A-Level Maths: Second-Order Differential Equations Exam-Focused Guide | A-Level 数学：二阶微分方程考点精讲

📚 A-Level Maths: Second-Order Differential Equations Exam-Focused Guide | A-Level 数学：二阶微分方程考点精讲

Second-order differential equations form a core part of the A-Level Maths syllabus, extending the ideas of integration and first-order ODEs to more complex physical systems. In this guide, we break down the structure of linear second-order differential equations with constant coefficients, the standard solution methods for both homogeneous and non-homogeneous cases, and the application of initial or boundary conditions. We also link the theory to the damped harmonic oscillator, a classic modelling scenario that frequently appears in exam questions.

二阶微分方程是 A-Level 数学核心内容之一，将积分和一阶常微分方程的思想延伸到更复杂的物理系统。本文详细拆解常系数线性二阶微分方程的结构、齐次与非齐次情形的标准解法，以及初始条件与边界条件的运用。我们还会将理论与典型考题中的阻尼谐振子模型联系起来，帮助你在考场上快速识别题型并准确作答。

1. What Is a Second-Order ODE? | 什么是二阶常微分方程？

A second-order ordinary differential equation (ODE) involves an unknown function y(x) together with its first and second derivatives, y′ and y″. In A-Level, we focus on linear equations with constant coefficients of the form a y″ + b y′ + c y = f(x), where a, b, c are constants and f(x) is a given function. If f(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous.

二阶常微分方程包含未知函数 y(x) 及其一阶导数 y′ 和二阶导数 y″。在 A-Level 中，我们主要讨论常系数线性方程 a y″ + b y′ + c y = f(x)，其中 a、b、c 为常数，f(x) 为已知函数。若 f(x) = 0，方程为齐次方程；否则为非齐次方程。

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

For the homogeneous case a y″ + b y′ + c y = 0, we assume a solution of the form y = e^λx. Substituting this into the ODE gives the characteristic equation a λ² + b λ + c = 0. The nature of the roots of this quadratic determines the form of the general solution.

对于齐次方程 a y″ + b y′ + c y = 0，假设解的形式为 y = e^λx，代入方程得到特征方程 a λ² + b λ + c = 0。该二次方程根的性质决定了通解的结构。

The characteristic equation can be solved using the quadratic formula λ = [ -b ± √(b² – 4ac) ] / (2a), and its discriminant Δ = b² – 4ac is the key indicator: Δ > 0 gives two distinct real roots; Δ = 0 gives a repeated real root; Δ < 0 gives a pair of complex conjugate roots.

利用求根公式 λ = [ -b ± √(b² – 4ac) ] / (2a) 可解特征方程，判别式 Δ = b² – 4ac 决定根的类型：Δ > 0 得到两相异实根；Δ = 0 得到重实根；Δ < 0 得到共轭复根。

3. Case 1: Distinct Real Roots | 情况一：两相异实根

When the characteristic equation has two distinct real roots λ₁ and λ₂, the general solution is y = A e^λ₁x + B e^λ₂x, where A and B are arbitrary constants. The solution is a linear combination of two independent exponential functions.

当特征方程有两个相异实根 λ₁ 和 λ₂ 时，通解为 y = A e^λ₁x + B e^λ₂x，其中 A、B 为任意常数。解是两个独立指数函数的线性组合。

Example: y″ – 5y′ + 6y = 0 ⇒ λ² – 5λ + 6 = 0 ⇒ (λ – 2)(λ – 3) = 0 ⇒ λ = 2, 3 ⇒ y = A e^2x + B e^3x.

示例：y″ – 5y′ + 6y = 0 ⇒ λ² – 5λ + 6 = 0 ⇒ (λ – 2)(λ – 3) = 0 ⇒ λ = 2, 3 ⇒ y = A e^2x + B e^3x。

4. Case 2: Repeated Real Root | 情况二：重实根

If the discriminant is zero, we have a repeated real root λ (twice). The general solution in this case is y = (A + B x) e^λx. The term x e^λx is needed to produce a second linearly independent solution.

若判别式为零，特征方程有重实根 λ，通解为 y = (A + B x) e^λx。必须引入 x e^λx 项才能生成第二个线性无关的解。

Example: y″ – 4y′ + 4y = 0 ⇒ λ² – 4λ + 4 = 0 ⇒ (λ – 2)² = 0 ⇒ λ = 2 (twice) ⇒ y = (A + B x) e^2x.

示例：y″ – 4y′ + 4y = 0 ⇒ λ² – 4λ + 4 = 0 ⇒ (λ – 2)² = 0 ⇒ λ = 2（二重根）⇒ y = (A + B x) e^2x。

5. Case 3: Complex Roots | 情况三：共轭复根

When Δ < 0, the characteristic equation gives a pair of complex conjugate roots λ = α ± iβ. The general solution is then expressed in trigonometric form: y = e^αx (A cos βx + B sin βx). This form is always real when A and B are real constants.

当 Δ < 0 时，特征方程给出共轭复根 λ = α ± iβ，通解可用三角函数表示为 y = e^αx (A cos βx + B sin βx)。只要 A、B 为实常数，该表达式即为实数。

Example: y″ + 4y′ + 13y = 0 ⇒ λ² + 4λ + 13 = 0 ⇒ λ = -2 ± 3i ⇒ y = e^-2x (A cos 3x + B sin 3x).

示例：y″ + 4y′ + 13y = 0 ⇒ λ² + 4λ + 13 = 0 ⇒ λ = -2 ± 3i ⇒ y = e^-2x (A cos 3x + B sin 3x)。

6. Summary of Homogeneous Solutions | 齐次方程通解总结

The three forms of the complementary function (CF) are summarised below for a homogeneous equation a y″ + b y′ + c y = 0:

以下总结齐次方程 a y″ + b y′ + c y = 0 的余函数（CF）的三种形式：

Δ > 0: y = A e^λ₁x + B e^λ₂x
Δ > 0：y = A e^λ₁x + B e^λ₂x
Δ = 0: y = (A + B x) e^λx
Δ = 0：y = (A + B x) e^λx
Δ < 0: y = e^αx (A cos βx + B sin βx)
Δ < 0：y = e^αx (A cos βx + B sin βx)

Always write the final CF in terms of real constants A and B. This part of the solution does not depend on the non-homogeneous term f(x).

始终用实常数 A、B 写出余函数。这部分解与非齐次项 f(x) 无关。

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

The general solution to a non-homogeneous ODE a y″ + b y′ + c y = f(x) is y = y_c + y_p, where y_c is the complementary function (solution to the homogeneous equation) and y_p is any particular integral (PI) that satisfies the full equation.

非齐次方程 a y″ + b y′ + c y = f(x) 的通解为 y = y_c + y_p，其中 y_c 为余函数（齐次方程的解），y_p 为满足完整方程的任意一个特解（PI）。

The method of undetermined coefficients is the standard technique for finding y_p when f(x) is a polynomial, exponential, sine/cosine, or a sum/product of these.

当 f(x) 为多项式、指数函数、正弦/余弦或其线性组合与乘积时，待定系数法是求 y_p 的标准方法。

8. Method of Undetermined Coefficients: Polynomial RHS | 待定系数法：多项式右端

If f(x) is an n-th degree polynomial, we propose a trial PI of the same degree with unknown coefficients. For example, if f(x) = x², try y_p = P x² + Q x + R. Substitute into the ODE and equate coefficients of like powers of x to find P, Q, R.

若 f(x) 为 n 次多项式，则试探特解设定为同次多项式，携带未知系数。例如 f(x) = x² 时，设 y_p = P x² + Q x + R，代入方程并比较 x 的同次幂系数，解出 P、Q、R。

Remember: if any term of the trial y_p already appears in the complementary function, you must multiply the trial function by x (or x² if necessary) to avoid duplication.

注意：若试探 y_p 中的某项已在余函数中出现，必须将试探函数乘以 x（必要时乘以 x²）以避免重复。

9. Method of Undetermined Coefficients: Exponential and Trigonometric RHS | 待定系数法：指数与三角右端

For f(x) = k e^px, try y_p = C e^px, unless e^px is already in the CF, in which case multiply by x (or x² for a double root). For f(x) = M cos ωx + N sin ωx, try y_p = P cos ωx + Q sin ωx; if these terms appear in the CF, multiply by x.

对于 f(x) = k e^px，设 y_p = C e^px，除非 e^px 已在 CF 中出现，此时乘以 x（重根则乘以 x²）。对于 f(x) = M cos ωx + N sin ωx，设 y_p = P cos ωx + Q sin ωx；若这些项出现在 CF 中，乘以 x。

Example: y″ + 3y′ + 2y = 5 e^-2x. CF: λ = -1, -2 ⇒ y_c = A e^-x + B e^-2x. Since e^-2x is in the CF, try y_p = C x e^-2x.

示例：y″ + 3y′ + 2y = 5 e^-2x。CF 为 λ = -1, -2 ⇒ y_c = A e^-x + B e^-2x。因 e^-2x 存在于 CF 中，设 y_p = C x e^-2x。

10. Using Initial or Boundary Conditions | 初始条件与边界条件的运用

Once the general solution y = y_c + y_p is found, arbitrary constants are determined by given conditions. Initial conditions specify y and y′ at a single point x = x₀, while boundary conditions specify y (or y′) at two different points. Plug the conditions into the general solution and its derivative to form simultaneous equations and solve for A and B.

求出通解 y = y_c + y_p 后，用已知条件确定任意常数。初始条件给出同一点 x = x₀ 处 y 与 y′ 的值，边界条件则给出两个不同点处的 y（或 y′）值。将条件代入通解及其导数，联立方程解出 A 与 B。

Typical exam question: solve y″ – 3y′ + 2y = 4x, given y(0) = 1, y′(0) = 0. Find y_c + y_p, then use the conditions to fix A and B. Always differentiate correctly and substitute carefully.

典型考题：解 y″ – 3y′ + 2y = 4x，已知 y(0) = 1，y′(0) = 0。先求出 y_c + y_p，再利用条件确定 A、B。务必正确求导并仔细代入。

11. Application: Damped Harmonic Oscillator | 应用：阻尼谐振子

A second-order ODE of the form m x″ + c x′ + k x = 0 models a mass-spring-damper system, where m is mass, c is damping coefficient, and k is spring constant. The characteristic equation m λ² + c λ + k = 0 leads to three damping regimes: underdamped (complex roots, oscillatory decay), critically damped (repeated root, fastest non-oscillatory return), and overdamped (distinct real roots, slow return without oscillation).

形如 m x″ + c x′ + k x = 0 的二阶常微分方程描述质量-弹簧-阻尼系统，其中 m 为质量，c 为阻尼系数，k 为弹簧常数。特征方程 m λ² + c λ + k = 0 对应三种阻尼状态：欠阻尼（复根，振荡衰减）、临界阻尼（重根，不发生振荡的最快回复）和过阻尼（相异实根，无振荡缓慢回复）。

Exam problems often provide initial displacement and velocity, then ask for the equation of motion. You first determine the type of damping from the discriminant, write the complementary function, and apply initial conditions. This is a direct application of everything covered in this guide.

考题中常给出初始位移和初始速度，要求写出运动方程。你需要先通过判别式判断阻尼类型，写出余函数，再代入初始条件。这是对本指南全部知识点的直接综合运用。

Published by TutorHao | Maths Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

A-Level Maths: Second-Order Differential Equations Exam-Focused Guide | A-Level 数学：二阶微分方程 考点精讲