Differential Equations: Key Exam Focus for AQA Maths | AQA 数学：微分方程 考点精讲

📚 Differential Equations: Key Exam Focus for AQA Maths | AQA 数学：微分方程 考点精讲

A solid grasp of differential equations is essential for AQA A-Level Mathematics, appearing in both pure and applied contexts. This article covers the key methods, common pitfalls, and exam strategies you need to master first-order separable equations, first-order linear equations with integrating factors, and second-order constant-coefficient linear ODEs, together with modelling applications.

扎实掌握微分方程是 AQA A-Level 数学的关键，出现在纯数和应用部分。本文涵盖你必须掌握的核心方法、常见错误和考试策略，包括一阶可分离方程、带积分因子的一阶线性方程、二阶常系数线性常微分方程以及建模应用。

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

A differential equation relates a function to its derivatives. The order is the highest derivative present. For AQA, we focus on first-order ODEs like dy/dx = f(x,y) and second-order linear ODEs with constant coefficients. A general solution contains arbitrary constants; a particular solution is obtained by applying initial or boundary conditions.

微分方程将函数与其导数联系起来。阶数是出现的最高阶导数。在 AQA 中，我们重点研究一阶常微分方程如 dy/dx = f(x,y) 和常系数二阶线性常微分方程。通解包含任意常数；通过应用初值或边界条件可得到特解。

2. Separable Differential Equations | 可分离变量的微分方程

If the equation can be written as dy/dx = g(x)h(y), separate variables so that all y terms are with dy and all x terms with dx. Then integrate both sides: ∫ 1/h(y) dy = ∫ g(x) dx. Always add a constant of integration (+C) on one side. Finally, rearrange to express y explicitly if required.

如果方程可写成 dy/dx = g(x)h(y)，分离变量使得所有含 y 的项与 dy 一起，所有含 x 的项与 dx 一起。然后两边积分：∫ 1/h(y) dy = ∫ g(x) dx。始终只在一边加上积分常数 (+C)。最后，若需要可重新排列以显式表示 y。

Example: dy/dx = (x+1)/y → ∫ y dy = ∫ (x+1) dx → ½y² = ½x² + x + C

例题：dy/dx = (x+1)/y → ∫ y dy = ∫ (x+1) dx → ½y² = ½x² + x + C

Common mistake: forgetting to write dy and dx correctly when separating. Always check that the equation is indeed separable; some AQA questions deliberately mix terms to test recognition.

常见错误：分离时忘记正确写出 dy 和 dx。始终检查方程是否确实可分离；AQA 的某些题目会故意混入别的项来考查识别能力。

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

A first-order linear ODE has the form dy/dx + P(x)y = Q(x). Multiply through by the integrating factor IF = e^(∫P(x)dx). The left-hand side becomes the derivative of y·IF, so integrate: y·IF = ∫ Q(x)·IF dx. Always evaluate ∫P(x)dx without the ‘+C’ when finding IF – a constant factor cancels later.

一阶线性常微分方程的形式为 dy/dx + P(x)y = Q(x)。乘上积分因子 IF = e^(∫P(x)dx)，左边就变成 y·IF 的导数，因此积分得：y·IF = ∫ Q(x)·IF dx。求 IF 时 ∫P(x)dx 切勿加 ‘+C’，因为常数因子稍后会被约掉。

IF = e^(∫Pdx) → d/dx(y·e^(∫Pdx)) = Q·e^(∫Pdx)

IF = e^(∫Pdx) → d/dx(y·e^(∫Pdx)) = Q·e^(∫Pdx)

After integration, include the +C and then use any given condition to find the particular solution. AQA often asks for the specific value of the constant.

积分之后，加上 +C，然后利用给定条件求出特解里的常数值。AQA 经常要求得到常数的具体取值。

4. Second-Order Homogeneous ODEs | 二阶齐次常微分方程

The standard form is a·d²y/dx² + b·dy/dx + c·y = 0, with constant real coefficients a, b, c. Assume a solution y = e^(λx), yielding the characteristic (auxiliary) equation aλ² + bλ + c = 0. The form of the complementary function depends on the discriminant Δ = b² – 4ac.

标准形式为 a·d²y/dx² + b·dy/dx + c·y = 0，其中 a,b,c 为实常数。设解为 y = e^(λx)，得到特征（辅助）方程 aλ² + bλ + c = 0。补函数的形式取决于判别式 Δ = b² – 4ac。

Roots of aλ²+bλ+c=0	Complementary Function y_c
Real distinct λ₁, λ₂	A e^(λ₁x) + B e^(λ₂x)
Repeated root λ	(A + Bx) e^(λx)
Complex λ = α ± iβ	e^(αx)(C cos(βx) + D sin(βx))

特征方程 aλ²+bλ+c=0 的根	补函数 y_c
不等实根 λ₁, λ₂	A e^(λ₁x) + B e^(λ₂x)
重根 λ	(A + Bx) e^(λx)
复根 λ = α ± iβ	e^(αx)(C cos(βx) + D sin(βx))

When dealing with complex roots, use Euler’s formula to convert e^(α±iβ)x into trigonometric form. AQA often tests the ability to write the solution without complex exponentials.

处理复根时，利用欧拉公式将 e^(α±iβ)x 转换为三角形式。AQA 经常会考查不使用复指数形式写出解的能力。

5. Non-Homogeneous Second-Order ODEs | 非齐次二阶常微分方程

The equation is a·d²y/dx² + b·dy/dx + c·y = f(x). The general solution is y = y_c + y_p, where y_c is the complementary function from the homogeneous equation and y_p is a particular integral. Find y_p by guessing a form similar to f(x) with undetermined coefficients.

方程为 a·d²y/dx² + b·dy/dx + c·y = f(x)。通解为 y = y_c + y_p，其中 y_c 是齐次方程的补函数，y_p 是特解。通过猜测与 f(x) 形式相似且带有待定系数的函数来求 y_p。

If f(x) = polynomial, try y_p = polynomial of same degree. If f(x) = k e^(px), try y_p = C e^(px), except when e^(px) is already in y_c, then multiply by x.

若 f(x) 为多项式，试设 y_p 为同次多项式。若 f(x) = k e^(px)，试设 y_p = C e^(px)，除非 e^(px) 已在 y_c 中出现，此时需乘以 x。

For f(x) = M cos(ωx) + N sin(ωx), try y_p = P cos(ωx) + Q sin(ωx). Substitute into the ODE, equate coefficients, and solve for P and Q.

对于 f(x) = M cos(ωx) + N sin(ωx)，试设 y_p = P cos(ωx) + Q sin(ωx)。代入原方程，比较系数，解出 P 和 Q。

6. Modelling with Differential Equations | 微分方程建模

Many AQA exam questions require constructing a differential equation from a word problem. Identify the rate of change (e.g., dT/dt, dN/dt) and express it in terms of the variable. Typical contexts include population growth, cooling, mixing, and mechanics.

许多 AQA 考题要求根据文字描述建立微分方程。找出变化率（如 dT/dt, dN/dt）并将其用变量表达出来。常见情境包括人口增长、冷却、混合以及力学问题。

The general principle: ‘rate of change = input rate – output rate’ or ‘rate proportional to difference’. Always define variables clearly and state the associated condition, then solve using the techniques from earlier sections.

一般原则：“变化率 = 输入率 – 输出率” 或 “速率与差值成正比”。始终明确定义变量并写出相关条件，然后用前面章节介绍的方法求解。

7. Exponential Growth and Decay | 指数增长与衰减

The model dN/dt = kN, with k constant, describes exponential growth (k>0) or decay (k<0). Separating variables gives N = N₀ e^(kt). The doubling time or half-life follows from setting N = 2N₀ or ½N₀ and solving for t.

模型 dN/dt = kN 中 k 为常数，描述指数增长（k>0）或衰减（k<0）。分离变量得 N = N₀ e^(kt)。加倍时间或半衰期可令 N = 2N₀ 或 ½N₀ 并解出 t。

Half-life t₁/₂ = (ln 2)/|k| for decay. Doubling time t_d = (ln 2)/k for growth.

衰减时半衰期 t₁/₂ = (ln 2)/|k|；增长时加倍时间 t_d = (ln 2)/k。

AQA sometimes embeds exponential growth inside a modified logistic context, but the core separable technique remains the same. Always check units and whether k is positive or negative.

AQA 有时会把指数增长嵌入修正的 logistic 情境，但核心的分离变量方法不变。务必检查单位以及 k 的正负。

8. Newton’s Law of Cooling | 牛顿冷却定律

Newton’s law states dT/dt = –k(T – T_env), where T is the object’s temperature and T_env is the constant ambient temperature. This is a first-order linear or separable equation. The solution is T = T_env + (T₀ – T_env) e^(–kt).

牛顿定律指出 dT/dt = –k(T – T_env)，其中 T 为物体温度，T_env 为恒定的环境温度。这是一阶线性方程，也可分离变量。解为 T = T_env + (T₀ – T_env) e^(–kt)。

Examiners often ask for the time taken to reach a certain temperature or the value of k from two data points. Remember that the temperature difference decays exponentially towards zero, so the object’s temperature asymptotically approaches T_env.

考官常要求计算达到某一温度所需的时间，或根据两个数据点求出 k 值。记住温度差随时间指数衰减趋于零，因此物体温度逐渐趋向 T_env。

9. Simple Harmonic Motion | 简谐运动

Simple harmonic motion (SHM) satisfies d²x/dt² = –ω²x, where ω is the angular frequency. The general solution is x = A cos(ωt) + B sin(ωt) or equivalently x = R cos(ωt – φ). AQA may link this to second-order ODEs: d²x/dt² + ω²x = 0.

简谐运动满足 d²x/dt² = –ω²x，其中 ω 为角频率。通解为 x = A cos(ωt) + B sin(ωt)，或等价地写为 x = R cos(ωt – φ)。AQA 可能将其与二阶常微分方程 d²x/dt² + ω²x = 0 关联。

The amplitude R = √(A²+B²) and phase angle φ = arctan(B/A). Velocity and acceleration follow from differentiation. Use initial displacement and velocity to find A and B.

振幅 R = √(A²+B²)，相位角 φ = arctan(B/A)。通过对位移求导可得速度和加速度。利用初始位移和初速度求出 A 和 B。

10. Exam Strategies & Common Pitfalls | 考试策略与常见错误

First, identify the type of ODE. Check if variables separate; if not, look for linear form with integrating factor; for second-order, write the characteristic equation. Always show the method clearly – marks are awarded for setting up the correct integral or trial function.

首先，识别常微分方程的类型。检查变量是否可分离；如果不是，寻找可用积分因子的线性形式；对于二阶，写出特征方程。务必清晰展示方法——建立正确积分或试函数的过程会得到步骤分。

Common slip: forgetting to apply initial conditions to find the constant in the general solution. Another is mishandling the sign when separating or moving terms. For second-order non-homogeneous equations, ensure the trial y_p does not duplicate terms in y_c; if it does, multiply by x or x² until linearly independent.

常见疏忽：忘记代入初值来求通解中的常数。另一错误是分离变量或移项时弄错符号。对于二阶非齐次方程，确保试设的 y_p 不与 y_c 中的项重复；若重复，需乘以 x 或 x² 直至线性无关。

When modelling, double-check that your differential equation matches the wording (e.g., ‘rate of decrease proportional to amount’ means dN/dt = –kN, not +kN). Keep track of units and whether quantities are increasing or decreasing.

进行建模时，反复确认你的微分方程符合题意（例如，“减少速率与数量成正比”意味着 dN/dt = –kN，而非 +kN）。注意单位以及量的增减。

Finally, practise past AQA papers; many differential equation questions follow a predictable pattern. Under timed conditions, allocate sufficient time to verify your particular solution satisfies both the ODE and the given conditions.

最后，练习 AQA 往年真题；许多微分方程题目均有套路可循。在限时条件下，留出充足时间检验特解是否满足常微分方程和所给条件。

11. Summary of Key Formulas | 关键公式总结

Separable: ∫ 1/h(y) dy = ∫ g(x) dx

可分离：∫ 1/h(y) dy = ∫ g(x) dx

Linear: IF = e^(∫P(x)dx), y·IF = ∫ Q·IF dx

线性：IF = e^(∫P(x)dx), y·IF = ∫ Q·IF dx

Second-order homogeneous: a·d²y/dx² + b·dy/dx + c·y = 0, roots λ from aλ²+bλ+c=0

二阶齐次：a·d²y/dx² + b·dy/dx + c·y = 0，根 λ 来自 aλ²+bλ+c=0

Exponential: N = N₀ e^(kt), cooling T = T_env + (T₀–T_env)e^(–kt)

指数：N = N₀ e^(kt)，冷却 T = T_env + (T₀–T_env)e^(–kt)

SHM: x = A cos(ωt) + B sin(ωt), ω² = k/m (mechanical)

简谐运动：x = A cos(ωt) + B sin(ωt), ω² = k/m（力学）

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