IGCSE WJEC Maths: Differential Equations Exam Focus | IGCSE WJEC 数学：微分方程考点精讲

📚 IGCSE WJEC Maths: Differential Equations Exam Focus | IGCSE WJEC 数学：微分方程考点精讲

Welcome to this focused revision guide on differential equations for the IGCSE WJEC Mathematics syllabus. Differential equations connect calculus to real-world rates of change, and mastering them is essential for success in both the core and extended papers. This article breaks down key concepts, common question types, and effective strategies step by step.

欢迎阅读这篇针对 IGCSE WJEC 数学大纲的微分方程复习指南。微分方程将微积分与现实世界的变化率联系起来，掌握它们对于在核心和扩展考试中取得成功至关重要。本文将逐步解析关键概念、常见题型和有效策略。

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

A differential equation is an equation that involves an unknown function and one or more of its derivatives. In IGCSE Mathematics, we primarily work with first-order differential equations, where the derivative dy/dx (or similar notation) is expressed in terms of the variables x and sometimes y. The goal is to find the original function y = f(x).

微分方程是包含未知函数及其导数的方程。在 IGCSE 数学中，我们主要处理一阶微分方程，其中导数 dy/dx（或类似记号）被表示为变量 x 甚至 y 的表达式。目标是求出原函数 y = f(x)。

For example, dy/dx = 3x² is a simple differential equation. The solution is not a single number but a family of functions, y = x³ + C, where C is a constant. This is a general solution because it includes all possible antiderivatives.

例如，dy/dx = 3x² 是一个简单的微分方程。它的解不是一个数值，而是一族函数 y = x³ + C，其中 C 为常数。这就是通解，因为它包含了所有可能的原函数。

2. Notation and Essential Terminology | 符号与基本术语

The derivative can be written in several ways: dy/dx, f'(x), or y’. All mean the rate of change of y with respect to x. A differential equation states a relationship between this derivative and the variables. The order of a differential equation refers to the highest derivative present; in IGCSE you will only encounter first-order ones.

导数可以写成多种形式：dy/dx、f'(x) 或 y’，都表示 y 关于 x 的变化率。而微分方程描述了这个导数与变量之间的关系。方程的阶是指最高阶导数的阶数；IGCSE 中只会出现一阶微分方程。

The process of solving a differential equation means finding all functions y = f(x) that satisfy it. The general solution contains an arbitrary constant C. If an additional condition is given, such as a point on the curve, we can determine C and obtain a particular solution.

求解微分方程意味着找到所有满足该方程的函数 y = f(x)。通解含有一个任意常数 C。若给出附加条件，如曲线上的某个点，我们就能确定 C 并得到特解。

3. Forming Differential Equations from Context | 从情境中建立微分方程

Many exam questions ask you to translate a written statement about a rate of change into a differential equation. Keywords like ‘rate’, ‘speed’, ‘growth’, ‘decay’, and ‘proportional’ are signs that you need to write something like dQ/dt = …

很多考题要求你将一段关于变化率的文字叙述转化为微分方程。遇到“速率”、“速度”、“增长”、“衰减”、“成正比”等关键词时，就需要写出形如 dQ/dt = … 的式子。

English: “The rate of change of the population P is proportional to P.” → dP/dt = kP
中文: “人口 P 的变化率与 P 成正比。” → dP/dt = kP
English: “The temperature T decreases at a rate proportional to the difference between T and room temperature 20°C.” → dT/dt = -k(T – 20)
中文: “温度 T 下降的速率与 T 和室温 20°C 之差成正比。” → dT/dt = -k(T – 20)
English: “The acceleration (rate of change of velocity) is constant, -9.8 m/s².” → dv/dt = -9.8
中文: “加速度（速度的变化率）为常数 -9.8 m/s²。” → dv/dt = -9.8

Always define your variables clearly: P for population, t for time, etc. This step is crucial for accurate solving later.

务必清晰地定义变量：P 代表人口，t 代表时间等。这一步对后续准确求解至关重要。

4. Solving dy/dx = f(x) by Direct Integration | 直接积分法求解 dy/dx = f(x)

If the differential equation is of the form dy/dx = f(x), where the right-hand side depends only on x, we can simply integrate both sides with respect to x. The solution is y = ∫ f(x) dx + C. Never forget the constant of integration!

如果微分方程形如 dy/dx = f(x)，右侧只依赖于 x，我们可以直接对 x 积分。解为 y = ∫ f(x) dx + C。一定不要忘记积分常数！

Example: Solve dy/dx = 4x³ – 6x + 2

Integrate: y = ∫ (4x³ – 6x + 2) dx = x⁴ – 3x² + 2x + C. That is the general solution.

积分得：y = ∫ (4x³ – 6x + 2) dx = x⁴ – 3x² + 2x + C。这就是通解。

If the equation uses other variables, the principle is the same. For ds/dt = 5t², we integrate to get s = (5/3)t³ + C. Always match the differential variable on both sides.

如果方程使用其他变量，原理同样。对于 ds/dt = 5t²，积分得 s = (5/3)t³ + C。务必保持两边微分变量一致。

5. Finding a Particular Solution Using Initial Conditions | 利用初始条件求特解

A particular solution is found when we are given extra information, often called initial conditions or boundary conditions, such as the value of y at a specific x. Substitute the given x and y into the general solution to solve for C.

当给出额外信息时，我们可以求出特解，这些信息通常称为初始条件或边界条件，如特定 x 下的 y 值。将给定的 x 和 y 代入通解即可解出 C。

Example: dy/dx = 6x² – 2, and y = 5 when x = 1. Find y in terms of x.

Step 1: Integrate: y = 2x³ – 2x + C. Step 2: Substitute x = 1, y = 5: 5 = 2(1)³ – 2(1) + C → 5 = 0 + C → C = 5. So the particular solution is y = 2x³ – 2x + 5.

第一步：积分得 y = 2x³ – 2x + C。第二步：代入 x = 1, y = 5：5 = 2(1)³ – 2(1) + C → 5 = 0 + C → C = 5。所以特解为 y = 2x³ – 2x + 5。

IGCSE questions often frame this in a geometrical way: “A curve passes through (2, 7) and its gradient is given by … Find the equation of the curve.” The gradient function is dy/dx, and you follow the same steps.

IGCSE 考题常以几何方式呈现：“一条曲线经过点 (2, 7) 且其斜率由……给出，求曲线方程。”斜率函数就是 dy/dx，求解步骤完全相同。

6. Solving Separable Equations: dy/dx = ky | 求解可分离方程：dy/dx = ky

Some IGCSE extended problems involve exponential growth or decay, modelled by dy/dx = ky where k is a constant. To solve, we separate the variables: bring all y terms to one side and all x terms to the other.

某些 IGCSE 扩展问题涉及指数增长或衰减，模型为 dy/dx = ky，其中 k 为常数。求解时，我们需分离变量：将所有含 y 的项移到一边，含 x 的项移到另一边。

Step-by-step: dy/dx = ky → (1/y) dy = k dx

Integrate both sides: ∫ (1/y) dy = ∫ k dx → ln|y| = kx + C. Then rewrite in exponential form: |y| = e^(kx + C) = e^C e^(kx). Let A = ±e^C, giving the general solution y = A e^(kx).

两边积分：∫ (1/y) dy = ∫ k dx → ln|y| = kx + C。再写成指数形式：|y| = e^(kx + C) = e^C e^(kx)。令 A = ±e^C，得到通解 y = A e^(kx)。

If an initial condition y(0) = y₀ is given, then A = y₀, so the particular solution is y = y₀ e^(kx). This formula describes exponential growth if k > 0 or decay if k < 0.

若给出初始条件 y(0) = y₀，则 A = y₀，特解为 y = y₀ e^(kx)。当 k > 0 时该公式描述指数增长，当 k < 0 时描述指数衰减。

In WJEC IGCSE questions, you may be asked to form the equation from a description like “the rate of increase of bacteria is proportional to the number present” and then solve it. Practice converting words into dy/dx = ky and then applying separation of variables.

在 WJEC IGCSE 考题中，你可能会被要求从描述“细菌的增加速率与现有数量成正比”建立方程并求解。请练习将文字转化为 dy/dx = ky，再使用分离变量法。

7. Motion and Kinematics Applications | 运动学应用

In kinematics, velocity v is the rate of change of displacement s: v = ds/dt. Acceleration a is the rate of change of velocity: a = dv/dt. Therefore, if acceleration is given as a function of time, we integrate to get velocity, then integrate again to get displacement.

在运动学中，速度 v 是位移 s 的变化率：v = ds/dt。加速度 a 是速度的变化率：a = dv/dt。因此，若加速度为时间的函数，我们先积分得速度，再积分得位移。

Example: a = dv/dt = 2t – 4, with initial velocity v(0) = 3. Find v(t).

Integrate: v = ∫ (2t – 4) dt = t² – 4t + C. Use v(0) = 3: 3 = 0 – 0 + C → C = 3. Therefore v = t² – 4t + 3.

积分：v = ∫ (2t – 4) dt = t² – 4t + C。利用 v(0) = 3：3 = 0 – 0 + C → C = 3。所以 v = t² – 4t + 3。

If you are also asked for displacement s(t) and given s(0) = 0, integrate v: s = ∫ (t² – 4t + 3) dt = (1/3)t³ – 2t² + 3t + D, then use s(0) = 0 to find D = 0.

若还需要求位移 s(t) 并已知 s(0) = 0，则对 v 积分：s = ∫ (t² – 4t + 3) dt = (1/3)t³ – 2t² + 3t + D，再利用 s(0) = 0 得到 D = 0。

Kinematic quantity	运动学量
Displacement s	位移 s
Velocity v = ds/dt	速度 v = ds/dt
Acceleration a = dv/dt	加速度 a = dv/dt

Always check the units and ensure you have included the constants of integration when moving from acceleration to velocity to displacement.

务必检查单位，并确保在从加速度到速度再到位移的每一步都加上了积分常数。

8. Rate of Change in Other Contexts | 其他情境中的变化率

Differential equations are widely used to model rates involving volume, area, temperature, and more. A classic IGCSE problem gives dV/dt, the rate of change of volume, and asks for V as a function of t, or for the amount at a specific time.

微分方程被广泛用于建立体积、面积、温度等变化率的模型。一道典型的 IGCSE 题会给出体积变化率 dV/dt，要求你求 V 关于 t 的函数，或特定时刻的数量。

For instance: “Water leaks from a tank at a rate proportional to the square root of the volume remaining, i.e. dV/dt = -k√V.” This is a separable equation: (1/√V) dV = -k dt. Integrate: 2√V = -kt + C. Rearrange to express V.

例如：“水从水箱中漏出的速率与剩余体积的平方根成正比，即 dV/dt = -k√V。”这是一个可分离方程：(1/√V) dV = -k dt，积分得 2√V = -kt + C，再整理出 V。

When given an initial volume V₀ at t=0, C = 2√V₀. The particular solution is 2√V = 2√V₀ – kt. Then you can answer questions like “find the time when the tank is half empty.”

若给出 t=0 时的初始体积 V₀，则 C = 2√V₀，特解为 2√V = 2√V₀ – kt。之后可以回答“求水箱半空所需时间”等问题。

9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Forgetting ‘+ C’: The most frequent error. After integrating, always write ‘+ C’ unless you are finding a definite integral for a particular reason. The constant accounts for all possible starting values.

忘记“+ C”：这是最常见的错误。积分后一定要写“+ C”，除非你有特殊理由在求定积分。常数 C 代表了所有可能的初始值。

Misinterpreting the derivative: Some students treat dy/dx as a fraction incorrectly when separating variables. Remember it is valid to multiply by dx only when the equation is separable, i.e. dy/dx = g(x)h(y). Do not randomly split it in other contexts.

误解导数：一些学生在分离变量时错误地处理 dy/dx。记住，只有当方程可分离即 dy/dx = g(x)h(y) 时，才可以将 dx 乘过去。在其他情境下不要随意拆分。

Losing the constant in exponential equations: When solving dy/dx = ky, after obtaining ln|y| = kx + C, do not forget to write y = A e^(kx) with A = ±e^C. A common mistake is writing y = e^(kx) + C, which is wrong.

在指数方程中丢失常数：求解 dy/dx = ky 时，得到 ln|y| = kx + C 后，别忘了写成 y = A e^(kx) 且 A = ±e^C。常见错误是写成 y = e^(kx) + C，这是错的。

Not substituting initial conditions correctly: Always substitute the given x and y into the integrated equation before simplifying. Check your final particular solution by differentiating to see if it yields the original dy/dx.

未正确代入初始条件：在化简通解之前，务必将给定的 x 和 y 代入积分后的方程。求出特解后，应对其求导，看是否能回到原来的 dy/dx，以此自查。

10. Exam-Style Questions with Worked Solutions | 考试风格题目与解答

Question 1: The gradient of a curve is given by dy/dx = 3x² – 4x + 1. The curve passes through the point (2, 5). Find the equation of the curve.

问题 1：一条曲线的斜率由 dy/dx = 3x² – 4x + 1 给出，该曲线经过点 (2, 5)。求曲线的方程。

Solution: Integrate: y = x³ – 2x² + x + C. Substitute x = 2, y = 5: 5 = 8 – 8 + 2 + C → C = 3. Equation: y = x³ – 2x² + x + 3.

解：积分得 y = x³ – 2x² + x + C。代入 x = 2, y = 5：5 = 8 – 8 + 2 + C → C = 3。曲线方程为 y = x³ – 2x² + x + 3。

Question 2: The rate of growth of a population P (in thousands) is proportional to P. Initially, P = 10, and after 2 years P = 15. Set up and solve the differential equation to find P in terms of t.

问题 2：某种群数量 P（单位为千）的增长率与 P 成正比。初始时 P

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IGCSE WJEC Maths: Differential Equations Exam Focus | IGCSE WJEC 数学：微分方程 考点精讲