Second Order Differential Equations: IB & CIE Exam Focus | IB CIE 数学：二阶微分方程考点精讲

📚 Second Order Differential Equations: IB & CIE Exam Focus | IB CIE 数学：二阶微分方程考点精讲

Second order differential equations form a core part of the IB and CIE A-Level Mathematics syllabuses, appearing in topics ranging from pure calculus to mechanics. Mastering the forms a y” + b y’ + c y = f(x), where a, b, c are constants, is essential for tackling both homogeneous and non-homogeneous cases. In this article we cover every exam-relevant detail, from the characteristic equation to the method of undetermined coefficients, and link the theory to real-world contexts like damped oscillations.

二阶微分方程是 IB 和 CIE A-Level 数学课程的核心内容，涵盖纯微积分与力学等多个板块。掌握形如 a y” + b y’ + c y = f(x) 的方程结构，是处理齐次和非齐次方程的基础。本文将从特征方程讲到待定系数法，并结合阻尼振动等实际背景，全面梳理常考题型与技巧。

1. Introduction to Second Order ODEs | 二阶常微分方程简介

A second order ordinary differential equation involves an unknown function y(x) and its derivatives up to the second order. The general linear form with constant coefficients is a y” + b y’ + c y = f(x), where f(x) is the forcing function. If f(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous.

二阶常微分方程包含未知函数 y(x) 及其一阶和二阶导数。常系数线性一般形式为 a y” + b y’ + c y = f(x)，其中 f(x) 为驱动函数。若 f(x) = 0，方程为齐次，否则为非齐次。

The solution y = y_c + y_p consists of the complementary function y_c (general solution of the homogeneous equation) and a particular integral y_p (any solution of the full non-homogeneous equation). This superposition principle is fundamental.

解的结构为 y = y_c + y_p，其中 y_c 是齐次方程的通解（补函数），y_p 是非齐次方程的一个特解（特积分）。这一叠加原理是整个求解框架的基础。

Both the IB Analysis & Approaches HL and CIE Further Mathematics or Pure Mathematics 3 modules test the ability to classify and solve such equations. Typical exam instructions ask for the general solution or a specific solution subject to boundary conditions.

IB 分析与方法 HL 以及 CIE 进阶数学/纯数 3 均会考查方程的分类与求解。常见考题要求写出通解，或给定边界条件求特解。

2. Homogeneous Linear Equations with Constant Coefficients | 常系数齐次线性方程

A homogeneous equation has the form a y” + b y’ + c y = 0, where a, b, c are real constants and a ≠ 0. The solution depends entirely on the roots of the auxiliary equation (also called the characteristic equation). We seek solutions of the form y = e^{rx}.

齐次方程形如 a y” + b y’ + c y = 0，a, b, c 为实常数且 a ≠ 0。求解全依赖于辅助方程（特征方程）的根。我们设解具有形式 y = e^{rx}。

Substituting y = e^{rx} yields a r² e^{rx} + b r e^{rx} + c e^{rx} = 0, giving a r² + b r + c = 0. This characteristic equation can be derived quickly by replacing y” with r², y’ with r, and y with 1.

代入 y = e^{rx} 得到 a r² + b r + c = 0。可快速记忆：用 r² 替换 y”，r 替换 y’，y 替换为 1。

For IB and CIE exams, you are expected to write the auxiliary equation, solve it, and construct the complementary function y_c according to the nature of the roots.

IB 和 CIE 考试要求写出辅助方程，求解并根据根的性质构建补函数 y_c。

3. The Characteristic Equation | 特征方程

Given a y” + b y’ + c y = 0, the characteristic equation is a r² + b r + c = 0. Its discriminant Δ = b² – 4ac determines the form of the complementary function. Always write it in standard form before factoring or applying the quadratic formula.

给定 a y” + b y’ + c y = 0，特征方程为 a r² + b r + c = 0。判别式 Δ = b² – 4ac 决定了补函数的形式。务必将方程写为标准形式后再因式分解或使用求根公式。

Worked example: For 2y” – 5y’ + 2y = 0, the characteristic equation is 2r² – 5r + 2 = 0. Factoring gives (2r – 1)(r – 2) = 0, so r = 1/2, 2. The complementary function is y_c = A e^{x/2} + B e^{2x}.

例如：2y” – 5y’ + 2y = 0，特征方程为 2r² – 5r + 2 = 0，因式分解得 (2r – 1)(r – 2) = 0，根为 r = 1/2, 2。补函数为 y_c = A e^{x/2} + B e^{2x}。

a r² + b r + c = 0 → Δ = b² – 4ac

4. Real and Distinct Roots | 相异实根情况

When Δ > 0 the characteristic equation has two distinct real roots r₁ and r₂. The complementary function is y_c = A e^{r₁x} + B e^{r₂x}, where A and B are arbitrary constants.

当 Δ > 0 时，特征方程有两个相异实根 r₁ 和 r₂。补函数为 y_c = A e^{r₁x} + B e^{r₂x}，A、B 为任意常数。

This is the simplest case and appears frequently in mechanics problems where the system is overdamped. Always ensure the exponential arguments are correct: r₁ and r₂ may be negative, leading to decaying solutions.

这是最简单的情形，常见于力学中过阻尼系统。务必注意指数参数的正负：r₁、r₂ 可能为负，导致解衰减。

Example: y” – 3y’ – 4y = 0 → r² – 3r – 4 = 0 → (r – 4)(r + 1) = 0 → r = 4, -1. General solution: y = A e^{4x} + B e^{-x}.

例：y” – 3y’ – 4y = 0 → r² – 3r – 4 = 0 → (r – 4)(r + 1) = 0 → r = 4, -1。通解：y = A e^{4x} + B e^{-x}。

5. Repeated Roots | 重根情况

If Δ = 0, there is a single repeated root r = -b/(2a). The complementary function must include an extra factor of x to ensure linear independence: y_c = (A + B x) e^{rx}.

若 Δ = 0，特征方程有重根 r = -b/(2a)。为保证线性无关，补函数须包含因子 x：y_c = (A + B x) e^{rx}。

This case is often overlooked by students. CIE mark schemes explicitly require the form (A + Bx)e^{rx}. For instance, y” – 4y’ + 4y = 0 gives r² – 4r + 4 = 0 → (r – 2)² = 0, so r = 2. Solution: y = (A + Bx) e^{2x}.

学生常忽略此情况。CIE 评分标准明确要求 (A + Bx)e^{rx} 形式。例如 y” – 4y’ + 4y = 0，特征方程 r² – 4r + 4 = 0 → (r – 2)² = 0，解为 y = (A + Bx) e^{2x}。

A common mistake is writing y = A e^{rx} + B e^{rx}, which collapses to a single constant. Always multiply one term by x.

常见错误是写成 y = A e^{rx} + B e^{rx}，这等于一个常数。务必在其中一个项乘以 x。

6. Complex Roots | 复根情况

When Δ < 0, the roots are complex conjugates r = α ± iβ, where i² = -1. The complementary function is y_c = e^{αx} [A cos(βx) + B sin(βx)]. This expresses oscillatory behaviour in mechanical systems.

当 Δ < 0 时，特征根为一对共轭复根 r = α ± iβ。补函数为 y_c = e^{αx} [A cos(βx) + B sin(βx)]，描述了力学系统中的振动行为。

Extract α and β correctly: if r = -2 ± 3i, then α = -2, β = 3. The real part α governs exponential growth or decay, while β gives the angular frequency.

正确提取 α 和 β：若 r = -2 ± 3i，则 α = -2，β = 3。实部 α 决定指数增减，β 给出角频率。

r = α ± iβ → y_c = e^{αx} (A cos βx + B sin βx)

Example: y” + 4y’ + 13y = 0. Characteristic: r² + 4r + 13 = 0 → r = [-4 ± √(16 – 52)]/2 = -2 ± 3i. Thus y = e^{-2x} (A cos 3x + B sin 3x).

例：y” + 4y’ + 13y = 0，特征方程 r² + 4r + 13 = 0 → r = -2 ± 3i。故 y = e^{-2x} (A cos 3x + B sin 3x)。

7. Non-Homogeneous Equations: Method of Undetermined Coefficients | 非齐次方程：待定系数法

For a non-homogeneous equation a y” + b y’ + c y = f(x), the particular integral y_p is found by assuming a form similar to f(x) and its independent derivatives. The method of undetermined coefficients works when f(x) is a polynomial, exponential, sine, cosine, or sums/products of these.

对于非齐次方程 a y” + b y’ + c y = f(x)，特解 y_p 通过假设与 f(x) 及其独立导数形式相似的函数求得。待定系数法适用于 f(x) 为多项式、指数、正余弦或其组合的情形。

Key rule: the assumed trial y_p must not match any part of the complementary function. If overlap occurs, multiply the trial function by x (or x² if needed) to remove duplication.

关键规则：假设的试探 y_p 不得与补函数有任何共同项。若出现重叠，需将试探函数乘以 x（必要时乘 x²）以消除重复。

8. Particular Integral for Polynomial, Exponential, Trigonometric Functions | 多项式、指数、三角函数的特解形式

Here are standard trial forms for common f(x). For a polynomial of degree n, try a full polynomial of degree n: y_p = P_n(x). For an exponential f(x) = k e^{px}, try y_p = C e^{px} unless p is a root of the characteristic equation.

以下是常见 f(x) 的标准试探形式：若 f(x) 为 n 次多项式，设 y_p 为同次完整多项式 P_n(x)。若 f(x) = k e^{px}，设 y_p = C e^{px}，除非 p 是特征方程的根。

For trig functions f(x) = M cos ωx + N sin ωx, try y_p = D cos ωx + E sin ωx. If iω is a root of the characteristic equation, multiply by x: y_p = x(D cos ωx + E sin ωx).

对于三角函数 f(x) = M cos ωx + N sin ωx，设 y_p = D cos ωx + E sin ωx。若 iω 是特征根，则乘以 x：y_p = x(D cos ωx + E sin ωx)。

f(x)	Trial y_p	Condition
3x² + 2	Ax² + Bx + C	None
5e^{3x}	Ce^{3x}	If 3 not a root
2 sin 4x	D cos 4x + E sin 4x	If no resonance

When dealing with sums, simply add the individual trial forms. Always substitute y_p into the ODE to find the coefficients by equating like terms.

处理函数相加时，只需将各试探形式相加。代入原方程，比较系数求解未知常数值。

9. Finding General Solutions and Applying Initial Conditions | 求通解并应用初始条件

The general solution is y = y_c + y_p. After obtaining the constants for y_p, leave A and B in the complementary function undetermined until initial or boundary conditions are applied.

通解为 y = y_c + y_p。求出 y_p 的系数后，保留补函数中的 A 和 B，直到代入初始条件或边界条件。

Conditions typically give y(0) and y'(0). First differentiate the full general solution, then substitute x = 0 and equate. This yields simultaneous equations for A and B.

条件通常给出 y(0) 和 y'(0)。先对完整通解求导，再代入 x = 0 建立关于 A、B 的方程组。

Example: y” – 3y’ + 2y = 4, with y(0) = 3, y'(0) = 2. y_c = A e^x + B e^{2x}. y_p = 2 (since constant). y = A e^x + B e^{2x} + 2. Using conditions gives A + B + 2 = 3, A + 2B = 2. Solving: A = 2, B = -1. Particular solution: y = 2e^x – e^{2x} + 2.

例：y” – 3y’ + 2y = 4，y(0) = 3，y'(0) = 2。y_c = A e^x + B e^{2x}，y_p = 2。y = A e^x + B e^{2x} + 2。代入条件得 A + B + 2 = 3，A + 2B = 2。解得 A = 2, B = -1。特解为 y = 2e^x – e^{2x} + 2。

10. Applications: Simple Harmonic Motion and Damped Oscillations | 应用：简谐运动与阻尼振动

Many mechanics problems reduce to a second order linear ODE. Simple harmonic motion (SHM) follows y” + ω² y = 0, giving y = A cos ωt + B sin ωt. Damped oscillations arise from my” + ky’ + cy = 0, with damping coefficient k.

许多力学问题可归结为二阶线性常微分方程。简谐运动遵循 y” + ω² y = 0，通解为 y = A cos ωt + B sin ωt。阻尼振动来自 my” + ky’ + cy = 0，其中 k 为阻尼系数。

Characteristic equation for mx” + cx’ + kx = 0: m r² + c r + k = 0. The damping condition (under, over, critical) depends on Δ = c² – 4mk. This directly maps to the real/complex root cases above.

对于 mx” + cx’ + kx = 0，特征方程为 m r² + c r + k = 0。阻尼状态（欠阻尼、过阻尼、临界阻尼）取决于 Δ = c² – 4mk，直接对应前述实根、复根情形。

IB exam questions may ask for the equation of motion and the long-term behaviour. Recognise that a negative real part of the roots implies the amplitude decays over time.

IB 考题可能要求写出运动方程并分析长期行为。需认识到，若特征根实部为负，振幅将随时间衰减。

11. Common Exam Pitfalls and Tips | 常见考试陷阱与技巧

One frequent error is forgetting to use the full polynomial for particular integrals. If f(x) = x², the trial must be Ax² + Bx + C, never just Ax². Missing lower-order terms will give an incomplete solution.

常见错误之一是特解试探缺少完整多项式。若 f(x) = x²，试探形式必须为 Ax² + Bx + C，不能只写 Ax²，遗漏低次项会导致求解失败。

Another pitfall is misidentifying when to multiply by x. Only multiply if the trial term already appears in y_c. Carefully compare the roots or the form of y_c with the trial.

另一个陷阱是错误判断何时需乘 x。只有当试探项出现在 y_c 中时才乘 x。务必仔细比对特征根或补函数形式与试探解。

Sign errors in the characteristic equation are common. Always rearrange into standard form a y” + b y’ + c y = … before extracting coefficients. Check the discriminant carefully.

特征方程中的符号错误频发。务必先将方程化为标准形式 a y” + b y’ + c y = …，再提取系数。仔细检查判别式。

For CIE, showing the substitution step for y_p earns method marks even if the algebra is incorrect. Always write down the derivatives of y_p clearly.

在 CIE 考试中，写出 y_p 及其导数的代入步骤可获得方法分，即使后续代数运算出错。务必清晰列出 y_p 的一阶和二阶导数。

Finally, always verify your final solution by differentiating and substituting back into the original equation if time permits.

最后，若时间允许，始终通过求导并代回原方程来验证最终解。

12. Summary and Final Checklist | 总结与检查列表

Write the ODE in standard form a y” + b y’ + c y = f(x). / 将方程写为标准形式 a y” + b y’ + c y = f(x)。
Find the homogeneous solution by solving a r² + b r + c = 0. / 通过解 a r² + b r + c = 0 求齐次解。
Classify roots: real distinct, repeated, or complex. / 区分根的类型：相异实根、重根、复根。
Select the correct complementary function y_c with arbitrary constants. / 选择正确的补函数 y_c，保留任意常数。
Determine a suitable trial y_p based on f(x), multiplying by x if overlap. / 根据 f(x) 确定合适试探 y_p，若重叠则乘 x。
Substitute y_p to find its coefficients. / 代入 y_p 求其系数。
Write the general solution y = y_c + y_p. / 写出通解 y = y_c + y_p。
Apply initial conditions to find A and B. / 应用初始条件求 A 和 B。
Check damping/resonance in applied problems. / 在应用题中检查阻尼/共振状态。
Avoid common errors: incomplete polynomials, missing x factors, sign mistakes. / 避免常见错误：多项式不完整、遗漏乘 x 因子、符号错误。

By following this structured approach and practicing with past paper questions, students can secure full marks on the second order ODE component of the IB and CIE examinations.

遵循这一结构化方法并多做历年真题，学生必能在 IB 和 CIE 考试的二阶常微分方程部分稳拿高分。

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Second Order Differential Equations: IB & CIE Exam Focus | IB CIE 数学：二阶微分方程考点精讲