📚 Mastering Differential Equations for Edexcel A-Level Maths | A-Level Edexcel 数学：微分方程 考点精讲
Differential equations lie at the heart of A-Level Mathematics, linking calculus with real-world modelling. In the Edexcel specification, you are expected to form, solve and interpret first-order and second-order differential equations. This article breaks down every key technique — from separating variables to finding particular integrals — and shows how exam questions test your understanding through contextual problems.
微分方程是 A-Level 数学的核心内容，它将微积分与现实世界建模联系在一起。根据 Edexcel 考试大纲，你需要掌握建立、求解和解释一阶与二阶微分方程。本文逐一拆解每一种关键方法——从分离变量到求特解——并展示考试题目如何通过情景问题检验你的理解。
1. What is a Differential Equation? | 什么是微分方程？
A differential equation is any equation that contains a derivative, such as dy/dx or d²y/dx². The order of a differential equation is determined by the highest derivative present. In Edexcel A-Level, you will work primarily with first-order and second-order ordinary differential equations.
微分方程是包含导数（如 dy/dx 或 d²y/dx²）的方程。微分方程的阶数由方程中出现的最高阶导数决定。在 Edexcel A-Level 中，你主要处理一阶和二阶常微分方程。
For example, dy/dx = 3x² is a first-order differential equation, while d²y/dx² + 4y = 0 is a second-order linear differential equation with constant coefficients. The general solution will contain arbitrary constants, and extra information (initial or boundary conditions) allows you to pin down a particular solution.
例如，dy/dx = 3x² 是一个一阶微分方程，而 d²y/dx² + 4y = 0 是一个常系数二阶线性微分方程。通解会包含任意常数，额外的信息（初始条件或边界条件）让你能够确定一个特解。
2. Forming Differential Equations from Context | 从情境中建立微分方程
Many Edexcel exam questions start by asking you to construct a differential equation from a written description. The key is to translate phrases like ‘rate of change is proportional to…’ into mathematical language.
许多 Edexcel 考题一开始会要求你根据文字描述建立一个微分方程。关键是把“变化率与……成正比”这类语句翻译成数学语言。
If a population P grows at a rate proportional to its size, we write dP/dt = kP. If a substance cools at a rate proportional to the difference between its temperature T and the ambient temperature Tₐ, we write dT/dt = −k(T − Tₐ). Always include a negative sign if the quantity is decreasing. Remember to define all variables clearly.
如果种群数量 P 以与其大小成正比的速率增长，我们写作 dP/dt = kP。如果物体冷却的速率与其温度 T 和环境温度 Tₐ 之差成正比，我们写作 dT/dt = −k(T − Tₐ)。若数量在减少，务必加上负号。记得清晰地定义所有变量。
3. Solving First-Order Separable Equations | 一阶可分离变量方程的解法
A first-order differential equation is separable if it can be written in the form g(y) dy/dx = f(x). The solution method involves rearranging so that all y terms appear on one side with dy and all x terms on the other side with dx, then integrating both sides.
若一阶微分方程可写成 g(y) dy/dx = f(x) 的形式，它就是可分离变量的。解法是移项使得所有含 y 的项与 dy 在一边，所有含 x 的项与 dx 在另一边，然后两边积分。
Example: solve dy/dx = xy. Separate: (1/y) dy = x dx. Integrate: ∫(1/y) dy = ∫x dx ⇒ ln|y| = ½ x² + C. Then exponentiate to get y = ± e^{C} e^{½x²}, which is usually written as y = A e^{½x²}, where A is an arbitrary constant.
例如：求解 dy/dx = xy。分离变量：(1/y) dy = x dx。积分：∫(1/y) dy = ∫x dx ⇒ ln|y| = ½ x² + C。然后取指数得到 y = ± e^{C} e^{½x²}，通常写作 y = A e^{½x²}，其中 A 为任意常数。
Always remember to include the constant of integration immediately. Many marks are lost by forgetting the ‘+ C’ before rearranging.
务必记得立刻加上积分常数。许多考生因为在移项之前忘记写“+ C”而丢分。
4. First-Order Linear Equations and Integrating Factor | 一阶线性方程与积分因子
A first-order linear differential equation takes the form dy/dx + P(x) y = Q(x). The integrating factor method is used here. The integrating factor I(x) is e^{∫P(x) dx}.
一阶线性微分方程的形式为 dy/dx + P(x) y = Q(x)。这里使用积分因子法。积分因子 I(x) = e^{∫P(x) dx}。
Multiply the whole equation by I(x). The left-hand side becomes the derivative of I(x) y. Then integrate both sides with respect to x. For instance, solve dy/dx + 2y = e^{−x}. Here P(x)=2, so I(x)=e^{∫2 dx}=e^{2x}. Multiply through: e^{2x} dy/dx + 2e^{2x} y = e^{x}. The left side is d/dx (e^{2x} y). Integrate: e^{2x} y = ∫ e^{x} dx = e^{x} + C. Thus y = e^{−x} + C e^{−2x}.
将整个方程乘以 I(x)，左边就变成 I(x) y 的导数，然后两边对 x 积分。例如，求解 dy/dx + 2y = e^{−x}。此处 P(x)=2，于是 I(x)=e^{∫2 dx}=e^{2x}。两边同乘：e^{2x} dy/dx + 2e^{2x} y = e^{x}。左边是 d/dx (e^{2x} y)。积分：e^{2x} y = ∫ e^{x} dx = e^{x} + C。因此 y = e^{−x} + C e^{−2x}。
This method is essential for A-Level and frequently appears in modelling contexts, such as tank mixing problems.
这一方法是 A-Level 的必考内容，常出现在建模情境中，例如水箱混合问题。
5. Second-Order Homogeneous Linear Equations | 二阶齐次线性方程
A second-order linear differential equation with constant coefficients has the form a d²y/dx² + b dy/dx + c y = 0. To solve it, you form the auxiliary equation: a m² + b m + c = 0, where m is a constant.
常系数二阶线性微分方程的形式为 a d²y/dx² + b dy/dx + c y = 0。为求其解，你需建立辅助方程：a m² + b m + c = 0，其中 m 为常数。
If the auxiliary equation has two distinct real roots m₁ and m₂, the general solution is y = A e^{m₁x} + B e^{m₂x}. If it has a repeated real root m, the form is y = (A + Bx) e^{mx}. If the roots are complex conjugate α ± iβ, the general solution is y = e^{αx} (C cos βx + D sin βx).
若辅助方程有两个相异实根 m₁ 和 m₂，通解为 y = A e^{m₁x} + B e^{m₂x}。若为重实根 m，形式为 y = (A + Bx) e^{mx}。若根为共轭复根 α ± iβ，通解为 y = e^{αx} (C cos βx + D sin βx)。
These forms match the physical behaviour of damped oscillations: overdamped, critically damped and underdamped cases all appear in exam modelling questions.
这些形式与阻尼振动的物理行为相匹配：过阻尼、临界阻尼和欠阻尼情形都会出现在考试建模题中。
6. Particular Integrals for Non-Homogeneous Equations | 非齐次方程的特解
When the equation is non-homogeneous, e.g. a d²y/dx² + b dy/dx + c y = f(x), the general solution is the sum of the complementary function (CF) and a particular integral (PI): y = y_CF + y_PI.
当方程为非齐次，例如 a d²y/dx² + b dy/dx + c y = f(x)，通解为余函数 (CF) 与特解 (PI) 之和：y = y_CF + y_PI。
The complementary function is found by solving the homogeneous equation (set f(x)=0). The particular integral is found using the method of undetermined coefficients. Try a form similar to f(x): for a polynomial, try a polynomial of the same degree; for e^{kx}, try λ e^{kx}; for sin kx or cos kx, try p cos kx + q sin kx. If the trial form appears in the CF, multiply by x (or x² if repeated).
余函数通过求解对应的齐次方程（令 f(x)=0）得到。特解使用待定系数法求得。尝试一个与 f(x) 形式相似的函数：若是多项式，则尝试同次数的多项式；对于 e^{kx}，尝试 λ e^{kx}；对于 sin kx 或 cos kx，尝试 p cos kx + q sin kx。若试解形式已出现在 CF 中，则乘以 x（若重复则乘以 x²）。
Example: y” − 3y’ + 2y = e^{x}. The auxiliary m²−3m+2=0 gives m=1,2, so CF = A e^{x} + B e^{2x}. For PI, try y = p x e^{x} (since e^{x} appears in CF). Substituting gives p=1/1 = 1, so PI = x e^{x}. General solution: y = A e^{x} + B e^{2x} + x e^{x}.
例：y” − 3y’ + 2y = e^{x}。辅助方程 m²−3m+2=0 得 m=1,2，故 CF = A e^{x} + B e^{2x}。对于特解 PI，尝试 y = p x e^{x}（因 e^{x} 已出现于 CF 中）。代入得 p=1，故 PI = x e^{x}。通解：y = A e^{x} + B e^{2x} + x e^{x}。
7. Applying Initial and Boundary Conditions | 应用初始条件与边界条件
Once you have the general solution, use given conditions to find the particular solution. Initial conditions provide the value of y and/or its derivatives at a single point, typically when x=0. Boundary conditions give values at two different points.
得到通解后，使用给定条件求特解。初始条件提供在某一点（通常是 x=0 时）y 和/或其导数的值。边界条件则给出在两个不同点的值。
For first-order ODEs, one condition determines the constant. For second-order, you need two conditions. Always substitute carefully and solve the resulting simultaneous equations if necessary.
对于一阶常微分方程，一个条件即可确定常数。对于二阶，你需要两个条件。务必仔细代入，必要时要解联立方程组。
In modelling, initial conditions often represent the starting population, starting temperature, initial displacement and velocity. These are key to making the mathematics match the real situation.
在建模中，初始条件通常代表初始种群数量、初始温度、初始位移和速度。这些是让数学符合实际情况的关键。
8. Modelling with First-Order ODEs | 一阶常微分方程建模
Edexcel expects you to interpret real-life problems such as Newton’s law of cooling, exponential growth/decay, and mixing problems. The steps are: 1) identify the rate of change and the relationship given; 2) form the differential equation; 3) solve using separation or integrating factor; 4) apply conditions to find constants; 5) answer the specific question, for example find the time when a quantity reaches a certain value.
Edexcel 要求你解释现实问题，如牛顿冷却定律、指数增长/衰减以及混合问题。步骤为：1) 找出变化率及给定的关系；2) 建立微分方程；3) 用分离变量或积分因子法求解；4) 应用条件求常数；5) 回答具体问题，例如求某量达到特定值的时间。
Newton’s Law of Cooling: dT/dt = −k (T − Tₐ). Solving yields T = Tₐ + (T₀ − Tₐ) e^{−kt}. This exponential decay type appears frequently, and you may need to use logs to find k from two data points.
牛顿冷却定律：dT/dt = −k (T − Tₐ)。解之得 T = Tₐ + (T₀ − Tₐ) e^{−kt}。这种指数衰减形式频繁出现，你可能需要利用两个数据点通过对数求 k。
9. Modelling with Second-Order ODEs: Simple Harmonic Motion | 二阶常微分方程建模：简谐运动
Simple harmonic motion (SHM) is modelled by d²x/dt² = −ω² x. The general solution is x = A cos ωt + B sin ωt, or equivalently x = R cos(ωt − φ). Here ω is the angular frequency, A and B are determined by initial displacement and velocity.
简谐运动 (SHM) 由 d²x/dt² = −ω² x 建模。其通解为 x = A cos ωt + B sin ωt，或等价地 x = R cos(ωt − φ)。其中 ω 为角频率，A 和 B 由初始位移和速度确定。
If the motion starts from rest at a displacement x₀, then x = x₀ cos ωt. If it starts from the equilibrium position with an initial velocity v₀, then x = (v₀/ω) sin ωt. Questions often ask for amplitude, period T = 2π/ω, or maximum speed.
若运动从位移 x₀ 处由静止开始，则 x = x₀ cos ωt。若从平衡位置以初速度 v₀ 开始，则 x = (v₀/ω) sin ωt。题目常要求振幅、周期 T = 2π/ω 或最大速率。
10. Damped and Forced Oscillations | 阻尼振动与受迫振动
Introducing a damping force proportional to velocity gives a d²x/dt² + b dx/dt + c x = 0. The auxiliary equation determines the nature of damping: heavy damping (two distinct real roots) gives no oscillation; critical damping (repeated real root) brings the system to rest fastest without oscillation; light damping (complex roots) yields decaying oscillations x = e^{−αt} (A cos βt + B sin βt).
引入与速度成正比的阻尼力，得到 a d²x/dt² + b dx/dt + c x = 0。辅助方程的根决定了阻尼的性质：过阻尼（两个相异实根）无振荡；临界阻尼（重实根）使系统在不振荡的情况下最快趋于静止；欠阻尼（复根）产生衰减振荡 x = e^{−αt} (A cos βt + B sin βt)。
For forced oscillations, the non-homogeneous term f(t) represents an external driving force. The complementary function describes the transient behaviour, and the particular integral gives the steady-state response. At resonance, the driving frequency equals the natural frequency, causing the amplitude to grow.
对于受迫振动，非齐次项 f(t) 代表外部驱动力。余函数描述暂态行为，特解给出稳态响应。在共振时，驱动频率等于固有频率，导致振幅不断增大。
11. Exam Tips and Common Pitfalls | 考试技巧与常见误区
Always check the order and type of the differential equation before choosing a method. For separable equations, make sure you can legitimately separate — factors only, no sums. When using integrating factor, simplify e^{∫P dx} correctly and remember to include the constant in the final integration.
在选择方法之前，务必检查微分方程的阶数和类型。对于可分离变量的方程，确保能够合理地分离——只能是乘法分解，不能有加法。使用积分因子时，要正确化简 e^{∫P dx}，并在最终积分时记得加上常数。
In second-order problems, writing down the auxiliary equation correctly is half the battle. When finding a particular integral, always look for duplication with the complementary function and multiply by x if needed. Never forget to apply both initial/boundary conditions and find all constants — leaving a solution with A and B without evaluating them will lose marks.
在二阶问题中，正确写出辅助方程就成功了一半。求特解时，务必检查与余函数的重叠情况，必要时乘以 x。绝不要忘记代入初始/边界条件求出所有常数——如果解中还留着未经计算的 A 和 B，必定会丢分。
Examiners also test your ability to interpret the final formula in context, so always re-read the question after solving to extract the required numerical answer.
出题人还会考察你在情境中解读最终公式的能力，因此解完之后务必重新审题，以得出题目要求的数值答案。
12. Summary and Key Formula Sheet | 总结与核心公式表
First-order separable: ∫(1/g(y)) dy = ∫f(x) dx.
Integrating factor: I(x) = e^{∫P(x) dx}, then d/dx (I y) = I Q(x).
Second-order homogeneous: a m² + b m + c = 0, and three cases for roots.
Particular integral: try form similar to f(x), adjust if overlap with CF.
Modelling: identify rate, form ODE, solve, apply conditions, interpret.
一阶可分离：∫(1/g(y)) dy = ∫f(x) dx。
积分因子：I(x) = e^{∫P(x) dx}，然后 d/dx (I y) = I Q(x)。
二阶齐次：a m² + b m + c = 0，分三种根的情形处理。
特解：尝试与 f(x) 相似的形式，若与 CF 重叠则调整。
建模：确定变化率，建立 ODE，求解，应用条件，解读结果。
Mastering these techniques will equip you to tackle any Edexcel differential equations question with confidence. Practice regularly with past papers, and focus on the transition from a verbal description to the correct mathematical equation — it is a skill that rewards careful reading and clear variable definitions.
掌握这些方法将使你能够自信地应对任何 Edexcel 微分方程考题。定期用真题练习，重点关注从文字描述到正确数学方程的转化——这是一项通过细心阅读和清晰变量定义便能获得回报的技能。
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