p·r − q = 0。务必直接从给定递推关系对应系数。
Forgetting n in the repeated-root case: Using only rⁿ when there is a double root fails to satisfy two independent initial conditions.

重根时忘记乘 n：重根时仅用 rⁿ 无法满足两个独立的初值条件。
Confusing R and θ in complex case: R is the modulus √q, not the real part. The angle θ is the argument, usually expressed in radians.

复根情形混淆 R 与 θ：R 是模 √q，不是实部。角 θ 是辐角，通常用弧度表示。
Not checking for resonance: If f(n) contains a term identical to the homogeneous solution, the particular solution must include an extra factor of n.

未检查共振：若 f(n) 中包含与齐次解相同的项，特解必须额外乘以因子 n。

On the exam, always verify your final formula by computing the first three terms both recursively and using your closed-form solution. This quick check catches most algebraic oversights and takes less than a minute.

考试时，务必通过递推和闭式公式分别计算前几项来验证最终公式。这个快速检查能发现大多数代数疏漏，耗时不到一分钟。

11. Higher-Order Equations (Extension) | 高阶方程（拓展）

The characteristic root method extends naturally to k-th order equations of the form uₙ₊ₖ + c₁·uₙ₊ₖ₋₁ + … + cₖ·uₙ = 0. The characteristic equation becomes a polynomial of degree k:

特征根法自然推广到 k 阶方程：uₙ₊ₖ + c₁·uₙ₊ₖ₋₁ + … + cₖ·uₙ = 0。特征方程变为 k 次多项式：

rᵏ + c₁·rᵏ⁻¹ + … + cₖ = 0

Distinct roots r₁, r₂, …, rₖ each contribute Aᵢ·rᵢⁿ to the solution. A root of multiplicity m contributes (A₁ + A₂n + … + Aₘnᵐ⁻¹)·rⁿ. Complex conjugate pairs are handled in the same way as for the second-order case.

每个相异根 rᵢ 对解贡献一项 Aᵢ·rᵢⁿ。m 重根贡献 (A₁ + A₂n + … + Aₘnᵐ⁻¹)·rⁿ。共轭复根按二阶情形相同方式处理。

Though IB assessments typically restrict explicit questions to second-order equations, understanding the extension deepens conceptual understanding and provides a safety margin for challenging problem-set questions.

虽然IB考试通常将显式问题限定在二阶方程，但理解推广形式能加深概念理解，并为应对高难度习题提供额外的安全边际。

12. Practice Problems with Selected Solutions | 练习题目及部分解答

Practice is essential for mastering this technique. Here we provide four problems of varying difficulty, followed by outline solutions.

练习是掌握这一技巧的关键。下面提供四道难度不同的练习，随后给出解题纲要。

Problem 1 (Basic): Solve uₙ₊₂ − 4uₙ₊₁ + 3uₙ = 0 with u₀ = 2, u₁ = 4.

题1（基础）：求解 uₙ₊₂ − 4uₙ₊₁ + 3uₙ = 0，u₀ = 2，u₁ = 4。

Outline: r² − 4r + 3 = (r − 1)(r − 3) = 0 → uₙ = A + B·3ⁿ. From u₀ = 2 and u₁ = 4, we get A = 1 and B = 1, so uₙ = 1 + 3ⁿ.

纲要：r² − 4r + 3 = (r − 1)(r − 3) = 0 → uₙ = A + B·3ⁿ。由 u₀ = 2 和 u₁ = 4 得 A = 1，B = 1，所以 uₙ = 1 + 3ⁿ。

Problem 2 (Repeated root): Solve uₙ₊₂ + 4uₙ₊₁ + 4uₙ = 0 with u₀ = 1, u₁ = 0.

题2（重根）：求解 uₙ₊₂ + 4uₙ₊₁ + 4uₙ = 0，u₀ = 1，u₁ = 0。

Outline: r² + 4r + 4 = (r + 2)² = 0 → uₙ = (A + Bn)(−2)ⁿ. From u₀ = 1, A = 1. From u₁ = 0, (A + B)(−2) = 0 → B = −1. Thus uₙ = (1 − n)(−2)ⁿ.

纲要：r² + 4r + 4 = (r + 2)² = 0 → uₙ = (A + Bn)(−2)ⁿ。由 u₀ = 1 得 A = 1。由 u₁ = 0 得 (A + B)(−2) = 0 → B = −1。故 uₙ = (1 − n)(−2)ⁿ。

Problem 3 (Complex roots): Solve uₙ₊₂ − 2uₙ₊₁ + 5uₙ = 0 with u₀ = 1, u₁ = 2.

题3（复根）：求解 uₙ₊₂ − 2uₙ₊₁ + 5uₙ = 0，u₀ = 1，u₁ = 2。

Outline: r² − 2r + 5 = 0 → r = 1 ± 2i. R = √5, θ = arctan(2). Solution: uₙ = (√5)ⁿ(A·cos(nθ) + B·sin(nθ)). Initial conditions yield A = 1 and B = 0, so uₙ = (√5)ⁿ·cos(nθ).

纲要：r² − 2r + 5 = 0 → r = 1 ± 2i。R = √5，θ = arctan(2)。解：uₙ = (√5)ⁿ(A·cos(nθ) + B·sin(nθ))。初值条件得 A = 1，B = 0，故 uₙ = (√5)ⁿ·cos(nθ)。

Problem 4 (Non-homogeneous): Solve uₙ₊₂ − uₙ₊₁ − 2uₙ = 3ⁿ with u₀ = 0, u₁ = 0.

题4（非齐次）：求解 uₙ₊₂ − uₙ₊₁ − 2uₙ = 3ⁿ，u₀ = 0，u₁ = 0。

Outline: Homogeneous: r² − r − 2 = (r − 2)(r + 1) = 0 → uₙʰ = A·2ⁿ + B·(−1)ⁿ. For the particular, try uₙᵖ = C·3ⁿ: C(9 − 3 − 2) = 4C = 1, so C = 1/4. General: uₙ = A·2ⁿ + B·(−1)ⁿ + (1/4)3ⁿ. Using the initial conditions give A = −1/20 and B = −4/5, so uₙ = (−1/20)2ⁿ − (4/5)(−

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Constant-Coefficient Linear Difference Equations: The Characteristic Root Method | 常系数线性差分方程的特征根解法