AQA Maths: Second Order Differential Equations Revision | AQA 数学：二阶微分方程考点精讲

📚 AQA Maths: Second Order Differential Equations Revision | AQA 数学：二阶微分方程考点精讲

Second order differential equations are a cornerstone of AQA Further Pure Mathematics. They model everything from simple harmonic motion to RLC circuits, and mastering them requires a clear understanding of auxiliary equations, complementary functions, and particular integrals. This guide breaks down every key concept, case by case, with linked worked examples to help you tackle any exam question with confidence.

二阶微分方程是 AQA 进阶纯数学的核心内容，从简谐运动到 RLC 电路均离不开它。要在考试中从容应对，必须透彻理解辅助方程、补函数和特解。本文逐一解析所有关键考点，配以关联例题，帮助你稳扎稳打，拿稳每一分。

1. Introduction to Second Order Differential Equations | 二阶微分方程简介

A second order ordinary differential equation (ODE) involves an unknown function y(x) and its derivatives up to the second order, d²y/dx². In AQA Further Maths, we focus on linear second order ODEs with constant coefficients, which take the general form a d²y/dx² + b dy/dx + c y = f(x), where a, b, c are constants and f(x) is a given function.

二阶常微分方程包含未知函数 y(x) 及其一阶、二阶导数。在 AQA 进阶数学中，我们研究的是常系数线性二阶方程，其一般形式为 a d²y/dx² + b dy/dx + c y = f(x)，其中 a, b, c 为常数，f(x) 是已知函数。

2. The Form of a Second Order Linear ODE | 二阶线性常微分方程的形式

We always assume a ≠ 0. The equation is classified as homogeneous if f(x) = 0, and non‑homogeneous (or inhomogeneous) if f(x) ≠ 0. The solution to a non‑homogeneous equation is the sum of two parts: the complementary function (CF), which solves the homogeneous equation, and a particular integral (PI), which accounts for f(x). So, the general solution is y = y_CF + y_PI.

我们始终设 a ≠ 0。若 f(x) = 0，方程是齐次的；若 f(x) ≠ 0，则为非齐次。非齐次方程的通解由两部分构成：满足齐次方程的补函数 (CF)，和反映 f(x) 影响的特解 (PI)。因此，通解为 y = y_CF + y_PI。

3. Homogeneous Equations: The Auxiliary Equation | 齐次方程：辅助方程

For a d²y/dx² + b dy/dx + c y = 0, we try a solution of the form y = e^mx. Substituting gives the auxiliary (or characteristic) equation a m² + b m + c = 0. Solving this quadratic provides the values of m that determine the form of the complementary function.

对于 a d²y/dx² + b dy/dx + c y = 0，尝试解 y = e^mx。代入后得到辅助方程 a m² + b m + c = 0。解此二次方程得到 m 的值，这些值决定了补函数的形式。

4. Case 1: Real and Distinct Roots | 情况1：实且不相等的根

If the auxiliary equation yields two distinct real roots m₁ and m₂, the complementary function is y_CF = A e^m₁x + B e^m₂x, where A and B are arbitrary constants. This is the simplest case and often arises when the discriminant b² − 4ac > 0.

若辅助方程有两相异实根 m₁ 和 m₂，则补函数为 y_CF = A e^m₁x + B e^m₂x，其中 A、B 为任意常数。这是最简单的情形，通常源自判别式 b² − 4ac > 0。

5. Case 2: Repeated Real Root | 情况2：重实根

When the discriminant is zero, we have one repeated real root m. The complementary function becomes y_CF = (A + Bx) e^mx. The extra factor of x is essential to ensure the two parts of the CF are linearly independent.

当判别式为零时，有一个重实根 m。补函数形式为 y_CF = (A + Bx) e^mx。添加的因子 x 是为了保证补函数的两个部分线性无关。

6. Case 3: Complex Conjugate Roots | 情况3：共轭复根

If the auxiliary equation has complex roots p ± iq, Euler’s formula lets us write the solution without complex exponentials. The complementary function is y_CF = e^px (A cos qx + B sin qx). This form appears frequently in oscillatory systems like damped harmonic motion.

若辅助方程有共轭复根 p ± iq，借助欧拉公式可避免复数指数。补函数为 y_CF = e^px (A cos qx + B sin qx)。这种形式经常出现在阻尼简谐振动等系统中。

7. Non‑Homogeneous Equations: Finding the Particular Integral | 非齐次方程：求特解

For a d²y/dx² + b dy/dx + c y = f(x), the particular integral y_PI is any solution that satisfies the full equation. We choose a trial function based on the form of f(x), substitute it into the left‑hand side, and equate coefficients to find the unknown parameters. Constants of integration are not included in y_PI.

对于 a d²y/dx² + b dy/dx + c y = f(x)，特解 y_PI 是满足完整方程的任意一个解。我们根据 f(x) 的形式选择一个试探函数，代入方程左边，通过比较系数确定未知参数。特解中不包含积分常数。

8. Standard Forms for the Particular Integral | 特解的标准形式

The choice of trial function follows a straightforward pattern, listed below. If the standard trial function already appears in the complementary function, multiply by x (or x² if necessary).

试探函数的选择遵循明确的规律，见下表。若标准试探函数已包含在补函数中，则乘以 x（必要时乘以 x²）。

f(x)	Trial y_PI
Polynomial of degree n	General polynomial of degree n
α e^kx	λ e^kx
α cos ωx + β sin ωx	p cos ωx + q sin ωx
Product, e.g. e^kx cos ωx	Include both sine and cosine multiplied by exponential

Always start with the most general form of the trial function, even if f(x) contains only a sine or cosine term. For example, if f(x) = 5 sin 3x, you must try p cos 3x + q sin 3x.

始终使用最一般的试探形式，即使 f(x) 只含正弦或余弦项。例如，若 f(x) = 5 sin 3x，也必须尝试 p cos 3x + q sin 3x。

9. Using Initial or Boundary Conditions | 使用初始条件或边界条件

After obtaining the general solution y = y_CF + y_PI, you may be given extra conditions such as y(0) = 2 and y'(0) = −1. Substitute these after differentiating the general solution to find the values of the arbitrary constants A and B, thus obtaining a particular solution to the ODE.

得到通解 y = y_CF + y_PI 后，题目常给出附加条件，如 y(0) = 2 和 y'(0) = −1。在对通解求导后代入这些条件，可求出任意常数 A、B 的值，从而得到满足初边值条件的特解。

10. Worked Example | 例题精讲

Example: Solve d²y/dx² − 5 dy/dx + 6y = 2e^4x, given y(0) = 1 and y'(0) = 0.

例题：求解 d²y/dx² − 5 dy/dx + 6y = 2e^4x，满足 y(0) = 1，y'(0) = 0。

Step 1 – Auxiliary equation: m² − 5m + 6 = 0 ⇒ m = 2, 3. Hence y_CF = A e^2x + B e^3x.

步骤1 – 辅助方程：m² − 5m + 6 = 0 ⇒ m = 2, 3。所以 y_CF = A e^2x + B e^3x。

Step 2 – Trial PI: since f(x)=2e^4x and 4 is not a root, let y_PI = λ e^4x. Substitute into LHS: (16λ − 20λ + 6λ) e^4x = 2λ e^4x. Setting this equal to 2e^4x gives 2λ = 2, so λ = 1. Hence y_PI = e^4x.

步骤2 – 试探特解：因 f(x)=2e^4x 且 4 不是特征根，设 y_PI = λ e^4x。代入方程左边得 (16λ − 20λ + 6λ) e^4x = 2λ e^4x。令其等于 2e^4x，得 2λ = 2，λ = 1。所以 y_PI = e^4x。

Step 3 – General solution: y = A e^2x + B e^3x + e^4x.

步骤3 – 通解：y = A e^2x + B e^3x + e^4x。

Step 4 – Apply conditions: y(0) = A + B + 1 = 1 ⇒ A + B = 0. y’ = 2A e^2x + 3B e^3x + 4e^4x, so y'(0) = 2A + 3B + 4 = 0 ⇒ 2A + 3B = −4. Solving gives A = 4, B = −4. Final answer: y = 4e^2x − 4e^3x + e^4x.

步骤4 – 代入条件：y(0) = A + B + 1 = 1 ⇒ A + B = 0。y’ = 2A e^2x + 3B e^3x + 4e^4x，故 y'(0) = 2A + 3B + 4 = 0 ⇒ 2A + 3B = −4。解得 A = 4，B = −4。最终答案：y = 4e^2x − 4e^3x + e^4x。

11. Common Mistakes and Exam Tips | 常见错误与应试技巧

Forgetting to check for overlap: If the trial PI duplicates a term in the CF, multiply by x. For repeated overlap, multiply by x².
忘记检查重叠：若试探特解与补函数中的项重复，必须乘以 x。若重复两次，乘以 x²。
Missing the full trial form: Even with f(x) = sin 2x, you must try p cos 2x + q sin 2x; otherwise the derivative terms may fail to match.
试探形式不完整：即使 f(x) = sin 2x，也必须尝试 p cos 2x + q sin 2x，否则导数项可能无法匹配。
Algebra errors in the auxiliary equation: Double‑check the discriminant and roots; mis‑identifying real vs complex roots changes the entire CF structure.
辅助方程的代数错误：务必仔细核对判别式和根；误判实根与复根会完全改变补函数结构。
Skipping derivative when applying conditions: Always differentiate the full general solution carefully before substituting x = 0 or boundary points.
代入条件时忘记求导：务必先将通解求导，再代入 x = 0 或边界点，不可跳过。
Never drop the ‘ + y_PI ‘: A common oversight is writing only the CF when a PI is required.
切勿遗漏特解：常见失误是当 f(x) 存在时只写了补函数，漏掉了 y_PI。

Cultivate a routine: (1) write the auxiliary equation and solve; (2) write the CF; (3) select and find the PI; (4) write the general solution; (5) apply conditions if given. Following these steps methodically eliminates most errors.

养成固定解题步骤：(1) 写出并解辅助方程；(2) 写出补函数；(3) 选择并求出特解；(4) 写出通解；(5) 若有条件则代入。有条不紊地执行这些步骤，能避开绝大多数错误。

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AQA Maths: Second Order Differential Equations Revision | AQA 数学：二阶微分方程 考点精讲