Introduction | 引言
First-order differential equations are a cornerstone topic in A-Level Mathematics (specifically Edexcel, AQA, OCR, and CAIE syllabi). They appear in Pure Mathematics Paper 2/3 and frequently surface in applied contexts such as mechanics, population dynamics, Newton’s law of cooling, and radioactive decay. This comprehensive guide walks you through every sub-topic, from separation of variables to integrating factors, with fully worked examples and examiner’s tips.
一阶微分方程是 A-Level 数学的核心内容（涵盖 Edexcel、AQA、OCR 及 CAIE 大纲），出现在纯数学卷2/卷3中，并在力学、人口动力学、牛顿冷却定律和放射性衰变等应用场景中频繁出现。本指南将带领你逐一攻克每个子主题——从分离变量法到积分因子法——配以完整例题解析和考官提分技巧。
1. What Is a Differential Equation? | 什么是微分方程？
A differential equation (DE) is an equation that relates a function with one or more of its derivatives. In A-Level, we focus on first-order ordinary differential equations (ODEs), meaning the highest derivative involved is the first derivative dy/dx.
微分方程是将一个函数与其一个或多个导数联系起来的方程。在 A-Level 阶段，我们主要研究一阶常微分方程，即所涉及的最高阶导数是一阶导数 dy/dx。
1.1 General Form | 一般形式
The general form of a first-order ODE is:
dy/dx = f(x, y)
Where f(x, y) is some expression involving x, y, or both. The goal is to find the function y = y(x) that satisfies this relationship — this is called the general solution. When we incorporate an initial condition (a specific pair of x and y values), we obtain the particular solution.
其中 f(x, y) 是包含 x、y 或两者的表达式。我们的目标是找到满足这一关系的函数 y = y(x)——这被称为通解。当引入初始条件（一组特定的 x 和 y 值）后，我们得到特解。
1.2 Syllabus Coverage by Exam Board | 各考试局大纲覆盖
- Edexcel: Differential Equations (Chapter 11, Pure 2) — 8 to 15 marks | 微分方程（第11章，纯数学卷2）—— 8至15分
- AQA: Solving Differential Equations (Section P, Pure) — 6 to 12 marks | 求解微分方程（P部分，纯数学）—— 6至12分
- OCR (MEI): First Order Differential Equations (Pure) — 5 to 10 marks | 一阶微分方程（纯数学）—— 5至10分
- CAIE (9709): Differential Equations (Paper 3, Pure 3) — 8 to 14 marks | 微分方程（卷3，纯数学3）—— 8至14分
2. Method 1: Separation of Variables | 方法一：分离变量法
This is the most straightforward technique and the one most frequently tested. It applies when the DE can be written in the form:
dy/dx = g(x) · h(y)
That is, the right-hand side factors into a product of a function of x only and a function of y only.
这是最直接也最常见的考试题型。当微分方程可以写成以上形式时适用——即右边可以分解为一个仅含 x 的函数与一个仅含 y 的函数的乘积。
2.1 The Method Step-by-Step | 逐步方法
Step 1: Separate the variables. Rearrange so that all terms involving y appear on the left with dy, and all terms involving x appear on the right with dx:
(1/h(y)) dy = g(x) dx
Step 2: Integrate both sides.
∫ (1/h(y)) dy = ∫ g(x) dx
Step 3: Solve for y if possible. Write your answer as y = f(x) + C, or leave it in implicit form if solving explicitly for y is too messy.
Step 4: Apply initial conditions (if given) to find the constant of integration C.
第1步：分离变量。 重新排列，使所有含 y 的项与 dy 一起出现在左边，所有含 x 的项与 dx 一起出现在右边。
第2步：两边积分。
第3步：求解 y。 尽可能写成 y = f(x) + C 的形式；若显式求解过于复杂，可保留隐式形式。
第4步：代入初始条件（若给出）求积分常数 C。
2.2 Worked Example 1 | 例题 1
Question: Solve dy/dx = 2xy, given that y(0) = 3.
Solution:
- Separate: (1/y) dy = 2x dx (assuming y ≠ 0)
- Integrate: ∫ (1/y) dy = ∫ 2x dx → ln|y| = x² + C
- Solve for y: |y| = e^(x² + C) = e^C · e^(x²). Let A = ±e^C, so y = A·e^(x²).
- Apply initial condition y(0) = 3: 3 = A·e⁰ → A = 3. Therefore, y = 3e^(x²).
题目： 求解 dy/dx = 2xy，已知 y(0) = 3。
解答：
- 分离变量： (1/y) dy = 2x dx（假设 y ≠ 0）
- 积分： ∫ (1/y) dy = ∫ 2x dx → ln|y| = x² + C
- 求解 y： |y| = e^(x²+C) = e^C · e^(x²)。令 A = ±e^C，得 y = A·e^(x²)。
- 代入初始条件 y(0) = 3： 3 = A·e⁰ → A = 3。因此 y = 3e^(x²)。
2.3 Worked Example 2 — Contextual (Newton’s Law of Cooling) | 例题2——应用题（牛顿冷却定律）
Question: A cup of coffee at 90°C is placed in a room at 20°C. The rate of cooling is proportional to the temperature difference between the coffee and the room. After 5 minutes, the coffee is 60°C. Find the temperature after 15 minutes.
题目： 一杯 90°C 的咖啡放在 20°C 的房间里。冷却速率与咖啡和室温的温差成正比。5 分钟后，咖啡温度为 60°C。求 15 分钟后的温度。
Solution | 解答：
Let T be the temperature at time t. Newton’s Law: dT/dt = -k(T – 20), where k > 0.
设 T 为时间 t 时的温度。牛顿冷却定律：dT/dt = -k(T – 20)，其中 k > 0。
- Separate: (1/(T – 20)) dT = -k dt
- Integrate: ln|T – 20| = -kt + C → T – 20 = Ae^(-kt) → T = 20 + Ae^(-kt)
- Apply T(0) = 90: 90 = 20 + A → A = 70. So T = 20 + 70e^(-kt).
- Apply T(5) = 60: 60 = 20 + 70e^(-5k) → e^(-5k) = 40/70 = 4/7 → -5k = ln(4/7) → k = (1/5)ln(7/4) ≈ 0.1119
- Find T(15): T(15) = 20 + 70e^(-15k) = 20 + 70(e^(-5k))³ = 20 + 70(4/7)³ = 20 + 70(64/343) = 20 + 13.06 ≈ 33.1°C
The coffee cools to approximately 33.1°C after 15 minutes. | 15 分钟后咖啡温度约为 33.1°C。
3. Method 2: Integrating Factor | 方法二：积分因子法
When a first-order DE is linear — i.e., it can be written in the standard form:
dy/dx + P(x)y = Q(x)
— we use the integrating factor method. This technique is required for Edexcel and CAIE, and is optional but useful for AQA and OCR.
当一阶微分方程是线性的——即可以写成标准形式 dy/dx + P(x)y = Q(x)——我们使用积分因子法。此方法为 Edexcel 和 CAIE 的必修内容，对 AQA 和 OCR 学生也非常实用。
3.1 The Integrating Factor Formula | 积分因子公式
The integrating factor (IF) is:
I(x) = e^(∫P(x)dx)
Multiply the entire DE by I(x), and the left-hand side becomes the derivative of a product:
d/dx [I(x)·y] = I(x)·Q(x)
Then integrate both sides:
I(x)·y = ∫ I(x)·Q(x) dx
将整个微分方程乘以 I(x)，左边变为乘积的导数；然后两边积分即可。
3.2 Worked Example 3 | 例题 3
Question: Solve dy/dx + 2y = e^(3x), with y(0) = 1.
题目： 求解 dy/dx + 2y = e^(3x)，已知 y(0) = 1。
Solution | 解答：
- Identify P(x) = 2, Q(x) = e^(3x).
- Compute IF: I(x) = e^(∫2 dx) = e^(2x).
- Multiply DE by IF: e^(2x)·dy/dx + 2e^(2x)·y = e^(5x). The LHS = d/dx[e^(2x)·y].
- Integrate: e^(2x)·y = ∫ e^(5x) dx = (1/5)e^(5x) + C.
- Solve for y: y = e^(-2x)[(1/5)e^(5x) + C] = (1/5)e^(3x) + Ce^(-2x).
- Apply y(0) = 1: 1 = 1/5 + C → C = 4/5.
- Final answer: y = (1/5)e^(3x) + (4/5)e^(-2x).
4. Contextual Applications | 实际应用场景
Differential equations are not just abstract mathematics — they model real-world phenomena. The A-Level exam will typically give you a word problem and ask you to (a) form the DE, (b) solve it, and (c) interpret the result. Here are the four most common contexts:
微分方程不仅仅是抽象数学——它们模拟现实世界的现象。A-Level 考试通常会给你一道文字题，要求你 (a) 建立微分方程，(b) 求解，(c) 解释结果。以下是四个最常见的应用场景：
4.1 Population Growth | 人口增长模型
Model: dP/dt = kP, where P is population and k is the growth rate. The solution is exponential: P = P₀e^(kt).
模型： dP/dt = kP，其中 P 为人口数量，k 为增长率。解为指数形式：P = P₀e^(kt)。
For more realistic models, the logistic equation dP/dt = kP(1 – P/M) is used, where M is the carrying capacity. Though this is beyond the standard A-Level syllabus, some exam boards (particularly Edexcel IAL) may touch on it in applied contexts.
更贴近实际的模型是逻辑斯谛方程 dP/dt = kP(1 – P/M)，其中 M 为环境承载量。虽然这在标准 A-Level 大纲之外，但一些考试局（特别是 Edexcel IAL）可能在应用背景中涉及。
4.2 Radioactive Decay | 放射性衰变
Model: dN/dt = -λN, where N is the number of undecayed nuclei and λ is the decay constant. Solution: N = N₀e^(-λt). The half-life T₁/₂ = ln(2)/λ.
模型： dN/dt = -λN，其中 N 为未衰变核的数量，λ 为衰变常数。解：N = N₀e^(-λt)。半衰期 T₁/₂ = ln(2)/λ。
4.3 Mixing Problems | 混合问题
A tank contains a salt solution. Pure water enters at a certain rate, and the well-mixed solution leaves at the same rate. Let S(t) be the amount of salt at time t:
dS/dt = (rate in) – (rate out) = 0 – (S/V)·(outflow rate)
一个水箱盛有盐水溶液。纯水以一定速率流入，充分混合后的溶液以相同速率流出。设 S(t) 为 t 时刻的盐量。
4.4 Mechanics: Velocity-Dependent Resistance | 力学：速度相关阻力
A body falling under gravity with air resistance proportional to velocity:
m(dv/dt) = mg – kv
This is a linear first-order DE solvable via integrating factor. The terminal velocity is v_terminal = mg/k, obtained by setting dv/dt = 0.
物体在重力作用下下落，空气阻力与速度成正比。这是一个线性一阶微分方程，可通过积分因子法求解。终端速度 v_terminal = mg/k，通过设 dv/dt = 0 获得。
5. Common Pitfalls and Examiner Tips | 常见失分点与考官建议
5.1 Forgetting the Constant of Integration | 忘记积分常数
The single most common mistake. After integrating, always add +C. Examiners typically deduct 1 mark for omitting it. Even if you are going to determine C from initial conditions immediately, you must show C first.
这是最常见的错误。积分后务必加上 +C。考官通常会为此扣 1 分。即使你打算立刻从初始条件求出 C，也必须先显示 C。
5.2 Division by Zero | 除以零
When separating variables involving 1/y or 1/(y – a), state that you are assuming y ≠ 0 or y ≠ a. A surprising number of students lose marks for not noting this assumption.
当分离涉及 1/y 或 1/(y – a) 的变量时，须声明你假设 y ≠ 0 或 y ≠ a。许多学生因未注明此假设而丢分。
5.3 Modulus Sign in Logarithms | 对数中的绝对值符号
When integrating 1/y, the correct antiderivative is ln|y|, not ln(y). Dropping the modulus can cost you an accuracy mark.
积分 1/y 时，正确的反导数是 ln|y| 而非 ln(y)。省略绝对值符号可能导致丢失准确性分数。
5.4 Mixing Up the Integrating Factor | 混淆积分因子
The IF is e^(∫P(x)dx). Do not confuse P(x) with Q(x). A common error is to set I(x) = e^(∫Q(x)dx), which is wrong.
积分因子是 e^(∫P(x)dx)。不要将 P(x) 与 Q(x) 混淆。常见错误是将 I(x) 设为 e^(∫Q(x)dx)，这是错误的。
5.5 Not Checking the Form Before Choosing a Method | 选择方法前未核对方程形式
Before diving in, ask yourself: is this separable, or is it linear? If it is linear, use the integrating factor. If it is separable, separate the variables. Trying to force the wrong method wastes time in the exam.
在开始解题之前，先问自己：这是可分离的，还是线性的？如果是线性的，用积分因子法；如果是可分离的，用分离变量法。强用错误方法只会浪费考试时间。
6. Exam-Style Practice Questions | 考试风格练习题
Q1 (Edexcel Style) | 第1题（Edexcel 风格）
Solve the differential equation dy/dx = y·cos(x), given that y(π/2) = e.
Answer: y = e^(sin(x) + 1 – 1) = e^(sin(x)). | 答案： y = e^(sin(x))。
Q2 (CAIE Style) | 第2题（CAIE 风格）
Find the general solution of x·dy/dx + y = x³.
Answer: y = (1/4)x³ + C/x. | 答案： y = (1/4)x³ + C/x。
Q3 (OCR Style) | 第3题（OCR 风格）
A radioactive substance decays at a rate proportional to its mass. Initially, the mass is 50 g, and after 10 days it is 40 g. Find its half-life.
Answer: T₁/₂ = 10·ln(2)/ln(1.25) ≈ 31.1 days. | 答案： T₁/₂ = 10·ln(2)/ln(1.25) ≈ 31.1 天。
Q4 (AQA Style) | 第4题（AQA 风格）
Water flows into a tank at 2 L/min and out at 2 L/min. The tank initially contains 100 L of water with 10 kg of dissolved salt. The incoming water is pure. Find the amount of salt after t minutes.
Answer: S(t) = 10e^(-t/50). | 答案： S(t) = 10e^(-t/50)。
7. Summary: How to Approach Any First-Order DE Question | 总结：如何应对任何一阶微分方程题目
Follow this decision tree in the exam:
- Read the question carefully. Is it a word problem? If so, extract the key mathematical relationship and form the DE. | 仔细读题。 是文字题吗？如果是，提取关键的数学关系并建立微分方程。
- Identify the type. Is the DE separable (dy/dx = g(x)·h(y)) or linear (dy/dx + P(x)y = Q(x))? | 判断类型。 方程是可分离的 (dy/dx = g(x)·h(y)) 还是线性的 (dy/dx + P(x)y = Q(x))？
- Choose the method. Separation of variables for separable DEs; integrating factor for linear DEs. | 选择方法。 可分离方程用分离变量法；线性方程用积分因子法。
- Solve systematically. Show every step — separate/integrate or find IF/multiply/integrate. Add +C. | 按步骤求解。 展示每一步——分离/积分或求IF/相乘/积分。记得加 +C。
- Apply initial conditions to find the particular solution. | 代入初始条件求特解。
- Check your answer. Does it satisfy the original DE? Does it satisfy the initial condition? A 30-second sanity check can save several marks. | 检查答案。 它满足原微分方程吗？满足初始条件吗？30 秒的复查可以挽救好几分。
8. Further Reading | 拓展阅读
Once you have mastered first-order ODEs, the next step is second-order differential equations — in particular, second-order linear ODEs with constant coefficients of the form a·d²y/dx² + b·dy/dx + cy = f(x), which appear in A-Level Further Mathematics (Edexcel FP2, CAIE Further Pure 2). These model damped harmonic oscillators, RLC circuits, and forced vibrations — powerful real-world applications that build directly on the techniques you have learned here.
掌握一阶常微分方程后，下一步是二阶微分方程——特别是形如 a·d²y/dx² + b·dy/dx + cy = f(x) 的常系数二阶线性常微分方程，这在 A-Level 进阶数学（Edexcel FP2、CAIE Further Pure 2）中出现。它们模拟阻尼谐振子、RLC 电路和受迫振动——这些强大的实际应用直接建立在你今天所学的技术之上。
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