GCSE CCEA Maths: Second Order Differential Equations – Key Points | GCSE CCEA 数学：二阶微分方程考点精讲

📚 GCSE CCEA Maths: Second Order Differential Equations – Key Points | GCSE CCEA 数学：二阶微分方程考点精讲

Second order differential equations form a key part of the CCEA GCSE Further Mathematics specification. They allow us to model systems involving acceleration, oscillations and damping. This article focuses on linear homogeneous second order ODEs with constant coefficients, covering the auxiliary equation method, the three types of general solution and how to apply initial conditions to find particular solutions.

二阶微分方程是 CCEA GCSE 进阶数学的核心考点之一。它们可以用来描述加速度、振动和阻尼等物理系统。本文将重点讲解常系数线性齐次二阶常微分方程，包括辅助方程法、三种通解形式以及如何利用初始条件求出特解。

1. Standard Form of a Second Order Linear ODE | 二阶线性常微分方程的标准形式

All the equations we deal with can be written in the standard form: a·d²y/dx² + b·dy/dx + c·y = 0, where a, b and c are real constants and a ≠ 0. The equation is linear because y and its derivatives appear only to the first power, and homogeneous because the right‑hand side is zero.

我们研究的所有方程都可以写成标准形式：a·d²y/dx² + b·dy/dx + c·y = 0，其中 a、b、c 为实常数且 a ≠ 0。方程是线性的，因为 y 及其导数都是一次幂；也是齐次的，因为右端为零。

2. The Auxiliary Equation | 辅助方程（特征方程）

We look for solutions of the form y = e^rx. Substituting y = e^rx, dy/dx = r e^rx and d²y/dx² = r² e^rx into the standard form and cancelling the non‑zero factor e^rx gives the auxiliary equation: a r² + b r + c = 0. The discriminant Δ = b² – 4ac determines the nature of the roots.

我们寻求形如 y = e^rx 的解。将 y = e^rx、dy/dx = r e^rx 以及 d²y/dx² = r² e^rx 代入标准形式，并约去非零因子 e^rx，得到辅助方程：a r² + b r + c = 0。判别式 Δ = b² – 4ac 决定了根的性质。

3. Case 1: Real and Distinct Roots | 情形一：相异实根

When Δ > 0, the auxiliary equation has two distinct real roots r₁ and r₂. The general solution is a linear combination of the two independent solutions: y = A e^r₁x + B e^r₂x, where A and B are arbitrary constants.

当 Δ > 0 时，辅助方程有两个不相等的实根 r₁ 和 r₂。通解是两个独立解的线性组合：y = A e^r₁x + B e^r₂x，其中 A 与 B 为任意常数。

4. Case 2: Repeated Roots | 情形二：重根

If Δ = 0, there is one repeated real root r = –b/(2a). One solution is e^rx, but a second linearly independent solution is needed. It is found to be x e^rx. The general solution takes the form y = (A + B x) e^{r x}.

若 Δ = 0，则有一个重实根 r = –b/(2a)。其中一个解为 e^rx，但需要另一个线性无关的解，我们选用 x e^rx。通解的形式为 y = (A + B x) e^{r x}。

5. Case 3: Complex Conjugate Roots | 情形三：共轭复根

When Δ < 0, the roots are complex conjugates. Write them as α ± iβ, where α = –b/(2a) and β = √(4ac – b²)/(2a) > 0. Using Euler’s formula, the real general solution is y = e^αx(A cos βx + B sin βx). The arbitrary constants A and B are real.

当 Δ < 0 时，根为一对共轭复数，记作 α ± iβ，其中 α = –b/(2a)，β = √(4ac – b²)/(2a) > 0。利用欧拉公式，实通解为 y = e^αx(A cos βx + B sin βx)，常数 A 和 B 为实数。

6. Writing the General Solution Clearly | 规范写出通解

Always state the general solution with clear labelling of the arbitrary constants A and B. Avoid using C₁ and C₂ unless instructed – A and B are standard in CCEA mark schemes. Also ensure the solution is expressed entirely in real terms, especially when roots are complex.

写出通解时，务必明确标注任意常数 A 和 B。除非题目要求，尽量不要用 C₁ 和 C₂，因为 CCEA 评分标准习惯使用 A 和 B。此外，确保解全部用实函数表达，尤其是在处理复根时。

7. Applying Initial Conditions to Find a Particular Solution | 应用初始条件求特解

To determine A and B, you need two initial conditions, usually y(x₀) = p and y'(x₀) = q. Differentiate your general solution, then substitute the given x-value into both y and y’ and solve the simultaneous equations. The result is a particular solution that satisfies the constraints.

为确定 A 和 B，你需要两个初始条件，通常为 y(x₀) = p 和 y'(x₀) = q。先对通解求导，再将给定的 x 值分别代入 y 与 y’，然后解联立方程组。所得结果即为满足约束的特解。

8. Worked Example 1: Distinct Real Roots | 例题一：相异实根

Solve y” – 5y’ + 6y = 0 with y(0) = 2 and y'(0) = 3. The auxiliary equation is r² – 5r + 6 = 0, giving r = 2, 3. The general solution is y = A e^2x + B e^3x. Using y(0)=2 gives A + B = 2. Differentiating: y’ = 2A e^2x + 3B e^3x; then y'(0) = 3 gives 2A + 3B = 3. Solving yields A = 3, B = –1. Hence the particular solution is y = 3 e^2x – e^3x.

求解 y” – 5y’ + 6y = 0，初始条件为 y(0) = 2，y'(0) = 3。辅助方程 r² – 5r + 6 = 0 得 r = 2, 3。通解为 y = A e^2x + B e^3x。由 y(0)=2 得 A + B = 2。求导得 y’ = 2A e^2x + 3B e^3x；代入 y'(0)=3 得 2A + 3B = 3。解得 A = 3，B = –1。故特解为 y = 3 e^2x – e^3x。

9. Worked Example 2: Repeated Roots | 例题二：重根

Solve y” – 6y’ + 9y = 0, given y(0) = 1 and y'(0) = 4. The auxiliary equation r² – 6r + 9 = 0 gives a repeated root r = 3. The general solution is y = (A + Bx) e^3x. At x = 0, y(0) = A = 1. Differentiate: y’ = B e^3x + 3(A + Bx) e^3x. Then y'(0) = B + 3A = 4, so B + 3 = 4, giving B = 1. The particular solution is y = (1 + x) e^3x.

求解 y” – 6y’ + 9y = 0，已知 y(0) = 1，y'(0) = 4。辅助方程 r² – 6r + 9 = 0 得重根 r = 3。通解为 y = (A + Bx) e^3x。当 x=0 时，y(0)=A=1。求导得 y’ = B e^3x + 3(A + Bx) e^3x。代入 y'(0) = B + 3A = 4，故 B+3=4，得 B=1。特解为 y = (1 + x) e^3x。

10. Worked Example 3: Complex Roots | 例题三：复根

Solve y” + 4y’ + 13y = 0 with y(0) = 2 and y'(0) = –2. The auxiliary equation r² + 4r + 13 = 0 has roots r = –2 ± 3i. Here α = –2, β = 3. The general solution is y = e^–2x(A cos 3x + B sin 3x). Using y(0) = 2 gives A = 2. Differentiate: y’ = –2 e^–2x(A cos 3x + B sin 3x) + e^–2x(–3A sin 3x + 3B cos 3x). Setting x = 0: y'(0) = –2A + 3B = –2 → –4 + 3B = –2 → B = 2/3. So the particular solution is y = e^–2x(2 cos 3x + (2/3) sin 3x).

求解 y” + 4y’ + 13y = 0，初始条件为 y(0) = 2，y'(0) = –2。辅助方程 r² + 4r + 13 = 0 的根为 r = –2 ± 3i。这里 α = –2，β = 3。通解为 y = e^–2x(A cos 3x + B sin 3x)。由 y(0)=2 得 A=2。求导：y’ = –2 e^–2x(A cos 3x + B sin 3x) + e^–2x(–3A sin 3x + 3B cos 3x)。代入 x=0：y'(0) = –2A + 3B = –2 → –4 + 3B = –2 → B = 2/3。特解为 y = e^–2x(2 cos 3x + (2/3) sin 3x)。

11. Physical Interpretation – Damping | 物理解释 – 阻尼振动

These equations frequently arise in mechanics as models of damped oscillatory motion, for example my” + k y’ + ω² y = 0. The three cases correspond to overdamping (distinct real roots, no oscillation), critical damping (repeated root, fastest return to equilibrium without oscillating) and underdamping (complex roots, decaying oscillations). Recognising the type of damping helps check whether your solution makes physical sense.

这类方程在力学中常作为阻尼振动模型出现，例如 my” + k y’ + ω² y = 0。三种情形分别对应过阻尼（相异实根，不振荡）、临界阻尼（重根，以最快速度回到平衡且不振荡）和欠阻尼（复根，振荡逐渐衰减）。识别阻尼类型有助于验证解是否具有物理合理性。

12. Common Mistakes and How to Avoid Them | 常见错误及对策

Watch out for these frequent errors: forgetting to divide through by a when the coefficient of d²y/dx² is not 1; mis‑applying the quadratic formula; using the wrong form for repeated roots (simply e^rx is insufficient); writing complex exponentials instead of real trig functions; and making algebraic slips when substituting initial conditions into the derivative. Always double‑check your differentiation and the signs in your simultaneous equations.

请注意以下常见错误：当 d²y/dx² 的系数不为 1 时忘记先除以 a；求根公式使用出错；在重根情形下只写出 e^rx 一个解；用复指数形式而非实数三角函数表达通解；以及在代入初始条件时出现代数运算或求导符号错误。务必仔细核对求导结果和联立方程中的正负号。

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GCSE CCEA Maths: Second Order Differential Equations – Key Points | GCSE CCEA 数学：二阶微分方程 考点精讲