📚 Second-Order Differential Equations | 二阶微分方程
Second-order differential equations are central to AQA A-level Mathematics and to many real-world models, from vibrating springs to electrical circuits. In this revision guide we build a systematic method for solving linear equations with constant coefficients, including both homogeneous and non-homogeneous cases.
二阶微分方程是 AQA A-Level 数学的核心内容，也是从弹簧振动到电路分析等众多现实模型的重要工具。本复习指南将为你建立一套求解常系数线性二阶方程的完整方法，涵盖齐次方程与非齐次方程。
1. The General Form | 一般形式
A second-order differential equation is an equation in which the highest-order derivative is the second derivative. In this topic we focus on linear equations with constant coefficients, written in the form
二阶微分方程是最高阶导数为二阶导数的方程。在本专题中，我们重点研究常系数线性方程，其形式为
a d²y/dx² + b dy/dx + c y = f(x)
where a, b and c are constants, a ≠ 0, and f(x) is a function of x. If f(x) = 0 the equation is called homogeneous; if f(x) is not zero, it is called non-homogeneous.
其中 a、b、c 为常数，a ≠ 0，f(x) 是 x 的函数。若 f(x) = 0，则称为齐次方程；若 f(x) 不为零，则称为非齐次方程。
2. Why the Exponential Works | 为什么用指数函数试探
For a homogeneous equation, a natural first guess is y = eλx. Differentiating gives dy/dx = λ eλx and d²y/dx² = λ² eλx.
对于齐次方程，最自然的尝试是 y = eλx。求导可得 dy/dx = λ eλx，d²y/dx² = λ² eλx。
Substituting these expressions into the homogeneous equation
将这些表达式代入齐次方程
a d²y/dx² + b dy/dx + c y = 0
gives
得到
eλx(aλ² + bλ + c) = 0
Because eλx is never zero, the product can equal zero only when the bracket is zero. This observation turns the differential equation into an algebraic equation.
因为 eλx 永不为零，乘积为零只能由括号部分为零决定。这一观察把微分方程转化成了代数方程。
3. The Auxiliary Equation | 辅助方程
The algebraic equation obtained from the exponential substitution is called the auxiliary equation, or characteristic equation:
通过指数替换得到的代数方程称为辅助方程，也称特征方程：
aλ² + bλ + c = 0
The nature of the general solution depends on the discriminant of this quadratic: b² − 4ac. There are three cases: real distinct roots, a repeated root, and complex conjugate roots.
通解的形式取决于该二次方程的判别式：b² − 4ac。共有三种情形：相异实根、重根、共轭复数根。
You should always solve the auxiliary equation before writing down the complementary function.
在写出通解之前，务必先求解辅助方程。
4. Real Distinct Roots | 相异实根
If the auxiliary equation has two real distinct roots λ₁ and λ₂, the general solution is
若辅助方程有两个相异实根 λ₁ 和 λ₂，则通解为
y = A eλ₁x + B eλ₂x
where A and B are arbitrary constants. These two exponential functions are linearly independent, so their combination spans all possible solutions.
其中 A 和 B 为任意常数。这两个指数函数线性无关，因此它们的线性组合能够表示所有可能的解。
For example, for y” − 3y’ + 2y = 0, the auxiliary equation is λ² − 3λ + 2 = 0, which factors as (λ − 1)(λ − 2) = 0. Hence λ₁ = 1 and λ₂ = 2, so the general solution is y = A ex + B e2x.
例如，对于 y” − 3y’ + 2y = 0，辅助方程为 λ² − 3λ + 2 = 0，可因式分解为 (λ − 1)(λ − 2) = 0。因此 λ₁ = 1，λ₂ = 2，所以通解为 y = A ex + B e2x。
5. Repeated Roots | 重根
If the discriminant is zero, the auxiliary equation has one repeated root λ = −b/(2a). In this case the two independent solutions cannot both be eλx; the second solution must be multiplied by x.
若判别式为零，辅助方程只有一个重根 λ = −b/(2a)。此时两个线性无关的解不能都是 eλx，第二个解必须乘以 x。
y = (A + Bx) eλx
For example, the equation y” + 6y’ + 9y = 0 has auxiliary equation λ² + 6λ + 9 = 0 = (λ + 3)², so λ = −3. The general solution is y = (A + Bx)e−3x.
例如，方程 y” + 6y’ + 9y = 0 的辅助方程为 λ² + 6λ + 9 = 0 = (λ + 3)²，因此 λ = −3。通解为 y = (A + Bx)e−3x。
6. Complex Roots | 复数根
If the discriminant is negative, the roots are complex conjugates λ = α ± βi, where α and β are real. The general solution can be written using Euler’s formula as
若判别式为负，则根为共轭复数 λ = α ± βi，其中 α 和 β 为实数。通解可利用欧拉公式写成
y = eαx(A cos βx + B sin βx)
Here A and B are real constants, β is taken as positive, and i² = −1. This form is essential when modelling oscillatory behaviour.
这里 A 和 B 是实常数，β 通常取正值，i² = −1。这一形式对描述振动行为至关重要。
For example, y” + 2y’ + 10y = 0 has auxiliary equation λ² + 2λ + 10 = 0, giving λ = −1 ± 3i. Hence y = e−x(A cos 3x + B sin 3x).
例如，y” + 2y’ + 10y = 0 的辅助方程为 λ² + 2λ + 10 = 0，得到 λ = −1 ± 3i。因此 y = e−x(A cos 3x + B sin 3x)。
7. Complementary Function and Particular Integral | 通解与特解
For a non-homogeneous equation of the form ay” + by’ + cy = f(x), the general solution is the sum of two parts:
对于形如 ay” + by’ + cy = f(x) 的非齐次方程，其一般解由两部分相加得到：
Complementary Function (CF): the general solution of the corresponding homogeneous equation ay” + by’ + cy = 0.
通解（CF）：对应齐次方程 ay” + by’ + cy = 0 的通解。
Particular Integral (PI): one specific solution of the full non-homogeneous equation.
特解（PI）：非齐次方程本身的一个具体解。
General solution = CF + PI
一般解 = CF + PI
The complementary function contains arbitrary constants; the particular integral contains no arbitrary constants.
通解中包含任意常数，而特解中不含任意常数。
8. Choosing a Particular Integral | 如何选择特解
The method of undetermined coefficients uses a trial form for the particular integral, based on the form of f(x).
待定系数法根据 f(x) 的形式来假设特解的形式。
| f(x) | Trial yₚ |
| Constant c | yₚ = C |
| Polynomial of degree n | yₚ = Cₙxⁿ + Cₙ₋₁xⁿ⁻¹ + … + C₀ |
| ekx | yₚ = A ekx |
| sin(kx) or cos(kx) | yₚ = A sin(kx) + B cos(kx) |
| ekxsin(kx) or ekxcos(kx) | yₚ = ekx(A sin(kx) + B cos(kx)) |
If the trial particular integral already appears in the complementary function, you must multiply the trial form by x. For example, if f(x) = e2x and the CF contains A e2x, use yₚ = Cx e2x.
如果假设的特解形式已经出现在通解中，则必须将试探形式乘以 x。例如，若 f(x) = e2x 且通解中已有 A e2x，则取 yₚ = Cx e2x。
9. Initial and Boundary Conditions | 初始条件与边界条件
Once the general solution is found, the arbitrary constants A and B are determined by using extra conditions. Initial conditions specify y and dy/dx at x = 0, while boundary conditions specify y at two different values of x.
在得到一般解后，利用额外条件确定任意常数 A 和 B。初始条件给出 x = 0 时的 y 与 dy/dx；边界条件给出两个不同 x 处的 y 值。
For example, consider y” − 4y = 0 with y(0) = 1 and y'(0) = 2. The auxiliary equation is λ² − 4 = 0, so λ = ±2 and y = A e2x + B e−2x. Then y’ = 2A e2x − 2B e−2x. Substituting x = 0 gives A + B = 1 and 2A − 2B = 2. Solving yields A = 1, B = 0, so y = e2x.
例如，考虑 y” − 4y = 0，且 y(0) = 1、y'(0) = 2。辅助方程为 λ² − 4 = 0，所以 λ = ±2，y = A e2x + B e−2x。则 y’ = 2A e2x − 2B e−2x。代入 x = 0 得 A + B = 1 和 2A − 2B = 2。联立解得 A = 1，B = 0，因此 y = e2x。
10. Damped Oscillations | 阻尼振动
Second-order equations with constant coefficients often appear in physics as models of damped oscillations. A typical equation is
常系数二阶方程在物理中常用来描述阻尼振动。典型方程为
m d²x/dt² + c dx/dt + kx = 0
where m is mass, c is the damping coefficient and k is the spring stiffness. The nature of the motion depends on the three cases:
其中 m 是质量，c 是阻尼系数，k 是弹簧刚度。运动性质取决于三种情形：
- Overdamping: real distinct roots, so the system returns to equilibrium slowly without oscillating.
- 过阻尼：相异实根，系统缓慢回到平衡位置，不发生振荡。
- Critical damping: a repeated root, so the system returns to equilibrium as fast as possible without oscillating.
- 临界阻尼：重根，系统以最快速度回到平衡位置而不发生振荡。
- Underdamping: complex roots, so the solution oscillates while the amplitude decays.
- 欠阻尼：复数根，解发生振荡且振幅逐渐衰减。
In AQA questions, you may be asked to interpret the mathematical solution in this physical context.
在 AQA 试题中，你可能会被要求从物理背景出发解释数学解的意义。
11. Full Worked Example | 完整例题
Solve y” + 4y’ + 5y = 2 sin x.
求解 y” + 4y’ + 5y = 2 sin x。
First find the complementary function. The auxiliary equation is λ² + 4λ + 5 = 0, so λ = −2 ± i. Hence
首先求通解。辅助方程为 λ² + 4λ + 5 = 0，故 λ = −2 ± i。因此
CF = e−2x(A cos x + B sin x)
Since f(x) = 2 sin x, try a particular integral of the form yₚ = p sin x + q cos x. Then yₚ’ = p cos x − q sin x and yₚ” = −p sin x − q cos x.
由于 f(x) = 2 sin x，可设特解 yₚ = p sin x + q cos x。则 yₚ’ = p cos x − q sin x，yₚ” = −p sin x − q cos x。
Substituting into y” + 4y’ + 5y gives
将上述表达式代入 y” + 4y’ + 5y 得
(4p − 4q) sin x + (4p + 4q) cos x = 2 sin x
Comparing coefficients, 4p − 4q = 2 and 4p + 4q = 0. Solving gives p = 1/4 and q = −1/4, so yₚ = ¼ sin x − ¼ cos x.
比较系数，得 4p − 4q = 2 和 4p + 4q = 0。解得 p = 1/4，q = −1/4，因此 yₚ = ¼ sin x − ¼ cos x。
The general solution is therefore
因此一般解为
y = e−2x(A cos x + B sin x) + ¼ sin x − ¼ cos x
12. AQA Exam Tips | AQA 考试提示
- Check whether the equation is homogeneous before deciding whether you need a particular integral.
- 先判断方程是否为齐次，再决定是否需要求特解。
- Write the auxiliary equation carefully: a d²y/dx² becomes aλ², b dy/dx becomes bλ, and cy becomes c.
- 认真写出辅助方程：a d²y/dx² 对应 aλ²，b dy/dx 对应 bλ，cy 对应 c。
- For a non-homogeneous equation, always find the complementary function first, then choose a particular integral using the table.
- 对于非齐次方程，先求通解，再利用表格选择特解形式。
- If the particular integral form overlaps with the complementary function, multiply by x.
- 如果特解形式与通解形式重叠，应乘以 x。
- When using initial or boundary conditions, differentiate
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