📚 IGCSE Maths: Differential Equations Exam Focus | IGCSE 数学：微分方程 考点精讲
Differential equations are a natural extension of differentiation and integration, allowing us to model a vast range of real-world phenomena. In IGCSE Maths, you are expected not only to solve simple differential equations by integration but also to apply initial conditions, recognise separable forms, and interpret solutions in contexts such as motion and growth. This article provides a thorough, exam‑focused walkthrough of every concept you need to master.
微分方程是微分的自然延伸，帮助我们建立大量现实世界的数学模型。在 IGCSE 数学中，你不仅要会用积分求解简单的微分方程，还要会代入初始条件、识别可分离变量的形式，并能在运动学、增长过程等问题中解释解的意义。本文将从考点出发，系统梳理每一个你需要掌握的知识点。
1. Understanding Differential Equations | 理解微分方程
A differential equation is an equation that contains an unknown function and one or more of its derivatives. The simplest type in IGCSE involves the first derivative dy/dx. For example, dy/dx = 3x² + 2 means ‘the rate of change of y with respect to x is given by 3x² + 2’.
微分方程是含有未知函数及其导数的方程。IGCSE 中最简单的一类只涉及一阶导数 dy/dx。例如 dy/dx = 3x² + 2 表示“y 关于 x 的变化率由 3x² + 2 给出”。
The order of a differential equation is the highest derivative it contains. Virtually all IGCSE questions deal with first‑order equations, though you may encounter second derivatives when linking acceleration, velocity and displacement in kinematics.
微分方程的阶由方程中出现的最高阶导数决定。IGCSE 的问题几乎全部围绕一阶方程，但在运动学中将加速度、速度与位移联系时，你可能会遇到二阶导数。
Recognising the form of a differential equation tells you which technique to use. The two main approaches you must know are direct integration and separation of variables.
识别微分方程的形式能帮助你判断该用什么方法。你必须掌握的两大方法是直接积分法和分离变量法。
2. Direct Integration: dy/dx = f(x) | 直接积分法：dy/dx = f(x)
When the right‑hand side depends only on x, the solution is obtained by 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.
当右边只含 x 时，可以直接对两边积分求解。若 dy/dx = f(x)，则 y = ∫ f(x) dx + C，其中 C 是积分常数。
For example, to solve dy/dx = 4x³ − sin x, integrate to get y = x⁴ + cos x + C. Never forget ‘+ C’ — a differential equation always has infinitely many solutions until an extra condition is supplied.
例如，求解 dy/dx = 4x³ − sin x，积分得 y = x⁴ + cos x + C。千万莫忘 ‘+ C’——在没有附加条件时，微分方程总有无穷多个解。
This technique is the foundation for all further work. Even when using separation of variables, you will eventually perform direct integration on both sides of the separated equation.
这一技巧是所有后续内容的基础。即使用分离变量法，你最终也要对分离后的等式两边进行直接积分。
3. General and Particular Solutions | 通解与特解
A solution containing the arbitrary constant C is called the general solution. To find a particular solution, you use an initial condition — a known point (x₀, y₀) that the curve must pass through.
含有任意常数 C 的解称为通解。要找出特解，你需要利用初始条件 —— 即曲线必须经过的一个已知点 (x₀, y₀)。
Suppose dy/dx = 2x and the curve passes through (1, 3). Integrate: y = x² + C. Substitute x = 1, y = 3 to obtain 3 = 1² + C, so C = 2. The particular solution is y = x² + 2.
假设 dy/dx = 2x 且曲线经过 (1, 3)。积分得 y = x² + C。代入 x = 1, y = 3 得到 3 = 1² + C，因此 C = 2。特解为 y = x² + 2。
Always write down the general solution first, then substitute the initial condition clearly. Examiners award marks for the method, even if the final constant is mis‑calculated.
务必先写出通解，再清晰地代入初始条件。即便最终常数算错，阅卷人也会按步骤给分。
4. Separation of Variables | 分离变量法
If the equation can be written in the form dy/dx = g(x)h(y), you can separate the variables: bring all y‑terms to the left with dy and all x‑terms to the right with dx. The equation becomes ∫ 1/h(y) dy = ∫ g(x) dx.
如果方程能写成 dy/dx = g(x)h(y)，你可以分离变量：将所有含 y 的项移到左边并配上 dy，所有含 x 的项移到右边配上 dx。方程变为 ∫ 1/h(y) dy = ∫ g(x) dx。
Consider dy/dx = (2x) / y. Multiply both sides by y dx to obtain y dy = 2x dx. Then integrate: y²/2 = x² + C, or y² = 2x² + 2C. You can rename the constant as a new letter, say k.
考虑 dy/dx = (2x) / y。两边乘以 y dx 得到 y dy = 2x dx。接着积分：y²/2 = x² + C，或写作 y² = 2x² + 2C。你可以将常数重命名，比如记作 k。
Watch out for divisions by zero when separating. If h(y) can be zero, you may lose a constant solution such as y = 0. In IGCSE, questions normally exclude these complications, but it is wise to remark that you assume h(y) ≠ 0.
分离时要注意是否除以零。若 h(y) 可能为零，你可能会丢失诸如 y = 0 这样的常数解。IGCSE 试题通常会排除这类复杂情况，但明智的做法是注明你假设 h(y) ≠ 0。
After integration, use the initial condition to find the particular solution. With some algebraic manipulation, you may be asked to express y explicitly in terms of x.
积分后利用初始条件求出特解。通过代数变形，有时会要求你将 y 表示成 x 的显函数。
5. Differential Equations in Kinematics | 运动学中的微分方程
Kinematics provides the most common applied setting for differential equations. If displacement s, velocity v and acceleration a are functions of time t, then v = ds/dt and a = dv/dt. Consequently, a = d²s/dt².
运动学为微分方程提供了最常见的应用背景。若位移 s、速度 v 和加速度 a 都是时间 t 的函数，那么 v = ds/dt 且 a = dv/dt。于是有 a = d²s/dt²。
Given acceleration as a function of time, you can find velocity by integrating a with respect to t and applying an initial velocity. For example, a = 6t, and at t = 0, v = 2. Then v = ∫ 6t dt = 3t² + C. Using v(0) = 2 gives C = 2, so v = 3t² + 2.
已知加速度是时间的函数，你可以对 a 关于 t 积分并代入初速度来求速度。例如 a = 6t，且 t = 0 时 v = 2。那么 v = ∫ 6t dt = 3t² + C。代入 v(0) = 2 得 C = 2，因此 v = 3t² + 2。
Integrating velocity yields displacement: s = ∫ v dt + D, where D is found from an initial position. This process illustrates how first‑order and second‑order differential equations naturally arise in physics.
对速度积分可得位移：s = ∫ v dt + D，其中 D 由初始位置确定。这一过程展示了一阶和二阶微分方程如何在物理中自然出现。
6. Exponential Growth and Decay | 指数增长与衰减
Many real‑world situations, such as population growth or radioactive decay, are modelled by dy/dt = k y, where k is a constant. This is a separable equation: ∫ 1/y dy = ∫ k dt gives ln|y| = kt + C, so y = A e^(kt) (where A = ±e^C).
许多现实情景（如种群增长或放射性衰变）可以用 dy/dt = k y 建模，其中 k 为常数。这是一个可分离变量的方程：∫ 1/y dy = ∫ k dt 给出 ln|y| = kt + C，因此 y = A e^(kt)（其中 A = ±e^C）。
If k > 0, the quantity grows exponentially; if k < 0, it decays. IGCSE questions sometimes give you the formula directly and ask you to find the time needed for a quantity to halve or double.
若 k > 0，量呈指数增长；若 k < 0，则呈指数衰减。IGCSE 有时会直接给出该公式，然后要求你计算数量减半或翻倍所需的时间。
Even when the formula is provided, you should be able to derive it using separation of variables. This demonstrates a deeper understanding and helps you spot errors if the given formula looks unusual.
即便公式是给出的，你也应能用分离变量法将其推导出来。这不仅能展示更深的理解，也有助于你在发现公式异常时识别错误。
7. Modelling Curves from Gradient Functions | 从斜率函数建模曲线
A classic IGCSE problem states: ‘The gradient of a curve at any point (x, y) is given by … Find the equation of the curve given that it passes through P.’ Here, the gradient function is exactly dy/dx, so you are solving a differential equation.
IGCSE 的经典问题会这样表述：“曲线上任一点 (x, y) 处的斜率由 …… 给出。已知曲线经过点 P，求曲线方程。” 这里的斜率函数就是 dy/dx，因此实际上你在解一个微分方程。
For instance, dy/dx = x / y, and the curve passes through (0, 2). Separate: y dy = x dx, integrate to get y²/2 = x²/2 + C. Using (0, 2) gives 2 = 0 + C, so y² = x² + 4. Because the point has y positive, you can write y = √(x² + 4).
例如，dy/dx = x / y，且曲线经过 (0, 2)。分离变量：y dy = x dx，积分得 y²/2 = x²/2 + C。代入 (0, 2) 得 2 = 0 + C，于是 y² = x² + 4。因为该点 y 为正，可写出 y = √(x² + 4)。
Always choose the appropriate branch (positive or negative) based on the given point. Marks are often allocated for stating the domain or the sign of y.
务必根据已知点选择恰当的分支（正或负）。卷面上常会留出分数给定义域的说明或 y 的符号判定。
8. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧
One of the most frequent mistakes is forgetting the constant of integration. Even if the question only asks for a particular solution, you must introduce ‘+ C’ first and then determine its value. Without it, you lose method marks.
最常见的错误之一就是忘掉积分常数。即便题目只要求特解，你也必须先引入 ‘+ C’，再确定其数值。没有这一步，你会失去方法分。
When separating variables, ensure you do not accidentally divide by an expression that could be zero. Check the context: a growth model with y > 0, for instance, avoids division by zero issues.
分离变量时，要确保没有不小心除以可能为零的表达式。根据上下文检查：例如 y > 0 的增长模型就能避开除零问题。
After integrating, check that your answer satisfies the original differential equation. A quick differentiation can save you from algebraic slips. In kinematics, pay close attention to units and the meaning of constants.
积分之后，检验你的答案是否满足原微分方程。快速求导一下能避免代数失误。在运动学中，要格外注意单位和常数的物理意义。
Finally, practise rewriting solutions in the form requested. Sometimes y² = 4x + 9 is acceptable; other times you must write y = ±√(4x + 9) and choose the correct sign.
最后，多练习按题目要求写出解的形式。有时 y² = 4x + 9 可以接受，有时则必须写成 y = ±√(4x + 9) 并选择正确的符号。
9. Worked Example from Past Papers | 真题实例解析
Question: Solve the differential equation dy/dx = (2x + 1) / y² given that y = 1 when x = 0. Express y in terms of x.
题目：解微分方程 dy/dx = (2x + 1) / y²，已知 x = 0 时 y = 1。将 y 表示为 x 的函数。
Separate variables: y² dy = (2x + 1) dx. Integrate both sides: ∫ y² dy = ∫ (2x + 1) dx → y³/3 = x² + x + C. Using x = 0, y = 1 gives 1/3 = 0 + 0 + C, so C = 1/3.
分离变量：y² dy = (2x + 1) dx。两边积分：∫ y² dy = ∫ (2x + 1) dx → y³/3 = x² + x + C。代入 x = 0, y = 1 得 1/3 = 0 + 0 + C，因此 C = 1/3。
Multiply by 3: y³ = 3x² + 3x + 1. Taking the cube root gives y = ∛(3x² + 3x + 1). Since the initial value is positive, we take the real cube root without a ± sign.
两边乘以 3：y³ = 3x² + 3x + 1。开立方根得 y = ∛(3x² + 3x + 1)。因为初值为正，我们取其三次实数根，不加 ± 号。
Always verify by differentiating your final y. If dy/dx matches the given expression, your solution is correct.
始终通过对最终 y 求导来验证。如果 dy/dx 与给定表达式吻合，则解正确无误。
10. Summary and Key Takeaways | 总结与关键要点
Differential equations in IGCSE link differentiation and integration. Master the two core techniques: direct integration when dy/dx = f(x), and separation of variables when dy/dx = f(x)g(y). Always include the constant of integration and use initial conditions to find particular solutions.
IGCSE 中的微分方程将微分与积分联系起来。掌握两大核心技巧：当 dy/dx = f(x) 时使用直接积分；当 dy/dx = f(x)g(y) 时使用分离变量。切勿遗漏积分常数，并利用初始条件求特解。
Apply these skills to kinematics (v = ds/dt, a = dv/dt), exponential models and curve‑sketching problems. Remember to check for domain issues and the sign of the dependent variable. With careful algebra and plenty of practice, differential equations become a reliable source of marks on your exam paper.
将这些技巧应用于运动学 (v = ds/dt, a = dv/dt)、指数模型和曲线求解问题。记住检查定义域及因变量的符号。通过仔细的代数操作和充分练习，微分方程将成为你试卷上可靠的得分点。
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