📚 IB AQA Mathematics: Second-Order Differential Equations | IB AQA 数学：二阶微分方程考点精讲
Second-order differential equations are a core topic in IB and AQA Mathematics, extending calculus techniques to model dynamic systems with acceleration, curvature and oscillations. Mastery of homogeneous and non-homogeneous forms, together with the method of undetermined coefficients, is essential for high marks. This guide unpacks every key concept, from the characteristic equation to practical modelling, with step‑by‑step clarity.
二阶微分方程是 IB 与 AQA 数学的核心内容，它把微积分方法推广到加速度、曲率和振荡等动态系统的建模中。熟练掌握齐次与非齐次形式，并结合待定系数法，是获得高分的关键。本文从特征方程到实际建模，逐层拆解每一个重要考点，步骤清晰易懂。
1. Introduction to Second-Order ODEs | 二阶常微分方程概述
A second-order ordinary differential equation (ODE) involves the second derivative d²y/dx² of an unknown function y(x). In the IB and AQA syllabi, we focus on linear equations with constant coefficients of the form a d²y/dx² + b dy/dx + c y = f(x), where a, b, c are constants. The equation is homogeneous when f(x) = 0, and non‑homogeneous otherwise. These equations appear in mechanics (spring‑mass systems), electronics (RLC circuits) and population dynamics.
二阶常微分方程包含未知函数 y(x) 的二阶导数 d²y/dx²。在 IB 与 AQA 课程中，我们主要讨论常系数线性方程 a d²y/dx² + b dy/dx + c y = f(x)，其中 a, b, c 为常数。当 f(x) = 0 时方程为齐次方程，否则为非齐次方程。这类方程广泛出现在力学（弹簧‑质量系统）、电子学（RLC 电路）和种群动力学中。
The general solution of a non‑homogeneous equation consists of two parts: the complementary function (CF), which solves the homogeneous equation, and a particular integral (PI) that accounts for f(x). Understanding this superposition principle is the foundation for all solution methods.
非齐次方程的通解由两部分组成：余函数 (CF) 满足齐次方程，特解 (PI) 则对应非齐次项 f(x)。理解这种叠加原理是掌握所有解法的根基。
2. Homogeneous Equations and the Characteristic Equation | 齐次方程与特征方程
To solve a y” + b y’ + c y = 0, assume a trial solution of the form y = e^(rx). Substituting yields the algebraic characteristic equation: a r² + b r + c = 0. The roots r determine the structure of the complementary function. Examiners expect you to write down the characteristic equation immediately and solve it, often by factorising or using the quadratic formula.
为求解 a y” + b y’ + c y = 0，假设试解 y = e^(rx)。代入后得到代数特征方程：a r² + b r + c = 0。根 r 决定了余函数的结构。考官希望你直接写出特征方程并求解，通常通过因式分解或使用求根公式完成。
The discriminant Δ = b² – 4ac dictates three cases: real distinct roots, a repeated real root, and complex conjugate roots. Each case generates a different family of functions for the complementary function.
判别式 Δ = b² – 4ac 决定了三种情形：相异实根、重实根、共轭复根。每种情形产生不同形式的余函数族。
3. Real Distinct Roots | 相异实根
When Δ > 0, the characteristic equation gives two distinct real roots r₁ and r₂. The complementary function is y_CF = A e^(r₁x) + B e^(r₂x), where A and B are arbitrary constants. This is the simplest case, and candidates must be able to write the solution directly after finding the roots.
当 Δ > 0 时，特征方程给出两个相异实根 r₁ 和 r₂。余函数为 y_CF = A e^(r₁x) + B e^(r₂x)，其中 A 与 B 为任意常数。这是最简单的情形，考生在求得根后必须能直接写出解。
For example, solving y” – 5y’ + 6y = 0 leads to r² – 5r + 6 = 0, giving r = 2, 3. The general solution is y = A e^(2x) + B e^(3x). Always check your roots by substituting back into the characteristic equation.
例如，求解 y” – 5y’ + 6y = 0 得到 r² – 5r + 6 = 0，解得 r = 2, 3。通解为 y = A e^(2x) + B e^(3x)。始终将根代回特征方程加以检验。
4. Repeated Roots | 重根情形
If Δ = 0, the characteristic equation has a single repeated root r = –b/(2a). In this case, the two linearly independent solutions are e^(rx) and x e^(rx). Hence the complementary function is y_CF = (A + Bx) e^(rx). This form is crucial because a simple sum of two identical exponentials would not satisfy the second‑order equation.
若 Δ = 0，特征方程有一个重根 r = –b/(2a)。此时两个线性无关解为 e^(rx) 与 x e^(rx)，因此余函数为 y_CF = (A + Bx) e^(rx)。这一形式至关重要，因为单纯将两个相同指数函数相加并不能满足二阶方程。
A typical example is y” – 4y’ + 4y = 0, giving r² – 4r + 4 = (r – 2)² = 0, so r = 2. The general solution is y = (A + Bx) e^(2x). Do not forget the constant B – it is easy to miss and lose marks.
典型例子：y” – 4y’ + 4y = 0，得 r² – 4r + 4 = (r – 2)² = 0，即 r = 2。通解为 y = (A + Bx) e^(2x)。不要遗漏常数 B，这是常见的失分点。
5. Complex Conjugate Roots | 共轭复根
When Δ < 0, the roots are complex conjugates α ± iβ, where α = –b/(2a) and β = √(4ac – b²)/(2a). The general solution of the homogeneous equation is y_CF = e^(αx) (C cos βx + D sin βx). This oscillatory behaviour appears constantly in physical applications such as damped harmonic motion.
当 Δ < 0 时，特征根为共轭复数 α ± iβ，其中 α = –b/(2a)，β = √(4ac – b²)/(2a)。齐次方程的通解为 y_CF = e^(αx) (C cos βx + D sin βx)。这种振荡行为频繁出现在阻尼简谐运动等物理应用中。
For instance, y” + 4y’ + 13y = 0 yields r² + 4r + 13 = 0, so r = –2 ± 3i. The solution is y = e^(–2x) (C cos 3x + D sin 3x). Be careful with the sign of α: it controls the exponential growth or decay.
例如，y” + 4y’ + 13y = 0 给出 r² + 4r + 13 = 0，得 r = –2 ± 3i。解为 y = e^(–2x) (C cos 3x + D sin 3x)。注意 α 的符号：它决定了指数增长或衰减。
6. Non‑Homogeneous Equations: The Particular Integral | 非齐次方程：特解的求法
For a y” + b y’ + c y = f(x), the general solution is y = y_CF + y_PI. The particular integral y_PI is any function that satisfies the non‑homogeneous equation, irrespective of constants. The method of undetermined coefficients provides a systematic way to guess a trial form based on the type of f(x).
对于 a y” + b y’ + c y = f(x)，通解为 y = y_CF + y_PI。特解 y_PI 是满足非齐次方程的任一函数，与常数无关。待定系数法提供了一种根据 f(x) 的类型进行试猜的系统方法。
The trial PI must be a linear combination of f(x) and its derivatives, with undetermined coefficients. After substituting into the ODE, equate coefficients to find these values. This method works for polynomials, exponentials, sines/cosines and sums of these.
试猜特解必须是 f(x) 及其导数的线性组合，并含有待定系数。代入原方程后，通过比较系数确定这些值。该方法适用于多项式、指数函数、正弦/余弦函数以及它们的和。
7. Undetermined Coefficients: Standard Trial Forms | 待定系数法：标准试解形式
The table below summarises the recommended trial particular integral y_PI for common forcing functions f(x). You must multiply the trial by x (or x²) if any term in the trial already appears in the complementary function – this avoids duplication.
下表总结了常见强迫函数 f(x) 的推荐试解特解形式 y_PI。若试解中的任一项已出现在余函数中，则必须乘以 x（或 x²），以避免重合。
| f(x) | Trial y_PI |
|---|---|
| Polynomial of degree n, e.g. 2x² + 3x – 1 | A xⁿ + B xⁿ⁻¹ + … (general polynomial of same degree) |
| k e^(px) | C e^(px) |
| p cos ωx + q sin ωx | C cos ωx + D sin ωx |
| Product e^(px) cos ωx (or sin) | e^(px)(C cos ωx + D sin ωx) |
For a polynomial of degree n, use a full polynomial of the same degree, even if lower‑order terms are missing in f(x). For example, if f(x) = x³, the trial PI is A x³ + B x² + C x + D. This ensures all derivative terms can be matched.
对于 n 次多项式，需使用同次的完整多项式，即使 f(x) 中缺少低次项。例如，若 f(x) = x³，试解为 A x³ + B x² + C x + D。这样才能匹配所有导数项。
8. Special Cases: Overlap with the Complementary Function | 特殊情况：与余函数重叠时的修正
If the standard trial PI contains a term that is already a solution of the homogeneous equation, the trial must be multiplied by x (or by x² if that term is a repeated root of the characteristic equation). This rule ensures linear independence of the particular integral from the complementary function.
若标准试解中含有已是齐次方程解的项，则必须将试解乘以 x（如果该项对应特征方程的重根，则乘以 x²）。这一规则保证了特解与余函数的线性无关。
Example: solve y” – 3y’ + 2y = e^(2x). The CF is y_CF = A e^(x) + B e^(2x). The normal trial for e^(2x) is C e^(2x), but e^(2x) already appears in the CF. Therefore try y_PI = C x e^(2x). After differentiation and substitution, you can determine C.
示例：求解 y” – 3y’ + 2y = e^(2x)。余函数为 y_CF = A e^(x) + B e^(2x)。e^(2x) 的标准试解为 C e^(2x)，但它已出现在 CF 中，因此试解应设为 y_PI = C x e^(2x)。求导并代入后可确定 C。
Similarly, for f(x) = e^(rx) cos ωx where r ± iω is a root, multiply the trial by x. This occurs in resonance problems and is a common exam pitfall.
类似地，若 f(x) = e^(rx) cos ωx 且 r ± iω 为特征根，则试解需乘以 x。这出现在共振问题中，是常见的考试陷阱。
9. Applying Initial Conditions to Determine Constants | 应用初始条件确定常数
Once the general solution y = y_CF + y_PI is written, two initial or boundary conditions (e.g. y(0) = y₀, y'(0) = v₀) are used to find the arbitrary constants A and B. Substitute the conditions into the expressions for y and its derivative, then solve the resulting linear system.
写出通解 y = y_CF + y_PI 后，利用两个初始条件或边界条件（如 y(0) = y₀, y'(0) = v₀）求出任意常数 A 与 B。将条件代入 y 及其导数的表达式，然后求解所得到的线性方程组。
Always differentiate carefully; errors in the derivative of the particular integral are a frequent source of lost marks. It is useful to compute y_PI’ separately and then add to y_CF’ before applying conditions.
务必仔细求导；特解导数计算错误是常见的失分原因。最好单独计算 y_PI’，再与 y_CF’ 相加，然后应用条件。
10. Modelling with Second‑Order ODEs and Exam Tips | 二阶常微分方程建模与应试技巧
Second‑order ODEs model numerous real‑world phenomena: simple harmonic motion (x” + ω² x = 0), damped oscillations (mx” + cx’ + kx = 0) and forced vibrations. In exam questions, you will often be given a contextual equation and asked to solve it under given initial conditions, interpreting the constants physically (amplitude, phase, damping factor).
二阶常微分方程可对许多现实现象建模：简谐运动 (x” + ω² x = 0)、阻尼振荡 (mx” + cx’ + kx = 0) 以及受迫振动。试题中常会给出一个实际情境的方程，要求你在给定初始条件下求解，并结合物理意义解释常数（振幅、相位、阻尼因子）。
Key revision strategies: memorise the three CF forms and the standard trial PI table; practise identifying overlap quickly; always write the full general solution before applying conditions; and double‑check the characteristic equation’s discriminant. With consistent practice, second‑order ODEs become one of the most predictable and rewarding topics on the paper.
关键的复习策略：熟记三种余函数形式与标准试解表格；练习快速识别重叠情形；在应用条件前一定要先写出完整的通解；并再次检查特征方程的判别式。经过持续练习，二阶常微分方程将成为试卷中规律性最强、最易得分的专题之一。
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