A-Level AQA Maths: Differential Equations Key Points | A-Level AQA 数学：微分方程考点精讲

📚 A-Level AQA Maths: Differential Equations Key Points | A-Level AQA 数学：微分方程考点精讲

Differential equations lie at the heart of A-Level Mathematics. They allow us to model how quantities change — from population growth to mechanical oscillations — and to solve for unknown functions given a rate of change. In the AQA specification, you encounter first-order separable equations, first-order linear equations solved via integrating factors, and second-order linear differential equations with constant coefficients. This guide walks you through the essential techniques, common pitfalls, and exam-ready strategies, all with paired English and Chinese explanations to strengthen your understanding.

微分方程是 A-Level 数学的核心内容。它们使我们能够模拟量是如何变化的——从人口增长到机械振动——并根据变化率解出未知函数。在 AQA 考纲中，你会遇到一阶可分离方程、用积分因子求解的一阶线性方程，以及常系数二阶线性微分方程。本指南将带你梳理关键技巧、常见错误和应试策略，并配有中英对照解释，以加深你的理解。

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

A differential equation relates a function to its derivatives. The order of a differential equation is the highest derivative that appears. For instance, dy/dx + 2y = eˣ is a first-order equation, while y” + 4y’ + 3y = 0 is a second-order equation. A solution is any function y = f(x) that satisfies the equation for all x in a given domain.

微分方程将函数与其导数联系起来。微分方程的阶数是出现的最高阶导数。例如，dy/dx + 2y = eˣ 是一阶方程，而 y” + 4y’ + 3y = 0 是二阶方程。解是使得方程在给定区间内对所有 x 都成立的函数 y = f(x)。

In AQA exams, you will often need to form a differential equation from a written description (the ‘rate of change is proportional to…’) and then solve it. Always identify the dependent variable, the independent variable, and any constants of proportionality.

在 AQA 考试中，你经常需要根据文字描述（“变化率与…成正比”）建立微分方程，然后求解。务必明确因变量、自变量以及任何比例常数。

2. Separable First-Order Equations | 可分离一阶微分方程

A first-order differential equation is called separable if it can be written in the form dy/dx = g(x)h(y). The method is to separate the variables: bring all y-terms to one side with dy, and all x-terms to the other with dx, then integrate both sides.

一阶微分方程如果能写成 dy/dx = g(x)h(y) 的形式，就称为可分离的。求解方法是分离变量：把所有 y 项和 dy 移到一边，所有 x 项和 dx 移到另一边，然后两边积分。

For example, to solve dy/dx = 3x²y, you rewrite it as (1/y) dy = 3x² dx. Integrating gives ln|y| = x³ + C, so y = A e^(x³) where A = ±e^C. Always write the constant of integration immediately after integrating, and be prepared to apply an initial condition to find the particular solution.

例如，求解 dy/dx = 3x²y，将其改写为 (1/y) dy = 3x² dx。积分得 ln|y| = x³ + C，所以 y = A e^(x³)，其中 A = ±e^C。积分后务必立即写上积分常数，并准备好利用初始条件求出特解。

Common exam tasks include finding general solutions and then using a point like (0, 2) to determine the arbitrary constant. Also watch for hidden separable equations where a simple substitution or rearrangement reveals the structure.

常见考题包括求通解，然后利用 (0, 2) 这样的点来确定任意常数。还要注意隐含的可分离方程，通过简单的代换或整理就能揭示其结构。

3. First-Order Linear Equations and the Integrating Factor | 一阶线性方程与积分因子

A first-order linear differential equation has the standard form dy/dx + P(x)y = Q(x). The integrating factor (IF) is given by I(x) = e^(∫ P(x) dx). Multiplying the entire equation by the IF transforms the left-hand side into the derivative of I(x)y.

一阶线性微分方程的标准形式为 dy/dx + P(x)y = Q(x)。积分因子 (IF) 为 I(x) = e^(∫ P(x) dx)。将整个方程乘以积分因子后，左边变成 I(x)y 的导数。

The steps are: (1) identify P(x) and calculate I(x); (2) multiply through by I(x) to obtain d/dx (I(x)y) = I(x)Q(x); (3) integrate both sides; (4) solve for y and apply initial conditions if given. For example, for dy/dx + 2xy = x, we have P(x) = 2x, so IF = e^(x²). Then d/dx (e^(x²) y) = x e^(x²), so e^(x²) y = (1/2) e^(x²) + C, giving y = 1/2 + C e^(-x²).

步骤为：(1) 确定 P(x) 并计算 I(x)；(2) 两边同乘 I(x)，得到 d/dx (I(x)y) = I(x)Q(x)；(3) 两边积分；(4) 解出 y，如有初始条件则代入。例如，对于 dy/dx + 2xy = x，P(x) = 2x，积分因子为 e^(x²)。于是 d/dx (e^(x²) y) = x e^(x²)，积分得 e^(x²) y = (1/2) e^(x²) + C，故 y = 1/2 + C e^(-x²)。

It is crucial to recognise the need for an integrating factor when the equation is not separable. Exam questions often mix both types, so check whether the equation can be rearranged into the linear standard form.

当方程不可分离时，识别是否需要使用积分因子至关重要。考试题常常混合两种类型，因此务必检查方程是否能转化为线性标准形式。

4. Second-Order Homogeneous Linear Equations with Constant Coefficients | 常系数二阶齐次线性微分方程

A homogeneous second-order linear differential equation with constant coefficients takes the form a y” + b y’ + c y = 0, where a, b, c are constants. To solve it, we assume a trial solution of the form y = e^(λx). Substituting this gives the auxiliary equation (or characteristic equation): a λ² + b λ + c = 0.

常系数二阶齐次线性微分方程的形式为 a y” + b y’ + c y = 0，其中 a, b, c 为常数。求解时，我们假设试解为 y = e^(λx)。代入后得到辅助方程（或特征方程）：a λ² + b λ + c = 0。

The roots of the auxiliary equation determine the form of the complementary function (CF), which is the general solution of the homogeneous equation. There are three cases: two distinct real roots, a repeated real root, and complex conjugate roots.

辅助方程根的性质决定了补函数 (CF) 的形式，即齐次方程的通解。共有三种情况：两个不同实根、一个重实根、以及共轭复根。

5. The Three Cases for the Complementary Function | 补函数的三种情况

Case 1: Two distinct real roots λ₁ and λ₂. The CF is y = A e^(λ₁ x) + B e^(λ₂ x). For example, y” – 3y’ + 2y = 0 gives λ² – 3λ + 2 = 0, so λ = 1 or 2. CF: y = A eˣ + B e^(2x).

情况 1: 两个不同实根 λ₁ 和 λ₂。补函数为 y = A e^(λ₁ x) + B e^(λ₂ x)。例如，y” – 3y’ + 2y = 0 得到 λ² – 3λ + 2 = 0，即 λ = 1 或 2。补函数：y = A eˣ + B e^(2x)。

Case 2: A repeated real root λ. The CF is y = (A + Bx) e^(λ x). Example: y” – 4y’ + 4y = 0 gives (λ – 2)² = 0, so λ = 2 (repeated). CF: y = (A + Bx) e^(2x).

情况 2: 一个重实根 λ。补函数为 y = (A + Bx) e^(λ x)。例如，y” – 4y’ + 4y = 0 得到 (λ – 2)² = 0，即 λ = 2（重根）。补函数：y = (A + Bx) e^(2x)。

Case 3: Complex conjugate roots λ = α ± iβ. The CF is y = e^(α x) (A cos βx + B sin βx). Example: y” + 2y’ + 5y = 0 gives λ = -1 ± 2i. CF: y = e^(-x) (A cos 2x + B sin 2x).

情况 3: 共轭复根 λ = α ± iβ。补函数为 y = e^(α x) (A cos βx + B sin βx)。例如，y” + 2y’ + 5y = 0 得到 λ = -1 ± 2i。补函数：y = e^(-x) (A cos 2x + B sin 2x)。

In exams, always write down the auxiliary equation explicitly, solve it, and then state the form of the complementary function with arbitrary constants A and B. Do not forget the arbitrary constants — they are essential for the full solution.

在考试中，务必明确写出辅助方程，求解，然后写出带有任意常数 A 和 B 的补函数形式。不要忘记任意常数——它们对于完整解至关重要。

6. Finding a Particular Integral for Non-Homogeneous Equations | 求非齐次方程的特解

When the equation is non-homogeneous, a y” + b y’ + c y = f(x), the general solution consists of the complementary function (CF) plus a particular integral (PI). The PI is any function y_p that satisfies the full equation. To find it, we use a trial function based on the form of f(x).

当方程为非齐次时，即 a y” + b y’ + c y = f(x)，通解由补函数 (CF) 加上特解 (PI) 组成。特解 y_p 是任何满足完整方程的函数。我们根据 f(x) 的形式设定试函数来求解。

The following table summarises common trial functions:

f(x)	Trial PI (yp)
Polynomial of degree n	General polynomial of degree n
e^(kx)	λ e^(kx)
cos ωx or sin ωx	λ cos ωx + μ sin ωx
Product (e.g., x eˣ)	Corresponding product with unknown coefficients

下表总结了常见的试函数：

f(x)	试特解 yp
n 次多项式	n 次一般多项式
e^(kx)	λ e^(kx)
cos ωx 或 sin ωx	λ cos ωx + μ sin ωx
乘积 (如 x eˣ)	含未知系数的相应乘积

If the standard trial function already appears in the CF, multiply the trial function by x (or x² for a repeated root). Substitute the trial PI into the differential equation and equate coefficients to find the values of the unknown constants. Then the general solution is y = CF + PI.

如果标准试函数已经出现在补函数中，则将试函数乘以 x（对于重根则乘以 x²）。将试特解代入微分方程，并比较系数求出未知常数的值。然后通解为 y = CF + PI。

7. Initial and Boundary Conditions | 初始条件与边界条件

After finding the general solution (CF + PI), you often need to determine the arbitrary constants A and B. In AQA problems, you may be given initial conditions (values of y and perhaps y’ at a single point) or boundary conditions (values of y at two different points). Both serve the same purpose: to pin down a unique particular solution.

求出通解 (CF + PI) 后，你通常需要确定任意常数 A 和 B。在 AQA 问题中，可能会给出初始条件（在某一点的 y 值，可能还有 y’ 值）或边界条件（在两个不同点的 y 值）。两者的目的相同：确定唯一的特解。

For a second-order equation, two conditions are needed. Plug the conditions into the general solution and its derivative if necessary; solve the resulting simultaneous equations for A and B. Always check your final solution by verifying that it satisfies the original differential equation and the given conditions.

对于二阶方程，需要两个条件。将条件代入通解及其导数（如果需要）；解出关于 A 和 B 的联立方程。务必检验最终解是否满足原微分方程和给定条件。

8. Modelling with Differential Equations: Exponential Growth and Decay | 微分方程建模：指数增长与衰减

Many real-world scenarios are modelled by the simple differential equation dN/dt = k N, or the more general dN/dt = k (N – a), where k is the growth/decay constant. Separation of variables leads to N = N₀ e^(k t) (or N = a + (N₀ – a) e^(k t)). The sign of k determines growth (k > 0) or decay (k < 0).

许多现实场景可以用简单的微分方程 dN/dt = k N 或更一般的 dN/dt = k (N – a) 来建模，其中 k 是增长/衰减常数。通过分离变量，得到 N = N₀ e^(k t)（或 N = a + (N₀ – a) e^(k t)）。k 的正负决定增长 (k > 0) 还是衰减 (k < 0)。

Exam questions often provide a context: population of bacteria, radioactive decay, cooling of an object (Newton’s Law of Cooling). You must interpret the wording to set up the correct differential equation. Remember that ‘rate of change is proportional to the current size’ translates to dN/dt ∝ N, hence dN/dt = k N.

考试常提供具体情境：细菌种群、放射性衰变、物体冷却（牛顿冷却定律）。你必须理解题干，建立正确的微分方程。记住，“变化率与当前大小成正比”转化为 dN/dt ∝ N，即 dN/dt = k N。

Also important is the concept of half-life for exponential decay: the time taken for N to halve is T = ln 2 / |k|. For exponential growth, you might be asked to find the doubling time.

对于指数衰减，半衰期的概念也很重要：N 减半所需的时间为 T = ln 2 / |k|。对于指数增长，则可能会要求计算倍增时间。

9. Modelling with Differential Equations: Mechanics and Mixtures | 微分方程建模：力学与混合问题

In mechanics problems, Newton’s second law can produce second-order differential equations. For example, for a particle falling under gravity with air resistance proportional to velocity, the equation might be m dv/dt = mg – kv, which is first-order. If resistance is proportional to velocity squared or involves position, it could lead to second-order or non-linear equations. At AQA, you mostly see linear first- or second-order ODEs.

在力学问题中，牛顿第二定律可以产生二阶微分方程。例如，一个受与速度成正比的空气阻力的下落物体，方程可能为 m dv/dt = mg – kv，这是一阶的。如果阻力与速度平方成正比或涉及位置，就可能得到二阶或非线性方程。在 AQA 考纲中，你主要遇到线性一、二阶常微分方程。

Mixture problems involve a tank with liquid entering and leaving; the rate of change of the amount of a substance is the rate in minus the rate out. This yields first-order linear equations, often solved via integrating factor. Carefully define variables and pay attention to concentrations and flow rates.

混合问题涉及有液体流入和流出的容器；某种物质数量的变化率为流入率减去流出率。这会得到一阶线性方程，通常用积分因子求解。仔细定义变量，注意浓度和流速。

10. Exam Technique and Common Mistakes | 应试技巧与常见错误

Avoid these typical errors: forgetting the constant of integration in separable equations; mishandling the integrating factor (e.g., missing the dx in the exponent integral); omitting the arbitrary constants in the complementary function; and picking the wrong trial form for the particular integral (especially when there is overlap with the CF).

避免以下典型错误：在可分离方程中遗忘积分常数；处理积分因子不当（例如，指数积分中漏掉 dx）；遗漏补函数中的任意常数；特解试函数选择错误（尤其是与补函数重叠时）。

When faced with a non-homogeneous second-order equation, always find the CF first, then examine f(x) and choose the PI trial accordingly. If your PI trial fails (usually because it duplicates a term in the CF), multiply by x. Always present your working step by step; examiners reward clear derivation of the auxiliary equation and identification of the complementary function.

遇到非齐次二阶方程时，始终先求补函数，然后检查 f(x) 并据此选择试特解。如果试特解失败（通常因为与补函数中的项重复），则乘以 x。务必逐步展示解题过程；考官会奖励清晰推导辅助方程和正确识别补函数的做法。

Finally, check your solution by differentiating and substituting back into the original equation. This quick validation can catch algebraic slips that might otherwise cost several marks.

最后，通过求导并代回原方程来检验你的解。这种快速验证可以避免因代数失误而丢失多分的遗憾。

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A-Level AQA Maths: Differential Equations Key Points | A-Level AQA 数学：微分方程 考点精讲