IGCSE CCEA Mathematics: Differential Equations Exam Guide | 微分方程考点精讲

📚 IGCSE CCEA Mathematics: Differential Equations Exam Guide | 微分方程考点精讲

Differential equations are a key topic in the IGCSE CCEA Mathematics Higher Tier syllabus. They allow us to model relationships involving rates of change and to find functions from information about their derivatives. In this guide, we break down the essential concepts, methods, and exam techniques you need to master this topic.

微分方程是 IGCSE CCEA 数学高等卷中的重点内容。它们用于建立涉及变化率的数学模型，并从导数信息反推原函数。本指南将拆解微分方程的核心概念、解题方法与应试技巧，帮助你全面掌握这一考点。

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

A differential equation is an equation that contains an unknown function and one or more of its derivatives. In the IGCSE CCEA course, the unknown function is usually y in terms of x, and the derivative is written as dy/dx. The equation describes how the rate of change of y relates to x, y, or both.

微分方程是包含未知函数及其一个或多个导数的方程。在 IGCSE CCEA 课程中，未知函数通常是关于 x 的 y，导数写作 dy/dx。该方程描述了 y 的变化率与 x、y 或两者之间的关系。

For example, dy/dx = 3x² is a simple differential equation. Solving it means finding the original function y = f(x) that satisfies this derivative relationship.

例如，dy/dx = 3x² 就是一个简单的微分方程。解这个方程意味着找到满足该导数关系的原函数 y = f(x)。

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

When the derivative is given purely as a function of x, solving the differential equation is just a matter of integrating both sides with respect to x. If dy/dx = f(x), then y = ∫ f(x) dx + C, where C is the constant of integration. This gives the general solution, which represents a family of curves.

当导数仅表示为关于 x 的函数时，解微分方程只需对两边关于 x 积分。若 dy/dx = f(x)，则 y = ∫ f(x) dx + C，其中 C 为积分常数。这得到的是通解，表示一簇曲线。

Example:

Solve dy/dx = 4x³ − 2x + 5.

Integrating: y = ∫ (4x³ − 2x + 5) dx = x⁴ − x² + 5x + C.

示例：

求解 dy/dx = 4x³ − 2x + 5。

积分得：y = ∫ (4x³ − 2x + 5) dx = x⁴ − x² + 5x + C。

Key point: Never forget the ‘+ C’. Without it, the solution is incomplete in an exam.

关键点：千万不能忘记 ‘+ C’。考试中若缺少它，解是不完整的。

3. Solving dy/dx = f(y) by Inversion and Integration | 倒数积分法求解 dy/dx = f(y)

When the derivative is given as a function of y only, we use the fact that dx/dy = 1 / (dy/dx). Rearranging gives dx/dy = 1/f(y). Then integrate with respect to y: x = ∫ 1/f(y) dy + C. You may need to rearrange afterwards to express y in terms of x, or leave it in implicit form if the question allows.

当导数仅以 y 的函数给出时，我们利用 dx/dy = 1 / (dy/dx) 这一性质。整理得 dx/dy = 1/f(y)。然后对 y 积分：x = ∫ 1/f(y) dy + C。之后可能需要重新整理，把 y 写成 x 的显函数，若题目允许也可保留隐函数形式。

Example:

Solve dy/dx = 6y².

We write dx/dy = 1/(6y²). Then x = ∫ (1/6) y⁻² dy = (1/6)(−y⁻¹) + C = −1/(6y) + C. Rearranging can give y in terms of x.

示例：

求解 dy/dx = 6y²。

我们写成 dx/dy = 1/(6y²)。然后 x = ∫ (1/6) y⁻² dy = (1/6)(−y⁻¹) + C = −1/(6y) + C。重新整理可将 y 用 x 表达。

4. Separable Differential Equations: The Core Method | 分离变量法：核心解法

The most common type in IGCSE CCEA is a separable differential equation of the form dy/dx = f(x)g(y). The method involves separating the variables so that all y terms (including dy) are on one side and all x terms (including dx) on the other: (1/g(y)) dy = f(x) dx. Then integrate both sides.

IGCSE CCEA 中最常见的类型是形如 dy/dx = f(x)g(y) 的可分离变量微分方程。该方法需要分离变量，让所有 y 项（包括 dy）在一边，所有 x 项（包括 dx）在另一边：(1/g(y)) dy = f(x) dx。然后两边同时积分。

Steps:

Rewrite the equation to isolate dy/dx if necessary.
Multiply both sides by dx and divide by g(y) to separate.
Integrate both sides. Remember one constant of integration on one side is enough.
Simplify and, if requested, solve for y explicitly.

步骤：

若需要，改写方程以分离出 dy/dx。
两边乘以 dx 并除以 g(y) 以分离变量。
两边积分。只需在一边加一个积分常数即可。
化简，若题目要求，解出 y 的显式表达式。

Example: Solve dy/dx = 2xy.

Separate: (1/y) dy = 2x dx. Integrate: ln|y| = x² + C. Then exponentiate: |y| = e^(x²+C) = e^C e^(x²). So y = A e^(x²), where A = ±e^C.

示例：解 dy/dx = 2xy。

分离变量：(1/y) dy = 2x dx。积分：ln|y| = x² + C。然后取指数：|y| = e^(x²+C) = e^C e^(x²)。所以 y = A e^(x²)，其中 A = ±e^C。

5. Finding Particular Solutions using Initial Conditions | 利用初始条件求特解

A general solution contains an arbitrary constant C. To find a particular solution, you need an initial condition, typically given as a pair of values (x₀, y₀) that satisfy the equation. Substitute these into the general solution and solve for C. Then rewrite the equation with this specific C value.

通解包含任意常数 C。为求特解，需要初始条件，通常以满足方程的一组值 (x₀, y₀) 给出。将这些值代入通解求出 C。然后用该特定 C 值重新写出方程。

Example: Given dy/dx = 2xy and y = 3 when x = 0, find the particular solution.

General solution: y = A e^(x²). Substitute: 3 = A e⁰ = A. Thus y = 3 e^(x²).

示例：已知 dy/dx = 2xy 且当 x = 0 时 y = 3，求特解。

通解：y = A e^(x²)。代入：3 = A e⁰ = A。因此 y = 3 e^(x²)。

Always box or clearly state the particular solution in the form y = f(x) if possible. This is often the final answer required.

若可能，始终将特解以 y = f(x) 的形式框出或明确写出。这通常是题目要求的最后答案。

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

One of the most important applications of differential equations in the CCEA syllabus is modelling exponential growth and decay. The basic form is dy/dx = k y, where k is a constant. If k > 0, it models growth; if k < 0, it models decay. The solution is y = A e^(k x), where A is the initial value when x = 0.

CCEA 教学大纲中微分方程最重要的应用之一是建立指数增长和衰减模型。基本形式为 dy/dx = k y，其中 k 为常数。若 k > 0，表示增长；若 k < 0，表示衰减。其解为 y = A e^(k x)，其中 A 为 x = 0 时的初始值。

In contextual problems, x often represents time t. For example, a population P(t) growing at a rate proportional to its current size: dP/dt = k P. Solution: P = P₀ e^(k t).

在实际问题中，x 常代表时间 t。例如，种群数量 P(t) 以与其当前大小成比例的速率增长：dP/dt = k P。解为：P = P₀ e^(k t)。

Be careful with units and interpretation: the constant k is the relative growth rate. Questions may ask you to find k from given data, or to predict a future value.

注意单位与解释：常数 k 为相对增长率。题目可能要求根据给定数据求 k，或预测未来值。

7. Rate of Change in Context: Forming Differential Equations | 结合情境建立微分方程

Sometimes you must construct a differential equation from a written description. Common phrases: ‘the rate of increase of y is proportional to y’ translates to dy/dt = k y. ‘The rate of decrease is proportional to the square of y’ becomes dy/dt = −k y². Linking sentences to mathematical symbols is a key skill.

有时你需要根据文字描述建立微分方程。常见表述：’y 的增长速率与 y 成正比’ 译作 dy/dt = k y。’减少速率与 y 的平方成正比’ 则变为 dy/dt = −k y²。将语句与数学符号对应是一项关键技能。

Steps to form a differential equation:

Identify the rate of change (dy/dt) and the quantity it depends on.
Determine if it is direct or inverse proportion, or a sum/difference.
Introduce a constant of proportionality k.
Write the equation and include any negative signs for decay.

建立微分方程的步骤：

确定变化率（dy/dt）及其依赖的量。
判断是正比、反比，还是和/差关系。
引入比例常数 k。
写出方程，衰减情况应包括负号。

Exam tip: Read the wording carefully. If the rate is proportional to the difference from a fixed value, you get equations like dT/dt = −k(T − 20) (Newton’s Law of Cooling type).

考试技巧：仔细读题。若变化率与某一固定值的差成正比，你会得到类似 dT/dt = −k(T − 20) 的方程（牛顿冷却定律型）。

8. Sketching Solution Curves and Slope Fields | 解曲线与斜率场草图

While not always heavily assessed, the ability to interpret a slope field (direction field) can appear. A slope field gives the value of dy/dx at various grid points. Drawing a solution curve means following the direction indicators smoothly, and if an initial point is given, the curve must pass through it.

虽然不一定重点考查，但解读斜率场（方向场）的能力可能出现在考题中。斜率场给出网格点处 dy/dx 的值。绘制解曲线意味着沿着方向指示平滑作曲线，若给出初始点，曲线必须穿过该点。

You might also be asked to show that a given function satisfies a differential equation by substituting it into both sides. This verifies it is a solution.

你也可能被要求通过代入给定函数到方程两边，证明该函数满足微分方程。这验证了它是一个解。

For the sketch, a rough curve following the arrows is sufficient. Focus especially on the behaviour where dy/dx = 0 (horizontal arrows) or where the slope is steep.

草图方面，沿着箭头大致画出曲线即可。尤其要关注 dy/dx = 0（水平箭头）和斜率陡峭的区域。

9. Common Mistakes and How to Avoid Them | 常见错误及如何避免

Mistake 1: Forgetting the constant of integration. Even if it disappears when finding a particular solution, you must include it initially. Always write y = … + C.

错误 1：忘记积分常数。即使求特解时常数会消去，初始步骤也必须包含。务必写上 y = … + C。

Mistake 2: Incorrect separation of variables. Ensure that after separation, one side contains only x and dx, the other only y and dy. If you have dy/dx = y/x, separating gives (1/y)dy = (1/x)dx — division by both y and x is needed.

错误 2：变量分离错误。确保分离后一边只有 x 和 dx，另一边只有 y 和 dy。如有 dy/dx = y/x，分离应得 (1/y)dy = (1/x)dx — 需同时除以 y 和乘以 dx、除以 x。

Mistake 3: Misapplying absolute values. When integrating 1/y, use ln|y|. When exponentiation removes ln, |y| = e^… , then introduce ± to drop absolute value properly. Most IGCSE contexts assume y > 0 so absolute signs can be simplified carefully.

错误 3：绝对值符号处理不当。积分 1/y 时用 ln|y|。取指数消去 ln 时得 |y| = e^…，然后通过 ± 正确去掉绝对值。大多数 IGCSE 场景可假设 y > 0，但需谨慎简化。

Mistake 4: Mixing up the roles of x and y when inverting dy/dx = f(y). Remember to write dx/dy and integrate with respect to y, then express y or x accordingly.

错误 4：在倒数法 dy/dx = f(y) 中混淆 x 与 y 的角色。记住要写成 dx/dy 并对 y 积分，然后相应表达 y 或 x。

10. Worked CCEA-Style Exam Question | CCEA 风格真题示例

Question:

The rate of increase of a population P, in thousands, t hours after the start of an experiment, is proportional to the population. Initially P = 2, and after 2 hours P = 3. Find an expression for P in terms of t.

问题：

实验开始 t 小时后，种群数量 P（以千计）的增长速率与种群数量成正比。初始时 P = 2，2 小时后 P = 3。求 P 关于 t 的表达式。

Solution:

Differential equation: dP/dt = k P. Separate: (1/P) dP = k dt. Integrate: ln|P| = k t + C. So P = A e^(k t), where A = e^C.

Using initial condition P=2 when t=0: 2 = A e⁰ = A. So A=2.

Using P=3 when t=2: 3 = 2 e^(2k). Thus e^(2k) = 1.5. Taking ln: 2k = ln(1.5), so k = ½ ln(1.5).

Final expression: P = 2 e^(½ ln(1.5) t) or simplified as P = 2 (1.5)^(t/2).

解答：

微分方程：dP/dt = k P。分离变量：(1/P) dP = k dt。积分：ln|P| = k t + C。故 P = A e^(k t)，其中 A = e^C。

利用初始条件 t=0 时 P=2：2 = A e⁰ = A，所以 A=2。

利用 t=2 时 P=3：3 = 2 e^(2k)。因此 e^(2k) = 1.5。取 ln：2k = ln(1.5)，得 k = ½ ln(1.5)。

最终表达式：P = 2 e^(½ ln(1.5) t) 或化简为 P = 2 (1.5)^(t/2)。

Exam marking highlights: Correct separation (1 mark), correct integration with constant (1 mark), finding A (1 mark), finding k (1 mark), final simplified expression (1 mark).

评分重点：正确分离变量（1 分），正确积分并带常数（1 分），求出 A（1 分），求出 k（1 分），最终简化表达式（1 分）。

11. Key Points for the Exam | 考试要点总结

Recognise the type of differential equation: dy/dx = f(x), dy/dx = f(y), or dy/dx = f(x)g(y).
For separable equations, the goal is to get all y’s on one side with dy and all x’s on the other with dx.
Only one constant of integration is needed, usually on the x-side after integration.
Initial conditions turn a general solution into a particular solution. Use them to find the constant.
Exponential models yield solutions of the form A e^(k t). Know how to find k from two data points.
Check your final answer by differentiating to see if you recover the original differential equation.

识别微分方程类型：dy/dx = f(x)、dy/dx = f(y) 或 dy/dx = f(x)g(y)。
对于可分离方程，目标是把所有 y 和 dy 放在一边，所有 x 和 dx 放在另一边。
只需一个积分常数，通常加在积分后的 x 一边。
初始条件将通解转化为特解。用它们求出常数值。
指数模型得出 A e^(k t) 形式的解。要掌握如何从两个数据点求 k。
通过求导检查最终答案，看是否能还原为原微分方程。

12. Final Tips & Exam Strategy | 最后提示与应试策略

Differential equation questions often carry high marks and are considered algebra-intensive. Manage your time wisely. Show every step clearly: separation, integration, constant determination, substitution of conditions. Even if you make a minor algebraic slip, you can still earn method marks. Always remember to write your final answer in the required form, and double-check that any given condition is fully used.

微分方程题常占据高分值，且代数运算量较大。合理分配时间。清晰展示每一步：分离变量、积分、确定常数、代入条件。即使出现小的代数失误，仍可获得方法分。务必按要求形式写出最终答案，并仔细检查是否已充分使用了所有给定条件。

Practice with past CCEA papers to become familiar with the typical wording, mark schemes, and the range of contexts (population, temperature, chemical concentration). With regular revision, these questions become a reliable source of marks.

多练习 CCEA 历年真题，熟悉典型措辞、评分方案以及各类情境（人口、温度、化学浓度）。通过定期复习，微分方程大题将成为你稳拿高分的题型。

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IGCSE CCEA Mathematics: Differential Equations Exam Guide | 微分方程考点精讲