📚 GCSE WJEC Mathematics: Differential Equations – A Complete Guide | GCSE WJEC 数学：微分方程 考点精讲
Differential equations might sound like an advanced topic, but at the GCSE level, particularly under the WJEC specification, they introduce the fundamental concepts of forming and solving simple first-order equations. These appear in contexts such as growth, decay, and motion, linking calculus with real-world problems.
微分方程听起来像是一个高深的课题，但在 GCSE 阶段，尤其是 WJEC 考试大纲下，它们引入了建立和求解简单一阶微分方程的基本概念。这些内容出现在增长、衰减和运动等情境中，将微积分与现实世界的问题联系起来。
1. Understanding the Basics of Differential Equations | 微分方程基础概念解析
A differential equation is an equation that contains a derivative. For WJEC GCSE, you will typically see dy/dx as part of an equation relating x and y. The goal is to find the original function y in terms of x.
微分方程是包含导数的方程。在 WJEC GCSE 中，你通常会看到 dy/dx 作为关联 x 和 y 的方程的一部分。目标是求出用 x 表示 y 的原始函数。
The simplest form you will encounter is dy/dx = f(x). To solve it, you integrate both sides with respect to x. This reverses differentiation.
你会遇到的最简单形式是 dy/dx = f(x)。要求解它，你对两边关于 x 进行积分。这是微分的逆运算。
For example, if dy/dx = 3x², then y = ∫ 3x² dx = x³ + C, where C is the constant of integration. Always remember to add ‘+ C’ unless initial conditions are given.
例如，如果 dy/dx = 3x²，则 y = ∫ 3x² dx = x³ + C，其中 C 是积分常数。除非给定了初始条件，否则一定要记得加上“+ C”。
2. Separating Variables to Solve Equations | 分离变量法求解方程
WJEC often includes differential equations where the variables can be separated. This means rearranging the equation so that all terms involving y are on one side with dy, and all terms involving x are on the other side with dx.
WJEC 经常包含可以分离变量的微分方程。这意味着重新排列方程，使所有包含 y 的项与 dy 在一侧，所有包含 x 的项与 dx 在另一侧。
Consider the equation dy/dx = 2x/y. Multiply both sides by y and dx to get y dy = 2x dx. Now integrate both sides: ∫ y dy = ∫ 2x dx.
考虑方程 dy/dx = 2x/y。将两边乘以 y 和 dx，得到 y dy = 2x dx。现在对两边积分：∫ y dy = ∫ 2x dx。
This yields ½ y² = x² + C. You can then rearrange to find y explicitly: y² = 2x² + 2C, so y = ±√(2x² + K) where K = 2C.
这得到 ½ y² = x² + C。然后你可以重新排列以显式求出 y：y² = 2x² + 2C，因此 y = ±√(2x² + K)，其中 K = 2C。
3. Applying Initial Conditions to Find Particular Solutions | 应用初始条件求特解
An initial condition is a known value of the function at a specific point, often given as y(a) = b. This allows you to determine the specific constant C and eliminate the ± ambiguity where possible.
初始条件是函数在某特定点的已知值，通常给出为 y(a) = b。这让你能够确定具体的常数 C，并在可能的情况下消除 ± 的歧义。
For instance, if dy/dx = 4x³ and you know y(1) = 3, integrate first: y = x⁴ + C. Substitute x = 1, y = 3: 3 = 1⁴ + C → C = 2.
例如，如果 dy/dx = 4x³ 且你知道 y(1) = 3，先积分：y = x⁴ + C。代入 x = 1, y = 3：3 = 1⁴ + C → C = 2。
Thus the particular solution is y = x⁴ + 2. Always write your final answer clearly, showing the substitution step for full marks.
因此特解为 y = x⁴ + 2。务必要清晰地写出最终答案，并展示代入步骤以获得满分。
4. Modelling Population Growth with Differential Equations | 用微分方程建模人口增长
A classic WJEC context is exponential growth, where the rate of change of a population P with respect to time t is proportional to P itself: dP/dt = kP.
一个经典的 WJEC 情境是指数增长，即人口 P 关于时间 t 的变化率与 P 本身成正比：dP/dt = kP。
To solve, separate variables: dP/P = k dt. Integrate: ln|P| = kt + C. Exponentiate: P = e^(kt + C) = Ae^(kt), where A = e^C.
求解时，分离变量：dP/P = k dt。积分：ln|P| = kt + C。取指数：P = e^(kt + C) = Ae^(kt)，其中 A = e^C。
If initially P₀ is the population at t = 0, then A = P₀, giving the well-known formula P = P₀e^(kt). This crops up in bacteria growth and compound interest problems.
如果初始时 P₀ 是 t = 0 时的人口，则 A = P₀，得到著名的公式 P = P₀e^(kt)。这出现在细菌生长和复利问题中。
5. Modelling Radioactive Decay and Cooling | 建模放射性衰变与冷却
Radioactive decay follows dN/dt = -λN, where N is the number of undecayed nuclei and λ is a positive decay constant. The negative sign shows decay.
放射性衰变遵循 dN/dt = -λN，其中 N 是未衰变原子核的数量，λ 是一个正的衰变常数。负号表示衰变。
Solving gives N = N₀e^(-λt). Similarly, Newton’s law of cooling uses dT/dt = -k(T – Tₐ), where T is temperature and Tₐ is ambient temperature.
求解得到 N = N₀e^(-λt)。类似地，牛顿冷却定律使用 dT/dt = -k(T – Tₐ)，其中 T 是温度，Tₐ 是环境温度。
Separate: dT/(T – Tₐ) = -k dt. Integrate: ln|T – Tₐ| = -kt + C → T – Tₐ = Ae^(-kt). Apply initial temperature to find A.
分离变量：dT/(T – Tₐ) = -k dt。积分：ln|T – Tₐ| = -kt + C → T – Tₐ = Ae^(-kt)。应用初始温度求出 A。
6. Interpreting the Rate of Change in Context | 在上下文中解读变化率
WJEC exam questions often ask you to interpret dy/dx or dP/dt in words. You must explain what the derivative represents at a given moment.
WJEC 考试题目经常要求你用文字解释 dy/dx 或 dP/dt。你必须解释在给定时刻导数代表什么。
For example, if V is volume of water in a tank and t is time, dV/dt = 5 means the volume is increasing at a rate of 5 units per unit time.
例如，如果 V 是水箱中水的体积，t 是时间，dV/dt = 5 表示体积以每单位时间 5 个单位的速度增加。
When dV/dt is negative, the tank is emptying. You may need to find when dV/dt = 0 (a turning point) to determine maximum or minimum volume.
当 dV/dt 为负时，水箱正在排水。你可能需要求出何时 dV/dt = 0（一个转折点）以确定最大或最小体积。
7. Forming Differential Equations from Descriptions | 根据描述建立微分方程
This skill is crucial: translating a written statement into a differential equation. Keywords like “rate of change”, “proportional to”, or “inversely proportional to” guide you.
这项技能至关重要：将书面陈述转化为微分方程。诸如“变化率”、“与……成正比”或“与……成反比”等关键词会指引你。
If “the rate of decrease of y is proportional to the square root of y”, write dy/dt = -k √y. The negative sign denotes decrease.
如果“y 的衰减速率与 y 的平方根成正比”，则写为 dy/dt = -k √y。负号表示衰减。
If “the gradient of a curve is inversely proportional to x” and the curve passes through (2, 5), set up dy/dx = k/x and substitute to find k.
如果“曲线的斜率与 x 成反比”且曲线经过 (2, 5)，建立 dy/dx = k/x 并代入求出 k。
8. Solving Differential Equations with Trigonometric Functions | 求解含三角函数的微分方程
WJEC also tests equations like dy/dx = sin x or dy/dx = cos² y. You must be confident integrating standard trigonometric forms.
WJEC 也测试像 dy/dx = sin x 或 dy/dx = cos² y 这样的方程。你必须对标准三角形式的积分有自信。
For dy/dx = sin x, integrate: y = -cos x + C. For dy/dx = sec² y, rearrange as cos² y dy = dx, then integrate: ∫ cos² y dy = x + C.
对于 dy/dx = sin x，积分：y = -cos x + C。对于 dy/dx = sec² y，重写为 cos² y dy = dx，然后积分：∫ cos² y dy = x + C。
Use the identity cos² y = (1 + cos 2y)/2 if needed. Then (1/2)∫ (1 + cos 2y) dy = x + C → (1/2)(y + ½ sin 2y) = x + C.
如果需要，使用恒等式 cos² y = (1 + cos 2y)/2。则 (1/2)∫ (1 + cos 2y) dy = x + C → (1/2)(y + ½ sin 2y) = x + C。
9. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
Leaving out the ‘+ C’ is the most frequent error. Every indefinite integration must include the constant. Only omit it if you are finding a definite integral or immediately substituting initial conditions.
遗漏“+ C”是最常见的错误。每一次不定积分都必须包含常数。仅当你求定积分或立即代入初始条件时才能省略。
Forgetting to separate variables correctly can trap you. Check that dy and dx are in the correct positions before integrating.
未能正确分离变量可能会困住你。在积分前检查 dy 和 dx 是否在正确的位置上。
Errors with algebraic manipulation, especially when exponentiating to remove logs. Remember: e^(ln A + kt) = e^(ln A) · e^(kt) = A·e^(kt).
代数操作错误，尤其是在取指数以去除对数时。记住：e^(ln A + kt) = e^(ln A) · e^(kt) = A·e^(kt)。
| Mistake (错误) | Correction (纠正) |
|---|---|
| Missing + C | Write + C immediately after integrating. |
| Incorrect separation | Multiply by dx and divide by the y-term. |
| Sign errors in decay | Use negative k for decay; check problem wording. |
| Not substituting initial conditions | Plug in x and y values to find C explicitly. |
10. Exam Strategy and Working Mark Scheme | 考试策略与分步评分方案
WJEC marks are awarded for method. Even if you make a numerical slip, showing separation, integration, and substitution steps can secure method marks.
WJEC 的分数是按方法给分的。即使你犯了数值错误，展示分离、积分和代入步骤也能确保拿到方法分。
Typical mark breakdown: 1 mark for correct separation, 2 marks for accurate integration (including correct trigonometric integration if present), 1 mark for constant C or finding A, 1 mark for final substitution.
典型的分值分配：正确分离得 1 分，准确积分得 2 分（如果涉及，包括正确的三角积分），常数 C 或求出 A 得 1 分，最终代入得 1 分。
Always check the units and whether the question asks for an explicit form (y = …) or an implicit form (equation relating x and y). Implicit answers can be accepted.
始终检查单位，以及问题是否要求显式形式（y = …）或隐式形式（关联 x 和 y 的方程）。隐式答案也可能被接受。
11. Practice Examples with Step-by-Step Solutions | 分步解题的练习示例
Example 1: Solve dy/dx = 6x² – 1 given y(1) = 4. Integrate: y = 2x³ – x + C. Substitute: 4 = 2(1)³ – 1 + C → C = 3. Solution: y = 2x³ – x + 3.
示例 1：求解 dy/dx = 6x² – 1，已知 y(1) = 4。积分：y = 2x³ – x + C。代入：4 = 2(1)³ – 1 + C → C = 3。解为：y = 2x³ – x + 3。
Example 2: The rate of change of temperature T of a coffee cup is dT/dt = -k(T – 20). If T(0) = 80 and T(5) = 50, find the particular solution.
示例 2：一杯咖啡的温度 T 的变化率为 dT/dt = -k(T – 20)。如果 T(0) = 80 且 T(5) = 50，求特解。
Separate: dT/(T – 20) = -k dt. Integrate: ln|T – 20| = -kt + C. T(0)=80: ln(60) = C. Then ln|T – 20| = -kt + ln 60.
分离变量：dT/(T – 20) = -k dt。积分：ln|T – 20| = -kt + C。T(0)=80：ln(60) = C。则 ln|T – 20| = -kt + ln 60。
Using T(5)=50: ln(30) = -5k + ln(60) → -5k = ln(30) – ln(60) = ln(0.5) → k = -0.2 ln(0.5) = 0.1386 (approx). Final solution: T = 20 + 60e^(-0.1386t).
使用 T(5)=50：ln(30) = -5k + ln(60) → -5k = ln(30) – ln(60) = ln(0.5) → k = -0.2 ln(0.5) = 0.1386（约）。最终解为：T = 20 + 60e^(-0.1386t)。
12. Linking Differential Equations to Other GCSE Topics | 微分方程与其他 GCSE 主题的联系
Differential equations are not isolated; they heavily rely on differentiation and integration skills from the calculus section. You must be able to differentiate polynomials, exponentials, and basic trig functions.
微分方程并非孤立存在；它们严重依赖于微积分部分的微分和积分技能。你必须能够对多项式、指数函数和基本三角函数进行微分。
They also connect to kinematics: velocity v = ds/dt, acceleration a = dv/dt. If a = g (constant), integrating gives v = gt + u and s = ½ gt² + ut.
它们还与运动学相关：速度 v = ds/dt，加速度 a = dv/dt。如果 a = g（常数），积分得到 v = gt + u 和 s = ½ gt² + ut。
In WJEC, a question might combine differential equations with curve sketching or area under a graph, so be prepared for cross-topic integration.
在 WJEC 中，一道题可能将微分方程与曲线草图或图下面积结合起来，因此要做好跨主题的综合准备。
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