A-Level OCR Maths: Differential Equations – Essential Exam Points | A-Level OCR 数学：微分方程考点精讲

📚 A-Level OCR Maths: Differential Equations – Essential Exam Points | A-Level OCR 数学：微分方程考点精讲

Differential equations are a cornerstone of the OCR A-Level Mathematics course, linking calculus to real-world phenomena such as population growth, cooling, and motion. This article distils every essential skill you need, from recognising a separable equation to applying initial conditions confidently. Whether you are preparing for the Pure paper or simply want a systematic recap, this guide covers all the exam-relevant points with clear examples.

微分方程是 OCR A-Level 数学课程的核心部分，它将微积分与现实世界中的现象（如人口增长、冷却定律和运动学）紧密联系在一起。本文提炼了所有必备技能，从识别可分离变量方程到熟练应用初始条件。无论你是在准备纯数学试卷还是只想系统地复习，本指南都将通过清晰的例子覆盖所有考点。

1. What Is a Differential Equation? | 什么是微分方程？

A differential equation is any equation that links an unknown function with one or more of its derivatives. In OCR, you will almost always meet first-order ordinary differential equations, where the derivative dy/dx is expressed in terms of x and y.

微分方程是联系未知函数与其一个或多个导数的方程。在 OCR 考试中，你几乎总会遇到一阶常微分方程，其中导数 dy/dx 用 x 和 y 表示。

The order of a differential equation is the highest derivative that appears. A first-order equation contains dy/dx but no higher derivatives. A solution to a differential equation is a function y = f(x) that satisfies the equation for all x in a given domain.

微分方程的阶数是指出现的最高阶导数。一阶方程包含 dy/dx 但不含更高阶导数。微分方程的解是使得方程在某个定义域内对所有 x 都成立的函数 y = f(x)。

General solutions contain an arbitrary constant. A particular solution is obtained when we use an extra condition, often called an initial condition or boundary condition, to fix that constant.

通解含有一个任意常数。当我们利用附加条件（常被称为初始条件或边界条件）确定该常数时，就得到了特解。

2. Forming Differential Equations from Context | 从实际背景建立微分方程

OCR exam questions often ask you to construct a differential equation from a word problem. The key is to translate statements about rates of change into derivatives. For example, ‘the rate of increase of a population P is proportional to P’ translates to dP/dt = kP.

OCR 试题经常要求你根据文字问题建立微分方程。关键是将关于变化率的描述转化为导数。例如，”人口 P 的增长速率与 P 成正比”可转化为 dP/dt = kP。

Watch for phrases such as ‘rate of decay’, ‘rate of cooling’, or ‘velocity is inversely proportional to displacement’. They all lead to relationships of the form dQ/dt = -kQ, dθ/dt = -k(θ – θ₀), or dx/dt = k/x. Always define your variables clearly before writing the equation.

注意”衰减速率””冷却速率”或”速度与位移成反比”等表述。它们都会导出形如 dQ/dt = -kQ, dθ/dt = -k(θ – θ₀) 或 dx/dt = k/x 的关系。在写出方程之前，一定要明确地定义变量。

Once the differential equation is formed, you can proceed to solve it. The phrasing of the question often guides you: if it says ‘find an expression for y in terms of x’, you are expected to solve the differential equation and apply any given conditions.

一旦建立起微分方程，你就可以着手求解。题目的措辞通常会给出提示：如果它说”求 y 关于 x 的表达式”，那么你就应该求解微分方程并结合给定的条件。

3. Separable Differential Equations – General Approach | 可分离变量微分方程 – 一般方法

The only class of first-order ODEs you need to solve analytically in OCR A-Level Maths is the separable equation. In its simplest form, the variables can be separated so that all terms involving y appear on one side with dy, and all terms involving x appear on the other side with dx.

在 OCR A-Level 数学中，你需要解析求解的唯一一类一阶常微分方程是可分离变量方程。在最简单的形式下，变量可以被分离，使得所有含 y 的项与 dy 在一侧，所有含 x 的项与 dx 在另一侧。

The general method is:

一般方法如下：

Rewrite dy/dx = g(x) · h(y) as (1/h(y)) dy = g(x) dx, provided h(y) ≠ 0.
Integrate both sides: ∫ (1/h(y)) dy = ∫ g(x) dx.
Add a single constant of integration, usually on the right-hand side.
Rearrange to make y the subject if possible, or leave the solution in implicit form.

将 dy/dx = g(x) · h(y) 改写为 (1/h(y)) dy = g(x) dx，前提是 h(y) ≠ 0。
两边积分：∫ (1/h(y)) dy = ∫ g(x) dx。
添加一个积分常数，通常放在右侧。
如果可能，将 y 写成显式函数，或保留隐式解。

Never forget the constant of integration. Omitting ‘+ c’ is one of the most common errors, and it costs marks even if all other steps are correct.

切勿忘记积分常数。忘记”+ c”是最常见的错误之一，即便其他步骤都正确也会丢分。

4. Solving Separable Equations: Basic Examples | 解可分离变量方程：基础例子

Consider the equation dy/dx = 3x²y. Separate the variables to get (1/y) dy = 3x² dx. Integrating gives ln|y| = x³ + c, so y = Ae^x³, where A = ±e^c or A = 0.

考虑方程 dy/dx = 3x²y。分离变量得到 (1/y) dy = 3x² dx。积分得 ln|y| = x³ + c，因此 y = Ae^x³，其中 A = ±e^c 或 A = 0。

Another common type is dy/dx = (2x)/(y+1). Cross-multiplying gives (y+1) dy = 2x dx. Integrating leads to (1/2)(y+1)² = x² + c. This can be left in implicit form or solved to give y = -1 ± √(2x² + 2c).

另一种常见类型是 dy/dx = (2x)/(y+1)。交叉相乘得 (y+1) dy = 2x dx。积分后得 (1/2)(y+1)² = x² + c。这既可以保留隐式形式，也可以解出 y = -1 ± √(2x² + 2c)。

Always check whether the original equation had any restrictions. For example, if y appeared in the denominator, the solution may exclude certain values.

一定要检查原方程是否有任何限制。例如，如果 y 出现在分母中，解可能会排除某些值。

5. Using Initial/Boundary Conditions | 使用初始/边界条件

An initial condition tells you the value of the dependent variable for a specific value of the independent variable, e.g. y = 2 when x = 0. Substituting these values into the general solution allows you to find the particular value of the constant.

初始条件告诉你当自变量取特定值时因变量的值，例如 x = 0 时 y = 2。将这些值代入通解即可求出常数的特定值。

Work systematically: after integrating, substitute the condition as early as possible, often right after writing the implicit equation containing ‘+ c’. This minimises algebraic manipulation and reduces sign errors.

要有条理地进行：积分后，尽早代入条件，通常是在写出含”+ c”的隐式方程之后立即代入。这样做可以尽量减少代数变形并降低符号错误。

Example: Given dy/dx = 2x(y–1) and y = 3 when x = 0. Separate to get (1/(y–1)) dy = 2x dx. Integrate: ln|y–1| = x² + c. Substitute: ln|3–1| = 0 + c ⇒ c = ln 2. Hence ln|y–1| = x² + ln 2, so y = 1 + 2e^x².

例题：已知 dy/dx = 2x(y–1) 且 x = 0 时 y = 3。分离变量得 (1/(y–1)) dy = 2x dx。积分：ln|y–1| = x² + c。代入：ln|3–1| = 0 + c ⇒ c = ln 2。因此 ln|y–1| = x² + ln 2，得 y = 1 + 2e^x²。

6. Exponential Growth and Decay Models | 指数增长与衰减模型

The equation dN/dt = kN describes exponential growth when k > 0 and exponential decay when k < 0. Separation gives ln|N| = kt + c, so N = N₀e^kt, where N₀ is the value of N at t = 0.

方程 dN/dt = kN 在 k > 0 时描述指数增长，在 k < 0 时描述指数衰减。分离变量得 ln|N| = kt + c，所以 N = N₀e^kt，其中 N₀ 是 t = 0 时 N 的值。

In OCR questions, you are often given two data points. Use one as the initial condition to find N₀, and the other to determine k. If N doubles in a certain time, set up the equation 2N₀ = N₀e^kT and solve for k.

在 OCR 考题中，通常会给你两个数据点。用其中一个作为初始条件求 N₀，再用另一个确定 k。如果 N 在一定时间内翻倍，则可建立方程 2N₀ = N₀e^kT 并求解 k。

Remember that the natural decay model for radioactive substances is dM/dt = –λM, which leads to M = M₀e^–λt. The half-life T_½ satisfies e^{–λT = ½, so λ = ln 2 / T_½.}

记住放射性物质的自然衰减模型为 dM/dt = –λM，从而得到 M = M₀e^–λt。半衰期 T_½ 满足 e^–λT = ½，因此 λ = ln 2 / T_½。

7. The Logistic-Like Equation: dx/dt = k(x – a) | 类逻辑斯谛方程 dx/dt = k(x – a)

OCR commonly tests equations of the form dx/dt = k(x – a), where a is a fixed constant. This typifies a situation where the rate of change of a quantity is proportional to the difference between the quantity and some limiting value, such as in Newton’s law of cooling.

OCR 经常考查形如 dx/dt = k(x – a) 的方程，其中 a 为固定常数。它代表这样一种情形：某量的变化率与该量与某个极限值之差成正比，例如牛顿冷却定律。

Separation gives ∫ 1/(x – a) dx = ∫ k dt ⇒ ln|x – a| = kt + c ⇒ x – a = Ae^kt. The sign of k determines whether x moves away from or towards a. In cooling contexts, k is negative, and x tends to a as t increases.

分离变量得 ∫ 1/(x – a) dx = ∫ k dt ⇒ ln|x – a| = kt + c ⇒ x – a = Ae^kt。k 的符号决定了 x 是远离还是趋近于 a。在冷却情境中，k 为负，随着 t 增大 x 趋近于 a。

Always identify the long-term (equilibrium) value. If the question states that a room warms up to 22°C, then a = 22 and you expect dθ/dt = –k(θ – 22) with k > 0. The solution is θ = 22 + Ae^–kt.

始终要找出长期（平衡）值。如果题目说房间升温至 22°C，那么 a = 22，你预期 dθ/dt = –k(θ – 22) 且 k > 0。解为 θ = 22 + Ae^–kt。

8. Applying to Motion (Velocity and Acceleration) | 应用于运动学（速度与加速度）

When a particle moves along a straight line, its velocity v satisfies v = dx/dt and its acceleration a = dv/dt = d²x/dt². OCR questions may give a differential equation linking v and x, such as dv/dt = –kv or v dv/dx = –g – kv².

当质点沿直线运动时，其速度 v 满足 v = dx/dt，加速度 a = dv/dt = d²x/dt²。OCR 题目可能会给出联系 v 和 x 的微分方程，例如 dv/dt = –kv 或 v dv/dx = –g – kv²。

In motion problems, you often need to separate variables with respect to v and x, using the identity a = v dv/dx. This is especially helpful if you are given acceleration as a function of x.

在运动问题中，你经常需要对 v 和 x 进行变量分离，使用恒等式 a = v dv/dx。如果加速度是以 x 的函数形式给出的，这将特别有用。

Typical example: a particle moves so that a = –4x. Write v dv/dx = –4x, separate: ∫ v dv = ∫ –4x dx, giving v²/2 = –2x² + c. Apply initial speed and position to find c.

典型例子：质点运动满足 a = –4x。写出 v dv/dx = –4x，分离变量：∫ v dv = ∫ –4x dx，得 v²/2 = –2x² + c。代入初速度和位置即可求出 c。

9. Common Mistakes and Tips | 常见错误与技巧

Losing the constant of integration is disastrous. Always write ‘+ c’ immediately after evaluating an indefinite integral. If you integrate both sides, a single constant on one side is sufficient.

遗漏积分常数是灾难性的。在计算不定积分后要立即写出”+ c”。如果两边都积分，仅在一侧添加一个常数就足够了。

Incorrect separation is another pitfall. Check that you have truly isolated all y-terms on one side and all x-terms on the other. Multiplying by dx or dy is a convenience, but you must ensure the equation is equivalent to the original.

错误的分离是另一个陷阱。务必确认已经将所有含 y 的项放在一边、所有含 x 的项放在另一边。乘以 dx 或 dy 只是简便做法，但必须确保方程与原方程等价。

Modulus signs matter. When integrating 1/y, use ln|y|. If the initial condition gives a positive value for y, you can drop the modulus, but justify it by stating y > 0.

绝对值符号很重要。积分 1/y 时要使用 ln|y|。如果初始条件给出 y 的正值，则可以去掉绝对值，但需要阐明 y > 0。

Finally, don’t forget to answer the specific question. If it asks for the time when a population reaches 1000, solve for t after finding the particular solution. Always express your final answer in the form requested, with appropriate units where necessary.

最后，不要忘记回答所提的具体问题。如果题目问人口达到 1000 的时间，在求出特解后要解出 t。最终答案应按照要求的格式给出，并在必要时附上适当单位。

10. Practice Question Walkthrough | 典型例题解析

Question: The rate of increase of a population P, in thousands, is proportional to (20 – P). When t = 0, P = 5, and when t = 2, P = 10. Find P in terms of t, and calculate the population after 5 hours.

题目：种群数量 P（以千计）的增长速率与 (20 – P) 成正比。t = 0 时 P = 5，t = 2 时 P = 10。求 P 关于 t 的表达式，并计算 5 小时后的种群数量。

Interpretation: dP/dt = k(20 – P). This is separable: ∫ 1/(20 – P) dP = ∫ k dt ⇒ –ln|20 – P| = kt + c. Rearranging: ln|20 – P| = –kt – c ⇒ 20 – P = Ae^–kt.

题意解读：dP/dt = k(20 – P)。该方程可分离：∫ 1/(20 – P) dP = ∫ k dt ⇒ –ln|20 – P| = kt + c。整理得 ln|20 – P| = –kt – c ⇒ 20 – P = Ae^–kt。

At t = 0, P = 5: 20 – 5 = A ⇒ A = 15, so 20 – P = 15e^–kt. At t = 2, P = 10: 20 – 10 = 15e^–2k ⇒ 10/15 = e^–2k ⇒ 2/3 = e^–2k. Take ln: –2k = ln(2/3) ⇒ k = –(1/2) ln(2/3) = (1/2) ln(3/2).

t = 0 时 P = 5：20 – 5 = A ⇒ A = 15，因此 20 – P = 15e^–kt。t = 2 时 P = 10：20 – 10 = 15e^–2k ⇒ 10/15 = e^–2k ⇒ 2/3 = e^–2k。取对数：–2k = ln(2/3) ⇒ k = –(1/2) ln(2/3) = (1/2) ln(3/2)。

Hence P = 20 – 15e^{–(t/2) ln(3/2)} = 20 – 15 (3/2)^–t/2. After 5 hours: P = 20 – 15(3/2)^–2.5. Evaluate numerically: (3/2)^–2.5 = (2/3)^2.5 ≈ (0.6667)^2.5 ≈ 0.363. So P ≈ 20 – 15×0.363 = 20 – 5.445 = 14.555 thousand, i.e. about 14,560.

因此 P = 20 – 15e^{–(t/2) ln(3/2)} = 20 – 15 (3/2)^–t/2。5 小时后：P = 20 – 15(3/2)^–2.5。数值计算：(3/2)^–2.5 = (2/3)^2.5 ≈ (0.6667)^2.5 ≈ 0.363。所以 P ≈ 20 – 15×0.363 = 20 – 5.445 = 14.555 千，即约 14,560。

This walkthrough illustrates the classic pattern: form the equation, separate variables, use two conditions to find constants, and finally interpret the result in context.

本例题展示了经典解题模式：建立方程、分离变量、利用两个条件求出常数，最后结合情境解释结果。

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A-Level OCR Maths: Differential Equations – Essential Exam Points | A-Level OCR 数学：微分方程 考点精讲