📚 Second Order Linear Differential Equations with Variable Coefficients | 变系数二阶线性微分方程
In A-level mathematics, you first meet second order linear differential equations with constant coefficients. However, many physical systems involve coefficients that change with position or time. This article extends your toolkit to variable coefficient equations, focusing on the Cauchy–Euler equation and the powerful methods of reduction of order and variation of parameters.
在 A-Level 数学中，你首先遇到的是常系数二阶线性微分方程。然而，许多物理系统中的系数会随位置或时间变化。本文将你的工具箱扩展到变系数方程，重点讨论柯西–欧拉方程，以及降阶法与参数变易法这两个强力方法。
1. The General Form and Linear Structure | 一般形式与线性结构
A second order linear differential equation with variable coefficients can be written in the standard form
变系数二阶线性微分方程可以写成标准形式
y″ + p(x)y′ + q(x)y = r(x)
where y′ and y″ denote the first and second derivatives of y with respect to x, and p, q, r are functions of x. If r(x) = 0, the equation is called homogeneous; otherwise it is non‑homogeneous.
其中 y′ 与 y″ 分别表示 y 对 x 的一阶与二阶导数，p、q、r 是 x 的函数。若 r(x) = 0，则方程称为齐次的；否则称为非齐次的。
The linearity of the equation means that if y₁ and y₂ are any two solutions of the homogeneous equation, then any linear combination c₁y₁ + c₂y₂ is also a solution. This structure is the foundation of all solution methods.
方程的线性意味着：若 y₁ 和 y₂ 是齐次方程的任意两个解，则任意的线性组合 c₁y₁ + c₂y₂ 仍是解。这一结构是所有求解方法的基础。
2. Existence and Uniqueness | 存在性与唯一性
For initial value problems, the standard theorem states that if p(x), q(x) and r(x) are continuous on an open interval containing x₀, and y(x₀) = α, y′(x₀) = β are given, then a unique solution exists on that interval.
对于初值问题，标准定理指出：若 p(x)、q(x) 和 r(x) 在包含 x₀ 的开区间上连续，并给定 y(x₀) = α、y′(x₀) = β，那么该区间上存在唯一解。
You do not need to memorise the proof for A-level, but the result justifies why two independent arbitrary constants appear in the general solution. For a homogeneous equation, the general solution must contain two linearly independent particular solutions.
你不需要为 A-Level 记住证明，但该结果解释了为什么通解中会出现两个独立的任意常数。对于齐次方程，通解必须包含两个线性无关的特解。
3. The Principle of Superposition | 叠加原理
Consider the homogeneous equation y″ + p(x)y′ + q(x)y = 0. If y₁ and y₂ are solutions, then for any constants A and B, the function
考虑齐次方程 y″ + p(x)y′ + q(x)y = 0。若 y₁ 和 y₂ 是解，则对任意常数 A 和 B，函数
y = Ay₁ + By₂
is also a solution. This is the superposition principle. To form the general solution, the two solutions must be linearly independent, meaning that one is not a constant multiple of the other. The Wronskian W = y₁y₂′ − y₂y₁′ is a convenient test: if W ≠ 0, then y₁ and y₂ are independent.
也是解。这就是叠加原理。要构成通解，两个解必须线性无关，即一个不是另一个的常数倍。用朗斯基行列式 W = y₁y₂′ − y₂y₁′ 来检验很方便：若 W ≠ 0，则 y₁ 与 y₂ 相互独立。
4. The Cauchy–Euler Equation | 柯西–欧拉方程
The most accessible variable coefficient equation is the Cauchy–Euler (or equidimensional) equation:
最容易处理的变系数方程是柯西–欧拉方程（又称等维方程）：
a x²y″ + b x y′ + c y = 0
where a, b, c are constants. The key idea is to try a solution of the form y = xᵐ, where m is a constant. Substituting y = xᵐ, y′ = m xᵐ⁻¹ and y″ = m(m−1)xᵐ⁻² into the equation gives
其中 a、b、c 为常数。关键思路是尝试形如 y = xᵐ 的解，m 为常数。将 y = xᵐ、y′ = m xᵐ⁻¹ 和 y″ = m(m−1)xᵐ⁻² 代入方程得
a m(m−1) xᵐ + b m xᵐ + c xᵐ = 0
Since xᵐ ≠ 0 for x ≠ 0, we obtain the auxiliary equation
因 xᵐ ≠ 0（x ≠ 0），便得到辅助方程
a m² + (b − a)m + c = 0
This quadratic equation in m determines the form of the solution.
这个关于 m 的二次方程决定了解的形式。
5. Solving the Cauchy–Euler Equation: Distinct Real Roots | 解柯西–欧拉方程：相异实根
If the auxiliary equation has two distinct real roots m₁ and m₂, then the general solution is
若辅助方程有两个相异实根 m₁ 与 m₂，则通解为
y = C₁xᵐ¹ + C₂xᵐ²
For example, solve x²y″ − 3xy′ + 3y = 0. The auxiliary equation is m² − 4m + 3 = 0, giving m = 1 or m = 3. Hence y = C₁x + C₂x³.
例如，求解 x²y″ − 3xy′ + 3y = 0。辅助方程为 m² − 4m + 3 = 0，得 m = 1 或 m = 3。故 y = C₁x + C₂x³。
This result resembles the constant coefficient case, but the powers of x replace exponentials.
这一结果与常系数情形相似，只是指数函数换成了 x 的幂。
6. Repeated and Complex Roots in the Cauchy–Euler Equation | 柯西–欧拉方程中的重根与复根
When the auxiliary equation has a repeated root m₀, the second independent solution is obtained by multiplying by ln x. The general solution becomes
当辅助方程有重根 m₀ 时，第二个线性无关解是将第一个解乘以 ln x。通解变为
y = (C₁ + C₂ ln x) xᵐ⁰
For complex roots m = α ± β i, we use xᵐ = e^(m ln x) and Euler’s formula. The solution is
若根为复数 m = α ± β i，利用 xᵐ = e^(m ln x) 和欧拉公式，解为
y = xᵅ [C₁ cos(β ln x) + C₂ sin(β ln x)]
These formulas are valid for x > 0; for x < 0, replace x by |x| in the logarithm.
以上公式在 x > 0 时成立；若 x < 0，则需将 x 换成 |x| 再取对数。
7. Reduction of Order | 降阶法
For a general homogeneous equation y″ + p(x)y′ + q(x)y = 0, suppose we know one non‑trivial solution y₁(x). We seek a second solution of the form y₂ = v y₁, where v is a function of x. Substituting into the equation and simplifying, the terms containing v cancel, leaving
对于一般的齐次方程 y″ + p(x)y′ + q(x)y = 0，假设我们知道一个非平凡解 y₁(x)。我们寻找形如 y₂ = v y₁ 的第二个解，其中 v 是 x 的函数。代入方程并化简后，含 v 的项相互抵消，只剩下
y₁ v″ + (2y₁′ + p y₁) v′ = 0
Let w = v′. Then this is a first order linear equation in w:
令 w = v′，则这是关于 w 的一阶线性方程：
y₁ w′ + (2y₁′ + p y₁) w = 0
Separating variables gives ∫ (1/w) dw = − ∫ (2y₁′/y₁ + p) dx, which can be integrated to find w. Then v = ∫ w dx.
分离变量得 ∫ (1/w) dw = − ∫ (2y₁′/y₁ + p) dx，积分求出 w，再积分 v = ∫ w dx。
Note that the reduction of order always produces a second solution containing a non‑zero constant multiple of y₁ plus the new independent part; we can ignore the constant multiple in the general solution.
注意，降阶法得到的第二个解总会包含 y₁ 的常数倍以及与 y₁ 无关的新部分；在通解中可以忽略这个常数倍。
8. Worked Example of Reduction of Order | 降阶法例题
Consider x y″ − y′ + 4x³y = 0 for x > 0. One given solution is y₁ = sin(x²). Use reduction of order to find the general solution.
考虑 x y″ − y′ + 4x³y = 0（x > 0）。已知一个解为 y₁ = sin(x²)。用降阶法求通解。
Step 1: Rewrite in standard form: y″ − (1/x)y′ + 4x²y = 0, so p(x) = −1/x.
第一步：写成标准形式：y″ − (1/x)y′ + 4x²y = 0，所以 p(x) = −1/x。
Step 2: Set y₂ = v sin(x²). We need to solve y₁ w′ + (2y₁′ + p y₁)w = 0. Here y₁′ = 2x cos(x²). Then
第二步：设 y₂ = v sin(x²)。我们需要解 y₁ w′ + (2y₁′ + p y₁)w = 0。这里 y₁′ = 2x cos(x²)，于是
2y₁′/y₁ + p = 4x cot(x²) − 1/x
Step 3: Integrate: ln|w| = − ∫(4x cot(x²) − 1/x) dx = −2 ln|sin(x²)| + ln|x|, so w = K x / sin²(x²).
第三步：积分：ln|w| = − ∫(4x cot(x²) − 1/x) dx = −2 ln|sin(x²)| + ln|x|，故 w = K x / sin²(x²)。
Step 4: v = ∫ w dx = K ∫ x csc²(x²) dx = −(K/2) cot(x²). Taking K = −2 gives v = cot(x²). Therefore
第四步：v = ∫ w dx = K ∫ x csc²(x²) dx = −(K/2) cot(x²)。取 K = −2 得 v = cot(x²)。因此
y₂ = cot(x²) sin(x²) = cos(x²)
The general solution is y = C₁ sin(x²) + C₂ cos(x²).
通解为 y = C₁ sin(x²) + C₂ cos(x²)。
9. Variation of Parameters | 参数变易法
For a non‑homogeneous equation y″ + p(x)y′ + q(x)y = f(x), if y₁ and y₂ are independent solutions of the homogeneous equation, we assume a particular solution
对于非齐次方程 y″ + p(x)y′ + q(x)y = f(x)，若 y₁ 与 y₂ 是齐次方程的两个独立解，我们设特解为
yₚ = u₁ y₁ + u₂ y₂
where u₁ and u₂ are functions of x. We impose the two conditions
其中 u₁ 和 u₂ 是 x 的函数。我们施加两个条件：
u₁′ y₁ + u₂′ y₂ = 0 and u₁′ y₁′ + u₂′ y₂′ = f(x)
Solving this linear system gives
解这个线性方程组得
u₁′ = − y₂ f(x) / W, u₂′ = y₁ f(x) / W
where W = y₁y₂′ − y₂y₁′ is the Wronskian. Finally integrate u₁′ and u₂′ with respect to x.
其中 W = y₁y₂′ − y₂y₁′ 是朗斯基行列式。最后对 u₁′ 和 u₂′ 关于 x 积分。
This method works for any continuous f(x), provided the integrals can be evaluated. It is often the easiest route for variable coefficient equations.
只要积分可求，该方法对任何连续 f(x) 都有效。对于变系数方程，这通常是最简便的途径。
10. Worked Example of Variation of Parameters | 参数变易法例题
Solve x²y″ − 2xy′ + 2y = x² ln x for x > 0, given that y₁ = x and y₂ = x² are solutions of the homogeneous equation.
求解 x²y″ − 2xy′ + 2y = x² ln x（x > 0），已知齐次方程的解为 y₁ = x 和 y₂ = x²。
First rewrite in standard form: y″ − (2/x)y′ + (2/x²)y = ln x, so f(x) = ln x.
首先写成标准形式：y″ − (2/x)y′ + (2/x²)y = ln x，所以 f(x) = ln x。
The Wronskian is W = x(2x) − x²(1) = x². Hence
朗斯基行列式 W = x(2x) − x²(1) = x²。因而
u₁′ = − x² ln x / x² = − ln x, u₂′ = x ln x / x² = (ln x)/x
Integrate: u₁ = ∫(−ln x) dx = −x ln x + x, and u₂ = ∫ (ln x)/x dx = ½ (ln x)².
积分：u₁ = ∫(−ln x) dx = −x ln x + x，u₂ = ∫ (ln x)/x dx = ½ (ln x)²。
Thus a particular solution is yₚ = (−x ln x + x)x + ½(ln x)² x² = −x² ln x + x² + ½ x²(ln x)².
因此特解为 yₚ = (−x ln x + x)x + ½(ln x)² x² = −x² ln x + x² + ½ x²(ln x)²。
The general solution is y = C₁x + C₂x² − x² ln x + ½ x²(ln x)²; the term x² can be absorbed into C₂x².
通解为 y = C₁x + C₂x² − x² ln x + ½ x²(ln x)²；其中的 x² 项可并入 C₂x²。
11. A Note on Series Solutions and Further Study | 关于级数解与进一步学习的说明
Many variable coefficient equations, such as Bessel’s equation or Legendre’s equation, cannot be solved by elementary functions. In such cases, we assume a power series solution y = Σₙ₌₀^∞ aₙ xⁿ and substitute it into the differential equation. Recurrence relations then determine the coefficients aₙ.
许多变系数方程，如贝塞尔方程或勒让德方程，不能用初等函数求解。此时我们假设幂级数解 y = Σₙ₌₀^∞ aₙ xⁿ，并将其代入微分方程。递推关系决定系数 aₙ。
For A-level and many first‑year university courses, the methods in this article—Cauchy–Euler, reduction of order, and variation of parameters—cover the most important solvable classes. They also illustrate the deep connection between linear algebra and differential equations.
对于 A-Level 和许多大学一年级课程，本文介绍的方法——柯西–欧拉、降阶法和参数变易法——已经涵盖最可解的重要类型。它们也体现了线性代数与微分方程之间的深刻联系。
Summary | 总结
Variable coefficient second order linear equations may seem challenging, but the three techniques you have studied here give a systematic route: identify an equidimensional equation and use the power substitution; or, if one solution is known, apply reduction of order; and for non‑homogeneous problems, use variation of parameters. Practice with the examples until you can apply each method fluently.
变系数二阶线性方程看似棘手，但这里学习的三种方法给出了系统思路：识别等维方程并采用幂替换；若已知一个解，则用降阶法；对于非齐次问题，则用参数变易法。反复练习例题，直到你能够熟练运用每种方法。
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