Formation of Second Order Differential Equations | 二阶微分方程的建立

📚 Formation of Second Order Differential Equations | 二阶微分方程的建立

In AQA A-Level Mathematics, second order differential equations appear in mechanics, oscillations and exponential growth models. This topic focuses on how such equations are formed, either by eliminating arbitrary constants from a given solution or by translating a physical law into a differential equation.

在 AQA A-Level 数学中，二阶微分方程出现在力学、振动和指数增长模型中。本主题重点介绍二阶微分方程是如何建立的，包括从已知解中消去任意常数，或将物理定律转化为微分方程。

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 and its second derivative d²y/dx², often together with y’ and y. The most common linear form at A-Level is shown below, where a, b, c are constants and f(x) is a function of x.

二阶微分方程包含未知函数 y 及其二阶导数 d²y/dx²，通常还包含 y’ 和 y。A-Level 中最常见的线性形式如下所示，其中 a、b、c 为常数，f(x) 是 x 的函数。

a d²y/dx² + b dy/dx + cy = f(x)

When f(x) = 0, the equation is homogeneous. Many formation questions lead to homogeneous equations because the given solution contains only exponential or trigonometric terms.

当 f(x) = 0 时，方程为齐次方程。许多建立方程的题目都会得到齐次方程，因为给定解只包含指数项或三角函数项。

2. Why Two Arbitrary Constants Appear | 为什么会出现两个任意常数

The general solution of a second order differential equation contains two arbitrary constants, usually written as A and B. This is because the equation involves a second derivative, so two integrations are needed to recover y from d²y/dx².

二阶微分方程的通解包含两个任意常数，通常写作 A 和 B。这是因为方程含有二阶导数，因此需要两次积分才能从 d²y/dx² 还原出 y。

For example, y = Ae^(2x) + Be^(-3x) is a general solution. The constants A and B can be determined only if two boundary or initial conditions are given.

例如，y = Ae^(2x) + Be^(-3x) 是一个通解。只有给定两个边界条件或初始条件，才能确定常数 A 和 B。

3. The Reverse Process: From Solution to Equation | 逆向过程：从解到方程

Normally we solve a differential equation to find its general solution. In this topic, the process is reversed: you are given a general solution and asked to find the second order differential equation that produced it.

通常我们求解微分方程以得到通解。本主题中的过程正好相反：给定一个通解，要求找出产生该通解的二阶微分方程。

This is useful because many physical systems are first observed through their solutions, such as oscillations, and the differential equation reveals the underlying relationship between displacement, velocity and acceleration.

这很有用，因为许多物理系统首先通过其解被观察到，例如振动，而微分方程揭示了位移、速度和加速度之间的基本关系。

4. Method: Eliminating Arbitrary Constants | 方法：消去任意常数

To form a second order differential equation from a general solution, follow these steps:

要从通解建立二阶微分方程，请遵循以下步骤：

Step 1: Differentiate the given solution once to obtain y’. | 步骤 1：对给定解求导一次，得到 y’。
Step 2: Differentiate again to obtain y”. | 步骤 2：再求导一次，得到 y”。
Step 3: Combine y, y’ and y” to eliminate the two arbitrary constants. | 步骤 3：将 y、y’ 和 y” 组合，消去两个任意常数。
Step 4: Simplify the result into the standard linear form. | 步骤 4：将结果化简为标准线性形式。

Because there are two constants, you must use at least the second derivative. The final equation should contain no A or B.

由于有两个常数，必须至少用到二阶导数。最终方程中不应含有 A 或 B。

5. Worked Example 1: Exponential Solutions | 例题 1：指数解

Let y = Ae^(2x) + Be^(-3x). Differentiate once and twice:

设 y = Ae^(2x) + Be^(-3x)。求导一次和两次：

y’ = 2Ae^(2x) − 3Be^(-3x), y” = 4Ae^(2x) + 9Be^(-3x)

To eliminate A and B, combine y”, y’ and y so that the coefficients of each exponential term become zero. The combination y” + y’ − 6y gives:

为了消去 A 和 B，将 y”、y’ 和 y 组合，使每个指数项的系数都变为 0。组合 y” + y’ − 6y 得到：

y” + y’ − 6y = (4+2−6)Ae^(2x) + (9−3−6)Be^(-3x) = 0

Hence the required second order differential equation is:

因此所求的二阶微分方程为：

y” + y’ − 6y = 0

This equation has auxiliary equation r² + r − 6 = 0, whose roots are r = 2 and r = −3, matching the original solution.

该方程的辅助方程为 r² + r − 6 = 0，其根为 r = 2 和 r = −3，与原始解一致。

6. Worked Example 2: Trigonometric Solutions | 例题 2：三角函数解

Consider the general solution y = A cos 3x + B sin 3x. Differentiate twice:

考虑通解 y = A cos 3x + B sin 3x。求导两次：

y’ = −3A sin 3x + 3B cos 3x

y” = −9A cos 3x − 9B sin 3x = −9y

Because y” = −9y, the differential equation is immediately obtained:

因为 y” = −9y，所以可以直接得到微分方程：

y” + 9y = 0

This is the standard form of simple harmonic motion with angular frequency 3. The constants A and B are set by initial displacement and velocity.

这是角频率为 3 的简谐运动标准形式。常数 A 和 B 由初始位移和速度决定。

7. Worked Example 3: Repeated Root Case | 例题 3：重根情形

When the auxiliary equation has a repeated root, the general solution contains the term (A + Bx)e^(2x). Let y = (A + Bx)e^(2x).

当辅助方程有重根时，通解包含 (A + Bx)e^(2x) 这一项。设 y = (A + Bx)e^(2x)。

Differentiate using the product rule:

使用乘法法则求导：

y’ = (B + 2A + 2Bx)e^(2x)

y” = (4A + 4B + 4Bx)e^(2x)

Now combine y” − 4y’ + 4y:

现在组合 y” − 4y’ + 4y：

y” − 4y’ + 4y = (4A+4B+4Bx − 8A−4B−8Bx + 4A+4Bx)e^(2x) = 0

Therefore the differential equation is:

因此微分方程为：

y” − 4y’ + 4y = 0

This has auxiliary equation r² − 4r + 4 = 0, giving the repeated root r = 2.

该方程的辅助方程为 r² − 4r + 4 = 0，得到重根 r = 2。

8. Forming Equations from Mechanics: Simple Harmonic Motion | 从力学建立方程：简谐运动

One of the most common physical applications is simple harmonic motion. For a particle of mass m attached to a spring with stiffness k, Hooke’s law gives the restoring force F = −kx, where x is displacement.

最常见的物理应用之一是简谐运动。对于质量为 m、连接在刚度为 k 的弹簧上的质点，胡克定律给出回复力 F = −kx，其中 x 为位移。

Newton’s second law states that F = ma, and acceleration a is the second derivative of displacement, a = d²x/dt². Equating forces gives:

牛顿第二定律指出 F = ma，加速度 a 是位移的二阶导数，a = d²x/dt²。令两力相等得到：

m d²x/dt² = −kx ⇒ d²x/dt² + (k/m)x = 0

This is usually written as d²x/dt² + ω²x = 0, where ω² = k/m. It is a second order differential equation formed directly from a physical law.

这通常写作 d²x/dt² + ω²x = 0，其中 ω² = k/m。这是一个直接由物理定律建立的二阶微分方程。

9. Newton’s Second Law and Damped Motion | 牛顿第二定律与阻尼运动

If a damping force proportional to velocity is also present, Newton’s second law becomes:

如果还存在与速度成正比的阻尼力，牛顿第二定律变为：

m d²x/dt² = −λ dx/dt − kx

Rearranging gives the standard damped motion equation:

整理后得到标准阻尼运动方程：

m x” + λ x’ + kx = 0

Here x” represents acceleration and x’ represents velocity. The negative signs show that both damping and spring forces oppose the motion.

这里 x” 表示加速度，x’ 表示速度。负号表示阻尼力和弹簧力都阻碍运动。

In exam questions, you may be asked to form this equation from a worded description of forces. Always link acceleration to d²x/dt² and velocity to dx/dt.

在考试题中，可能会要求你根据力的文字描述建立该方程。务必始终将加速度与 d²x/dt² 联系，将速度与 dx/dt 联系。

10. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Students often lose marks by making small algebraic slips when eliminating constants. The following points can help avoid these errors.

学生在消去常数时常因小的代数失误而失分。以下几点有助于避免这些错误。

Differentiate carefully, especially with the product rule when x appears in the solution. | 求导要仔细，尤其是当解中出现 x 时，要正确使用乘法法则。
Remember that two arbitrary constants require you to find y” as well as y’. | 记住两个任意常数需要同时求出 y” 和 y’。
Do not leave A or B in the final equation; they should cancel completely. | 最终方程中不要留下 A 或 B；它们应完全消去。
Check your answer by solving the differential equation or by substituting the original solution back. | 通过求解微分方程或将原解代回方程来检查答案。
In mechanics problems, use x for displacement and t for time, not y and x. | 在力学问题中，用 x 表示位移、t 表示时间，不要用 y 和 x。

11. Practice Questions | 练习题

Question 1: Find the second order differential equation whose general solution is y = Ae^(4x) + Be^(-x).

问题 1：求通解为 y = Ae^(4x) + Be^(-x) 的二阶微分方程。

Answer: y” − 3y’ − 4y = 0. | 答案：y” − 3y’ − 4y = 0。

Question 2: Find the second order differential equation with general solution y = A cos 2x + B sin 2x.

问题 2：求通解为 y = A cos 2x + B sin 2x 的二阶微分方程。

Answer: y” + 4y = 0. | 答案：y” + 4y = 0。

Question 3: A particle of mass m on a spring of stiffness k moves under Hooke’s law. Show that its displacement x satisfies d²x/dt² + (k/m)x = 0.

问题 3：质量为 m 的质点在刚度为 k 的弹簧上按胡克定律运动。证明其位移 x 满足 d²x/dt² + (k/m)x = 0。

Answer: Use F = ma with F = −kx and a = d²x/dt². | 答案：使用 F = ma，其中 F = −kx，a = d²x/dt²。

12. Summary | 总结

A second order differential equation has a general solution containing two arbitrary constants. To form the equation from a given solution, differentiate twice and combine y, y’ and y” to eliminate the constants.

二阶微分方程的通解包含两个任意常数。要从给定解建立方程，需对其求导两次，并将 y、y’ 和 y” 组合以消去常数。

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Formation of Second Order Differential Equations | 二阶微分方程的建立