Mastering Separation of Variables for First-Order Differential Equations | 掌握一阶微分方程的变量分离法

📚 Mastering Separation of Variables for First-Order Differential Equations | 掌握一阶微分方程的变量分离法

Differential equations lie at the heart of A-Level Maths, especially within the Edexcel Pure Mathematics specification. One of the most accessible yet powerful techniques you must master is ‘separation of variables’. This method allows you to solve first-order ordinary differential equations where the rate of change of a quantity can be expressed as a product of a function of the independent variable and a function of the dependent variable. Many real-world problems – from population growth and radioactive decay to Newton’s law of cooling – can be modelled and solved using this approach. In this article, we will unpack the method step by step, work through carefully chosen examples, highlight typical exam pitfalls, and equip you with the confidence to tackle any separation of variables question on your Edexcel paper.

微分方程是A-Level数学的核心内容，尤其在Edexcel纯数学考纲中占有重要地位。你必须掌握的一种最直观且功能强大的技巧就是“变量分离法”。当一阶常微分方程的变化率可以表示为只含自变量的函数与只含因变量的函数的乘积时，就可以用这种方法求解。从人口增长、放射性衰变到牛顿冷却定律，许多实际问题都可以借助这个工具建模并求解。本文将逐步拆解这一方法，通过精心设计的例题进行演练，指出考试中常见的失分陷阱，帮助你充满信心地应对Edexcel试卷中任何一道变量分离法题目。

1. Recognising When to Separate | 识别何时可以使用变量分离法

The separation of variables technique can be applied to any first-order differential equation that can be written in the form dy/dx = g(x)·h(y), where g(x) is a function of x only and h(y) is a function of y only. In the Edexcel exam, the equation is often given in a slightly disguised manner, such as dy/dx = f(x,y) where you need to factorise the right-hand side. You may also encounter differential equations accompanied by initial conditions, allowing you to determine the constant of integration and find a particular solution.

变量分离法适用于任何可以写成 dy/dx = g(x)·h(y) 形式的一阶微分方程，其中 g(x) 仅包含 x，h(y) 仅包含 y。在Edexcel考试中，方程往往会给得稍微隐蔽一些，比如 dy/dx = f(x,y)，此时你需要对右边进行因式分解。题目还常常会给出初始条件，要求你确定积分常数并求出特解。

• Standard form: dy/dx = g(x) · h(y)
• 中文理解： 标准形式为 dy/dx = g(x)·h(y)
• Key check: Can you isolate all y terms (including dy) on one side and all x terms (including dx) on the other?
• 关键检查： 能否将所有含 y 的项（包括 dy）移到方程一边，所有含 x 的项（包括 dx）移到另一边？

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2. The Mechanics of the Technique | 变量分离法的运算机制

Once a differential equation is written in the form dy/dx = g(x)h(y), the separation procedure involves treating dy/dx as a fraction (which is justified by the chain rule) and rewriting the equation so that every term involving y is on the side of dy, and every term involving x is on the side of dx. More precisely, you rearrange to obtain 1/h(y) dy = g(x) dx, then integrate both sides independently. Remember to add a constant of integration on one side only – usually on the side that is more convenient. This gives a general solution in implicit form, which often needs to be rearranged to express y explicitly in terms of x.

将微分方程写成 dy/dx = g(x)h(y) 的形式后，分离过程就是把 dy/dx 当作分数来处理（这可以通过链式法则严格证明），然后重新整理，使得所有含 y 的项都跟随 dy，所有含 x 的项都跟随 dx。更准确地说，你要移项变成 1/h(y) dy = g(x) dx，然后对两边分别积分。别忘了只在一侧加上积分常数——通常加在较方便的那一侧。这样就会得到隐式通解，往往还需要进一步整理，把 y 表示成 x 的显函数。

Step: ∫ 1/h(y) dy = ∫ g(x) dx

The rigorous justification lies in the inverse relationship between differentiation and integration. By applying the chain rule, the derivative of ∫ 1/h(y) dy with respect to x gives (1/h(y))·dy/dx, which matches the separated equation.

严格的依据在于微分与积分的互逆关系。利用链式法则，∫ 1/h(y) dy 对 x 求导恰好得到 (1/h(y))·dy/dx，这与分离后的方程一致。

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3. Step-by-Step Guided Example: Exponential Growth | 分步指导例题：指数增长

Consider the differential equation dy/dx = 3y, with the initial condition y(0) = 5. This is a classic exponential growth model. We identify that g(x)=3 and h(y)=y, so we can separate variables.

考虑微分方程 dy/dx = 3y，初始条件为 y(0) = 5。这是一个经典的指数增长模型。我们可以识别出 g(x)=3、h(y)=y，所以可以分离变量。

Step 1: Separate: (1/y) dy = 3 dx.

步骤1：分离变量：(1/y) dy = 3 dx。

Step 2: Integrate both sides: ∫ 1/y dy = ∫ 3 dx → ln|y| = 3x + C.

步骤2：两边积分：∫ 1/y dy = ∫ 3 dx → ln|y| = 3x + C。

Step 3: Exponentiate to solve for y: |y| = e^(3x+C) = e^C · e^(3x). Let A = ±e^C, so y = A e^(3x).

步骤3：取指数解出 y：|y| = e^(3x+C) = e^C · e^(3x)。令 A = ±e^C，于是 y = A e^(3x)。

Step 4: Apply initial condition: 5 = A e^(0) → A = 5. Therefore, the particular solution is y = 5e^(3x).

步骤4：代入初始条件：5 = A e^(0) → A = 5。因此特解为 y = 5e^(3x)。

y = 5 e³ˣ

This clean result illustrates a pattern you will encounter frequently: exponential solutions emerge naturally when the rate of change is proportional to the quantity itself.

这个简洁的结果展示了一种常见的模式：当变化率与量本身成正比时，自然就会得到指数解。

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4. Handling More Complex Functions | 处理更复杂的函数形式

In an Edexcel exam, the simple y and x separation is often replaced by expressions like dy/dx = (x+2)/y² or dy/dx = e^(x) · cos y. The principle remains the same, but careful algebraic manipulation is required. For dy/dx = (x+2)/y², multiply both sides by y² dx to get y² dy = (x+2) dx. Then integrate: ∫ y² dy = ∫ (x+2) dx → y³/3 = (1/2)x² + 2x + C. Multiplying through by 3 gives y³ = (3/2)x² + 6x + D, where D = 3C.

在Edexcel考试中，简单的 y 和 x 分离往往会被类似 dy/dx = (x+2)/y² 或 dy/dx = e^(x)·cos y 这样的表达式取代。原理不变，但需要谨慎的代数操作。对于 dy/dx = (x+2)/y²，两边同乘 y² dx 得到 y² dy = (x+2) dx。然后积分：∫ y² dy = ∫ (x+2) dx → y³/3 = (1/2)x² + 2x + C。两边同乘3后得到 y³ = (3/2)x² + 6x + D，其中 D = 3C。

When trigonometric functions appear, such as dy/dx = tan y · sec² x, the separated form is cot y dy = sec² x dx. Integration yields ln|sin y| = tan x + C. It is essential to be fluent with standard integrals of trigonometric functions – a non-negotiable for Edexcel success.

当出现三角函数时，例如 dy/dx = tan y · sec² x，分离后变成 cot y dy = sec² x dx。积分可得 ln|sin y| = tan x + C。对三角函数的标准积分公式滚瓜烂熟，是拿下Edexcel考试的必备条件。

Separable form	Integral to evaluate
sec² x tan y	∫ cot y dy = ∫ sec² x dx
x√(y) / (1+ x²)	∫ y⁻¹/² dy = ∫ x/(1+x²) dx

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5. Applying Initial Conditions to Find Particular Solutions | 使用初始条件求特解

The general solution obtained from integration always contains an arbitrary constant. In most exam questions, you will be given an initial or boundary condition – for instance, y = 2 when x = 0 – to determine the value of that constant. It is best practice to substitute the condition immediately after integration, before attempting to rearrange the implicit equation into an explicit form. This approach often simplifies the algebra and reduces sign errors.

积分得到的通解总含有一个任意常数。在大多数试题中，题目会给出初始或边界条件——例如 x=0 时 y=2——以确定常数的值。最佳做法是在积分之后立即代入条件，再尝试将隐式方程整理为显式形式。这样往往能简化代数运算，减少符号错误。

For example, after solving dy/dx = (2x+1)/y and obtaining ½y² = x² + x + C, you could be given y(1)=3. Substituting gives ½(9) = 1 + 1 + C → 4.5 = 2 + C → C = 2.5. The particular solution is then y² = 2x² + 2x + 5. Leaving the answer in a clean implicit form is often accepted unless the question explicitly asks for y in terms of x.

例如，在求解 dy/dx = (2x+1)/y 并得到 ½y² = x² + x + C 之后，可能给出 y(1)=3。代入得到 ½(9)=1+1+C → 4.5=2+C → C=2.5。于是特解为 y² = 2x² + 2x + 5。除非题目明确要求写成 y 关于 x 的显式，保留简洁的隐式形式通常也能被接受。

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6. Rearranging into the Required Form | 整理为题目要求的形式

Edexcel exam questions frequently ask for the solution in a specific format, such as y = f(x) or an expression like ln|y+1| = 2x + k. Always read the wording carefully. When exponentiation is required, remember the absolute value sign: e^(ln|y|) = |y|. You can remove the absolute value by introducing a ± constant, which can then be absorbed into a new constant A. The new constant can take any non-zero real value, and sometimes zero is also permitted depending on the context.

Edexcel试题经常要求将答案写成特定形式，比如 y = f(x) 或者 ln|y+1| = 2x + k 这样的表达式。务必仔细审题。当需要取指数时，别忘了绝对值符号：e^(ln|y|) = |y|。你可以通过引入 ± 常数来去掉绝对值，并将其吸收为一个新常数 A。新常数可以取任意非零实数，有时根据上下文，零也是允许的。

For the equation dy/dx = y + 1, separation leads to ∫ 1/(y+1) dy = ∫ 1 dx, giving ln|y+1| = x + C. Then |y+1| = e^(x+C) = e^C e^x, so y+1 = A e^x, where A = ±e^C. Finally y = A e^x − 1. It is very common for students to lose marks by forgetting the −1 or mishandling the constant.

对于方程 dy/dx = y + 1，分离变量得到 ∫ 1/(y+1) dy = ∫ 1 dx，从而 ln|y+1| = x + C。那么 |y+1| = e^(x+C) = e^C e^x，因此 y+1 = A e^x，其中 A = ±e^C。最终 y = A e^x − 1。学生经常因为忘记 −1 或者常数处理不当而失分。

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7. Common Mistakes and How to Avoid Them | 常见错误与规避方法

Even confident students can stumble on separation of variables under exam pressure. Here are the most frequent pitfalls and their remedies.

即便是自信满满的学生，在考试压力下也可能在变量分离上栽跟头。以下是最常见的陷阱及其应对方法。

• Forgetting the constant of integration: Always write ‘+ C’ immediately after the integration step. Many mark schemes award a method mark for including the constant.
• 忘记积分常数： 每次积分后立即写上“+ C”。多数评分标准会给包含常数的步骤方法分。
• Misplacing the constant: Adding C to both sides unnecessarily. One side is sufficient; adding to both simply combines into a new constant.
• 常数位置不对： 没必要两边都加 C；加在一侧即可，加两次无非是合并成一个新常数。
• Incorrect handling of absolute values: When integrating 1/(y-a), remember to use ln|y-a|. The modulus can be lifted if the initial condition guarantees a positive argument.
• 处理绝对值不当： 积分 1/(y-a) 时，记住要用 ln|y-a|。如果初始条件能保证表达式为正，就可以去掉绝对值。
• Dividing by a function that could be zero: If h(y) could equal zero, the solution y = constant is a singular solution that should be considered separately.
• 除以可能为零的函数： 如果 h(y) 可能为零，那么 y = 常数 就是一个奇异解，应该单独考虑。
• Algebra slip when exponentiating: e^(ln|y|+C) is not |y| + e^C; it is |y|·e^C. Use the law e^(A+B) = e^A · e^B.
• 取指数时代数出错： e^(ln|y|+C) 不是 |y| + e^C，而是 |y|·e^C。要使用 e^(A+B)=e^A·e^B。

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8. Connecting to Real-World Modelling Contexts | 联系实际建模场景

Edexcel loves to embed differential equations in a contextual problem. You might see a chemical reaction where the rate of concentration change is proportional to the product of reactants, or a temperature change according to Newton’s law: dθ/dt = −k(θ − θₛ), where θₛ is the surrounding temperature. For this equation, separation gives ∫ 1/(θ − θₛ) dθ = ∫ −k dt, leading to ln|θ − θₛ| = −kt + C, and then θ = A e⁻ᵏᵗ + θₛ. Physics and engineering scenarios often expect you to interpret the meaning of the constant A and the long-term behaviour as t → ∞.

Edexcel喜欢把微分方程嵌入实际情境中。你可能会看到化学反应速率与反应物浓度的乘积成正比，或者根据牛顿定律描述温度变化：dθ/dt = −k(θ − θₛ)，其中 θₛ 是环境温度。对此方程进行分离，得到 ∫ 1/(θ − θₛ) dθ = ∫ −k dt，从而 ln|θ − θₛ| = −kt + C，进而 θ = A e⁻ᵏᵗ + θₛ。物理和工程情境常常要求你解释常数 A 的意义以及 t→∞ 时的长期行为。

Make sure you can confidently link the mathematical solution back to the context. For example, when solving a population model dP/dt = kP, the general solution P = P₀ eᵏᵗ predicts exponential growth. If a question asks ‘how long until the population reaches 1000?’, you need to set P = 1000, substitute the known P₀, and solve for t using logarithms.

务必能够自信地将数学解与情境联系起来。例如，求解人口模型 dP/dt = kP 时，通解 P = P₀ eᵏᵗ 预测指数增长。如果题目问“人口何时达到1000？”，就需要令 P=1000，代入已知的 P₀，并用对数求出 t。

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9. Tackling an Exam-Style Question | 攻克一道考试风格题目

Question: Find the particular solution of the differential equation dy/dx = x√(1 − y²), given that y = 0 when x = 0. Give your answer in the form y = f(x).

题目： 求微分方程 dy/dx = x√(1 − y²) 的特解，已知 x=0 时 y=0。答案写成 y = f(x) 的形式。

Step 1: Separate variables: 1/√(1 − y²) dy = x dx.

步骤1：分离变量：1/√(1 − y²) dy = x dx。

Step 2: Integrate both sides: arcsin(y) = ½x² + C. (Recall ∫ 1/√(1 − y²) dy = arcsin(y) + constant.)

步骤2：两边积分：arcsin(y) = ½x² + C。（记住 ∫ 1/√(1 − y²) dy = arcsin(y) + 常数。）

Step 3: Use initial condition y(0)=0: arcsin(0) = 0 + C → C = 0.

步骤3：利用初始条件 y(0)=0：arcsin(0) = 0 + C → C = 0。

Step 4: Solve for y: y = sin(½x²). Since arcsin gives outputs in [−π/2, π/2], the sine function returns the corresponding y.

步骤4：解出 y：y = sin(½x²)。因为 arcsin 的取值范围是 [−π/2, π/2]，正弦函数自然返回对应的 y。

Final answer: y = sin(x²/2)

This example highlights the need to recall the integral of 1/√(1−y²) as arcsin y, a standard result found on the Edexcel formula sheet. It also elegantly shows how the initial condition can drastically simplify the answer.

这个例子突显了必须牢记 1/√(1−y²) 的积分是 arcsin y，这是 Edexcel 公式表上的标准结果。它也优雅地展示了初始条件如何能大幅简化答案。

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10. Verifying Your Solution | 验证你的解

A quick check that many top-performing students use is to differentiate their final answer and substitute it back into the original differential equation. For the previous example, y = sin(½x²). Then dy/dx = cos(½x²)·x. The original right-hand side is x√(1 − sin²(½x²)) = x|cos(½x²)|. For the given domain around zero, cos(½x²) is positive, so the equation holds true. This verification step can catch algebraic slips and deepen understanding.

很多高分学生会采用一个快速检查方法：对最终答案求导并代回原微分方程。以刚才的例子为例，y = sin(½x²)，dy/dx = cos(½x²)·x。原方程右边是 x√(1 − sin²(½x²)) = x|cos(½x²)|。在零附近的定义域内，cos(½x²) 为正，所以方程成立。这种验证步骤可以发现代数错误并加深理解。

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11. Connecting to Other Integration Techniques | 与其他积分技巧的联系

Separation of variables frequently requires competent integration skills beyond the basic polynomial and trigonometric functions. You may need partial fractions when separating expressions like dy/dx = (y+2)/(x²−1) after rewriting as 1/(y+2) dy = 1/(x²−1) dx. The right-hand side integrates using the decomposition 1/(x²−1) = ½[1/(x−1) − 1/(x+1)], giving log terms. Similarly, substitution or integration by recognition may be needed when encountering forms such as dy/dx = e^(x) sin² y, where you must integrate cosec² y dy. A solid foundation in pure integration is therefore unavoidable.

变量分离法常常需要扎实的积分功底，远不止基本的多项式和三角函数积分。当碰到类似 dy/dx = (y+2)/(x²−1) 这样的方程，改写为 1/(y+2) dy = 1/(x²−1) dx 后，右边需要使用部分分式分解：1/(x²−1) = ½[1/(x−1) − 1/(x+1)]，从而得到对数项。同样，如果遇到 dy/dx = e^(x) sin² y，就必须积分 cosec² y dy，这时可能会用到代换法或观察识别法。因此，扎实的纯积分基础是无论如何绕不开的。

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12. Summary and Revision Checklist | 总结与复习清单

To master separation of variables for your Edexcel A-Level, ensure you can consistently carry out the following actions: recognise the separable form dy/dx = g(x)h(y); rearrange and integrate each side correctly, adding a constant; handle trigonometric, exponential and rational integrands; apply initial conditions before or after algebraic manipulation depending on which is simpler; understand how to treat absolute values and singular solutions; and finally, interpret solutions in context. Practise with past paper questions under timed conditions, and always verify your final answer by differentiation. With disciplined practice, you will turn this topic into a reliable source of marks.

要想在 Edexcel A-Level 中驾驭变量分离法，你必须能够熟练完成以下动作：识别可分离形式 dy/dx = g(x)h(y)；正确移项并积分左右两边，加上常数；处理三角函数、指数和有理被积函数；根据繁简程度，在代数整理之前或之后代入初始条件；理解如何处理绝对值与奇异解；最后，在情境中解释解的意义。在限时条件下练习历年真题，并始终通过求导来验证最终答案。通过有素的训练，你定能把这一专题变成稳稳的得分来源。

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