Complementary Functions and Particular Integrals | 互补函数与特解积分

📚 Complementary Functions and Particular Integrals | 互补函数与特解积分

When studying second-order differential equations in AQA Further Mathematics, the main goal is to solve equations of the form a d²y/dx² + b dy/dx + c y = f(x), where a, b and c are constants. The solution has two parts: the complementary function, which comes from the homogeneous equation, and a particular integral, which is a specific solution to the full non-homogeneous equation. This article explains how to find both parts and how to combine them into a general solution.

在 AQA 进阶数学中，学习二阶微分方程的主要目标是求解形如 a d²y/dx² + b dy/dx + c y = f(x) 的方程，其中 a、b、c 为常数。方程的解由两部分组成：互补函数来自齐次方程，特解积分则是非齐次方程的一个特定解。本文将解释如何求出这两部分，以及如何将它们组合成通解。

1. The Structure of the Problem | 问题的结构

Consider a second-order linear differential equation with constant coefficients written as a d²y/dx² + b dy/dx + c y = f(x), where a ≠ 0. The left-hand side is linear because y, dy/dx and d²y/dx² appear only to the first power and are not multiplied together. The right-hand side f(x) is called the forcing term or non-homogeneous term. If f(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous and requires both a complementary function and a particular integral.

考虑一个常系数二阶线性微分方程，写作 a d²y/dx² + b dy/dx + c y = f(x)，其中 a ≠ 0。左边是线性的，因为 y、dy/dx 和 d²y/dx² 都只以一次方出现，并且没有彼此相乘。右边的 f(x) 称为强迫项或非齐次项。如果 f(x) = 0，方程是齐次的；否则该方程是非齐次的，需要同时求出互补函数和特解积分。

a d²y/dx² + b dy/dx + c y = f(x)

2. Complementary Function and the Auxiliary Equation | 互补函数与辅助方程

The complementary function solves the associated homogeneous equation a d²y/dx² + b dy/dx + c y = 0. Since the coefficients are constant, we look for trial solutions of the form y = e^mx. Differentiating gives dy/dx = m e^mx and d²y/dx² = m² e^mx. Substituting these into the homogeneous equation produces (a m² + b m + c) e^mx = 0. Because e^mx is never zero, the auxiliary equation must be a m² + b m + c = 0. The roots of this quadratic determine the form of the complementary function.

互补函数求解相应的齐次方程 a d²y/dx² + b dy/dx + c y = 0。由于系数是常数，我们可以尝试形如 y = e^mx 的解。对其求导得到 dy/dx = m e^mx 和 d²y/dx² = m² e^mx。将这些代入齐次方程，得到 (a m² + b m + c) e^mx = 0。因为 e^mx 永不为零，所以辅助方程必然是 a m² + b m + c = 0。这个二次方程的根决定了互补函数的形式。

a m² + b m + c = 0

3. Case 1: Real and Distinct Roots | 情况一：相异实根

If the discriminant b² – 4ac is positive, the auxiliary equation has two distinct real roots, say m₁ and m₂. The complementary function is y_c = A e^m₁x + B e^m₂x, where A and B are arbitrary constants. These constants are not fixed until initial or boundary conditions are applied. For example, the equation d²y/dx² – 3 dy/dx + 2y = 0 has auxiliary equation m² – 3m + 2 = 0, which factorises as (m – 1)(m – 2) = 0, giving roots m₁ = 1 and m₂ = 2. Hence y_c = A e^x + B e^2x.

如果判别式 b² – 4ac 为正，辅助方程有两个相异实根，记为 m₁ 和 m₂。互补函数为 y_c = A e^m₁x + B e^m₂x，其中 A 和 B 是任意常数。这些常数要等到代入初始条件或边界条件后才能确定。例如，方程 d²y/dx² – 3 dy/dx + 2y = 0 的辅助方程是 m² – 3m + 2 = 0，因式分解为 (m – 1)(m – 2) = 0，得到根 m₁ = 1 和 m₂ = 2。因此 y_c = A e^x + B e^2x

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