Second-Order Differential Equations | IB 数学：二阶微分方程考点精讲

📚 Second-Order Differential Equations | IB 数学：二阶微分方程考点精讲

Second-order differential equations form a core part of the IB Mathematics: Analysis and Approaches HL syllabus. They appear in contexts ranging from pure calculus to modelling real-world phenomena such as mechanical vibrations, electrical circuits, and population dynamics. Mastering this topic requires a clear understanding of the homogeneous case, the characteristic equation, the three types of roots, and the method of undetermined coefficients for non-homogeneous equations. This article covers all essential points you need for your IB exam, with step-by-step explanations, examples, and common pitfalls.

二阶微分方程是 IB 数学：分析与方法 HL 课程的核心内容之一。它们不仅出现在纯微积分题目中，还广泛用于建模现实世界中的振动、电路和人口动态等现象。掌握这一主题需要清晰理解齐次情形、特征方程、三种特征根类型以及非齐次方程的待定系数法。本文涵盖 IB 考试所需的所有关键知识点，并配有分步讲解、示例和常见错误分析。

1. What Is a Second-Order Differential Equation? | 什么是二阶微分方程？

A second-order linear differential equation has the general form a d²y/dx² + b dy/dx + c y = f(x), where a, b, c are constants (for IB purposes) and f(x) is a function of x. If f(x) = 0, the equation is called homogeneous; otherwise, it is non-homogeneous. The solution y = y_c + y_p consists of the complementary function (general solution of the homogeneous equation) and a particular integral (any solution of the non-homogeneous equation).

二阶线性微分方程的一般形式为 a d²y/dx² + b dy/dx + c y = f(x)，其中 a, b, c 为常数（在 IB 范围内），f(x) 是关于 x 的函数。当 f(x) = 0 时，方程称为齐次方程；否则为非齐次方程。解的结构为 y = y_c + y_p，其中 y_c 为补函数（齐次方程的通解），y_p 为特解（非齐次方程的任意一个解）。

2. Homogeneous Equations and the Characteristic Equation | 齐次方程与特征方程

For a homogeneous equation a d²y/dx² + b dy/dx + c y = 0, we assume a solution of the form y = e^mx. Substituting gives the auxiliary (characteristic) equation: a m² + b m + c = 0. The roots m₁ and m₂ determine the form of the complementary function. This method works because the exponential function reproduces itself under differentiation.

对于齐次方程 a d²y/dx² + b dy/dx + c y = 0，我们假设解的形式为 y = e^mx。代入后得到辅助（特征）方程：a m² + b m + c = 0。根 m₁ 和 m₂ 决定了补函数的形式。该方法之所以有效，是因为指数函数在求导后保持相同形式。

3. Case 1: Two Distinct Real Roots | 情形一：两个不等实根

If b² – 4ac > 0, the characteristic equation has two distinct real roots m₁ and m₂. The general solution is y = A e^m₁x + B e^m₂x, where A and B are arbitrary constants. This case often arises in overdamped systems. For example, solve y” – 5y’ + 6y = 0. The characteristic equation is m² – 5m + 6 = 0, with roots m = 2 and m = 3, so y = A e^2x + B e^3x.

若 b² – 4ac > 0，特征方程有两个不同的实根 m₁ 和 m₂。通解为 y = A e^m₁x + B e^m₂x，其中 A、B 为任意常数。这种情况常见于过阻尼系统。例如，求解 y” – 5y’ + 6y = 0。特征方程为 m² – 5m + 6 = 0，根为 m = 2 和 m = 3，因此通解为 y = A e^2x + B e^3x。

4. Case 2: Repeated Real Root | 情形二：重实根

When b² – 4ac = 0, the characteristic equation has a repeated root m = -b/(2a). In this case, the two independent solutions are e^mx and x e^mx. The general solution is y = (A + B x) e^mx. For instance, y” – 4y’ + 4y = 0 gives m² – 4m + 4 = 0, so m = 2 (twice). The solution is y = (A + B x) e^2x. This represents critical damping in physical applications.

当 b² – 4ac = 0 时，特征方程有一个重根 m = -b/(2a)。此时两个线性无关的解为 e^mx 和 x e^mx。通解为 y = (A + B x) e^mx。例如，y” – 4y’ + 4y = 0 导出 m² – 4m + 4 = 0，则 m = 2（二重根）。解为 y = (A + B x) e^2x。这在实际应用中表示临界阻尼。

5. Case 3: Complex Conjugate Roots | 情形三：共轭复根

If b² – 4ac < 0, the characteristic equation yields complex conjugate roots m = α ± i β, where α = -b/(2a) and β = √(4ac - b²)/(2a). The general solution is y = e^αx (A cos βx + B sin βx). For example, y” + 4y’ + 13y = 0 gives m = -2 ± 3i, so y = e^-2x (A cos 3x + B sin 3x). This solution describes damped oscillatory motion.

如果 b² – 4ac < 0，特征方程产生共轭复根 m = α ± i β，其中 α = -b/(2a)，β = √(4ac - b²)/(2a)。通解为 y = e^αx (A cos βx + B sin βx)。例如，y” + 4y’ + 13y = 0 得到 m = -2 ± 3i，因此 y = e^-2x (A cos 3x + B sin 3x)。这种解描述了阻尼振荡运动。

6. Constructing the Complementary Function Summary | 补函数构建方法总结

The complementary function y_c is entirely determined by the roots of the characteristic equation. IB students must be able to quickly identify the discriminant sign and write down the correct form. A useful table helps:

补函数 y_c 完全由特征方程的根决定。IB 学生必须能够快速判断判别式的符号并写出正确形式。以下表格可供参考：

Roots (根)	Form of y_c (补函数形式)
Distinct real m₁, m₂	A e^m₁x + B e^m₂x
Repeated real m	(A + B x) e^mx
Complex α ± iβ	e^αx (A cos βx + B sin βx)

7. Non-Homogeneous Equations and the Particular Integral | 非齐次方程与特解

A non-homogeneous equation has the form a y” + b y’ + c y = f(x). The general solution is y = y_c + y_p, where y_p is any particular solution. To find y_p, we use the method of undetermined coefficients, which works when f(x) is a polynomial, exponential, sine, cosine, or a combination of these. The key idea is to guess a form for y_p with unknown coefficients and substitute into the equation to determine them.

非齐次方程形式为 a y” + b y’ + c y = f(x)。通解为 y = y_c + y_p，其中 y_p 为任一特解。为求 y_p，我们使用待定系数法，该方法适用于 f(x) 为多项式、指数函数、正弦、余弦或其组合的情况。核心思路是预设含待定系数的 y_p 形式，代入方程确定系数。

8. Guessing Rules for y_p | 特解 y_p 的设定规则

The trial form of y_p depends on f(x). Basic guidelines: if f(x) is a polynomial of degree n, try a general polynomial of degree n; if f(x) = p e^kx, try C e^kx; if f(x) = p cos ωx + q sin ωx, try C cos ωx + D sin ωx. If any term in the trial y_p already appears in y_c, multiply the trial y_p by x (or x² if necessary) to avoid duplication.

y_p 的尝试形式取决于 f(x)。基本规则：若 f(x) 是 n 次多项式，尝试 n 次一般多项式；若 f(x) = p e^kx，尝试 C e^kx；若 f(x) = p cos ωx + q sin ωx，尝试 C cos ωx + D sin ωx。若 y_p 尝试式中任一项已出现在 y_c 中，则将尝试式乘以 x（必要时乘以 x²），避免重复。

Example: Solve y” – 3y’ + 2y = 4e^3x. The complementary function from roots 1 and 2 is y_c = A e^x + B e^2x. Since f(x) = 4e^3x does not appear in y_c, try y_p = C e^3x. Substituting gives 9C e^3x – 9C e^3x + 2C e^3x = 4e^3x → 2C = 4 → C = 2. Thus y_p = 2e^3x and general solution y = A e^x + B e^2x + 2e^3x.

示例：求解 y” – 3y’ + 2y = 4e^3x。由根 1 和 2 得补函数 y_c = A e^x + B e^2x。因 f(x) = 4e^3x 未出现在 y_c 中，尝试 y_p = C e^3x。代入得 9C e^3x – 9C e^3x + 2C e^3x = 4e^3x → 2C = 4 → C = 2。因此 y_p = 2e^3x，通解为 y = A e^x + B e^2x + 2e^3x。

9. Handling Duplication and Modification | 重复项处理与修正

When f(x) contains a term that solves the homogeneous equation, the standard trial form must be multiplied by x (or x²). For instance, solve y” – 4y’ + 4y = e^2x. The repeated root m = 2 gives y_c = (A + B x) e^2x. Trying y_p = C e^2x fails because both e^2x and x e^2x are solutions of the homogeneous equation. Therefore, try y_p = C x² e^2x. Substitution will yield C = 1/2, so y_p = ½ x² e^2x.

当 f(x) 含齐次方程的解时，标准尝试式必须乘以 x（或 x²）。例如，求解 y” – 4y’ + 4y = e^2x。重根 m = 2 给出 y_c = (A + B x) e^2x。尝试 y_p = C e^2x 会失败，因为 e^2x 和 x e^2x 均为齐次方程的解。因此，尝试 y_p = C x² e^2x。代入可得 C = 1/2，故 y_p = ½ x² e^2x。

This modification ensures the particular integral is linearly independent from the complementary function. Always compare the trial form with y_c before solving for coefficients.

这种修正确保特解与补函数线性无关。务必在求解系数前比较尝试式与 y_c。

10. Initial Conditions and Unique Solutions | 初始条件与唯一解

To find the unique solution of an IB problem, you will often be given initial conditions, such as y(0) = a and y'(0) = b. After obtaining the general solution y = y_c + y_p, substitute x = 0 into y and y’ to form simultaneous equations and solve for the arbitrary constants A and B. Always differentiate carefully and remember that these constants come from the complementary function only, as y_p has no arbitrary constants.

为求得 IB 问题的唯一解，通常会给出初始条件，例如 y(0) = a 和 y'(0) = b。在得到通解 y = y_c + y_p 后，将 x = 0 代入 y 和 y’，联立方程求解任意常数 A 和 B。请仔细求导，并牢记这些常数仅来自补函数，因为 y_p 不含任意常数。

Example: Given y” + 9y = 0, y(0) = 2, y'(0) = 6. The characteristic equation m² + 9 = 0 gives m = ±3i, so y = A cos 3x + B sin 3x. Then y(0) = A = 2. y’ = -3A sin 3x + 3B cos 3x, so y'(0) = 3B = 6 → B = 2. Thus unique solution is y = 2 cos 3x + 2 sin 3x.

示例：给定 y” + 9y = 0，y(0) = 2，y'(0) = 6。特征方程 m² + 9 = 0 得 m = ±3i，故 y = A cos 3x + B sin 3x。则 y(0) = A = 2。y’ = -3A sin 3x + 3B cos 3x，故 y'(0) = 3B = 6 → B = 2。因此唯一解为 y = 2 cos 3x + 2 sin 3x。

11. Applications: Simple Harmonic Motion and Damped Oscillations | 应用：简谐运动与阻尼振荡

In mechanics, the equation for a mass-spring system with damping is m d²x/dt² + c dx/dt + k x = 0, where m is mass, c is damping constant, and k is spring stiffness. This leads to characteristic equation m r² + c r + k = 0. The behaviour depends on the discriminant: overdamped (real distinct roots), critically damped (repeated root), and underdamped (complex roots), which gives oscillatory motion with decaying amplitude. For forced oscillations, the right-hand side becomes a driving force F(t), and the particular integral represents the steady-state solution.

在力学中，带阻尼的质量-弹簧系统方程为 m d²x/dt² + c dx/dt + k x = 0，其中 m 为质量，c 为阻尼常数，k 为弹簧刚度。这导出特征方程 m r² + c r + k = 0。系统行为取决于判别式：过阻尼（不等实根）、临界阻尼（重根）和欠阻尼（复根），后者产生振幅衰减的振荡运动。对于受迫振动，右侧变为驱动力 F(t)，特解则代表稳态解。

IB exam questions often model a pendulum or a spring. You may need to interpret the physical meaning of the constants, or find the time when the amplitude falls to a certain value. Knowing the link between the sign of the discriminant and the motion type is essential.

IB 试题经常模拟钟摆或弹簧。你可能需要解释常数的物理意义，或找出振幅降至某值的时间。理解判别式符号与运动类型之间的联系至关重要。

12. Common Mistakes and Exam Tips | 常见错误与应试技巧

Forgetting the second solution for repeated roots. Always include the x e^mx term.
忘记重根时的第二个解。 务必包含 x e^mx 项。
Incorrect trial function for y_p when f(x) is a sum. Treat each term separately and take the sum. If f(x) = e^x + sin x, try C e^x + D cos x + E sin x.
当 f(x) 为和式时尝试式不正确。 分别处理每一项并求和。若 f(x) = e^x + sin x，尝试 C e^x + D cos x + E sin x。
Missing the duplication rule. Always check whether any part of the trial y_p appears in y_c. If yes, multiply by x.
遗漏重复项规则。 始终检查 y_p 尝试式是否出现在 y_c 中。若是，则乘以 x。
Arithmetic errors in differentiating y_p. Write y’_p and y”_p carefully, especially when product rule applies for terms like x e^mx.
求导 y_p 时计算错误。 仔细写出 y’_p 和 y”_p，尤其当涉及 x e^mx 等项需用乘法法则时。
Solving characteristic equation quickly. Recognise that the discriminant determines the solution form; factorisation or the quadratic formula should be second nature.
快速求解特征方程。 认识到判别式决定解的形式；因式分解或求根公式应熟练自如。

Mastering second-order differential equations requires systematic practice. Begin by classifying the equation, solving the characteristic equation, writing y_c, then determining y_p using the correct trial form and modification rule. Finally, apply initial conditions. This structured approach works for every IB problem and builds confidence. Practice with past paper questions to become fluent in the algebraic manipulations and to recognise the common function types.

掌握二阶微分方程需要系统性的练习。从方程分类开始，求解特征方程，写出 y_c，再通过正确的尝试式和修正规则确定 y_p，最后应用初始条件。这种结构化方法适用于每道 IB 题目，并能建立信心。通过练习历年真题来熟悉代数运算并识别常见函数类型。

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Second-Order Differential Equations | IB 数学：二阶微分方程 考点精讲