📚 4. Constant-coefficient Linear Difference Equations | 4. 常系数线性差分方程
Difference equations, also known as recurrence relations, describe sequences in which each term is defined as a function of preceding terms. In IB Mathematics, a key topic is the solution of linear difference equations with constant coefficients, where the relationship between terms involves fixed multipliers and additions. Mastering these equations unlocks the ability to model population growth, financial loans, and the famous Fibonacci sequence, making them a powerful tool for both pure and applied problems.
差分方程，也称递推关系，描述的是每一项都由前几项定义的数列。在IB数学中，常系数线性差分方程是一个核心主题，这类方程中各项之间的关系包含固定的乘数和加数。掌握这些方程能够帮助你建立人口增长、金融贷款以及著名的斐波那契数列等模型，是纯粹数学和应用问题中的一个有力工具。
1. Introduction to Difference Equations | 差分方程简介
A difference equation is an equation that relates a term in a sequence to its predecessors. For a sequence un, a first‑order relation might be un+1 = 2un + 1, while a second‑order one connects un+2, un+1 and un. The 'order' refers to the number of previous terms involved, and 'constant‑coefficient' means that the multipliers of those terms are fixed numbers, independent of n.
差分方程是联系数列某项与其前几项的等式。对于一个序列 un，一阶关系可以写作 un+1 = 2un + 1，而二阶关系则联系 un+2、un+1 和 un。“阶”表示所涉及前项的数量，“常系数”意味着这些项的乘数是固定的、不依赖于 n 的数值。
2. Linear Constant‑Coefficient Form | 线性常系数差分方程的标准形式
The general linear constant‑coefficient difference equation of order k can be written as:
xn+k + c1xn+k-1 + c2xn+k-2 + ... + ckxn = f(n)
where c1, ..., ck n) is a known function of n. If f(n) = 0, the equation is homogeneous; otherwise it is non‑homogeneous. In IB examinations, you will mainly encounter first‑ and second‑order equations of this type.
一般形式的 k 阶线性常系数差分方程可写为：
xn+k + c1xn+k-1 + c2xn+k-2 + … + ckxn = f(n)
其中 c1 到 ck 是常数，f(n) 是已知的关于 n 的函数。若 f(n) = 0，则为齐次方程；否则为非齐次方程。在IB考试中，你主要会遇到一阶和二阶的此类方程。
3. First‑Order Homogeneous Equations | 一阶齐次方程
Consider the simplest case: un+1 = a un, where a is a constant. By repeated substitution, the solution is un = u0 an or, more generally, un = C an for any constant C determined by an initial condition. This exponential form is the building block for all higher‑order equations.
考虑最简单的情形：un+1 = a un，其中 a 为常数。通过反复代入，解为 un = u0 an，或更一般地，un = C an，C 由初始条件决定。这种指数形式是所有高阶方程的基础模块。
4. First‑Order Non‑Homogeneous Equations | 一阶非齐次方程
For un+1 = a un + g(n), the general solution is the sum of the complementary solution C an ives k = a k + p, so k = p/(1 − a) provided a ≠ 1. If a = 1, the solution is arithmetic: un = u0 + n p.
对于 un+1 = a un + g(n)，通解为齐次解 C an 与一个特解之和。当 g(n) 为常数 p 时，尝试常数特解 k。代入得 k = a k + p，因此当 a ≠ 1 时，k = p/(1 − a)。如果 a = 1，解为等差形式：un = u0 + n p。
) is a polynomial, try a polynomial of the same degree. For exponential forcing g(n) = b rn with r ≠ a, try a particular solution of the form A rn; if r = a, use A n an. These choices mirror the method of undetermined coefficients for differential equations.
当 g(n) 是多项式时，尝试同次数的多项式作为特解。对于指数型驱动项 g(n) = b rn，若 r ≠ a，尝试形如 A rn 的特解；若 r = a，则使用 A n an。这些选择与微分方程中的待定系数法十分相似。
5. Second‑Order Homogeneous Equations and the Characteristic Equation | 二阶齐次方程与特征方程
A second‑order linear homogeneous equation with constant coefficients has the form
a un+2 + b un+1 + c un = 0, a ≠ 0
Assuming a solution of the form un = rn and substituting yields the characteristic equation a r2 + b r + c = 0. The roots of this quadratic determine the nature of the general solution.
二阶常系数线性齐次差分方程的形式为
a un+2 + b un+1 + c un = 0, a ≠ 0
假设解形如 un = rn 并代入，即得特征方程 a r2 + b r + c = 0。这个二次方程的根决定了通解的性质。
6. Distinct Real Roots | 相异实根
If the characteristic equation has two distinct real roots r1 and r2, the general solution is
un = A r1n + B r2n
Constants A and B are found from two initial conditions, typically u0 and u1. This case arises in many classic sequences, such as the difference equation for the Fibonacci numbers.
若特征方程有两个相异实根 r1 和 r2，通解为
un = A r1n + B r2n
常数 A 和 B 由两个初始条件（通常为 u0 和 u1）求出。很多经典数列都属于这种情况，比如斐波那契数列的差分方程。
7. Repeated Roots | 重根
When the characteristic equation has a repeated root r, both rn and n rn satisfy the recurrence. The general solution becomes
un = (A + B n) rn
This can be verified by direct substitution. Always check that the two terms are linearly independent; the linear function in n provides this independence.
当特征方程有重根 r 时，rn 和 n rn 都满足递推关系。通解变为
un = (A + B n) rn
这可以通过直接代入来验证。一定要确保两项线性无关；函数中的 n 因子恰好提供了这种无关性。
8. Complex Roots | 复根
acteristic equation has complex conjugate roots α ± iβ, we express them in polar form: r e±iθ, where r = √α2 + β2 and tan θ = β/α. The real general solution is written as
un = rn (C cos(nθ) + D sin(nθ))
This oscillatory behaviour appears in models with damping or cyclical patterns. Always use the real form unless instructed otherwise.
若特征方程有共轭复根 α ± iβ，我们将其写成极坐标形式：r e±iθ，其中 r = √α2 + β2，tan θ = β/α。实通解可以写成
un = rn (C cos(nθ) + D sin(nθ))
这种振荡行为出现在含阻尼或周期模式的模型中。除非另有说明，答题时一律使用实函数形式。
9. Second‑Order Non‑Homogeneous Equations | 二阶非齐次方程
The equation a un+2 + b un+1 + c un = f(n) has general solution un = un(h) + un(p), where un(h) is the general solution of the associated homogeneous equation and un(p) is any particular solution of the non‑homogeneous equation. The form of un(p) depends on f(n).
方程 a un+2 + b un+1 + c un = f(n) 的通解为 un = un(h) + un(p)，其中 un(h) 是对应齐次方程的通解，而 un(p) 是非齐次方程的任意一个特解。un(p) 的形式由 f(n) 决定。
10. Method of Undetermined Coefficients | 待定系数法
For common f(n), we guess a particular solution with unknown coefficients and substitute into the equation to determine them. The main rules are:
对于常见的 f(n)，我们猜测一个含待定系数的特解形式，并代入方程加以确定。主要规则如下：
| f(n) | Trial un(p) |
|---|---|
| Constant k | D |
| Polynomial P(n) of degree m | Polynomial of degree m |
| k · sn, s not a root of char. eq. | C sn |
| k · sn, s a single root of char. eq. | C n sn |
| A cos(ωn) or A sin(ωn) | C cos(ωn) + D sin(ωn) |
If f(n) is a product of these functions, multiply the corresponding trial forms. If any term in the trial solution already appears in the complementary solution, multiply by an additional factor of n (or n2 for double roots) until it is independent.
如果 f(n) 是这些基本形式的乘积，则将对应的试探形式相乘。如果试探解中的某项已经出现在齐次解中，就额外乘以 n（对重根可乘 n2），直到线性无关为止。
11. Applications: Fibonacci Sequence and Finance | 应用：斐波那契数列与金融模型
The Fibonacci recurrence Fn+2 = Fn+1 + Fn, with F0=0, F1=1, gives characteristic equation r2 − r − 1 = 0. The roots are φ = (1+√5)/2 and ψ = (1−√5)/2, leading to Binet's formula Fn = (φn – ψn)/√5. This classical result shows how irrational numbers combine to produce integer sequences.
斐波那契递推 Fn+2 = Fn+1 + Fn，其中 F0=0，F1=1，给出特征方程 r2 − r − 1 = 0。根为 φ = (1+√5)/2 和 ψ = (1−√5)/2，由此导出 Binet 公式 Fn = (φn – ψn)/√5。这一经典结果展示了无理数如何组合成整数数列。
In finance, the repayment of a loan with fixed periodic payments can be modeled by a first‑order linear difference equation. If Pn is the outstanding balance after n periods, we often have Pn+1 = (1+i) Pn − R, where i is the interest rate per period and R is the repayment. Solving gives Pn = (P0 − R/i)(1+i)n + R/i, which determines the number of periods needed to pay off the loan.
在金融中，固定分期偿还的贷款可以用一阶线性差分方程建模。若 Pn 表示 n 期后的未偿还余额，我们常有 Pn+1 = (1+i) Pn − R，其中 i 为每期利率，R 为还款额。求解可得 Pn = (P0 − R/i)(1+i)n + R/i，从而确定还清贷款所需的期数。
12. IB Exam Tips and Summary | IB考试技巧与总结
Always begin by writing the characteristic equation for homogeneous equations and find its roots. Use the initial conditions to determine the constants at the very end – never before forming the general solution. For non‑homogeneous equations, find the complementary solution first, then add a correctly chosen particular solution. Check your answer by verifying the recurrence for a few early terms.
解题时，首先要写出齐次方程的特征方程并求根。在最后阶段才利用初始条件确定常数——切勿在写出通解之前进行。对于非齐次方程，先求出齐次通解，再叠加一个正确选定的特解。最后，通过验证递推的前几项来检查答案的正确性。
ution, miswriting the polar form of complex roots, and mishandling the case a = 1 in first‑ ts, exponentials, and trigonometric functions, to build confidence for the IB examination.
常见错误包括：试探特解与齐次解重叠时忘记乘以 n；复根的极坐标形式书写
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