📚 Solving Second-Order Linear Homogeneous Differential Equations with Constant Coefficients | 二阶常系数齐次线性微分方程的解法
In IB Mathematics HL, solving differential equations is a core skill. This article focuses on second-order linear homogeneous differential equations with constant coefficients, a standard topic in Paper 3 or Calculus options.
在 IB 数学高阶水平中，解微分方程是核心技能。本文聚焦于二阶常系数齐次线性微分方程，这是 Paper 3 或微积分选项中的标准内容。
1. Standard Form | 标准形式
A second-order linear homogeneous differential equation with constant coefficients has the general form
二阶常系数齐次线性微分方程的一般形式为
a y″ + b y′ + c y = 0
where a, b, c are real constants and a ≠ 0. The term “homogeneous” means the right-hand side is zero.
其中 a, b, c 为实常数且 a ≠ 0。“齐次”指等号右侧为零。
For example, 2y″ − 3y′ + 5y = 0 is homogeneous, while 2y″ − 3y′ + 5y = eˣ is non-homogeneous.
例如，2y″ − 3y′ + 5y = 0 是齐次的，而 2y″ − 3y′ + 5y = eˣ 是非齐次的。
2. The Characteristic Equation | 特征方程
To solve the equation, we assume a solution of the form y = e^(rx), where r is a constant to be determined.
为求解该方程，我们假设解的形式为 y = e^(rx)，其中 r 是待定常数。
Substituting y = e^(rx), y′ = r e^(rx), y″ = r² e^(rx) into a y″ + b y′ + c y = 0 gives
将 y = e^(rx)、y′ = r e^(rx)、y″ = r² e^(rx) 代入 a y″ + b y′ + c y = 0，得到
a r² + b r + c = 0
This quadratic equation is called the characteristic (or auxiliary) equation. Its roots determine the general solution.
这个二次方程称为特征方程（或辅助方程）。它的根决定通解的形式。
3. Two Distinct Real Roots | 两个不同实根
If the characteristic equation has two distinct real roots r₁ and r₂, then the general solution is
若特征方程有两个不同的实根 r₁ 和 r₂，则通解为
y = C₁ e^(r₁x) + C₂ e^(r₂x)
This works because both e^(r₁x) and e^(r₂x) satisfy the differential equation, and the linear combination covers all possible solutions.
这是因为 e^(r₁x) 和 e^(r₂x) 都满足该微分方程，而线性组合涵盖所有可能的解。
For example, for y″ − 3y′ + 2y = 0, the characteristic equation r² − 3r + 2 = 0 gives r = 1 and r = 2, so y = C₁eˣ + C₂e^(2x).
例如，对于 y″ − 3y′ + 2y = 0，特征方程 r² − 3r + 2 = 0 给出 r = 1 和 r = 2，因此 y = C₁eˣ + C₂e^(2x)。
4. Repeated Real Root | 重根
If the characteristic equation has a repeated real root r, then the general solution is not simply C e^(rx) because that only gives one independent solution.
若特征方程有重实根 r，则通解不能简单地写成 C e^(rx)，因为那只能给出一个线性无关的解。
In this case, the second independent solution is x e^(rx), and the general solution is
此时，第二个线性无关的解为 x e^(rx)，通解为
y = (C₁ + C₂ x) e^(rx)
For instance, y″ − 4y′ + 4y = 0 has characteristic equation (r − 2)² = 0, so y = (C₁ + C₂x)e^(2x).
例如，y″ − 4y′ + 4y = 0 的特征方程为 (r − 2)² = 0，所以 y = (C₁ + C₂x)e^(2x)。
5. Complex Conjugate Roots | 共轭复根
If the discriminant b² − 4ac is negative, the roots are complex numbers of the form r = α ± iβ, where α = −b/(2a) and β = √(4ac − b²)/(2a).
若判别式 b² − 4ac 为负，则根为共轭复数 r = α ± iβ，其中 α = −b/(2a)，β = √(4ac − b²)/(2a)。
Using Euler’s formula, the general solution can be written in real form as
利用欧拉公式，通解可以写成实数形式：
y = e^(αx) (C₁ cos(βx) + C₂ sin(βx))
For example, y″ + 4y′ + 13y = 0 has roots r = −2 ± 3i, so y = e^(−2x)(C₁ cos 3x + C₂ sin 3x).
例如，y″ + 4y′ + 13y = 0 的根为 r = −2 ± 3i，所以 y = e^(−2x)(C₁ cos 3x + C₂ sin 3x)。
6. Initial Conditions and Particular Solutions | 初始条件与特解
The general solution contains two arbitrary constants C₁ and C₂. To find a particular solution, we need two initial conditions, typically y(x₀) and y′(x₀).
通解中包含两个任意常数 C₁ 和 C₂。为了求特解，需要两个初始条件，通常是 y(x₀) 和 y′(x₀)。
Substitute the initial values into the general solution and its derivative, then solve the resulting system of linear equations for C₁ and C₂.
将初始值代入通解及其导数，然后解出关于 C₁ 和 C₂ 的线性方程组。
For example, for y″ + y = 0 with y(0) = 2 and y′(0) = 1, the general solution is y = C₁ cos x + C₂ sin x. From y(0) = 2 we get C₁ = 2; from y′(0) = C₂ we get C₂ = 1. Thus y = 2 cos x + sin x.
例如，对于 y″ + y = 0，且 y(0) = 2、y′(0) = 1，通解为 y = C₁ cos x + C₂ sin x。由 y(0) = 2 得 C₁ = 2；由 y′(0) = C₂ 得 C₂ = 1。因此 y = 2 cos x + sin x。
7. Euler’s Formula and Complex Exponential Form | 欧拉公式与复指数形式
Euler’s formula states e^(iθ) = cos θ + i sin θ. It connects the complex exponential with trigonometric functions.
欧拉公式表明 e^(iθ) = cos θ + i sin θ。它将复指数与三角函数联系起来。
When the characteristic roots are α ± iβ, the complex solutions e^((α+iβ)x) and e^((α−iβ)x) can be combined into real solutions via linear combinations.
当特征根为 α ± iβ 时，复解 e^((α+iβ)x) 和 e^((α−iβ)x) 可以通过线性组合化为实解。
e^(αx) cos(βx) = (e^((α+iβ)x) + e^((α−iβ)x)) / 2
e^(αx) sin(βx) = (e^((α+iβ)x) − e^((α−iβ)x)) / (2i)
This explains why sin and cos appear in the real general solution.
这解释了为什么实数通解中会出现 sin 和 cos。
8. Linear Combination and Structure of the General Solution | 线性组合与通解结构
The set of all solutions of a homogeneous linear differential equation forms a vector space of dimension two. Any two linearly independent solutions form a fundamental set.
齐次线性微分方程的所有解构成一个二维向量空间。任意两个线性无关的解构成一个基本解组。
For distinct real roots, the fundamental set is {e^(r₁x), e^(r₂x)}; for a repeated root, it is {e^(rx), x e^(rx)}; for complex roots, it is {e^(αx) cos(βx), e^(αx) sin(βx)}.
对于不同实根，基本解组为 {e^(r₁x), e^(r₂x)}；对于重根，为 {e^(rx), x e^(rx)}；对于复根，为 {e^(αx) cos(βx), e^(αx) sin(βx)}。
The general solution is the linear combination of these two independent solutions with arbitrary constants.
通解就是这两个线性无关解的带有任意常数的线性组合。
9. Step-by-Step Solving Procedure | 解题步骤
Follow these steps to solve any second-order homogeneous equation with constant coefficients.
按照以下步骤求解任何二阶常系数齐次方程。
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Step 1: Write the equation in the form a y″ + b y′ + c y = 0.
第一步：将方程写成 a y″ + b y′ + c y = 0 的形式。
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Step 2: Form the characteristic equation a r² + b r + c = 0.
第二步：写出特征方程 a r² + b r + c = 0。
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Step 3: Solve for r using factorisation or the quadratic formula.
第三步：用因式分解或求根公式解出 r。
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Step 4: Based on the discriminant, choose the appropriate general solution form.
第四步：根据判别式，选择相应的通解形式。
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Step 5: If initial conditions are given, determine the constants.
第五步：若给定初始条件，确定常数。
Remember the sign of the discriminant: if b² − 4ac > 0, distinct real roots; if = 0, repeated root; if < 0, complex roots.
记住判别式的符号：若 b² − 4ac > 0，为不同实根；若 = 0，为重根；若 < 0，为复根。
10. Common IB Exam Questions and Typical Pitfalls | 常见IB题型与易错点
IB questions often ask you to solve an equation, find the constant given initial conditions, or interpret the behaviour of the solution as x → ∞.
IB 常考题型包括：解方程、根据初始条件求常数，或分析当 x → ∞ 时解的行为。
Common mistakes include writing the wrong characteristic equation, forgetting the factor x for a repeated root, or incorrectly simplifying complex roots.
常见错误包括：写错特征方程、在重根时漏掉因子 x，或错误化简复根。
Another pitfall: when the coefficient a is not 1, divide the whole equation by a first if helpful, but keep the characteristic equation consistent with the original coefficients.
另一个易错点：当系数 a 不为 1 时，可以先将整个方程除以 a，但特征方程必须与原系数一致。
Always check the discriminant before writing the final answer.
在写出最终答案前，务必检查判别式。
11. Worked Example | 完整例题
Solve the initial value problem: y″ − 6y′ + 25y = 0, y(0) = 1, y′(0) = 5.
求解初值问题：y″ − 6y′ + 25y = 0，y(0) = 1，y′(0) = 5。
The characteristic equation is r² − 6r + 25 = 0. Applying the quadratic formula,
特征方程为 r² − 6r + 25 = 0。应用求根公式，
r = (6 ± √(36 − 100)) / 2 = (6 ± √(−64)) / 2 = 3 ± 4i
Thus α = 3 and β = 4, so the general solution is
因此 α = 3，β = 4，所以通解为
y = e^(3x)(C₁ cos 4x + C₂ sin 4x)
Using y(0) = 1 gives C₁ = 1. Differentiating:
由 y(0) = 1 得 C₁ = 1。求导：
y′ = e^(3x)(3(C₁ cos 4x + C₂ sin 4x) + (−4C₁ sin 4x + 4C₂ cos 4x))
At x = 0, y′(0) = 3C₁ + 4C₂ = 5. Since C₁ = 1, 4C₂ = 2, so C₂ = 1/2. The particular solution is
在 x = 0 处，y′(0) = 3C₁ + 4C₂ = 5。由于 C₁ = 1，得 4C₂ = 2，所以 C₂ = 1/2。特解为
y = e^(3x)(cos 4x + ½ sin 4x)
12. Connection to IB Assessment Objectives | 与IB评估目标的联系
In IB exams, you are expected to demonstrate the ability to choose the correct method, justify the form of the solution, and interpret the result in contexts such as simple harmonic motion.
在 IB 考试中，你需要展示选择正确方法、证明解形式的合理性，并在简谐运动等情境中解释结果的能力。
This topic extends your knowledge of differentiation and integration, and it directly supports the study of systems of differential equations and real-world modelling.
该主题扩展了微积分知识，并直接支持微分方程组和实际建模的学习。
Mastering this technique also prepares you for solving non-homogeneous equations using particular integrals, which often appears in HL Paper 3.
掌握这一技巧也为使用特积分求解非齐次方程做好准备，这常出现在 HL Paper 3 中。
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