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 never a single function: it is an entire family of functions controlled by two arbitrary constants. That family is built from two pieces - the complementary function (CF), which solves the associated homogeneous equation, and the particular integral (PI), which is one specific solution of the full non-homogeneous equation.

当你求解常系数二阶线性微分方程时，答案从来不是单一函数，而是由两个任意常数控制的整个函数族。这个函数族由两部分构成：补函数（CF）求解对应的齐次方程，特积分（PI）则是完整非齐次方程的一个特解。

This revision guide walks through the whole method for the AQA A-level Further Mathematics specification: forming the auxiliary equation, reading off the correct CF for real, repeated and complex roots, choosing a sensible trial function for the PI, handling the awkward resonance case, and finishing the job with boundary or initial conditions.

本复习指南系统讲解 AQA A-level 进阶数学大纲要求的完整方法：写出辅助方程、根据不同类型的根写出正确的补函数、为特积分选择合适的试探函数、处理麻烦的共振情形，最后用边界条件或初始条件定出常数。

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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 ordinary differential equation with constant coefficients has the standard form below, where a, b and c are constants with a ≠ 0, and f(x) is the forcing term, sometimes called the right-hand side or the input.

常系数二阶线性常微分方程具有下面的标准形式，其中 a、b、c 为常数且 a ≠ 0，f(x) 称为强迫项，有时也叫右端项或输入项。

a d²y/dx² + b dy/dx + c y = f(x)

= 0 the equation is homogeneous, and its general solution is the complementary function y_c. If f(x) is not zero the equation is non-homogeneous, and its general solution is the sum of the complementary function and any one particular integral y_p.

若 f(x) = 0，方程为齐次方程，其通解就是补函数 y_c。若 f(x) 不为零，方程为非齐次方程，其通解等于补函数与任意一个特积分 y_p 之和。

y = y_c + y_p

Two structural facts matter for marks. First, y_c always carries exactly two arbitrary constants, because the equation is second order. Second, the particular integral carries none: it is a single fixed function that satisfies the whole equation.

解答结构有两点直接关系到得分。第一，y_c 恰好含两个任意常数，因为方程是二阶的。第二，特积分不含任何任意常数，它是满足整个方程的一个固定函数。

Linearity is what makes this split legal. If y_c solves the homogeneous equation and y_p solves the full equation, then y_c + y_p solves the full equation, and in fact every solution of the full equation can be written in that form.

这种拆分之所以合法，来自方程的线性性：若 y_c 满足齐次方程、y_p 满足完整方程，则 y_c + y_p 也满足完整方程，而且完整方程的每一个解都可以写成这种形式。

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

side to zero and look for an exponential-type solution by assuming a trial solution y = e^mx, where m is a constant still to be found.

为求补函数，先把右端项置零，并假设方程有形如 y = e^mx 的指数型解，其中 m 是待定常数。

Differentiating gives dy/dx = m e^mx and d²y/dx² = m² e^mx. Substituting into a d²y/dx² + b dy/dx + c y = 0 gives a m² e^mx + b m e^mx + c e^mx = 0. Since e^mx is never zero, it cancels and leaves the auxiliary equation.

求导得 dy/dx = m e^mx，d²y/dx² = m² e^mx。代入 a d²y/dx² + b dy/dx + c y = 0 得 a m² e^mx + b m e^mx + c e^mx = 0。由于 e^mx 恒不为零，可以约去，剩下的就是辅助方程。

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a m² + b m + c = 0

The auxiliary equation is a quadratic, so it has two roots counted with multiplicity. Its discriminant b² − 4ac tells you immediately which of the three cases you are in.

辅助方程是二次方程，按重数计有两个根。它的判别式 b² − 4ac 立即告诉你属于三种情形中的哪一种。

m = [ −b ± √ b² − 4ac ] / 2a

Discriminant 判别式	Nature of roots 根的性质	Complementary function 补函数
b² − 4ac > 0	two distinct real roots m₁, m₂	yc = A em₁x + B em₂x
b² − 4ac = 0	one repeated real root m	yc = (A + Bx) emx
b² − 4ac < 0	complex conjugates α ± iβ	yc = eαx(A cos βx + B sin βx)

Always write the auxiliary equation down before choosing the form of the complementary function. Examiners award method marks for a correctly formed auxiliary equation even when later algebra slips.

一定要先写出辅助方程，再选择补函数的形式。即使后面的代数出错，只要辅助方程写对，阅卷时仍会给出方法分。

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

When b² − 4ac > 0, the auxiliary equation has two distinct real roots m₁ and m₂. The general solution of the homogeneous equation is the sum of two independent exponential terms.

当 b² − 4ac > 0 时，辅助方程有两个不同实根 m₁ 和 m₂。齐次方程的通解是两个相互独立的指数项之和。

y_c = A e^m₁x + B e^m₂x

A common slip is to write the two constants as one, or to forget one exponential term entirely. Both constants are needed: two initial conditions or two boundary conditions will later determine both A and B.

常见错误是把两个常数写成一个，或者干脆漏掉一个指数项。两个常数都必须保留：后面要用两个初始条件或两个边界条件分别定出 A 和 B。

le, if the auxiliary equation factors as (m + 3)(m − 1) = 0, then m₁ = −3 and m₂ = 1, so y_c = A e^−3x + B e^x. Notice that a negative root produces a decaying term, which is exactly what you would expect in a damped physical system.

例如，若辅助方程分解为 (m + 3)(m − 1) = 0，则 m₁ = −3、m₂ = 1，故 y_c = A e^−3x + B e^x。注意负根给出衰减项，这正符合阻尼物理系统的预期。

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4. Repeated Real Roots | 辅助方程有重根

ponential solution appears. A second independent solution is obtained by multiplying by x.

当 b² − 4ac = 0 时，两根重合为 m = −b / 2a，此时只得到一个指数解。第二个独立解需要乘以 x 来获得。

y_c = (A + Bx) e^mx

Writing only A e^mx for the repeated case is a guaranteed loss of marks, because it cannot supply the two independent constants that a second-order equation demands.

在重根情形下只写 A e^mx 必定失分，因为它无法提供二阶方程所要求的两个独立常数。

For instance, the auxiliary equation m² − 6m +

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