GCSE Maths: Differential Equations Essentials | GCSE 数学：微分方程考点精讲

📚 GCSE Maths: Differential Equations Essentials | GCSE 数学：微分方程考点精讲

Differential equations appear in GCSE Further Maths and some higher-tier courses, linking calculus to real-world change. This guide covers direct integration, separable equations, exponential models, and common pitfalls—everything you need to build a solid foundation and tackle exam questions with confidence.

微分方程出现在 GCSE 进阶数学和部分高阶课程中，将微积分与现实世界的变化联系起来。本指南涵盖直接积分法、可分离变量方程、指数模型及常见易错点——助你打好扎实基础，从容应对考试题目。

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

A differential equation is any equation that involves a derivative, such as dy/dx. It describes how one quantity changes with respect to another. For example, dy/dx = 3x² tells us the rate of change of y with respect to x.

微分方程是任何包含导数的方程，例如 dy/dx。它描述一个量相对于另一个量如何变化。例如，dy/dx = 3x² 告诉我们 y 随 x 变化的速率。

In GCSE-level problems, you will mostly see first-order differential equations. These can often be solved by direct integration or by separating variables. The general solution contains an arbitrary constant, while a particular solution is found when an initial condition is given.

在 GCSE 级别的题目中，你遇到的大多是一阶微分方程。它们通常可以通过直接积分或分离变量法求解。通解含有一个任意常数，而给出初始条件后可求得特解。

2. Solving by Direct Integration | 直接积分法求解

If the differential equation is of the form dy/dx = f(x), you can simply integrate both sides with respect to x. The result is y = ∫ f(x) dx + C, where C is the constant of integration. This is the general solution.

若微分方程形如 dy/dx = f(x)，你可以直接对 x 积分两边。结果是 y = ∫ f(x) dx + C，其中 C 是积分常数。这便是通解。

For example, given dy/dx = 4x³ − 2x + 1, integrating gives y = x⁴ − x² + x + C. Always remember to add ‘+ C’. Leaving out the constant is one of the most common mistakes.

例如，给定 dy/dx = 4x³ − 2x + 1，积分得 y = x⁴ − x² + x + C。务必记得加上“+ C”。遗漏常数是最常见的错误之一。

English: Step 1: Write dy/dx = f(x). Step 2: Integrate both sides: y = ∫ f(x) dx. Step 3: Add the constant C to get the general solution.
中文：第一步：写出 dy/dx = f(x)。第二步：两边积分：y = ∫ f(x) dx。第三步：加上常数 C 得到通解。

3. Finding Particular Solutions | 求特解

If you are given a boundary condition—such as a point (x₀, y₀) on the curve—you can find the specific value of C. Substitute the known x and y values into the general solution and solve for C. This gives the particular solution.

如果给出边界条件——例如曲线上一点 (x₀, y₀)——你可以求出 C 的具体值。将已知的 x 和 y 值代入通解并解出 C。这样就得到了特解。

For instance, if dy/dx = 6x and the curve passes through (1, 5), integrate to get y = 3x² + C. Substitute x=1, y=5: 5 = 3(1)² + C → C = 2. Hence, the particular solution is y = 3x² + 2.

例如，dy/dx = 6x 且曲线经过 (1, 5)，积分得 y = 3x² + C。代入 x=1, y=5：5 = 3(1)² + C → C = 2。因此特解为 y = 3x² + 2。

Always present the final answer without the constant, because you have used the condition to determine it. This is a common requirement in GCSE Further Maths exam questions.

最终答案中不要保留常数，因为你已利用条件确定了它。这是 GCSE 进阶数学考题中的常见要求。

4. Separable Differential Equations | 可分离变量的微分方程

Some equations are not immediate integration problems. A separable equation looks like dy/dx = g(x)h(y). You can separate the variables by rewriting it so that all y-terms are on one side with dy, and all x-terms on the other with dx.

有些方程不能直接积分。可分离变量的方程形式如 dy/dx = g(x)h(y)。你可以通过重写来分离变量，使所有含 y 的项与 dy 在一边，所有含 x 的项与 dx 在另一边。

After separation, the equation becomes (1/h(y)) dy = g(x) dx. Then you integrate both sides independently. This technique is fundamental and appears in many modelling scenarios.

分离后方程变为 (1/h(y)) dy = g(x) dx。然后分别对两边积分。这一技巧是基础，出现在许多建模场景中。

Example: dy/dx = 2xy. Separate to get (1/y) dy = 2x dx. Integrate: ln|y| = x² + C. Then rewrite in explicit form if needed: y = Ae^(x²), where A = ±e^C.

示例：dy/dx = 2xy。分离得 (1/y) dy = 2x dx。积分：ln|y| = x² + C。若需要可改写为显式形式：y = Ae^(x²)，其中 A = ±e^C。

5. Step-by-Step Separation of Variables | 分离变量法步骤详解

Here is a reliable routine: (1) Write dy/dx = f(x, y). (2) If the right-hand side can be factorised as g(x)h(y), separate to (1/h(y)) dy = g(x) dx. (3) Integrate both sides, adding a single constant on one side. (4) Simplify using algebraic and logarithmic rules. (5) If an initial condition is given, substitute to find the constant.

以下是一个可靠流程：(1) 写出 dy/dx = f(x, y)。(2) 若右边可因式分解为 g(x)h(y)，分离为 (1/h(y)) dy = g(x) dx。(3) 对两边积分，仅在一边添加一个常数。(4) 利用代数和 log 运算法则化简。(5) 如果给出初始条件，代入求常数。

Always check that h(y) ≠ 0 when dividing. Some solutions may be lost or need separate consideration. However, at GCSE level, the equations are designed to avoid such complications.

做除法时务必检查 h(y) ≠ 0。某些解可能会丢失或需单独讨论。但在 GCSE 级别，题目会设计得避开这些复杂情况。

Worked example: dy/dx = (x+1)/(y²). Separate: y² dy = (x+1) dx. Integrate: (1/3)y³ = (1/2)x² + x + C. Multiply by 3: y³ = (3/2)x² + 3x + 3C. You may leave the constant as a single letter, say, K.

范例：dy/dx = (x+1)/(y²)。分离：y² dy = (x+1) dx。积分：(1/3)y³ = (1/2)x² + x + C。乘以 3：y³ = (3/2)x² + 3x + 3C。你可用单个字母 K 代替常数。

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

A particularly important differential equation is dy/dt = ky, where k is a constant. If k > 0, it models exponential growth; if k < 0, it models exponential decay. The general solution is y = Ae^(kt), where A is the initial value when t=0.

一个特别重要的微分方程是 dy/dt = ky，其中 k 为常数。若 k > 0，它模拟指数增长；若 k < 0，模拟指数衰减。通解为 y = Ae^(kt)，其中 A 是 t=0 时的初值。

Derivation: separate variables as (1/y) dy = k dt, then integrate to get ln|y| = kt + C, so y = e^(kt + C) = Ae^(kt). This simple model appears in population growth, radioactive decay, and compound interest.

推导：分离变量得 (1/y) dy = k dt，积分得 ln|y| = kt + C，因此 y = e^(kt + C) = Ae^(kt)。这个简单模型出现在种群增长、放射性衰变和复利中。

Given an initial quantity P₀ and a growth rate k, the amount at time t is P = P₀ e^(kt). The same structure describes depreciation and cooling. Understanding this pattern is key to solving contextual problems.

给定初始量 P₀ 和增长率 k，时间 t 时的量为 P = P₀ e^(kt)。同样的结构可描述贬值与冷却。理解这一模式是解决情境问题的关键。

7. Modelling Population Growth | 人口增长模型

Suppose a population P(t) grows at a rate proportional to its current size: dP/dt = kP. If the initial population is 500 and the population doubles in 4 years, find k. First write P = 500 e^(kt). After 4 years, 1000 = 500 e^(4k) → e^(4k) = 2 → 4k = ln 2 → k = (ln 2)/4.

假设人口 P(t) 的增长速率与其当前规模成正比：dP/dt = kP。若初始人口为 500，且 4 年后翻倍，求 k。首先写出 P = 500 e^(kt)。4 年后，1000 = 500 e^(4k) → e^(4k) = 2 → 4k = ln 2 → k = (ln 2)/4。

Then you can predict the population at any time t. The doubling time is always (ln 2)/k. This is a standard result that often appears in exams, so memorising the formula T_double = ln2 / k is useful.

然后你可以预测任意时间 t 的人口。倍增时间永远是 (ln 2)/k。这是一个常在考试中出现的标准结论，记住公式 T_double = ln2 / k 很有用。

Be careful with units: if time is in years, k is in 1/years. Always state units where applicable. A clear diagram showing the exponential curve can help verify your answer, though not required for the calculation.

注意单位：若时间以年计，则 k 的单位为 1/年。只要适用都要注明单位。虽然计算不要求，画出指数曲线示意图有助于验证答案。

8. Newton’s Law of Cooling | 牛顿冷却定律

Newton’s Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. This gives the differential equation dT/dt = −k(T − T_env), where T_env is the constant surrounding temperature.

牛顿冷却定律指出，物体温度的变化速率与其温度和环境温度之差成正比。这给出微分方程 dT/dt = −k(T − T_env)，其中 T_env 是恒定的环境温度。

Separate variables: (1/(T − T_env)) dT = −k dt. Integrate to get ln|T − T_env| = −k t + C. Exponentiate: T − T_env = Ae^(−kt), so T = T_env + Ae^(−kt). The constant A is the initial temperature difference.

分离变量：(1/(T − T_env)) dT = −k dt。积分得 ln|T − T_env| = −k t + C。取指数：T − T_env = Ae^(−kt)，所以 T = T_env + Ae^(−kt)。常数 A 是初始温度差。

This model often appears in context, for example a hot drink cooling in a room. You might be asked to find k given two temperature readings, or to predict the temperature after a certain time. Set up the equations carefully and solve using logs.

该模型常用于情境题，例如热饮在房间中冷却。题目可能要求根据两个温度读数求 k，或预测某段时间后的温度。仔细建立方程，运用对数求解。

9. Applications in Physics | 物理中的应用

Differential equations link seamlessly with kinematics. Velocity v is the derivative of displacement s, and acceleration a is the derivative of v. Thus, if acceleration is given as a function of time, a = dv/dt = f(t), integrate to get velocity, then integrate again to get displacement.

微分方程与运动学无缝连接。速度 v 是位移 s 的导数，加速度 a 是 v 的导数。因此，若加速度以时间的函数给出，a = dv/dt = f(t)，积分得速度，再积分得位移。

For example, a particle moves such that a = 6t. With initial conditions v(0)=2, s(0)=5. Integrate: v = ∫ 6t dt = 3t² + C. Using v(0)=2 gives C=2, so v = 3t² + 2. Then s = ∫ (3t²+2) dt = t³ + 2t + D. Using s(0)=5 gives D=5, so s = t³ + 2t + 5.

例如，质点运动满足 a = 6t。初值为 v(0)=2，s(0)=5。积分：v = ∫ 6t dt = 3t² + C。由 v(0)=2 得 C=2，故 v = 3t² + 2。然后 s = ∫ (3t²+2) dt = t³ + 2t + D。由 s(0)=5 得 D=5，故 s = t³ + 2t + 5。

This two-step integration mirrors solving a second-order differential equation. Though GCSE doesn’t formally cover second-order, the kinematic reasoning is well within the syllabus and bridges into A-level mechanics.

这种两步积分类似于求解二阶微分方程。虽然 GCSE 不正式涉及二阶方程，但这种运动学推理完全在大纲范围内，并为 A-level 力学架起桥梁。

10. Common Mistakes to Avoid | 常见错误

English: Forgetting the constant of integration ‘C’. Always add it after indefinite integration, then find it using initial conditions.
中文：忘记积分常数“C”。不定积分后务必加上 C，再使用初始条件求出具体值。
English: Misapplying separation of variables: ensure you move all y-terms to the left with dy before integrating, and that the expression is factorisable.
中文：分离变量不当：确保积分前将所有含 y 的项连同 dy 移至左边，并确保表达式可因式分解。
English: Losing the absolute value when integrating 1/y. The result is ln|y|. However, in many GCSE contexts, y is positive, so you can write ln y directly.
中文：积分 1/y 时丢失绝对值。结果是 ln|y|。但在很多 GCSE 情境中 y 为正，可直接写 ln y。
English: Not simplifying constants properly. You can combine constants into a single letter when exponentiating, but do not forget to redefine the constant correctly.
中文：未合理简化常数。取指数时可将常数合并为一个字母，但切勿忘记正确定义新的常数。

11. Exam Tips and Tricks | 考试技巧

When a question asks you to “form and solve” a differential equation, read the wording carefully to identify the rate and any proportionality. Write down the equation in mathematical symbols before attempting to solve. Explicitly state any assumptions.

当题目要求“建立并求解”微分方程时，仔细阅读题干，识别速率和比例关系。在尝试求解前先用数学符号写出方程。明确列出任何假设。

Always show the separation step even if it feels simple. Marks are awarded for method. For exponential growth, writing “ln|y| = kt + C” explicitly earns method marks before the final e-form. Never jump straight to the final solution.

即使看似简单，也要展示分离步骤。解法有分。对于指数增长，写出“ln|y| = kt + C”可在得到最终 e 形式前获得步骤分。绝不要直接跳到最终解。

Check your answer by differentiating it and plugging it back into the original differential equation. This verification takes seconds and can catch algebraic errors. Also make sure your final solution satisfies the initial condition.

通过对答案求导并代回原微分方程进行检验。这种验证只需几秒钟，能发现代数错误。还要确保最终解满足初始条件。

12. Practice Question Walkthrough | 典型例题精讲

Question: A container of water leaks at a rate proportional to the volume V of water remaining. Initially the volume is 2000 cm³, and after 10 minutes it is 1500 cm³. Find V(t) and determine the volume after 20 minutes.

题目：一个容器漏水，速率与剩余水量 V 成正比。初始体积为 2000 cm³，10 分钟后为 1500 cm³。求 V(t)，并确定 20 分钟后的体积。

Step	English Action	中文操作
1	Set up equation: dV/dt = −kV, with k > 0 since volume decreases.	建立方程：dV/dt = −kV，k > 0 因为体积在减少。
2	Separate: (1/V) dV = −k dt.	分离变量：(1/V) dV = −k dt。
3	Integrate: ln V = −kt + C.	积分：ln V = −kt + C。
4	Exponentiate: V = e^(−kt + C) = Ae^(−kt). At t=0, V=2000 ⇒ A=2000.	取指数：V = e^(−kt + C) = Ae^(−kt)。t=0 时 V=2000 ⇒ A=2000。
5	Use t=10, V=1500: 1500 = 2000 e^(−10k) ⇒ e^(−10k) = 0.75 ⇒ −10k = ln 0.75 ⇒ k = −(ln 0.75)/10 ≈ 0.02877.	利用 t=10, V=1500：1500 = 2000 e^(−10k) ⇒ e^(−10k) = 0.75 ⇒ −10k = ln 0.75 ⇒ k ≈ 0.02877。
6	Thus V(t) = 2000 e^(−0.02877t).	因此 V(t) = 2000 e^(−0.02877t)。
7	At t=20: V = 2000 e^(−0.02877×20) ≈ 2000 × 0.5625 ≈ 1125 cm³.	t=20 时：V ≈ 2000 × 0.5625 ≈ 1125 cm³。

This structured approach earns full marks. Remember to give answers to a sensible degree of accuracy and include units.

这种结构化解法可获得满分。记得以合理的精度给出答案并注明单位。

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GCSE Maths: Differential Equations Essentials | GCSE 数学：微分方程 考点精讲