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

📚 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 + b dy/dx + c y = f(x), where a, b, c are constants 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。

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 = e^rx. Substituting this into the equation gives a r² e^rx + b r e^rx + c e^rx = 0, which factors to e^rx(a r² + b r + c) = 0. Since e^rx 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 = e^rx 的解。将其代入方程得 a r² e^rx + b r e^rx + c e^rx = 0，可因式分解为 e^rx(a r² + b r + c) = 0。因为 e^rx 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 y_c.

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 y_c 的不同通解。

• Real and distinct roots (Δ > 0, r₁ ≠ r₂): y_c = A e^r₁x + B e^r₂x. 实相异根：y_c = A e^r₁x + B e^r₂x.
• Repeated real root (Δ = 0, r = r₁ = r₂): y_c = (A + Bx) e^rx. 重实

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