Second Order Linear Differential Equations with Variable Coefficients | 二阶线性变系数微分方程

📚 Second Order Linear Differential Equations with Variable Coefficients | 二阶线性变系数微分方程

In AQA A-Level Mathematics, second order linear differential equations usually appear with constant coefficients. However, the syllabus also includes linear second order equations whose coefficients are functions of x. These variable coefficient equations require special techniques: trying a power solution, reducing the order once one solution is known, or using a given substitution to transform the equation.

在 AQA A-Level 数学中，二阶线性微分方程通常以常系数形式出现。但考纲也包含系数为 x 的函数的线性二阶方程。这类变系数方程需要特殊方法：尝试幂函数解、已知一个解时降阶、或使用给定代换转化方程。

1. General Form and Linearity | 一般形式与线性性质

A second order linear differential equation in standard form is y'' + p(x)y' q(x)y = r(x). Here p, q and r are functions of x, and the equation is linear because y, y' and y'' appear only to the first power and are not multiplied together.

标准形式的二阶线性微分方程是 y'' + p(x)y' + q(x)y = r(x)。其中 p、q、r 是 x 的函数，该方程是线性的，因为 y、y' 和 y'' 都只出现一次幂，且不相乘。

n is homogeneous; otherwise it is non-homogeneous. The general solution has the form y = y_c + y_p, where y_c is the complementary function and y_p is a particular integral.

若 r(x) = 0，方程为齐次方程；否则为非齐次方程。通解形式为 y = y_c + y_p，其中 y_c 为余函数，y_p 为特积分。

2. Why Variable Coefficients Need New Methods | 为什么变系数方程需要新方法

With constant coefficients, we solve the auxiliary equation and use exponential, trigonometric or polynomial forms. When coefficients are functions such as x or x², those standard guesses often fail because the equation is not invariant under a shift in x.

对于常系数方程，我们求解辅助方程并使用指数、三角或多项式形式。当系数为 x 或 x² 等函数时，这些标准猜测往往失效，因为方程在 x 平移下不再保持不变。

The key AQA techniques are: use the Cauchy-Euler substitution y = x^m, change variable x = e^t to produce a constant coefficient equation, or use reduction of order when one solution is given.

AQA 的关键方法是：使用柯西-欧拉代换 y = x^m；令 x = e^t 转化为常系数方程；或在给定一个解时使用降阶法。

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3. The Cauchy-Euler Equation | 柯西-欧拉方程

The most common variable coefficient equation in A-Level is the Cauchy-Euler equation: x²y'' + axy' + by = 0, where a and b are constants. For x > 0, try y = x^m.

A-Level 中最常见的变系数方程是柯西-欧拉方程：x²y'' + axy' + by = 0，其中 a、b 为常数。当 x > 0 时，尝试 y = x^m。

Differentiate: y' = m x^m−1, y'' = m(m−1) x^m−2. Substituting gives the indicial equation m(m−1) + am + b = 0.

求导得 y' = m x^m−1，y'' = m(m−1) x^m−2。代入后得到指数方程 m(m−1) + am + b = 0。

For distinct real roots m₁ and m₂, the complementary function is y = A x^m₁ + B x^m₂. For a repeated root m, it becomes y = (A + B ln x) x^m.

若指数方程有两个不等实根 m₁ 和 m₂，则余函数为 y = A x^m₁ + B x^m₂。若为二重根 m，则为 y = (A + B ln x) x^m。

For complex roots m = α ± βi, the real form is y = x^α [ A cos(β ln x) + B sin(β ln x) ].

若为复根 m = α ± βi，则实形式为 y = x^α [ A cos(β ln x) + B sin(β ln x) ]。

4. Transforming by x = e^t | 用代换 x = e^t 转化

Put x = e^t t = ln x. The chain rule gives dy/dx = (1/x) dy/dt and d²y/dx² = (1/x²)(d²y/dt² − dy/dt).

令 x = e^t，则 t = ln x。链式法则给出 dy/dx = (1/x) dy/dt 和 d²y/dx² = (1/x²)(d²y/dt² − dy/dt)。

Substitution into x²y'' + axy' 0 yields d²y/dt² + (a − 1) dy/dt + by = 0, which is a constant coefficient equation in t.

代入 x²y'' + axy' + by = 0 得到 d²y/dt² + (a − 1) dy/dt + by = 0，这是关于 t 的常系数方程。

Solve this using the auxiliary equation, then replace t by ln x. This method explains why the Cauchy-Euler solution involves x^m and ln x.

用辅助方程求解该方程，再将 t 替换为 ln x。这一方法解释了柯西-欧拉方程的解为何含有 x^m 和 ln x。

5. Reduction of Order | 降阶法

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