Second Order Differential Equations | 二阶微分方程

📚 Second Order Differential Equations | 二阶微分方程

Second order differential equations are a core part of AQA A-level mathematics, especially in the pure and mechanics strands. They model many real-world systems such as oscillations, electrical circuits and mechanical vibrations.

二阶微分方程是 AQA A-level 数学的核心内容之一，尤其是在纯数学和力学部分。它们可以描述许多现实系统，例如振动、电路和机械振荡。

In this revision guide, we will work through the standard methods for solving linear second order differential equations with constant coefficients, including homogeneous and non-homogeneous cases.

在这份复习指南中，我们将系统讲解常系数线性二阶微分方程的标准解法，包括齐次与非齐次情形。

1. What is a Second Order Differential Equation? | 什么是二阶微分方程？

A second order differential equation involves an unknown function y(x) and its derivatives up to the second derivative, d²y/dx².

二阶微分方程包含未知函数 y(x) 及其最高为二阶导数的导数 d²y/dx²。

In this topic we focus on linear equations with constant coefficients, written in the general form

在本主题中，我们重点研究常系数线性方程，其一般形式为

a d²y/dx² + b dy/dx + c y = f(x)

where a, b and c are constants, with a ≠ 0.

其中 a、b 和 c 为常数，且 a ≠ 0。

If f(x) = 0, the equation is called homogeneous; otherwise it is called non-homogeneous.

如果 f(x) = 0，则该方程称为齐次方程；否则称为非齐次方程。

The general solution of a non-homogeneous equation is found in two parts: the complementary function and a particular integral.

非齐次方程的通解由两部分组成：余函数和一个特解。

2. The Auxiliary Equation | 辅助方程

For the homogeneous equation a d²y/dx² + b dy/dx + c y = 0, we try a solution of the form y = e^(λx).

对于齐次方程 a d²y/dx² + b dy/dx + c y = 0，我们尝试形如 y = e^(λx) 的解。

Substituting this into the equation gives the characteristic equation, also called the auxiliary equation:

将其代入方程可得到特征方程，也称为辅助方程：

aλ² + bλ + c = 0

The roots of this quadratic equation determine the form of the complementary function (CF).

这个二次方程的根决定了余函数（CF）的形式。

There are three possible cases for the roots: two distinct real roots, one repeated real root, or a pair of complex conjugate roots.

根有三种可能的情形：两个不相等的实根、一个重实根，或一对共轭复根。

3. Case 1: Real Distinct Roots | 情形一：实且不相等根

If the auxiliary equation has two distinct real roots λ₁ and λ₂, the complementary function is

如果辅助方程有两个不相等的实根 λ₁ 和 λ₂，则余函数为

y = A e^(λ₁x) + B e^(λ₂x)

where A and B are arbitrary constants.

其中 A 和 B 为任意常数。

For example, the equation y” – 5y’ + 6y = 0 has auxiliary equation λ² – 5λ + 6 = 0, so λ = 2 and λ = 3.

例如，方程 y” – 5y’ + 6y = 0 的辅助方程为 λ² – 5λ + 6 = 0，因此 λ = 2 和 λ = 3。

Therefore the complementary function is y = A e^(2x) + B e^(3x).

因此余函数为 y = A e^(2x) + B e^(3x)。

The constants A and B are found later using initial or boundary conditions.

常数 A 和 B 将在之后利用初始条件或边界条件求出。

4. Case 2: Repeated Real Roots | 情形二：实重根

If the auxiliary equation has a repeated real root λ, so that the roots are λ₁ = λ₂ = λ, the complementary function is

如果辅助方程有一个实重根 λ，即 λ₁ = λ₂ = λ，则余函数为

y = (A + Bx) e^(λx)

where A and B are arbitrary constants.

其中 A 和 B 为任意常数。

The extra factor x is necessary because y = A e^(λx) alone would only give one independent solution.

额外的因子 x 是必要的，因为仅有 y = A e^(λx) 只能给出一个独立解。

For example, y” – 6y’ + 9y = 0 gives λ² – 6λ + 9 = 0, so λ = 3 is a repeated root.

例如，y” – 6y’ + 9y = 0 对应 λ² – 6λ + 9 = 0，因此 λ = 3 为重根。

The complementary function is y = (A + Bx) e^(3x).

余函数为 y = (A + Bx) e^(3x)。

5. Case 3: Complex Conjugate Roots | 情形三：共轭复根

If the auxiliary equation has complex conjugate roots of the form λ = p ± iq, where p and q are real and q ≠ 0, the complementary function is

如果辅助方程有形如 λ = p ± iq 的共轭复根，其中 p 和 q 为实数且 q ≠ 0，则余函数为

y = e^(px) (A cos qx + B sin qx)

where A and B are arbitrary constants.

其中 A 和 B 为任意常数。

This form comes from Euler’s formula e^(ix) = cos x + i sin x, and it avoids the use of complex exponentials in the final answer.

这种形式来源于欧拉公式 e^(ix) = cos x + i sin x，它避免了在最终答案中使用复指数。

For example, y” + 4y’ + 13y = 0 gives λ² + 4λ + 13 = 0, which has roots λ = -2 ± 3i.

例如，y” + 4y’ + 13y = 0 对应 λ² + 4λ + 13 = 0，其根为 λ = -2 ± 3i。

So p = -2 and q = 3, and the complementary function is y = e^(-2x) (A cos 3x + B sin 3x).

因此 p = -2，q = 3，余函数为 y = e^(-2x) (A cos 3x + B sin 3x)。

6. Non-Homogeneous Equations and Particular Integrals | 非齐次方程与特解

For the non-homogeneous equation a d²y/dx² + b dy/dx + c y = f(x), the general solution is

对于非齐次方程 a d²y/dx² + b dy/dx + c y = f(x)，通解为

y = complementary function + particular integral

The complementary function solves the associated homogeneous equation, while the particular integral is any one function that satisfies the full non-homogeneous equation.

余函数满足相应的齐次方程，而特解是满足完整非齐次方程的任意一个函数。

To find the particular integral, we use a trial function whose form depends on the type of f(x).

为了求特解，我们根据 f(x) 的类型使用不同形式的试探函数。

If f(x) is a sum of different types, the particular integral is the sum of the individual trial functions.

如果 f(x) 是不同类型函数之和，则特解是各个试探函数之和。

7. Trial Functions for Common f(x) | 常见 f(x) 的试探函数

The table below summarises the standard trial functions used for a particular integral.

下表总结了求特解时使用的标准试探函数。

f(x)	Trial function \| 试探函数	Notes \| 说明
Polynomial of degree n, e.g. ax² + bx + c	General polynomial of degree n, e.g. Px² + Qx + R	Include all lower powers even if some are missing.
k e^(rx)	C e^(rx)	Use C, not k.
k cos ωx or k sin ωx	C cos ωx + D sin ωx	Include both cosine and sine terms.
Product, e.g. e^(rx) cos ωx	e^(rx) (C cos ωx + D sin ωx)	Keep exponential factor unchanged.

If any term of the trial function already appears in the complementary function, multiply the trial function by x until no term overlaps with the CF.

如果试探函数中的任何一项已经出现在余函数中，则应将试探函数乘以 x，直到与余函数没有重复项为止。

8. Finding a Particular Integral: Examples | 求特解：例题

Consider the equation y” – 5y’ + 6y = 12 e^(4x).

考虑方程 y” – 5y’ + 6y = 12 e^(4x)。

The complementary function is y = A e^(2x) + B e^(3x), as found earlier.

如前所述，余函数为 y = A e^(2x) + B e^(3x)。

Since f(x) = 12 e^(4x), we try a particular integral of the form y_p = C e^(4x).

因为 f(x) = 12 e^(4x)，我们尝试使用特解 y_p = C e^(4x)。

Substituting y_p into the equation gives 16C e^(4x) – 20C e^(4x) + 6C e^(4x) = 12 e^(4x), so 2C = 12 and C = 6.

将 y_p 代入方程得到 16C e^(4x) – 20C e^(4x) + 6C e^(4x) = 12 e^(4x)，因此 2C = 12，C = 6。

The general solution is y = A e^(2x) + B e^(3x) + 6 e^(4x).

通解为 y = A e^(2x) + B e^(3x) + 6 e^(4x)。

For a trigonometric example, consider y” + 9y = 5 sin 2x.

对于三角函数的例子，考虑 y” + 9y = 5 sin 2x。

The complementary function is y = A cos 3x + B sin 3x, since the auxiliary equation λ² + 9 = 0 gives λ = ±3i.

余函数为 y = A cos 3x + B sin 3x，因为辅助方程 λ² + 9 = 0 给出 λ = ±3i。

For the particular integral we try y_p = C cos 2x + D sin 2x.

对于特解，我们尝试 y_p = C cos 2x + D sin 2x。

Substituting and equating coefficients gives C = 0 and D = 1, so y_p = sin 2x.

代入并比较系数可得 C = 0，D = 1，因此 y_p = sin 2x。

The general solution is y = A cos 3x + B sin 3x + sin 2x.

通解为 y = A cos 3x + B sin 3x + sin 2x。

9. Initial and Boundary Conditions | 初始条件与边界条件

To determine the arbitrary constants A and B, we need extra information, usually given as initial or boundary conditions.

为了确定任意常数 A 和 B，我们需要额外信息，通常以初始条件或边界条件给出。

Common conditions specify the value of y and the value of dy/dx at a particular point, often x = 0.

常见条件会给出在某一点（通常是 x = 0）处 y 的值和 dy/dx 的值。

For example, if y = A e^(2x) + B e^(3x) + 6 e^(4x) and we are given y(0) = 8 and y'(0) = 30, we first substitute x = 0.

例如，如果 y = A e^(2x) + B e^(3x) + 6 e^(4x)，且已知 y(0) = 8 和 y'(0) = 30，我们先代入 x = 0。

This gives A + B + 6 = 8, so A + B = 2.

由此得到 A + B + 6 = 8，因此 A + B = 2。

Differentiating, y’ = 2A e^(2x) + 3B e^(3x) + 24 e^(4x), so at x = 0 we get 2A + 3B + 24 = 30, hence 2A + 3B = 6.

求导得 y’ = 2A e^(2x) + 3B e^(3x) + 24 e^(4x)，因此在 x = 0 处有 2A + 3B + 24 = 30，即 2A + 3B = 6。

Solving the simultaneous equations A + B = 2 and 2A + 3B = 6 gives A = 0 and B = 2.

解联立方程 A + B = 2 与 2A + 3B = 6 得到 A = 0，B = 2。

The particular solution is therefore y = 2 e^(3x) + 6 e^(4x).

因此特解为 y = 2 e^(3x) + 6 e^(4x)。

10. Exam Strategy and Common Mistakes | 考试策略与常见错误

When answering a second order differential equation question, follow a clear step-by-step method.

在解答二阶微分方程题目时，应遵循清晰的逐步方法。

First, write the equation in standard form and identify a, b, c and f(x).

首先，将方程写成标准形式，并确定 a、b、c 和 f(x)。

Second, solve the auxiliary equation to find the complementary function.

其次，解辅助方程以求出余函数。

Third, choose an appropriate trial function for the particular integral and substitute it carefully.

第三，为特解选择合适的试探函数，并仔细代入。

Fourth, assemble the general solution and apply any given conditions to find the constants.

第四，组合出通解，并应用给定条件求出常数。

A common mistake is forgetting to include both the complementary function and the particular integral in the final answer.

常见错误之一是最终答案中忘记同时包含余函数和特解。

Another common mistake is using a trial function that overlaps with the complementary function without multiplying by x.

另一个常见错误是试探函数与余函数出现重叠，却没有乘以 x。

Also be careful when differentiating products such as (A + Bx) e^(λx) or e^(px) cos qx, as small slip-ups often lead to incorrect constants.

此外，在求导如 (A + Bx) e^(λx) 或 e^(px) cos qx 这类乘积时要格外小心，因为小失误常常导致常数错误。

Always check your final solution by substituting back into the original differential equation if time allows.

如果时间允许，应始终将最终解代回原微分方程进行检查。

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Second Order Differential Equations | 二阶微分方程