A-Level WJEC Mathematics: Mastering Second Order Differential Equations | A-Level WJEC 数学：二阶微分方程考点精讲

📚 A-Level WJEC Mathematics: Mastering Second Order Differential Equations | A-Level WJEC 数学：二阶微分方程考点精讲

Second order differential equations appear across mechanics, circuits and population models. In WJEC A‑Level Mathematics, you must be confident solving linear equations with constant coefficients — both homogeneous and non‑homogeneous — and interpreting solutions in context. This guide breaks down every essential technique, from the characteristic equation to forced oscillations, so you can tackle exam questions with clarity.

二阶微分方程在力学、电路和种群模型中均有出现。在 WJEC A‑Level 数学中，你们需要熟练掌握常系数线性微分方程的解法，包括齐次和非齐次情形，并能在实际情境中解读解。本文逐一拆解所有核心技巧，从特征方程到受迫振动，助你清晰应对考试题目。

1. Introduction to Second Order ODEs | 二阶常微分方程简介

A second order ordinary differential equation (ODE) involves an unknown function y(x) and its derivatives up to y”. The general linear form with constant coefficients is a d²y/dx² + b dy/dx + c y = f(x). When f(x)=0 the equation is homogeneous; otherwise it is non‑homogeneous. WJEC papers expect you to classify the equation and then solve systematically.

二阶常微分方程包含未知函数 y(x) 及其最高至二阶的导数。常系数线性方程的一般形式为 a d²y/dx² + b dy/dx + c y = f(x)。当 f(x)=0 时，方程为齐次的；否则为非齐次。WJEC 试卷要求你能够对方程进行分类，然后系统求解。

2. Homogeneous Equations with Constant Coefficients | 常系数齐次方程

Start with the homogeneous case: a y” + b y’ + c y = 0. Because the coefficients are constant, we look for solutions of the form y = eʳᵗ (if the independent variable is t) or eʳˣ. Substituting this trial solution transforms the ODE into an algebraic equation called the characteristic equation. Any linear combination of two independent solutions forms the complementary function (CF).

从齐次情形入手：a y” + b y’ + c y = 0。由于系数为常数，我们寻找形如 y = eʳᵗ（若自变量为 t）或 eʳˣ 的解。代入试验解后，微分方程转化为一个代数方程，即特征方程。两个独立解的任意线性组合构成余函数。

3. The Characteristic Equation | 特征方程

Substituting y = eʳˣ into a y” + b y’ + c y = 0 gives a r² eʳˣ + b r eʳˣ + c eʳˣ = 0, so eʳˣ (a r² + b r + c) = 0. Since eʳˣ ≠ 0, we obtain the characteristic equation:

a r² + b r + c = 0

This quadratic determines the type of solution. The discriminant Δ = b² − 4ac tells you whether the roots are real and distinct, repeated, or complex conjugates. Solving the quadratic is always your first step.

将 y = eʳˣ 代入 a y” + b y’ + c y = 0 得到 a r² eʳˣ + b r eʳˣ + c eʳˣ = 0，即 eʳˣ (a r² + b r + c) = 0。由于 eʳˣ ≠ 0，得到特征方程。

a r² + b r + c = 0

此二次方程决定解的类型。判别式 Δ = b² − 4ac 告诉你根是相异实根、重根还是共轭复根。求解二次方程始终是你的第一步。

4. Case 1: Real Distinct Roots | 情形一：两个不相等实根

If the characteristic equation has two distinct real roots r₁ and r₂, the general solution of the homogeneous equation is:

y = A eʳ¹ˣ + B eʳ²ˣ

where A and B are arbitrary constants. For example, y” − 3y’ + 2y = 0 gives r² − 3r + 2 = 0 → r₁ = 1, r₂ = 2. Hence CF: y = A eˣ + B e²ˣ. This is the simplest case and frequently appears in WJEC C4/FP2.

若特征方程有两个相异实根 r₁ 与 r₂，齐次方程的通解为：

y = A eʳ¹ˣ + B eʳ²ˣ

其中 A、B 为任意常数。例如 y” − 3y’ + 2y = 0 得出 r² − 3r + 2 = 0 → r₁ = 1, r₂ = 2，故通解 y = A eˣ + B e²ˣ。这是最简单的情形，在 WJEC C4/FP2 中频繁出现。

5. Case 2: Repeated Roots | 情形二：重根

When b² − 4ac = 0, the characteristic equation yields one repeated real root r = −b/(2a). A single exponential eʳˣ is one solution; a second independent solution is found by multiplying by x. The general solution becomes:

y = (A + Bx) eʳˣ

For instance, y” + 4y’ + 4y = 0 gives (r + 2)² = 0, so r = −2 (repeated). Therefore y = (A + Bx) e⁻²ˣ. Always check you haven’t missed the x‑factor in the exam.

当 b² − 4ac = 0 时，特征方程产生一个重实根 r = −b/(2a)。单一的指数函数 eʳˣ 是一个解；第二个独立解通过乘以 x 得到。通解为：

y = (A + Bx) eʳˣ

例如 y” + 4y’ + 4y = 0 给出 (r + 2)² = 0，故 r = −2（重根），因而 y = (A + Bx) e⁻²ˣ。考试时务必检查有无遗漏 x 项。

6. Case 3: Complex Roots | 情形三：复根

If the roots are complex, they come as a conjugate pair r = α ± iβ. The two real‑valued solutions are eᵅˣ cos βx and eᵅˣ sin βx. The general solution is:

y = eᵅˣ (A cos βx + B sin βx)

Equivalently, using the harmonic form R eᵅˣ cos(βx − φ). Example: y” + 2y’ + 5y = 0 gives r = −1 ± 2i, so y = e⁻ˣ (A cos 2x + B sin 2x). This case models damped oscillations and is crucial for mechanics contexts.

若根为复数，它们以共轭对 r = α ± iβ 出现。两个实值解为 eᵅˣ cos βx 与 eᵅˣ sin βx。通解为：

y = eᵅˣ (A cos βx + B sin βx)

等价地，可用谐波形式 R eᵅˣ cos(βx − φ) 表示。例如 y” + 2y’ + 5y = 0 给出 r = −1 ± 2i，故 y = e⁻ˣ (A cos 2x + B sin 2x)。此种情形模拟阻尼振动，在力学情境中至关重要。

7. Non‑Homogeneous Equations: Finding Particular Integrals | 非齐次方程：求特解

When the right‑hand side f(x) ≠ 0, the full solution splits into two parts: the complementary function (CF), which solves the homogeneous equation, and a particular integral (PI) that fits the forcing term. WJEC syllabuses focus on f(x) being a polynomial, exponential, sine/cosine, or a combination. Your goal is to propose a PI of a similar form, then determine any unknown coefficients.

当右边 f(x) ≠ 0 时，全解分为两部分：解齐次方程的余函数，以及匹配强迫项的特解。WJEC 大纲侧重 f(x) 为多项式、指数、正弦/余弦或其组合的情形。你的目标是提出形式类似的试验特解，然后确定未知系数。

8. Method of Undetermined Coefficients | 待定系数法

Choose a trial PI depending on f(x):

f(x) = polynomial of degree n → trial PI = general polynomial of degree n.
f(x) = p eᵏˣ → trial PI = λ eᵏˣ (if eᵏˣ is not already in the CF).
f(x) = p sin kx + q cos kx → trial PI = λ sin kx + μ cos kx.
For products, use the product of the corresponding trial forms.

If the trial PI overlaps with the CF, multiply the trial PI by x (or x² if necessary). Substitute trial PI into the ODE, equate coefficients, and solve for the constants. This systematic comparison of coefficients is a regular WJEC exam technique.

试特解的选择依 f(x) 而定：

f(x) 为 n 次多项式 → 试特解为 n 次一般多项式。
f(x) = p eᵏˣ → 试特解为 λ eᵏˣ（若 eᵏˣ 不在余函数中）。
f(x) = p sin kx + q cos kx → 试特解为 λ sin kx + μ cos kx。
乘积的情形则使用相应试验形式的乘积。

若试特解与余函数重合，需乘以 x（必要时乘 x²）。将试特解代入原方程，比较系数，求出常数。这种系统的系数比较法是 WJEC 常见考法。

9. Complete Solution = Complementary Function + Particular Integral | 通解 = 余函数 + 特解

The general solution of a non‑homogeneous linear ODE is:

y = y_CF + y_PI

First find the CF by solving the homogeneous equation (using the characteristic equation), then determine the PI with undetermined coefficients. Add them together and do not forget the arbitrary constants from the CF — they will be fixed later by initial conditions. A common mistake is to omit the CF and present only the PI.

非齐次线性常微分方程的通解为：

y = y_CF + y_PI

首先通过解齐次方程（利用特征方程）求出余函数，然后用待定系数法确定特解。将两者相加，且不可遗漏余函数中的任意常数——它们随后将由初始条件确定。常见错误是只给出特解而丢失余函数。

10. Applying Initial/Boundary Conditions | 应用初始/边界条件

Once the general solution is written, use given conditions (e.g. y(0)=2, y'(0)=−1) to find the constants A, B. Plug the condition into the expression for y and, if necessary, differentiate y to obtain y’, then substitute the second condition. This produces simultaneous equations for A and B. In WJEC questions, these final constants often simplify nicely and allow you to write the particular solution answering the physical problem.

写出通解后，利用给定条件（如 y(0)=2, y'(0)=−1）确定常数 A 与 B。将条件代入 y 的表达式，必要时求出 y’ 后再代入第二个条件。这样会得到关于 A 和 B 的联立方程组。在 WJEC 试题中，这些最终常数往往会简洁地化简，从而写出回应物理问题的特解。

11. Applications: Damped Harmonic Motion | 应用：阻尼简谐运动

Many WJEC mechanics questions model a mass‑spring‑damper system: m x” + c x’ + k x = 0. The characteristic equation m r² + c r + k = 0 determines the damping regime:

c² > 4mk: overdamping (real distinct roots, no oscillation).
c² = 4mk: critical damping (repeated root, fastest return to equilibrium).
c² < 4mk: underdamping (complex roots, decaying oscillations).

The solution form eᵅᵗ (A cos ωt + B sin ωt) describes decaying vibrations. Be ready to sketch graphs or interpret the amplitude envelope eᵅᵗ.

许多 WJEC 力学题模拟质量‑弹簧‑阻尼系统：m x” + c x’ + k x = 0。特征方程 m r² + c r + k = 0 决定阻尼类型：

c² > 4mk：过阻尼（相异实根，无振荡）。
c² = 4mk：临界阻尼（重根，最快回到平衡位置）。
c² < 4mk：欠阻尼（复根，衰减振荡）。

解的形式 eᵅᵗ (A cos ωt + B sin ωt) 描述衰减振动。备考时要会描图或解读幅值包络线 eᵅᵗ。

12. Exam Tips & Common Mistakes | 考试技巧与常见错误

Six points to sharpen your technique:

Always write the characteristic equation clearly and check the discriminant.
When a CF term appears in the trial PI, remember to multiply by x (or x²).
Do not drop ‘A’ and ‘B’ before applying conditions — the CF must retain its constants.
For trigonometric non‑homogeneous terms use both sin and cos in the trial PI, even if only one appears on the right.
After finding the complete solution, verify by differentiation that it satisfies the original ODE.
In mechanics problems, link initial displacement and velocity directly to y(0) and y'(0).

Avoiding these slips will secure method marks, which carry substantial weight in WJEC marking schemes.

六个要点提升解题技巧：

务必清晰写出特征方程并检查判别式。
当试特解中包含余函数的项时，记得乘以 x（或 x²）。
在施加条件之前不要丢掉常数 A 和 B——余函数必须保留其常数。
对于三角型非齐次项，试特解中要同时使用 sin 和 cos，即使右边只出现其中一个。
求出全解后，可通过求导验证它是否满足原微分方程。
在力学题中，要直接将初位移和初速度与 y(0) 和 y'(0) 联系起来。

避免这些疏漏将确保得到方法分，而方法分在 WJEC 评分方案中占很大比重。

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A-Level WJEC Mathematics: Mastering Second Order Differential Equations | A-Level WJEC 数学：二阶微分方程 考点精讲