A-Level Edexcel 数学：分离变量法求解一阶微分方程 / Solving First-Order Differential Equations by Separation of Variables

引言 / Introduction

微分方程是 A-Level 数学中最迷人也是最具挑战性的主题之一。它们不仅是纯数学的核心组成部分，也是物理、工程、经济和生物等应用数学领域的基础工具。在 Edexcel A-Level 数学大纲中，一阶微分方程构成了微分方程模块的入门部分，而分离变量法是学生需要掌握的第一种求解技巧。本文将系统性地讲解分离变量法的原理、步骤和常见变体，并通过详细例题帮助你建立扎实的解题能力。

Differential equations are among the most fascinating yet challenging topics in A-Level Mathematics. They are not only a core component of pure mathematics but also a foundational tool in applied fields such as physics, engineering, economics, and biology. In the Edexcel A-Level Mathematics syllabus, first-order differential equations form the entry point to the differential equations module, and separation of variables is the first solving technique students must master. This article systematically explains the principles, steps, and common variations of separation of variables, with detailed worked examples to build your problem-solving confidence.

什么是一阶微分方程？ / What Is a First-Order Differential Equation?

一阶微分方程是包含一个未知函数及其一阶导数的方程。一般形式为：

dy/dx = f(x, y)

其中 y 是 x 的未知函数，dy/dx 表示 y 关于 x 的变化率。方程中只出现一阶导数（dy/dx），不涉及二阶或更高阶导数，因此称为“一阶”（first-order）。一阶微分方程描述了系统状态随一个变量的变化规律，例如人口增长速率、放射性衰变速率、物体冷却速率等，都可用一阶微分方程建模。

A first-order differential equation is an equation involving an unknown function and its first derivative. The general form is:

dy/dx = f(x, y)

where y is an unknown function of x, and dy/dx represents the rate of change of y with respect to x. Only the first derivative (dy/dx) appears — no second-order or higher derivatives — hence the name “first-order.” First-order differential equations describe how a system’s state changes with respect to a single variable: population growth rate, radioactive decay rate, cooling rate of an object — all can be modelled with first-order differential equations.

什么是分离变量法？ / What Is Separation of Variables?

分离变量法（Separation of Variables）是求解一阶微分方程最基本、最直观的方法。当一个微分方程可以写成以下形式时，就能使用分离变量法：

dy/dx = g(x) · h(y)

也就是说，方程右侧可以分解为“仅含 x 的函数”与“仅含 y 的函数”的乘积。如果能做到这一点，我们就可以将含 y 的项移到等号一边（与 dy 在一起），将含 x 的项移到等号另一边（与 dx 在一起），然后对两边分别积分。这个思想朴素却强大，是求解许多实际问题的第一选择。

Separation of Variables is the most fundamental and intuitive method for solving first-order differential equations. When a differential equation can be written in the form:

dy/dx = g(x) · h(y)

i.e., the right-hand side can be factored into a product of “a function of x only” and “a function of y only,” we can apply this method. Move all terms involving y to one side (together with dy), and all terms involving x to the other side (together with dx), then integrate both sides. The idea is simple yet powerful — it is often the first approach to try for many real-world problems.

分离变量法的标准步骤 / Standard Steps of Separation of Variables

掌握以下五个步骤，就能应对绝大多数分离变量法的题目：

1. 识别可分离性 / Identify Separability：检查方程是否能写成 dy/dx = g(x) · h(y) 的形式。如果不能，考虑其他方法（如积分因子法）。
2. 分离变量 / Separate Variables：将方程改写为 (1/h(y)) dy = g(x) dx。注意 h(y) ≠ 0 的情况需要单独讨论。
3. 两边积分 / Integrate Both Sides：对等式两边分别积分：∫ (1/h(y)) dy = ∫ g(x) dx。不要忘记加积分常数！
4. 求解 y / Solve for y：将积分结果整理成 y = f(x) + C 或隐式形式 F(x, y) = C。如果题目给出了初始条件（initial condition），代入求出具体的 C 值。
5. 验证 / Verify：将结果代回原方程，检查是否满足。这一步在考试中能帮你发现符号错误。

Master these five steps and you can handle the vast majority of separation-of-variables questions:

1. Identify Separability: Check whether the equation can be written as dy/dx = g(x) · h(y). If not, consider alternative methods (e.g., integrating factor).
2. Separate Variables: Rewrite as (1/h(y)) dy = g(x) dx. Pay attention to cases where h(y) = 0 — these may require separate treatment.
3. Integrate Both Sides: Integrate each side: ∫ (1/h(y)) dy = ∫ g(x) dx. Do not forget the constant of integration!
4. Solve for y: Rearrange the result to y = f(x) + C or implicit form F(x, y) = C. If an initial condition is given, substitute to find the specific value of C.
5. Verify: Substitute your solution back into the original equation. This step can catch sign errors in the exam.

例题一：基础分离变量 / Example 1: Basic Separation

题目 / Problem：求解微分方程 dy/dx = 2xy，并给出通解。

解答 / Solution：

第1步 — 识别：方程已经是 dy/dx = g(x) · h(y) 的形式，其中 g(x) = 2x，h(y) = y。可以直接分离。

第2步 — 分离：将 y 移到左边，x 移到右边：

(1/y) dy = 2x dx

注意：假设 y ≠ 0。y = 0 是否是解？代回原方程：若 y = 0，则 dy/dx = 0，左边 = 0，右边 = 2x · 0 = 0，是解。但通常通解已涵盖此退化情况，考试中注明即可。

第3步 — 积分：

∫ (1/y) dy = ∫ 2x dx

ln|y| = x² + C

第4步 — 解出 y：

|y| = e^(x² + C) = e^C · e^(x²)

令 A = ±e^C（A 为非零常数），则：

y = A · e^(x²)

这就是通解（general solution）。注意 A 可以是任意实常数（包括零，对应 y = 0 的平凡解）。

Problem: Solve the differential equation dy/dx = 2xy and give the general solution.

Solution:

Step 1 — Identify: The equation is already in the form dy/dx = g(x) · h(y), with g(x) = 2x, h(y) = y. We can separate directly.

Step 2 — Separate: Move y to the left, x to the right:

(1/y) dy = 2x dx

Note: we assume y ≠ 0. Is y = 0 a solution? Substitute back: if y = 0, then dy/dx = 0, LHS = 0, RHS = 2x · 0 = 0 — it is a solution. But the general solution usually covers this degenerate case; just mention it in the exam.

Step 3 — Integrate:

∫ (1/y) dy = ∫ 2x dx

ln|y| = x² + C

Step 4 — Solve for y:

|y| = e^(x² + C) = e^C · e^(x²)

Let A = ±e^C (A is a non-zero constant), then:

y = A · e^(x²)

This is the general solution. Note that A can be any real constant (including zero, corresponding to the trivial solution y = 0).

例题二：含初始条件的特解 / Example 2: Particular Solution with Initial Condition

题目 / Problem：求解 dy/dx = (x + 1) / y，满足 y(0) = 2。

解答 / Solution：

分离：

y dy = (x + 1) dx

积分：

∫ y dy = ∫ (x + 1) dx

y²/2 = x²/2 + x + C

乘以 2：

y² = x² + 2x + 2C

令 K = 2C（为方便）：

y² = x² + 2x + K

代入初始条件 y(0) = 2：

(2)² = 0² + 2(0) + K → 4 = K

特解：

y² = x² + 2x + 4

y = √(x² + 2x + 4)（取正根因为 y(0) = 2 > 0）

这是一个典型的初始值问题（Initial Value Problem, IVP），初始条件确定了积分常数，从而从一族曲线中选出唯一满足条件的特解。

Problem: Solve dy/dx = (x + 1) / y, given that y(0) = 2.

Solution:

Separate:

y dy = (x + 1) dx

Integrate:

∫ y dy = ∫ (x + 1) dx

y²/2 = x²/2 + x + C

Multiply by 2:

y² = x² + 2x + 2C

Let K = 2C for convenience:

y² = x² + 2x + K

Apply initial condition y(0) = 2:

(2)² = 0² + 2(0) + K → 4 = K

Particular solution:

y² = x² + 2x + 4

y = √(x² + 2x + 4) (take the positive root since y(0) = 2 > 0)

This is a classic Initial Value Problem (IVP). The initial condition pins down the integration constant, selecting the unique solution curve from an entire family.

例题三：指数型微分方程（自然增长与衰减）/ Example 3: Exponential Differential Equations (Natural Growth & Decay)

题目 / Problem：一个细菌培养皿中，细菌数量 N 的增长速率与当前数量成正比。已知初始有 100 个细菌，2 小时后增加到 400 个。求任意时刻 t 的细菌数量表达式，并计算 5 小时后的细菌数量。

解答 / Solution：

根据题意建立微分方程：

dN/dt = kN（其中 k 为正常数，即增长率）

分离变量：

(1/N) dN = k dt

积分：

∫ (1/N) dN = ∫ k dt

ln|N| = kt + C

N = Ae^(kt)，其中 A = e^C

代入 N(0) = 100：

100 = Ae^0 → A = 100

N = 100e^(kt)

代入 N(2) = 400 求 k：

400 = 100e^(2k)

4 = e^(2k)

2k = ln 4 → k = (ln 4)/2 = ln 2

最终模型：

N(t) = 100e^(t ln 2) = 100 · 2^t

5 小时后的数量：

N(5) = 100 · 2^5 = 100 · 32 = 3200 个细菌

指数增长模型是分离变量法最经典的应用之一。同样的模型也适用于放射性衰变（k 为负）、复利计算、药物代谢等场景。

Problem: In a bacterial culture, the growth rate of the bacterial population N is proportional to the current population. Initially there are 100 bacteria, and after 2 hours the population increases to 400. Find the expression for N at any time t, and calculate the population after 5 hours.

Solution:

Formulate the differential equation from the description:

dN/dt = kN (where k is a positive constant — the growth rate)

Separate variables:

(1/N) dN = k dt

Integrate:

∫ (1/N) dN = ∫ k dt

ln|N| = kt + C

N = Ae^(kt), where A = e^C

Apply N(0) = 100:

100 = Ae^0 → A = 100

N = 100e^(kt)

Apply N(2) = 400 to find k:

400 = 100e^(2k)

4 = e^(2k)

2k = ln 4 → k = (ln 4)/2 = ln 2

Final model:

N(t) = 100e^(t ln 2) = 100 · 2^t

Population after 5 hours:

N(5) = 100 · 2^5 = 100 · 32 = 3200 bacteria

The exponential growth model is one of the most classic applications of separation of variables. The same model applies to radioactive decay (k negative), compound interest, drug metabolism, and many other scenarios.

常见易错点与考试技巧 / Common Pitfalls & Exam Tips

1. 忘记积分常数 / Forgetting the Constant of Integration

这是 A-Level 考试中最常见的失分点。每次积分都必须加上常数 C，即使你认为可以“两边抵消”。积分常数代表一族解（family of solutions），是微分方程通解的核心特征。如果题目有初始条件，也必须先写出含 C 的通解再代入求值，不能跳过这一步。

This is the single most common mark-losing mistake in A-Level exams. You must add the constant C every time you integrate — even if you think it will “cancel out on both sides.” The integration constant represents a family of solutions and is the defining feature of a general solution. Even when an initial condition is given, you must first write the general solution with C, then substitute to find its value — never skip this step.

2. 绝对值处理不当 / Mishandling Absolute Values

积分 1/y 得到 ln|y| 而非 ln y。当后续步骤通过指数函数消除 ln 时，绝对值符号转化为 ± 号，最终被吸收进常数 A。许多学生在这一步犯错，直接写成 ln y 而丢失了负值解。虽然在最终答案中常数 A 的任意性能覆盖正负情况，但推导过程中省略绝对值是不严谨的，可能被扣分。

Integrating 1/y gives ln|y|, not ln y. When the logarithm is later eliminated via exponentiation, the absolute value transforms into a ± sign, which is eventually absorbed into the constant A. Many students make mistakes here, writing ln y directly and losing the negative solution branch. Although the arbitrariness of constant A in the final answer covers both positive and negative cases, omitting the absolute value in the derivation is mathematically imprecise and may lose marks.

3. h(y) = 0 的奇异解 / Singular Solutions Where h(y) = 0

分离变量时除以 h(y)，必须考虑 h(y) = 0 的情况。例如 dy/dx = y²，分离后 1/y² dy = dx，但 y = 0 也是原方程的解（0 的导数是 0，右边 y² = 0² = 0）。这种“丢失的解”称为奇异解（singular solution），在 Edexcel 考试中通常需要注明。

When dividing by h(y) during separation, you must consider cases where h(y) = 0. For example, in dy/dx = y², after separation we get 1/y² dy = dx, but y = 0 is also a solution to the original equation (the derivative of 0 is 0, and the RHS y² = 0² = 0). Such “lost solutions” are called singular solutions, and they usually need to be noted in Edexcel exams.

4. 将 x 和 y 混在同一积分中 / Mixing x and y in the Same Integral

分离变量后，左边积分仅涉及 y，右边积分仅涉及 x。不能出现 ∫ (y + x) dx 这种混在一起的情况。分离的彻底性是方法的前提。

After separation, the left-hand integral involves only y, and the right-hand integral involves only x. You must not have mixed integrals like ∫ (y + x) dx. Thorough separation is the prerequisite for the method to work.

分离变量法的扩展 / Extensions of Separation of Variables

掌握了基础分离变量法后，Edexcel A-Level 还会考查以下变体：

可化为可分离形式的方程 / Equations Reducible to Separable Form：某些方程初看不可分离，但通过代换（substitution）可以转化。例如齐次方程 dy/dx = f(y/x)，令 v = y/x，则 y = vx，dy/dx = v + x(dv/dx)，代入后往往可以分离变量。

部分分式辅助积分 / Partial Fractions in Integration：有时分离后的积分 ∫ 1/h(y) dy 需要借助部分分式法（partial fractions）来计算，尤其是在 h(y) 为二次多项式时。例如 ∫ 1/(y² − 1) dy = ∫ 1/[(y−1)(y+1)] dy = (1/2)∫ [1/(y−1) − 1/(y+1)] dy。

隐式通解 / Implicit General Solutions：有时积分后无法显式解出 y = f(x)，此时保留隐式形式 F(x, y) = C 是完全可接受的答案。Edexcel 评分标准明确允许隐式解。

After mastering basic separation of variables, Edexcel A-Level also tests these variants:

Equations Reducible to Separable Form: Some equations do not appear separable at first glance but can be transformed via substitution. For example, for homogeneous equations dy/dx = f(y/x), let v = y/x, then y = vx and dy/dx = v + x(dv/dx). Substituting often yields a separable equation in v and x.

Partial Fractions in Integration: Sometimes the separated integral ∫ 1/h(y) dy requires partial fractions, especially when h(y) is a quadratic polynomial. For instance, ∫ 1/(y² − 1) dy = ∫ 1/[(y−1)(y+1)] dy = (1/2)∫ [1/(y−1) − 1/(y+1)] dy.

Implicit General Solutions: Sometimes after integration you cannot solve explicitly for y = f(x). In such cases, leaving the answer in implicit form F(x, y) = C is perfectly acceptable. Edexcel mark schemes explicitly allow implicit solutions.

练习题目 / Practice Problems

尝试独立完成以下题目，然后对照答案检验：

1. 求 dy/dx = y·cos x 的通解。 / Find the general solution of dy/dx = y·cos x.
2. 求 dy/dx = x²/y³，满足 y(0) = 1 的特解。 / Solve dy/dx = x²/y³, given y(0) = 1.
3. 一个放射性样品以与其当前质量成正比的速率衰变。初始质量为 50g，10 天后减少到 40g。求半衰期。 / A radioactive sample decays at a rate proportional to its current mass. Initial mass is 50g, and after 10 days it has reduced to 40g. Find the half-life.
4. 求 (1 + x²)dy/dx = xy 的通解。 / Find the general solution of (1 + x²)dy/dx = xy.
5. 牛顿冷却定律：物体冷却速率正比于物体温度与环境温度之差。一杯 90°C 的咖啡放在 20°C 的房间中，5 分钟后降至 60°C。求再过 5 分钟后的温度。 / Newton’s Law of Cooling: the cooling rate is proportional to the temperature difference between the object and its surroundings. A 90°C cup of coffee is placed in a 20°C room and cools to 60°C in 5 minutes. Find the temperature after a further 5 minutes.

答案速查 / Quick Answer Check

1. y = Ae^(sin x)
2. y⁴ = (4/3)x³ + 1 → y = ⁴√((4/3)x³ + 1)
3. k = (1/10)ln(0.8)，半衰期 t₁/₂ = (ln 2)/|k| ≈ 31.1 天 / half-life ≈ 31.1 days
4. y = A√(1 + x²)
5. 约 38.6°C / approximately 38.6°C（提示：T(t) = 20 + 70e^(kt)，先求 k，再代入 t = 10 / Hint: find k first, then substitute t = 10）

总结 / Summary

分离变量法是 A-Level Edexcel 数学微分方程模块的核心方法。掌握它的关键在于：(1) 准确识别可分离形式；(2) 严格按步骤分离、积分、求解；(3) 不遗漏积分常数和奇异解。熟练后，无论是纯数学题目还是应用题（增长率、衰变、冷却等），你都能从容应对。建议至少练习 20–30 道不同类型的题目，形成肌肉记忆，确保考试中能做到零失误。

Separation of variables is the core method in the Edexcel A-Level Mathematics differential equations module. The keys to mastery are: (1) accurately identifying separable forms; (2) rigorously following the steps — separate, integrate, solve; (3) never omitting the integration constant or singular solutions. Once proficient, you will confidently handle both pure mathematics problems and application problems (growth rate, decay, cooling, etc.). We recommend practising at least 20–30 questions of varied types to build muscle memory, ensuring zero mistakes in the exam.