📚 GCSE CCEA Maths: Differential Equations Revision Notes | CCEA 数学：微分方程考点精讲
Differential equations are a key bridge between differentiation and integration in the CCEA GCSE Mathematics syllabus. They often appear in questions involving motion, growth or the process of working backwards from a derivative to find the original function. Understanding how to solve simple first‑order differential equations and apply initial conditions gives you a powerful tool for tackling higher‑tier problems.
微分方程是 CCEA GCSE 数学课程中连接微分与积分的重要桥梁。它们经常出现在涉及运动、增长或从导数反求原函数的问题中。掌握如何求解简单的一阶微分方程并应用初始条件，将为你解决高阶问题提供强有力的工具。
1. What Is a Differential Equation? | 什么是微分方程？
rivative is given as a function of x. Solving the differential equation means finding the original function y = F(x) by integration.
微分方程是指包含导数的方程，例如 dy/dx 或 dv/dt。在 GCSE 阶段，最常见的形式是 dy/dx = f(x)，即导数被表示为 x 的函数。求解微分方程就是通过积分找出原函数 y = F(x)。
dt, where s is displacement, v is velocity and a is acceleration. These are differential equations that relate rates of change.
我们在运动学中也会遇到微分方程：v = ds/dt 和 a = dv/dt，其中 s 是位移，v 是速度，a 是加速度。这些都是关联变化率的微分方程。
2. Review of Differentiation | 微分复习
y = xⁿ, then dy/dx = n xⁿ⁻¹. You also need to handle constant multiples and sums of terms.
在求解微分方程之前，你需要熟练掌握微分。GCSE 的核心法则是幂法则：若 y = xⁿ，则 dy/dx = n xⁿ⁻¹。同时你还需要处理常数倍与多项式的和。
| f(x) | f'(x) |
|---|---|
| x³ | 3x² |
| 5x² | 10x |
| 7x | 7 |
| 4 (constant) | 0 |
This table shows basic examples. When we solve a differential equation, we are given the derivative and need to reverse the process.
上表展示了基本例子。当我们求解微分方程时，我们已知导数，需要逆向操作。
3. Review of Integration | 积分复习
add the constant of integration. The rule is: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, valid for n ≠ -1.
积分是微分的逆运算。要从 dy/dx = xⁿ 求出 y，我们将指数加 1 并除以新的指数，再加上积分常数。其法则为：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，当 n ≠ -1 时成立。
| dy/dx | y = ∫ (dy/dx) dx |
|---|---|
| 2x | x² + C |
| 3x² | x³ + C |
| 6x² + 4x | 2x³ + 2x² + C |
The constant C appears because differentiation of a constant is zero. We will determine C when extra information is provided.
常数 C 的出现是因为常数的导数为零。当题目提供额外信息时，我们将确定 C 的值。
4. Solving dy/dx = f(x) | 求解 dy/dx = f(x)
equation is of the form dy/dx = f(x). To solve, we integrate both sides with respect to x:
y = ∫ f(x) dx
最简形式的微分方程是 dy/dx = f(x)。我们可以对 x 两边积分来求解：y = ∫ f(x) dx。
For example, solve dy/dx = 4x³ − 2x + 1.
例如，求解 dy/dx = 4x³ − 2x + 1。
y = ∫ (4x³ − 2x + 1) dx = x⁴ − x² + x + C
The solution is a family of curves differing only by the constant C. We call this the general solution.
其解为一族曲线，仅相差常数 C。我们称之为通解。
5. The Constant of Integration | 积分常数
The constant of integration, usually written as C, must always be included when performing indefinite integration. In the context of differential equations, omitting C loses marks and gives an incomplete description of all possible solutions.
积分常数通常写作 C，在进行不定积分时必须始终包含。在微分方程的情境中，遗漏 C 会丢分，并且无法描述所有可能的解。
Think of C as shifting the entire graph up or down. Without it, we would have only one specific function instead of the full set.
可以把 C 想象成将整个图像上下平移。没有它，我们就只能得到一个特定的函数，而非完整集合。
6. Using Initial Conditions | 利用初始条件
ce y = 5 when x = 2. We substitute these values into the general solution and solve for C.
初始条件能够让我们求出 C 的特定值。它通常以曲线上某点的形式给出，例如当 x = 2 时 y = 5。我们将这些值代入通解并解出 C。
Example: Given dy/dx = 6x² and the point (1, 4) lies on the curve, find y in terms of x.
例子：已知 dy/dx = 6x²，且曲线经过点 (1,4)，求 y 关于 x 的表达式。
ubstitute x = 1, y = 4 → 4 = 2(1)³ + C → C = 2. So the particular solution is y = 2x³ + 2.
积分得：y = ∫ 6x² dx = 2x³ + C。代入 x=1, y=4 → 4 = 2(1)³ + C → C = 2。因此特解为 y = 2x³ + 2。
7. Differential Equations in Kinematics | 运动学中的微分方程
In kinematics problems, displacement s, velocity v and acceleration a are linked by differential equations:
v = ds/dt and a = dv/dt
在运动学问题中，位移 s、速度 v 和加速度 a 由微分方程联系：v = ds/dt 以及 a = dv/dt。
If we know acceleration as a function of time, we can integrate to find velocity, then integrate again to find displacement. Each integration introduces a constant which can be found using initial values for velocity or displacement.
若我们知道加速度是时间的函数，就可以通过积分求速度，再积分求位移。每次积分都会引入一个常数，可利用初速度或初位移求出这些常数。
For example, a car accelerates from rest with a = 3 m/s². Find v and s after time t.
例如，一辆汽车从静止开始以 a = 3 m/s² 加速。求经过时间 t 后的速度和位移。
Integrate a: v = ∫ 3 dt = 3t + C₁. At t=0, v=0 ⇒ C₁ = 0, so v = 3t.
对 a 积分：v = ∫ 3 dt = 3t + C₁。在 t=0 时，v=0 ⇒ C₁=0，因此 v=3t。
Integrate v: s = ∫ 3t dt = (3/2)t² + C₂. At t=0, assume s=0 ⇒ C₂ = 0, so s = 1.5 t².
对 v 积分：s = ∫ 3t dt = (3/2)t² + C₂。在 t=0 时，设 s=0 ⇒ C₂=0，所以 s = 1.5 t²。
8. Forming Differential Equations | 建立微分方程
ons require you to construct a differential equation from a verbal description. Words like "the rate of change of y with respect to x is proportional to …" indicate a differential equation. Translate "proportional to" into "= k ×", where k is a constant.
有些题目会要求你根据文字描述建立微分方程。诸如“y 随 x 的变化率与……成正比”这类表述就暗示了微分方程。将“成正比”翻译为“= k ×”，其中 k 为常数。
Example: "The velocity of a particle changes at a rate that is inversely proportional to time." This gives dv/dt = k/t.
例子：“某粒子的速度变化率与时间成反比。”便得到 dv/dt = k/t。
Then you may be asked to solve the equation, using integration and a given condition to find k.
随后你可能会被要求求解该方程，借助积分和给定条件求出 k。
9. Dealing with Cases Where dy/dx Depends on y | 当 dy/dx 依赖于 y 时
here k is a constant. This is solved by recognising that the exponential function y = A eᵏˣ ted). For CCEA GCSE, such cases are rare but could appear as a direct recognition problem.
在 GCSE 进阶层次偶尔会出现 dy/dx = ky，其中 k 为常数。这可以通过识别指数函数 y = A eᵏˣ 是解来求解，或者通过分离变量法（尽管不一定要求）。对 CCEA GCSE 而言，这类题目很少见，但可能作为直接识别题出现。
If told that dy/dx = 0.2y and y=100 when x=0, you can state y = 100 e⁰·²ˣ using knowledge of exponential growth.
若已知 dy/dx = 0.2y 且 x=0 时 y=100，你可以利用指数增长的知识写出 y = 100 e⁰·²ˣ。
10. Common Mistakes to Avoid | 常见错误
- Forgetting the +C: Always add the constant of integration unless the integral is definite.
- Mishandling the power rule: Remember to add one to the power and divide by the new power, not multiply.
- Ignoring initial conditions: Once you have the general solution, use the given x and y values to find C.
- Confusing s, v, and a: Keep clear that v = ds/dt and a = dv/dt. Integrating a gives v, not s directly.
- 忘记 +C： 除非是定积分，否则总要加上积分常数。
- 幂法则运用错误： 记住指数加 1，再除以新指数，而非乘以。
- 忽略初始条件： 得到通解后，务必用所给的 x 和 y 值求出 C。
- 混淆 s、v 和 a： 明确 v = ds/dt，a = dv/dt。对 a 积分得到 v，而不是直接得到 s。
11. Worked Example | 例题详解
asses through (2, 11) and its gradient is dy/dx = 3x² + 2. Find the equation of the curve.
一条曲线经过点 (2, 11)，且其斜率为 dy/dx = 3x² + 2。求该曲线的方程。
Integrate: y = ∫ (3x² + 2) dx = x³ + 2x + C.
积分得：y = ∫ (3x² + 2) dx = x³ + 2x + C。
Use the point: 11 = (2)³ + 2(2) + C → 11 = 8 + 4 + C → C = -1.
利用该点：11 = (2)³ + 2(2) + C → 11 = 8 + 4 + C → C = -1。
Thus the equation is y = x³ + 2x − 1.
因此曲线方程为 y = x³ + 2x − 1。
12. Exam Tips | 考试技巧
In the exam, show clear steps: write the integral, include the +C, substitute the given values neatly and state the particular solution. Even if you make an arithmetic slip, method marks are awarded for a correct integration approach and proper use of conditions.
在考试中，要展示清晰步骤：写出积分、包含 +C、整齐地代入已知数值并写出特解。即使出现计算错误，正确的积分方法以及恰当使用条件也能为你赢得方法分。
Check your final answer by differentiating it to see if you recover the original dy/dx. This is a quick verification that can prevent careless mistakes.
通过对最终答案求导检查是否得到原 dy/dx。这是一种快速验证方法，能避免粗心错误。
eep track of units (m, s, m/s, m/s²) and make sure you know when t=0 conditions are given.
对于运动学问题，要注意单位（m、s、m/s、m/s²），并确保明确知道 t=0 时的条件。
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