📚 Second Order Linear Differential Equations with Variable Coefficients | 变系数二阶线性微分方程
In AQA Further Mathematics, second order linear differential equations with variable coefficients appear when the coefficients of y”, y’ and y are functions of the independent variable x, rather than constants. These equations are more general than the constant-coefficient case and require special techniques such as Cauchy–Euler substitution, reduction of order, and power series methods.
在 AQA 进阶数学中，变系数二阶线性微分方程是指 y”、y’ 和 y 的系数为自变量 x 的函数，而非常数。这类方程比常系数情形更一般，需要使用特殊技巧，例如柯西-欧拉代换、降阶法和幂级数解法。
1. Standard Form and Overview | 标准形式与概览
The general form of a second order linear differential equation with variable coefficients is
y” + P(x) y’ + Q(x) y = R(x)
When R(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous. The methods you use depend on the structure of P(x) and Q(x).
一般的变系数二阶线性微分方程可写为
y” + P(x) y’ + Q(x) y = R(x)
当 R(x) = 0 时方程为齐次方程，否则为非齐次方程。具体采用哪种解法取决于 P(x) 与 Q(x) 的结构。
2. The Cauchy–Euler Equation | 柯西-欧拉方程
The Cauchy–Euler equation (also called Euler–Cauchy equation) has the form
a x² y” + b x y’ + c y = f(x)
Here a, b and c are constants, and the powers of x match the order of the derivative: x² multiplies y”, x multiplies y’, and a constant multiplies y.
柯西-欧拉方程（又称欧拉-柯西方程）具有如下形式
a x² y” + b x y’ + c y = f(x)
其中 a、b、c 为常数，且 x 的幂次与导数阶数对应：x² 乘 y”，x 乘 y’，常数乘 y。
3. Substituting x = eᵗ: From Variable to Constant Coefficients | 代换 x = eᵗ：从变系数到常系数
To solve a homogeneous Cauchy–Euler equation, make the substitution x = eᵗ, so that t = ln x. Then the derivatives transform as
dy/dx = (1/x) dy/dt, d²y/dx² = (1/x²)(d²y/dt² − dy/dt)
Substituting into a x² y” + b x y’ + c y = 0 gives constant-coefficient equation
a d²y/dt² + (b − a) dy/dt + c y = 0
The corresponding auxiliary equation is a m² + (b − a) m + c = 0.
解齐次柯西-欧拉方程时，使用代换 x = eᵗ，即 t = ln x。此时导数变换为
dy/dx = (1/x) dy/dt, d²y/dx² = (1/x²)(d²y/dt² − dy/dt)
代入 a x² y” + b x y’ + c y = 0 后得到常系数方程
a d²y/dt² + (b − a) dy/dt + c y = 0
对应的特征方程为 a m² + (b − a) m + c = 0。
4. Worked Example: Homogeneous Cauchy–Euler | 例题：齐次柯西-欧拉方程
Example: solve x² y” − 3x y’ + 4y = 0 for x > 0.
例题：求解 x² y” − 3x y’ + 4y = 0（x > 0）。
Let x = eᵗ. Then a = 1, b = −3, c = 4. The transformed equation is
d²y/dt² + (−3 −1) dy/dt + 4y = 0
so d²y/dt² − 4 dy/dt + 4y = 0. The characteristic equation is
m² − 4m + 4 = 0, i.e. (m − 2)² = 0
Thus there is a repeated root m = 2, giving y = (A + Bt)e²ᵗ. Since t = ln x, the general solution is
y = (A + B ln x) x²
代换 x = eᵗ，此时 a = 1，b = −3，c = 4。变换后的方程为
d²y/dt² + (−3 −1) dy/dt + 4y = 0
即 d²y/dt² − 4 dy/dt + 4y = 0。特征方程为
m² − 4m + 4 = 0， 即 (m − 2)² = 0
于是有重根 m = 2，得到 y = (A + Bt)e²ᵗ。因 t = ln x，通解为
y = (A + B ln x) x²
5. Worked Example: Non-homogeneous Cauchy–Euler | 例题：非齐次柯西-欧拉方程
Example: solve x² y” − 3x y’ + 4y = ln x.
例题：求解 x² y” − 3x y’ + 4y = ln x。
Using x = eᵗ, the non-homogeneous term ln x becomes t, and the transformed equation is
d²y/dt² − 4 dy/dt + 4y = t
For the particular integral, try y = At + B. Then y’ = A and y” = 0, so
−4A + 4(At + B) = t
Comparing coefficients gives A = 1/4 and B = 1/4. Hence the general solution is
y = (A + B ln x)x² + (ln x + 1)/4
使用 x = eᵗ 后，非齐次项 ln x 变为 t，变换后的方程为
d²y/dt² − 4 dy/dt + 4y = t
求特解时可设 y = At + B，则 y’ = A，y” = 0，因此
−4A + 4(At + B) = t
比较系数得 A = 1/4，B = 1/4。所以通解为
y = (A + B ln x)x² + (ln x + 1)/4
6. Reduction of Order: The Basic Idea | 降阶法：基本思想
For a general equation y” + P(x)y’ + Q(x)y = 0, if one non-zero solution y₁(x) is known, we can find a second linearly independent solution y₂(x) by writing y₂ = v(x) y₁(x). This is called reduction of order because substituting this form reduces the equation to a first order equation for v’.
对于一般方程 y” + P(x)y’ + Q(x)y = 0，若已知一个非零解 y₁(x)，则可令 y₂ = v(x) y₁(x) 来求第二个线性无关解。这种方法称为降阶法，因为代入后方程会化为关于 v’ 的一阶方程。
7. Deriving the Reduction-of-Order Formula | 降阶法公式推导
Set y = v y₁. Then
y’ = v’y₁ + v y₁’, y” = v”y₁ + 2v’y₁’ + v y₁”
Substitute into y” + P y’ + Q y = 0. Since y₁ is a solution, the terms involving v cancel, leaving
y₁ v” + (2y₁’ + P y₁) v’ = 0
Let u = v’. Then the equation becomes separable:
u’ / u = −(2y₁’/y₁ + P)
Integrating once gives the standard formula
y₂ = y₁ ∫ ( e^{−∫ P dx} / y₁² ) dx
令 y = v y₁，则
y’ = v’y₁ + v y₁’, y” = v”y₁ + 2v’y₁’ + v y₁”
代入 y” + P y’ + Q y = 0 后，含有 v 的项因 y₁ 是解而抵消，余下
y₁ v” + (2y₁’ + P y₁) v’ = 0
令 u = v’，则方程变为可分离变量形式：
u’ / u = −(2y₁’/y₁ + P)
积分一次即得标准公式
y₂ = y₁ ∫ ( e^{−∫ P dx} / y₁² ) dx
8. Worked Example: Reduction of Order | 例题：降阶法
Example: given that y₁ = eˣ is a solution of x y” + (1 − 2x)y’ + (x − 1)y = 0, find the general solution for x > 0.
例题：已知 y₁ = eˣ 是方程 x y” + (1 − 2x)y’ + (x − 1)y = 0 的一个解，求其通解（x > 0）。
First divide by x to get P(x) = 1/x − 2. Then
e^{−∫P dx} = e^{−(ln x − 2x)} = e^{2x}/x
Using the formula,
y₂ = eˣ ∫ ( (e^{2x}/x) / e^{2x} ) dx = eˣ ∫ (1/x) dx = eˣ ln x
Hence the general solution is
y = (A + B ln x) eˣ
先将方程除以 x，得 P(x) = 1/x − 2。于是
e^{−∫P dx} = e^{−(ln x − 2x)} = e^{2x}/x
代入公式得
y₂ = eˣ ∫ ( (e^{2x}/x) / e^{2x} ) dx = eˣ ∫ (1/x) dx = eˣ ln x
因此通解为
y = (A + B ln x) eˣ
9. Power Series Solutions | 幂级数解法
When P(x) and Q(x) are complicated but analytic at x₀, assume a power series solution
y = Σₙ₌₀^∞ aₙ xⁿ
Differentiate term by term and substitute into the differential equation. Then compare coefficients of like powers of x to obtain a recurrence relation for aₙ.
当 P(x) 和 Q(x) 较复杂但在 x₀ 处解析时，可假设幂级数解
y = Σₙ₌₀^∞ aₙ xⁿ
逐项求导并代入微分方程，然后比较同次幂的系数，得到关于 aₙ 的递推关系。
10. Airy’s Equation: A Series Example | 艾里
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