📚 Second Order Differential Equations | 二阶微分方程
Second order differential equations are one of the most rewarding topics in AQA A-Level Mathematics: a single algebraic routine solves an entire family of problems about springs, pendulums, damped oscillations and electrical circuits. This revision guide covers the general form, the three cases of the auxiliary equation, the method of undetermined coefficients for particular integrals, the special failure case that produces resonance, and the exam technique that turns a long calculation into reliable marks.
二阶微分方程是 AQA A-Level 数学中最有回报的考点之一：一套代数流程就能解决弹簧、单摆、阻尼振动和电路这一整类问题。本复习指南将依次讲解方程的一般形式、辅助方程的三种情形、用待定系数法求特解、会产生共振的特殊失效情形，以及把冗长计算变成稳定得分的应试技巧。
1. General Form and Terminology | 方程的一般形式与术语
A linear second order differential equation with constant coefficients takes the form shown below, where a, b and c are real constants with a ≠ 0, and f(x) is a given function of x. The word linear means that y, dy/dx and d²y/dx² each appear only to the first power and are never multiplied together; second order means that the highest derivative present is d²y/dx².
常系数线性二阶微分方程的形式如下，其中 a、b、c 为实常数且 a ≠ 0，f(x) 是给定的 x 的函数。所谓线性，是指 y、dy/dx 与 d²y/dx² 都只出现一次方，且彼此不相乘；所谓二阶，是指方程中出现的最高阶导数为 d²y/dx²。
a d²y/dx² + b dy/dx + c y = f(x)
, where the complementary function CF solves the homogeneous version and the particular integral PI is any one solution of the full equation. Because the CF contains two arbitrary constants A and B, two extra conditions — usually values of y and dy/dx at a single point, or values of y at two different points — are needed to single out one unique solution.
若 f(x) = 0，方程称为齐次方程；若 f(x) 不为零，则为非齐次方程。这类方程的每一个解都可以写成 y = CF + PI，其中余函数 CF 是齐次方程的解，特解 PI 是原方程的任意一个解。由于 CF 中含有两个任意常数 A 与 B，还需要两个附加条件——通常是在同一点处给出 y 与 dy/dx 的值，或在两个不同点处给出 y 的值——才能确定唯一的解。
2. The Auxiliary Equation | 辅助方程与余函数
To solve the homogeneous equation, try a solution of the form y = emx, where m is a constant still to be found. Differentiating gives dy/dx = m emx and d²y/dx² = m² emx. Substituting these into a d²y/dx² + b dy/dx + c y = 0 and factoring out the factor emx, which is never zero for real m, leaves a simple quadratic in m.
求解齐次方程时，先尝试形如 y = emx 的解，其中 m 是待定常数。求导得到 dy/dx = m emx，d²y/dx² = m² emx。把它们代入 a d²y/dx² + b dy/dx + c y = 0，并约去恒不为零的因子 emx，就剩下一个关于 m 的二次方程。
a m² + b m + c = 0
This quadratic is called the auxiliary equation, or characteristic equation. Its roots completely determine the complementary function, and the sign of the discriminant b² − 4ac tells you which of the three possible shapes the answer will take. Writing down the auxiliary equation correctly in the first line of working is therefore the single most important step of the whole method, because an error here ruins every later line.
这个二次方程称为辅助方程（或特征方程）。它的根完全决定余函数的形式，而判别式 b² − 4ac 的符号则决定答案属于三种可能形状中的哪一种。因此，在解题第一行就正确写出辅助方程，是整个方法中最关键的一步，因为这里出错会让后面每一行都作废。
3. The Three Cases of the Auxiliary Equation | 辅助方程的三种情形
Solving the auxiliary equation always produces one of three outcomes: two distinct real roots, one repeated real root, or a pair of complex conjugate roots. Each case gives a different complementary function, and examiners expect you to identify the case before you write the answer.
解辅助方程只会得到三种结果：两个不等的实根、一个重实根，或一对共轭复根。每种情形对应不同的余函数，考官希望你在写出答案之前先判断属于哪一种情形。
| Discriminant / Roots | 判别式与根 | Roots m | 根 m | Complementary Function | 余函数 CF |
|---|---|---|
| b² − 4ac > 0 | m₁, m₂ real and distinct | y = A em₁x + B em₂x |
| b² − 4ac = 0 | m = −b / 2a (repeated) | y = (A + Bx) emx |
| b² − 4ac < 0 | m = α ± iβ | y = eαx(A cos βx + B sin βx) |
Notice that in every case exactly two arbitrary constants appear, and no more. If your answer contains three constants, or only one, something has gone wrong earlier in the solution.
注意每种情形都恰好出现两个任意常数，既不多也不少。如果你的答案里有三个常数，或者只有一个，说明前面的步骤出了问题。
4. Case 1: Distinct Real Roots | 情形一：两个不等的实根
These are real and distinct, so the general solution is the sum of two exponential terms.
求解 d²y/dx² − 5 dy/dx + 6y = 0。辅助方程为 m² − 5m + 6 = 0，因式分解得 (m − 2)(m − 3) = 0，所以 m = 2 与 m = 3。两实根不相等，因此通解是两个指数项之和。
y = A e2x + B e3x
You can always check by differentiation: dy/dx = 2A e2x + 3B e3x, and d²y/dx² = 4A e2x + 9B e3x. Substituting gives 4A e2x + 9B e3x − 5(2A e2x + 3B e3x) + 6(A e2x + B e3x) = (4A − 10A + 6A)e2x + (9B − 15B + 6B)e3x = 0, confirming the solution.
你随时可以用求导来检验：dy/dx = 2A e2x + 3B e3x，d²y/dx² = 4A e2x + 9B e3x。代入原方程得 4A e2x + 9B e3x − 5(2A e2x + 3B e3x) + 6(A e2x + B e3x) = (4A − 10A + 6A)e2x + (9B − 15B + 6B)e3x = 0，解得到验证。
5. Case 2: A Repeated Real Root | 情形二：重实根
+ 4 = 0 factorises as (m − 2)² = 0, giving the single repeated root m = 2. Writing y = A e2x + B e2x produce one independent solution, so the second solution is obtained by multiplying by x.
求解 d²y/dx² − 4 dy/dx + 4y = 0。辅助方程 m² − 4m + 4 = 0 分解为 (m − 2)² = 0，得到唯一的重根 m = 2。若写成 y = A e2x + B e2x，两个项其实是一个解，所以第二个解要乘上 x 得到。
y = (A + Bx) e2x
Checking the extra term x e2x: its derivatives are dy/dx = e2x(1 + 2x) and d²y/dx² = e2x(4 + 4x). Then d²y/dx² − 4 dy/dx + 4y = e2x[(4 + 4x) − 4(1 + 2x) + 4x] = e2x(0) = 0, so x e2x really is a solution, and it is independent of e2x because of the factor x.
检验多出来的那一项 x e2x：它的导数为 dy/dx = e2x(1 +
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