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Second Order Differential Equations | 二阶微分方程 考点精讲

2026-07-12 23:21 · 本文部分内容由AI辅助创作

📚 Second Order Differential Equations | 二阶微分方程 考点精讲

Second order differential equations are a key topic in AQA IGCSE Further Mathematics. They model many physical phenomena and require a systematic approach to find general solutions. Mastering the characteristic equation, types of roots, and the method of undetermined coefficients is essential for exam success.

二阶微分方程是AQA IGCSE进阶数学中的一个关键主题。它们可以模拟许多物理现象，需要系统性的方法来求解通解。掌握特征方程、根的类型以及待定系数法对于考试成功至关重要。

1. Introduction to Second Order ODEs | 二阶常微分方程简介

A second order ordinary differential equation (ODE) involves an unknown function y(x) and its derivatives up to the second order, d2y/dx2. The most general linear second order ODE with constant coefficients is a d2y/dx2 stants and a ≠ 0. If f(x) = 0, the equation is homogeneous; otherwise it is inhomogeneous.

二阶常微分方程（ODE）涉及未知函数y(x)及其最高二阶导数d2y/dx2。最常见的常系数线性二阶ODE形式为 a d2y/dx2 + b dy/dx + c y = f(x)，其中a, b, c为常数且a ≠ 0。如果f(x)=0，方程为齐次的；否则为非齐次的。

In AQA exams, you will mainly work with constant coefficient equations and be asked to find general solutions, particular solutions, or interpret solutions in context.

在AQA考试中，你主要处理常系数方程，并被要求求通解、特解或在具体情境下解读解。

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2. Homogeneous Linear Equations with Constant Coefficients | 常系数齐次线性方程

For a homogeneous equation a y'' + b y' + c y = 0, we seek solutions of the form y = erx. Substituting this into the equation gives a r2 erx + b r erx + c erx = 0, which factors to erx(a r2 + b r + c) = 0. Since erx is never zero, we obtain the characteristic equation a r2 + b r + c = 0. The roots of this quadratic determine the form of the general solution.

对于齐次方程 a y'' + b y' + c y = 0，我们寻找形如 y = erx 的解。将其代入方程得 a r2 erx + b r erx + c erx = 0，可因式分解为 erx(a r2 + b r + c) = 0。因为 erx 永不为零，我们得到特征方程 a r2 + b r + c = 0。该二次方程的根决定了通解的形式。

3. The Characteristic Equation | 特征方程

The characteristic equation is a quadratic: a r2 + b r + c = 0. Its discriminant Δ = b2 - 4ac governs the nature of the roots. Three cases arise, each yielding a distinct general solution for the complementary function yc.

特征方程是一个二次方程：a r2 + b r + c = 0。其判别式 Δ = b2 - 4ac 决定了根的性质。共有三种情况，每种情况给出补函数 yc 的不同通解。