Second-Order Differential Equations for AS Mathematics | AS 数学：二阶微分方程考点精讲

📚 Second-Order Differential Equations for AS Mathematics | AS 数学：二阶微分方程考点精讲

Mastering second-order differential equations is a crucial step in AS Mathematics, enabling you to solve problems involving oscillations, forces, and rates of change. This revision guide covers the essential techniques: solving homogeneous equations via the characteristic equation, classifying the three types of roots, forming the complementary function, and handling inhomogeneous cases with the method of undetermined coefficients, and applying boundary or initial conditions to obtain a particular solution. Every method is presented in the order you should apply it in an examination.

掌握二阶微分方程是 AS 数学的关键一步，使你能够解决涉及振动、力和变化率的问题。本复习指南涵盖必备技巧：通过特征方程求解齐次方程、判别三类根、构建补函数、用待定系数法处理非齐次情形，以及应用边界条件或初始条件求出特解。所有方法都按照考试中应当使用的顺序呈现。

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1. What Are Second-Order Differential Equations? | 什么是二阶微分方程？

A second-order differential equation is an equation that relates a function y(x) to its first derivative dy/dx and its second derivative d²y/dx². The word 'second-order' simply means that the highest derivative appearing in the equation is the second derivative. These equations appear naturally in physics and mechanics: acceleration is the second derivative of displacement, so Newton's second law applied to a moving particle, a mass on a spring, or a pendulum produces a second-order ODE. They also describe rates of change in population and cooling models once the rate itself changes.

二阶微分方程是把函数 y(x) 与它的一阶导数 dy/dx 及二阶导数 d²y/dx² 联系起来的方程。"二阶"就是说方程中出现的最高阶导数是二阶导数。这类方程在物理和力学中自然出现：加速度是位移的二阶导数，所以把牛顿第二定律用于运动质点、弹簧振子或单摆，就会得到二阶常微分方程。当变化率本身还在变化时，人口模型和冷却模型也会用到它。

In AS Mathematics we do not study every possible type. We restrict attention to linear equations with constant coefficients, because these have a completely systematic solution method based on a quadratic equation. This restriction is what makes the topic so examinable: once you can classify the roots of one quadratic, you can write down the whole family of solutions.

在 AS 数学中，我们并不研究所有可能的类型，而是只研究常系数线性方程，因为这类方程有一套完全系统的解法，核心只是一个二次方程。正是这个限制使本考点特别适合考试：只要能判别一个二次方程的根，你就能写出整个解族。

a d²y/dx² + b dy/dx + c y = f(x)

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2. The Standard Form of a Linear Constant-Coefficient ODE | 常系数线性常微分方程的标准形式

constants with a ≠ 0. The function f(x) on the right-hand side is called the forcing function, because in applications it represents an external driving force or input. Keeping the equation in this layout matters: the coefficients a, b, c are read off directly to build the characteristic equation, so a sign slip here ruins every later step.

我们总是先把方程写成标准形式 a d²y/dx² + b dy/dx + c y = f(x)，其中 a、b、c 是实数常数且 a ≠ 0。右端的函数 f(x) 称为强迫函数，因为在应用中它代表外部驱动力或输入。保持这种排列很重要：系数 a、b、c 会被直接读出以构造特征方程，所以这一步的符号错误会毁掉后面所有步骤。

us; if f(x) is not identically zero, the equation is inhomogeneous (also called non-homogeneous). A second classification concerns the coefficients: if a, b and c are constants the equation has constant coefficients, but if any of them depends on x the equation has variable coefficients and lies beyond AS level. If the leading coefficient a is not 1, it is often helpful to divide the whole equation by a first, so that the characteristic equation is monic.

若 f(x) = 0，方程是齐次的；若 f(x) 不恒为零，方程是非齐次的。第二种分类针对系数：若 a、b、c 都是常数，则方程为常系数；若其中任何一个依赖于 x，则为变系数方程，超出 AS 范围。若首项系数 a 不是 1，通常先把整个方程除以 a，使特征方程成为首一形式。

Homogeneous: a y″ + b y′ + c y = 0   |   Inhomogeneous: a y″ + b y′ + c y = f(x)

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3. The Structure of the General Solution: CF + PI | 通解的结构：补函数加特解

which is the general solution of the corresponding homogeneous equation a y″ + b y′ + c y = 0. The second is a particular integral (PI), which is any single function that satisfies the full inhomogeneous equation. The general solution is their sum.

非齐次线性方程的通解总是分成两个互相独立的部分。第一部分是补函数（CF），即对应齐次方程 a y″ + b y′ + c y = 0 的通解；第二部分是特解（PI），即满足完整非齐次方程的任意一个函数。通解就是二者之和。

y = y_c + y_p

Two facts explain why this structure works. First, a second-order equation needs exactly two arbitrary constants in its general solution, and all of them come from the CF, because the PI contains no arbitrary constants. Second, if y_c solves the homogeneous equation and y_p solves the inhomogeneous one, then substituting y = y_c + y_p id solution. Understanding this structure is the foundation for every method that follows.

有两点可以解释这个结构为何成立。第一，二阶方程的通解必须恰好包含两个任意常数，而它们全部来自 CF，因为 PI 不含任意常数。第二，若 y_c 满足齐次方程、y_p 满足非齐次方程，则代入 y = y_c + y_p 得到 f(x) + 0 = f(x)，因此这个和确实是解。理解这一结构是所有后续方法的基础。

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4. The Characteristic Equation | 特征方程

To find the CF we look for solutions of the homogeneous equation of the form y = e^mx, where m is a constant still to be determined. Differentiating gives y′ = m e^mx and y″ = m² e^mx. Substituting these into a y″ + b y′ + c y = 0 produces a m² e^mx + b m e^mx + c e^mx = 0. Since e^mx is never zero, we may divide through by it and obtain a quadratic equation in m.

求 CF 时，我们先假设齐次方程有形如 y = e^mx 的解，其中 m 是待定常数。求导得 y′ = m e^mx，y″ = m² e^mx。把它们代入 a y″ + b y′ + c y = 0，得到 a m² e^mx + b m e^mx + c e^mx = 0。由于 e^mx 永远不为零，可以两边约去它，从而得到关于 m 的二次方程。

a m² + b m + c = 0

s applies: distinct real roots when Δ > 0, one repeated real root when Δ = 0, and a complex conjugate pair when Δ < 0. Classifying the roots first is the single most important habit in this topic, because the shape of the CF depends entirely on that classification.

这就是特征方程，也叫辅助方程。它的判别式 Δ = b² − 4ac 立刻告诉你属于三种情形中的哪一种：Δ > 0 时有两个不同实根；Δ = 0 时有一个重实根；Δ < 0 时为一对共轭复根。先对根分类是本考点最重要的习惯，因为 CF 的形式完全取决于这个分类。

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5. Case 1: Distinct Real Roots | 情形一：两个不同实根

If the characteristic equation has two distinct real roots m₁ and m₂, then both e^m₁x and e^m₂x are solutions of the homogeneous equation, and they are independent of each other. The complementary function is therefore a linear combination of them, with two arbitrary constants A and B.

若特征方程有两个不同实根 m₁ 和 m₂，则 e^m₁x 与 e^m₂x 都是齐次方程的解，而且彼此线性无关。因此补函数是它们的线性组合，含两个任意常数 A 和 B。

y_c = A e^m₁x + B e^m₂x

Notice the three typical exam situations that produce this case. When both roots are positive, both exponential terms grow. When both are negative, both terms decay to zero as x increases, which models a damping or cooling process. When one root is positive and the other negative, the solution contains both a growing and a decaying term, and the initial conditions decide which one dominates at small x. Never write the sum with the same constant in both terms: the two constants are independent and must be kept separate.

注意产生本情形的三种典型考题。两根都为正时，两个指数项都增长；两根都为负时，两项随 x 增大都衰减到零，可用来描述阻尼或冷却过程；一根正、一根负时，解中既有增长项也有衰减项，初始条件决定在小 x 处哪一项占主导。千万不要在两个项里写同一个常数：这两个常数互相独立，必须分开。

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6. Case 2: A Repeated Real Root | 情形二：一个重实根

o, the characteristic equation has a single repeated root m = −b / (2a). One solution is e^mx, but a second-order equation needs two independent solutions, and simply writing A e^mx + B e^mx collapses into one term. The standard remedy is to multiply the second copy by x, which produces a genuinely independent second solution.

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若判别式为零，特征方程只有一个重根 m = −b / (2a)。一个解是 e^mx，但二阶方程需要两个无关解，而写成 A e^mx + B e^mx 会合并成一项。标准做法是把第二个解乘以 x，从而得到一个真正独立的第二解。

y_c = (A + Bx) e^mx

You can verify that x e^mx conditions to find A and B. Remember that a repeated root can be positive, negative or zero; if m = 0 the CF becomes simply A + Bx, a straight line, which is a useful sanity check when the original equation has no y′ term and no y term left in the homogeneous part.

你可以对 x e^mx 求两次导再代入来验证它确实是解，但考试中直接引用标准形式，然后用给定条件求 A 和 B 即可。注意重根可正、可负、也可为零；若 m = 0，CF 就变成 A + Bx，即一条直线，当齐次部分既不剩 y′ 项也不剩 y 项时，这是很好的检验。

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7. Case 3: Complex Conjugate Roots | 情形三：一对共轭复根

discriminant is negative, the roots take the form m = α ± iβ, where α = −b / (2a) and β = √4ac − b² / (2a) with β > 0. Using Euler's relation, the exponential of a complex argument can be written in terms of sine and cosine, which removes all imaginary numbers from the final answer and produces the most important CF in applied mathematics.

若判别式为负，两根形如 m = α ± iβ，其中 α = −b / (2a)，β = √4ac − b² / (2a) 且 β > 0。利用欧拉关系，复指数可以写成正弦与余弦的组合，从而让最终答案中不出现虚数，并得到应用数学中最重要的 CF。

y_c = e^αx (A cos βx + B sin βx)

This case models oscillation. The factor e^αx s simple harmonic, with period 2π / β. Two exam traps live here. First, if the roots are purely imaginary, α = 0, so the exponential factor disappears and must not be written as e^0x in the final answer. Second, β must be taken as positive; writing −β merely swaps the roles of A and B and is not wrong, but it can confuse the marking.

本情形描述振动。因子 e^αx 控制振幅：α < 0 时振动衰减（阻尼运动）；α > 0 时振幅增长；α = 0 时振幅恒定，为简谐运动，周期为 2π / β。这里有两个考试陷阱。第一，若根为纯虚数，则 α = 0，指数因子消失，最终答案中不能写成 e^0x。第二，β 应取正值；写成 −β 只是交换了 A 与 B 的角色，本身不算错，但容易让阅卷产生困惑。

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8. Finding the Particular Integral | 求特解（PI）

Once the CF is known, we look for a particular integral, that is, any one function that satisfies the full inhomogeneous equation. The method of undetermined coefficients says: try a function of the same general shape as f(x), substitute it into the equation, and match coefficients to determine the unknown constants. The table below lists the standard trials for AS level.

求出 CF 之后，我们要找一个特解，即满足完整非齐次方程的任意一个函数。待定系数法告诉我们：先试一个与 f(x) 形状相同的函数，代入方程，再比较系数以确定未知常数。下表列出 AS 阶段的标准试解形式。

f(x)	Trial PI	中文说明
Polynomial of degree n, e.g. 4x + 1	px + q (full polynomial of degree n)	多项式：完整写出同次多项式，不能省略常数项
ekx	p ekx	指数型：试 p ekx
sin kx or cos kx	p cos kx + q sin kx	三角型：正余弦必须成对出现
Sum of the above	Sum of the corresponding trials	叠加原理：把各项试解相加

After substituting, compare the coefficient of each independent function on both sides, for example the coefficients of x, of the constant term, or of cos kx and sin kx. This yields simultaneous equations for p, q and any other unknowns. Always add the CF back at the end: the PI alone is never the general solution.

代入后比较两端各个独立函数的系数，例如 x 的系数、常数项，或者 cos kx 与 sin kx 的系数，从而得到关于 p、q 等待定常数的联立方程。最后一定要把 CF 加回去：单独的 PI 永远不是通解。

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9. When the Trial PI Fails: Resonance | 试解失效：共振与重复情形

There is one situation in which the standard trial does not work. If the function you try is already part of the complementary function, substituting it can never balance f(x), because the left-hand side collapses to zero. In that case multiply the trial by x, and substitute again. If it still fails, which happens with a repeated root, multiply by x² instead.

有一种情况标准试解会失效。如果你试的函数已经包含在补函数中，代入后左端会化为零，永远无法与 f(x) 平衡。此时应把试解乘以 x 再代入；若仍然失效（这在重根时出现），就改为乘以 x²。

tes the CF: multiply by x &nbsp;&nbsp;→&nbsp;&nbsp; if it still duplicates: multiply by x²

equency of the system, and the amplitude grows without bound; engineers call this resonance. For example, the equation d²y/dx² + 4y = sin 2x has CF A cos 2x + B sin 2x, so sin 2x is already present and the correct trial is x(p cos 2x + q sin 2x). Failing to apply this rule is one of the most common ways to lose marks.

这条规则有物理含义。当 f(x) 本身是齐次方程的解时，驱动力正好以系统的固有频率振动，振幅会无限增大，工程上称为共振。例如方程 d²y/dx² + 4y = sin 2x 的 CF 是 A cos 2x + B sin 2x，sin 2x 已经出现，因此正确的试解是 x(p cos 2x + q sin 2x)。不应用这条规则是最常见的失分点之一。

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10. Applying Boundary and Initial Conditions | 应用边界条件与初始条件

as initial conditions, such as y = 3 and dy/dx = 0 when x = 0, or as boundary conditions, such as y = 0 when x = 0 and y = 1 when x = 2. Always substitute after the CF and PI have been combined, never before.

通解含两个任意常数，因此需要两个额外信息才能确定唯一的解。它们可以是初始条件，例如 x = 0 时 y = 3 且 dy/dx = 0；也可以是边界条件，例如 x = 0 时 y = 0、x = 2 时 y = 1。务必在 CF 与 PI 合并之后再代入，而不是在之前。

The procedure is mechanical but unforgiving. First differentiate the general solution to obtain dy/dx. Then substitute the first condition into y and the second into either y or dy/dx, according to what is given, producing two simultaneous linear equations in A and B. Solve them by elimination or substitution, then quote the final solution with the constants replaced by their numerical values. A quick check: put the values back into your final answer and confirm both conditions are satisfied.

步骤机械但不容出错。先对通解求导得到 dy/dx；再把第一个条件代入 y，第二个条件按题目给定代入 y 或 dy/dx，得到关于 A 与 B 的两个线性联立方程；用消元或代入法求解，最后写出用数值替换常数后的最终解。快速检验：把数值代回最终答案，确认两个条件都满足。

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11. A Worked Exam-Style Example | 典型例题精讲

f(x) = 4x, so the first step is to write down the characteristic equation m² − 3m + 2 = 0.

求解 d²y/dx² − 3 dy/dx + 2y = 4x，已知 x = 0 时 y = 0 且 dy/dx = 0。方程已是标准形式，a = 1、b = −3、c = 2、f(x) = 4x，因此第一步写出特征方程 m² − 3m + 2 = 0。

= 2. These are distinct real roots, so case 1 applies and the complementary function is y_c = A e^x + B e^2x. Because f(x) = 4x is linear, we try a full linear PI of the form y_p = px + q.

因式分解得 (m − 1)(m − 2) = 0，所以 m₁ = 1、m₂ = 2。这是两个不同实根，属于情形一，补函数为 y_c = A e^x + B e^2x。由于 f(x) = 4x 是一次式，我们试完整的线性特解 y_p = px + q。

Differentiating gives y_p′ = p and y_p nstant terms gives 2q − 3p = 0, so 2q = 6 and q = 3. The particular integral is therefore y_p = 2x + 3, and the general solution is y = A e^x + B e^2x + 2x + 3.

求导得 y_p′ = p、y_p″ = 0。代入方程得 0 − 3p + 2(px + q) = 4x，即 2px + (2q − 3p) = 4x。比较 x 的系数得 2p = 4，故 p = 2；比较常数项得 2q − 3p = 0，故 2q = 6、q = 3。因此特解为 y_p = 2x + 3，通

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