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

A second order differential equation involves a function and its first and second derivatives. In many mathematical and physical problems, such equations arise naturally from geometry or from fundamental laws of physics and engineering. Learning how to ‘form’ these equations — rather than just solve them — is an essential skill for A-Level Further Mathematics.

二阶微分方程涉及一个函数及其一阶和二阶导数。在许多数学与物理问题中，这类方程会自然地产生于几何条件或物理/工程基本定律。学习如何“建立”这些方程——而不仅仅是求解它们——是 A-Level 进阶数学中一项关键技能。

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

An ordinary differential equation (ODE) is said to be of second order if it contains the second derivative of the unknown function but no higher derivatives. Its standard linear form is shown below.

如果一个常微分方程（ODE）包含未知函数的二阶导数，而不含更高阶导数，则称其为二阶微分方程。其标准线性形式如下。

a(x) d²y/dx² + b(x) dy/dx + c(x)y = f(x)

Here a(x), b(x) and c(x) are coefficient functions, and f(x) is the forcing term. If f(x) = 0, the equation is homogeneous; otherwise it is non-homogeneous.

其中，a(x)、b(x) 和 c(x) 是系数函数，f(x) 是强迫项。当 f(x) = 0 时，方程为齐次的；否则为非齐次方程。

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2. General Solution and Arbitrary Constants | 通解与任意常数

A second order ODE generally has a general solution containing two arbitrary constants. This is because integration is performed twice in obtaining the solution, introducing two independent constants of integration.

二阶微分方程的通解通常包含两个任意常数。这是因为求解过程中需要积分两次，每次都会引入一个独立的积分常数。

For example, the simple equation d²y/dx² = 0 has general solution y = Ax + B, where A and B are arbitrary constants.

例如，简单方程 d²y/dx² = 0 的通解为 y = Ax + B，其中 A 和 B 为任意常数。

Conversely, if a family of curves is defined by an equation with two arbitrary constants, we can eliminate these constants to obtain the corresponding second order ODE. This process is the central idea of ‘formation’.

反过来，如果一个曲线族由含有两个任意常数的方程定义，我们可以通过消去这些常数来得到相应的二阶微分方程。这一过程正是“建立方程”的核心思想。

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3. Method 1: Eliminating Arbitrary Constants | 方法一：消去任意常数

Given a relation y = F(x, A, B) that contains two arbitrary constants A and B, we differentiate twice with respect to x:

给定一个含有两个任意常数 A 和 B 的关系式 y = F(x, A, B)，我们对 x 求两次导数：

dy/dx = F₁(x, A, B), d²y/dx² = F₂(x, A, B)

The three equations (for y, y’ and y”) are then combined by eliminating A and B. The result is a second order differential equation that is satisfied by every curve in the family.

然后联立这 3 个方程（y、y’、y”），消去 A 和 B，得到该曲线族中每条曲线都满足的二阶微分方程。

In practice, this elimination often requires solving for A and B in terms of y and y’, then substituting into y”. Alternatively, if the form of the general solution is known, the equation can be written directly from the characteristic roots.

实际操作中，往往需要先用 y 和 y’ 表示 A 和 B，再将它们代入 y”。另一种方法是，如果已知通解的形式，可以依据特征根直接写出方程。

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4. Worked Example: Exponential Form | 例题：指数型通解

Form the differential equation whose general solution is y = A e^(2x) + B e^(-3x).

建立以 y = A e^(2x) + B e^(-3x) 为通解的微分方程。

Differentiate the given solution twice:

对给定通解求两次导数：

y’ = 2A e^(2x) – 3B e^(-3x), y” = 4A e^(2x) + 9B e^(-3x)

From the first two equations we can solve for A e^(2x) and B e^(-3x):

由前两个方程，我们解出 A e^(2x) 与 B e^(-3x)：

A e^(2x) = (3y + y’)/5, B e^(-3x) = (2y – y’)/5

Substituting these into the expression for y” gives:

将它们代入 y” 的表达式：

y” = 4[(3y + y’)/5] + 9[(2y – y’)/5] = 6y – y’

Thus the required equation is y” + y’ – 6y = 0.

因此所求方程为 y” + y’ – 6y = 0。

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5. Worked Example: Trigonometric Form | 例题：三角型通解

Form the differential equation from y = A cos 3x + B sin 3x.

由 y = A cos 3x + B sin 3x 建立微分方程。

Differentiating once and twice gives:

一次和二次求导得到：

y’ = -3A sin 3x + 3B cos

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