Solving Second-Order Linear Unforced Equations | 二阶线性无强迫微分方程解法

📚 Solving Second-Order Linear Unforced Equations | 二阶线性无强迫微分方程解法

Second-order linear differential equations lie at the heart of modelling oscillatory systems, from mass-spring dampers to RLC circuits. This article concentrates on unforced (homogeneous) equations, where no external driving term appears on the right-hand side. Mastering the technique of solving such equations via the characteristic equation is essential for IB Mathematics: Analysis and Approaches (HL) and further work in physics and engineering.

二阶线性微分方程是描述振动系统（从弹簧振子到 RLC 电路）的核心工具。本文专注于无外力（齐次）方程，即方程右端没有外部驱动项。通过特征方程求解此类方程是 IB 数学分析与方法 (HL) 的重要技能，也是进一步学习物理和工程的基础。

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1. Introduction to Second-Order Linear ODEs | 二阶线性常微分方程导论

A second-order linear ordinary differential equation (ODE) has the general form a(x) y'' + b(x) y' estrict our attention to constant-coefficient equations where a, b, c are real constants and a ≠ 0. The term 'linear' able y and its derivatives appear only to the first power and are not multiplied together.

二阶线性常微分方程的一般形式为 a(x) y'' + b(x) y' + c(x) y = f(x)。在 IB 高阶课程中，我们仅研究常系数方程，即 a、b、c 为实常数且 a ≠ 0。“线性”是指因变量 y 及其导数只以一次幂出现，并且不会彼此相乘。

When the forcing term f(x) is identically zero, the equation is called unforced or homogeneous. The solution strategy relies on exploiting the fact that exponential functions are eigenfunctions of the derivative operator, leading to an algebraic auxiliary equation.

当强迫项 f(x) 恒为零时，方程称为无外力方程或齐次方程。求解策略基于指数函数是导数算子的特征函数这一事实，从而导出一个代数辅助方程。

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2. The Concept of Unforced (Homogeneous) Equations | 无外力（齐次）方程的概念

An unforced second-order linear equation is written as a y'' + b y' ut means the system evolves solely from its initial conditions. Any linear combination of two independent solutions is again a solution – this is the principle of superposition, which only holds for homogeneous linear ODEs.

无外力二阶线性方程写作 a y'' + b y' + c y = 0。没有外部输入意味着系统的演化完全由初始条件决定。两个独立解的任意线性组合仍然是解——这就是叠加原理，它只对齐次线性常微分方程成立。

The general solution will contain two arbitrary constants, reflecting the two degrees of freedom of a second-order equation. To determine these constants uniquely, two pieces of information, typically initial values y(0) and y'(0), are required.

通解将包含两个任意常数，这反映了二阶方程的两个自由度。为了唯一确定这些常数，需要两个条件，通常是初值 y(0) 和 y'(0)。

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3. Standard Forms and Classification | 标准形式和分类

Without loss of generality we can divide through by a to obtain the normalised form y'' + p y' aviour of solutions is entirely determined by the nature of the roots of the associated characteristic equation.

不失一般性，我们可以将方程除以 a 得到标准形式 y'' + p y' + q y = 0，其中 p = b/a 且 q = c/a。解的行为完全由对应特征方程的根的性质决定。

d form). These are: two distinct real roots, one repeated real root, and a pair of complex conjugate roots.

根据判别式 Δ = p² − 4q（或等价地使用非标准化形式中的 Δ = b² − 4ac），我们将解分为三种互斥的情形：两个相异实根、一个重实根和一对共轭复根。

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4. The Characteristic Equation | 特征方程

We seek solutions of the form y = e^rx. Substituting y' = r e^rx and y'' = r² e^rx into a y'' + b y' + c y = 0 and cancelling the common factor e^rx (which is never zero) yields the characteristic equation:

我们寻找形如 y = e^rx 的解。将 y' = r e^rx 和 y'' = r² e^rx 代入 a y'' + b y' + c y = 0，并消去非零公因子 e^rx，得到特征方程：

a r² + b r + c = 0

ion. The discriminant Δ = b² − 4ac acts as the classifier for the three solution families.

这是一个关于 r 的二次方程。它的根 r₁ 和 r₂ 决定了通解的结构。判别式 Δ = b² − 4ac 充当三种解族的分类器。

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5. Case 1: Distinct Real Roots | 情形1：相异实根

0, the characteristic equation has two distinct real roots r₁ and r₂. The functions y₁ = e^r₁x and y₂ = e^r₂x are linearly independent, so the general solution of the homogeneous equation is their linear combination:

当 Δ > 0 时，特征方程有两个相异实根 r₁ 和 r₂。函数 y₁ = e^r₁x 和 y₂ = e^r₂x 线性无关，因此齐次方程的通解为它们的线性组合：

y = A e^r₁x + B e^r₂x

Because r₁ and r₂ are real, the solutions exhibit purely exponential growth or decay (or a combination of both) without oscillation. This case often arises in overdamped mechanical systems or in first-order chemical kinetics combined with a second integrating factor.

由于 r₁ 和 r₂ 均为实数，解呈现纯粹的指数增长或衰减（或两者的组合），没有振荡。这种情形常出现在过阻尼机械系统或与二阶积分因子相结合的一级化学动力学中。

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6. Case 2: Repeated Real Roots | 情形2：重实根

the characteristic equation has a single repeated root r = −b/(2a). One solution is y₁ = e^rx. A second, linearly independent solution is obtained by multiplying by x, giving y₂ = x e^rx. The general solution therefore is:

当 Δ = 0 时，特征方程有一个重根 r = −b/(2a)。一个解为 y₁ = e^rx，而另一个线性无关解可通过乘 x 得到，即 y₂ = x e^rx。因此通解为：

y = (A + B x) e^rx

The presence of the term x e^rx means that the solution can exhibit a subtle interplay between polynomial and exponential behaviour, characteristic of critically damped systems where the system returns to equilibrium as quickly as possible without oscillating.

x e^rx 项的出现意味着解可以在多项式与指数行为之间展现出微妙的相互作用，这是临界阻尼系统的特征——系统在不振荡的前提下以尽可能快的速度返回平衡位置。

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

When Δ < 0, the roots are a complex conjugate pair: r = α ± iβ, where α = −b/(2a) and β = √4ac − b²/(2a). Euler's formula e^iθ = cos θ + i sin θ allows us to express the real-valued general solution as:

当 Δ < 0 时，根为一对共轭复数：r = α ± iβ，其中 α = −b/(2a)，β = √4ac − b²/(2a)。利用欧拉公式 e^iθ = cos θ + i sin θ，可以将实数值通解表示为：

y = e^αx (C cos(βx) + D sin(βx))

The exponential factor e^αx equency β. If α = 0, the solution is purely sinusoidal, describing simple harmonic motion.

指数因子 e^αx 控制振幅的增长（α > 0）或衰减（α < 0），而三角函数部分以角频率 β 产生振荡行为。若 α = 0，解为纯正弦曲线，描述简谐运动。

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8. Working Examples – Case 1 | 实例解析 – 情形1

Consider the equation y'' − 5y' es as (r − 2)(r − 3) = 0. The roots are r₁ = 2, r₂ = 3, so the general solution is y = A e^2x + B e^3x.

考虑方程 y'' − 5y' + 6y = 0。特征方程为 r² − 5r + 6 = 0，可分解为 (r − 2)(r − 3) = 0。根为 r₁ = 2、r₂ = 3，因此通解为 y = A e^2x + B e^3x。

If initial conditions are y(0) = 4 and y'(0) = 9, we substitute to find A and B. From y(0) = A + B = 4, and differentiating gives y' = 2A e^2x + 3B e^3x so y' A + 3B = 9. Solving the system yields A = 3, B = 1, giving the particular solution y = 3 e^2x + e^3x.

若初值条件为 y(0) = 4 和 y'(0) = 9，我们代入求 A 和 B。由 y(0) = A + B = 4，求导得 y' = 2A e^2x + 3B e^3x 故 y'(0) = 2A + 3B = 9。解此方程组得 A = 3、B = 1，特解为 y = 3 e^2x + e^3x。

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9. Working Examples – Case 2 | 实例解析 – 情形2

Solve y'' − 6y' = 36 − 36 = 0, giving a repeated root r = 3. Hence the general solution is y = (A + B x) e^3x.

求解 y'' − 6y' + 9y = 0。特征方程 r² − 6r + 9 = 0 的判别式 Δ = 36 − 36 = 0，求得重根 r = 3。因此通解为 y = (A + B x) e^3x。

With initial values y(0) = 2 and y'(0) = 1, we compute y(0) = A = 2. Differentiating: y' = B e^3x + 3(A + B x) e^3x, so y'(0) = B + 3A = 1. Substituting A = 2 gives B + 6 = 1 ⇒ B = −5. The particular solution is y = (2 − 5x) e^3x.

给定初值 y(0) = 2、y'(0) = 1，计算得 y(0) = A = 2。求导：y' = B e^3x + 3(A + B x) e^3x，故 y'(0) = B + 3A = 1。代入 A = 2 得 B + 6 = 1 ⇒ B = −5。特解为 y = (2 − 5x) e^3x。

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10. Working Examples – Case 3 | 实例解析 – 情形3

Solve y'' + 4y' + 2)² + 9 = 0, so r = −2 ± 3i. Here α = −2 and β = 3, yielding the general solution y = e^−2x (C cos(3x) + D sin(3x)).

求解 y'' + 4y' + 13y = 0。特征方程为 r² + 4r + 13 = 0。配平方得 (r + 2)² + 9 = 0，故 r = −2 ± 3i。此处 α = −2、β = 3，通解为 y = e^−2x (C cos(3x) + D sin(3x))。

If y(0) = 1 and y'(0) = 1, we immediately have C = 1. Differentiate: y' = e^−2x(−2C cos3x − 2D sin3x − 3C sin3x + 3D cos3x). Evaluating at x = 0 gives y'(0) = −2C + 3D = 1 ⇒ 3D = 3 ⇒ D = 1. Thus the particular solution is y = e^−2x (cos(3x) + sin(3x)).

若 y(0) = 1 且 y'(0) = 1，立刻得 C = 1。求导：y' = e^−2x(−2C cos3x − 2D sin3x − 3C sin3x + 3D cos3x)。在 x = 0 处计算得 y'(0) = −2C + 3D = 1 ⇒ 3D = 3 ⇒ D = 1。因此特解为 y = e^−2x (cos(3x) + sin(3x))。

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11. Initial Value Problems | 初值问题

Initial value problems (IVPs) for second-order linear homogeneous ODEs require specifying y(x₀) and y'(x₀). These conditions determine the two arbitrary constants uniquely. In the IB examination, x₀ is almost always zero, but the method extends seamlessly to any starting point.

二阶线性齐次常微分方程的初值问题要求指定 y(x₀) 和 y'(x₀)。这些条件可唯一确定两个任意常数。在 IB 考试中，x₀ 几乎总是零，但该方法可无缝推广至任意起始点。

The procedure is straightforward: first write down the general solution in its appropriate form (distinct real, repeated, or complex). Then differentiate to obtain the velocity-like expression y'(x). Substitute the initial values to form a system of two linear equations in the two constants, and solve.

求解步骤非常直接：首先根据情形写出适当形式的通解（相异实根、重根或复根）；然后求导得到类似速度的表达式 y'(x)；代入初值建立关于两个常数的线性方程组并求解。

A useful check is to verify the dimensions or order of magnitude of the constants. In oscillatory cases, the coefficients C and D relate directly to the initial amplitude and phase, and can be rewritten in the form R e^αx cos(βx − δ) if required.

一个有用的检验是校核常数的量级或大小。在振荡情形中，系数 C 和 D 直接与初始振幅和相位相关，必要时可改写为 R e^αx cos(βx − δ) 的形式。

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12. Summary and Key Points | 总结与考点

The solution of a y'' + b y' three possible scenarios and their general solutions. Recognising which case applies and writing the correct trial solution is the central skill tested in IB Mathematics HL.

求解 a y'' + b y' + c y = 0 完全取决于特征方程 a r² + b r + c = 0 的根。下表总结了三种可能出现的情况及其通解。识别适用哪种情形并写出正确的试探解是 IB 数学 HL 考查的核心技能。

Case / 情形	Discriminant Δ / 判别式	Roots / 根	General Solution / 通解
1. Distinct real / 相异实根	Δ > 0	r₁, r₂	y = A er₁x + B er₂x
2. Repeated real / 重实根	Δ = 0	r = −b/(2a)	y = (A + B x) erx
3. Complex conjugate / 共轭复根	Δ < 0	α ± iβ	y = eαx (C cos βx + D sin βx)

Memorising these three forms is essential, but more important is understanding why they arise from the superposition of fundamental solutions. In an exam setting, always check the discriminant first, then write the general solution, and finally apply any given initial conditions to fix the constants.

记忆这三种形式至关重要，但更重要的理解它们为何由基本解的叠加产生。在考试中，务必先计算判别式，再写出通解，最后应用给定的初值条件确定常数。

The unforced equation is the gateway to understanding forced oscillations and resonance, which appear in further topics. A confident command of the homogeneous case will make advanced applications much more accessible.

无外力方程是理解受迫振动和共振的入口，这些内容将在进阶主题中出现。对齐次情形充满信心的掌握将使高阶应用变得容易得多。

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