📚 Solving Differential Equations: Separation of Variables | 微分方程求解：变量分离法
In A-Level Mathematics, differential equations describe relationships involving rates of change. This article focuses on first-order differential equations solvable by separating variables, a key technique for Edexcel Pure Mathematics exams. We will explore the method, illustrate with worked examples, and offer exam strategies.
在A-Level数学中，微分方程描述涉及变化率的关系。本文重点讨论可通过变量分离法求解的一阶微分方程，这是Edexcel纯数考试的关键技巧。我们将探索该方法，通过实例展示，并提供考试策略。
1. What is a Differential Equation? | 什么是微分方程？
e of y with respect to x equals 3x². Such equations arise naturally in physics, biology, and economics whenever a quantity's growth or decay depends on its current value.
包含未知函数导数的方程称为微分方程。例如，dy/dx = 3x² 表示 y 关于 x 的变化率等于 3x²。这类方程自然地出现在物理、生物和经济学中，只要一个量的增长或衰减依赖于其当前值。
e. The solution to a differential equation is a function y = f(x) that satisfies the equation, often involving an arbitrary constant.
在 Edexcel A-Level 中，你主要处理一阶常微分方程，最高阶导数为一次导数。微分方程的解是一个满足方程的函数 y = f(x)，通常包含任意常数。
2. Notation and Terminology | 记号和术语
The most common notation in A-Level is Leibniz' as a function of both x and y. The order of a differential equation is the highest derivative present; here we focus on first-order equations.
A-Level 中最常用的记号是莱布尼茨记号 dy/dx，其清楚地显示了变量。形如 dy/dx = f(x,y) 的方程将导数表示为 x 和 y 的函数。微分方程的阶是其中出现的最高阶导数；这里我们关注一阶方程。
(0) = 2, to determine the constant. Edexcel exam questions frequently ask you to find both forms.
通解包含任意常数（通常记作 C 或 A）。特解通过使用初始条件（如 y(0) = 2）确定该常数而得到。Edexcel 的考题经常要求你求出这两种形式。
3. The Separation of Variables Technique | 变量分离法
(x) dx, then integrate both sides. This transforms the problem into two simpler integrals.
形如 dy/dx = g(x)h(y) 的微分方程可通过分离变量求解。你将其改写为 (1/h(y)) dy = g(x) dx，然后对两边积分。这将问题转化为两个更简单的积分。
The key steps are: (i) identify g(x) and h(y); (ii) rearrange so all y‑terms are with dy and all x‑terms with dx; (iii) integrate both sides, adding the constant of integration on one side; (iv) solve for y if possible, and apply any given condition.
关键步骤为：(i) 确定 g(x) 和 h(y)；(ii) 重排使所有含 y 的项与 dy 在一起，所有含 x 的项与 dx 在一起；(iii) 对两边积分，并在其中一边加上积分常数；(iv) 尽可能解出 y，并应用给定的条件。
∫ (1/h(y)) dy = ∫ g(x) dx
Always remember that the constant of integration is crucial; omitting it will lose marks in exams.
务必记住积分常数至关重要；漏写它将在考试中失分。
4. Worked Example 1: Exponential Growth | 例1：指数增长
. This models unrestricted population growth or compound interest. Separate the variables:
考虑微分方程 dy/dx = ky，其中 k 为常量。它模拟无限制的人口增长或复利。分离变量：
(1/y) dy = k dx
Integrate both sides: ∫ (1/y) dy = ∫ k dx ⇒ ln|y| = kx + C. To remove the logarithm, exponentiate: |y| = eᵏˣ⁺ᶜ = eᶜ·eᵏˣ. Since eᶜ is a positive constant, we can write y = A eᵏˣ, where A = ± eᶜ (A ≠ 0). This is the general solution.
两边积分：∫ (1/y) dy = ∫ k dx ⇒ ln|y| = kx + C。为去掉对数，取指数：|y| = eᵏˣ⁺ᶜ = eᶜ·eᵏˣ。由于 eᶜ 是正常数，可写作 y = A eᵏˣ，其中 A = ± eᶜ（A ≠ 0）。这就是通解。
cal contexts, y is positive, so we can directly drop the absolute value and write y = A eᵏˣ tial condition y(0) = y₀ is given, then A = y₀, yielding the particular solution y = y₀ eᵏˣ.
在许多物理情境中 y 为正，因此可直接去掉绝对值并写成 y = A eᵏˣ。若给定初始条件 y(0) = y₀，则 A = y₀，得到特解 y = y₀ eᵏˣ。
5. Worked Example 2: Exponential Decay with Initial Condition | 例2：带初始条件的指数衰减
e equation dy/dx = −λy (λ > 0) describes radioactive decay or cooling. Separate variables:
方程 dy/dx = −λy（λ > 0）描述放射性衰变或冷却过程。分离变量：
(1/y) dy = −λ dx
Integrate: ln|y| = −λx + C. Let y(0) = y₀ be the initial amount. Substituting x = 0, y = y₀ gives ln|y₀| = C. Therefore ln|y| = −λx + ln|y₀| ⇒ ln|y/y₀| = −λx. Exponentiating gives y = y₀ e⁻⁽λˣ⁾ (assuming y₀ > 0).
积分：ln|y| = −λx + C。令 y(0) = y₀ 为初始量。代入 x = 0, y = y₀ 得 ln|y₀| = C。因此 ln|y| = −λx + ln|y₀| ⇒ ln|y/y₀| = −λx。取指数得 y = y₀ e⁻⁽λˣ⁾（假设 y₀ > 0）。
This particular solution shows that the quantity decays exponentially. In exams, showing the step from the general solution to the particular solution using initial conditions is essential.
该特解表明量呈指数衰减。考试中，要展示从通解利用初始条件导出特解的步骤，这很关键。
6. Finding General and Particular Solutions | 求通解和特解
The general solution always includes one arbitrary constant for a first-order ODE. This constant reflects the family of curves that satisfy the equation. When an initial condition (x₀,y₀) is supplied, you substitute to solve for the constant, obtaining a unique particular solution.
对一阶常微分方程，通解总含有一个任意常数。该常数反映了满足方程的曲线族。当提供初始条件 (x₀,y₀) 时，代入求解常数，得到唯一的特解。
ple, if dy/dx = 2x/y, the separated form is y dy = 2x dx. Integrating yields y²/2 = x² + C ⇒ . Often we merge constants: let D = 2C, so y² = 2x² + D. If given y(1) = 2, then 4 = 2 + D ⇒ D = 2, giving y² = 2x² + 2.
例如，若 dy/dx = 2x/y，分离得 y dy = 2x dx。积分得 y²/2 = x² + C ⇒ y² = 2x² + 2C。通常合并常数：令 D = 2C，则 y² = 2x² + D。若给定 y(1) = 2，则 4 = 2 + D ⇒ D = 2，得 y² = 2x² + 2。
Always present your constant in the neatest form, e.g. A, C, or K. Avoid leaving expressions like 2C unnecessarily; rename it so the final answer looks tidy.
总是用最简洁的形式表示常数，如 A、C 或 K。避免不必要地保留如 2C 的表达式；重命名常数使最终答案整洁。
7. Modelling Real-World Scenarios | 现实世界建模
Differential equations model numerous real phenomena: population growth (dy/dt ∝ y), radioactive decay (dy/dt ∝ −y), Newton’s law of cooling (dT/dt ∝ T − Tₐ), and simple chemical reactions. The Edexcel specification expects you to interpret the rate in context and translate words into d y/d x form.
微分方程可模拟众多现实现象：人口增长（dy/dt ∝ y）、放射性衰变（dy/dt ∝ −y）、牛顿冷却定律（dT/dt ∝ T − Tₐ）以及简单的化学反应。Edexcel 大纲要求你能根据上下文解释变化率，并将文字转化为 dy/dx 形式。
For instance, “the rate of increase of a population is proportional to the population size” converts to dP/dt = kP. “The temperature of a body falls at a rate proportional to the excess over room temperature” becomes dT/dt = −k(T − 20) if room is 20°C.
例如，“人口增长率与人口数量成正比”转化为 dP/dt = kP。“物体温度下降速率与超出室温的温差成正比”变为 dT/dt = −k(T − 20)（假设室温 20°C）。
Once the equation is set up, you apply separation of variables and use any given data to find constants. The final step often involves substituting back to answer a specific question, like finding the population after 5 years.
一旦建立方程，你应用变量分离法并利用给定数据求出常数。最后一步通常需要代回以解答具体问题，比如求 5 年后的人口数量。
8. Common Pitfalls and How to Avoid Them | 常见错误及如何避免
One frequent mistake is forgetting the constant of integration and writing y = eᵏˣ instead of y = A eᵏˣ particular case, not the general solution. Always insert +C immediately after integrating.
一个常见错误是忘记积分常数，写成 y = eᵏˣ 而不是 y = A eᵏˣ。常数并非可有可无；漏掉它会使解成为特例而非通解。务必在积分后立即加上 +C。
ple, y = 0 might also satisfy the original DE, and Edexcel sometimes expects you to state this.
另一个陷阱是除以可能为零的函数，例如当 y 可能为零时除以 y。此时应检查是否有遗漏的常数解。例如 y = 0 也可能满足原微分方程，Edexcel 有时要求你指出这一点。
When using logarithms, remember the absolute value. After integrating 1/y, you get ln|y|, not ln y. This becomes important when the solution may take negative values, though many exam contexts assume y > 0.
使用对数时，记住绝对值。对 1/y 积分得到 ln|y|，而非 ln y。当解可能取负值时这一点很重要，尽管许多考试情境假设 y > 0。
9. Practice Problem | 练习题
/y, given that y = 3 when x = 1. Find the particular solution and express y in terms of x.
尝试这道考试风格的问题：求解微分方程 dy/dx = (x+1)/y，已知当 x = 1 时 y = 3。求特解并将 y 用 x 表示。
Solution: Separate variables: y dy = (x+1) dx. Integrate: ½ y² = ½ x² + x + C ⇒ y² = x² + 2x + D (where D = 2C). Using y(1)=3: 9 = 1 + 2 + D ⇒ y² = x² + 2x + 6. Since y = 3 when x=1 is positive, we take the positive square root: y = √x² + 2x + 6. This is the required particular solution.
解答：分离变量：y dy = (x+1) dx。积分：½ y² = ½ x² + x + C ⇒ y² = x² + 2x + D（其中 D = 2C）。利用 y(1)=3：9 = 1 + 2 + D ⇒ D = 6。因此 y² = x² + 2x + 6。由于 x=1 时 y=3 为正，取正平方根：y = √x² + 2x + 6。这就是要求的特解。
10. Summary and Exam Tips | 总结与考试技巧
teps: separate, integrate, and apply initial conditions. Practice with a variety of functions, including those requiring integration of rational and radical expressions.
变量分离法是求解形如 dy/dx = g(x)h(y) 的一阶常微分方程的强大且直接的方法。掌握三个核心步骤：分离、积分并
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