Complementary Functions and Particular Integrals | 补函数与特积分

📚 Complementary Functions and Particular Integrals | 补函数与特积分

When you solve a second-order linear differential equation with constant coefficients, the answer is not a single function but a family of functions. It is made of two parts: the complementary function (CF), which solves the associated homogeneous equation, and the particular integral (PI), which is one specific solution of the full equation. This revision guide explains how to find both parts for the AQA A-level specification.

当你求解常系数二阶线性微分方程时，答案不是单一函数，而是一个函数族。它由两部分组成：补函数（CF）求解对应的齐次方程，特积分（PI）则是完整方程的一个特解。本复习指南说明如何在 AQA A-level 考试大纲下求出这两部分。

1. The General Second-Order Linear ODE | 二阶线性常微分方程的一般形式

A second-order linear ODE with constant coefficients has the form shown below, where a, b and c are constants and f(x) is the forcing term. If f(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous.

常系数二阶线性常微分方程的形式如下，其中 a、b、c 为常数，f(x) 为强迫项。若 f(x) = 0，则方程为齐次的；否则为非齐次的。

a d²y/dx² + b dy/dx + c y = f(x)

The solution is split as y = yc + yp. The complementary function yc contains two arbitrary constants and solves the homogeneous equation; the particular integral yp is any single solution of the full equation.

解拆分为 y = yc + yp。补函数 yc 含有两个任意常数，求解齐次方程；特积分 yp 是完整方程的任意一个特解。

2. The Homogeneous Equation and Auxiliary Equation | 齐次方程与辅助方程

To find yc, set f(x) = 0 and assume a trial solution y = e^(mx). Substituting into a d²y/dx² + b dy/dx + c y = 0 gives the auxiliary equation.

为求 yc，令 f(x) = 0 并假设试探解 y = e^(mx)。代入 a d²y/dx² + b dy/dx + c y = 0 后得到辅助方程。

a m² + b m + c = 0

The roots of the auxiliary equation determine the form of the complementary function. Always write the auxiliary equation before choosing the CF.

辅助方程的根决定补函数的形式。在选择补函数之前，必须先写出辅助方程。

3. Distinct Real Roots of the Auxiliary Equation | 辅助方程有两个不同实根

If the auxiliary equation has two distinct real roots m₁ and m₂, the complementary function is the sum of two independent exponential

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