Exercise 26B: Solving First-Order Differential Equations by Separation of Variables | 练习26B：用分离变量法解一阶微分方程

📚 Exercise 26B: Solving First-Order Differential Equations by Separation of Variables | 练习26B：用分离变量法解一阶微分方程

Exercise 26B in the IB Mathematics curriculum focuses on a fundamental technique for solving first-order differential equations: separation of variables. This method is essential for tackling problems involving growth, decay, and many physical models. In this article, we will explore the theory behind separable equations, walk through worked examples, highlight common mistakes, and provide practical tips for mastering Exercise 26B.

IB数学课程中的练习26B重点讲解求解一阶微分方程的基本技巧：分离变量法。这一方法对于处理增长、衰减以及许多物理模型问题至关重要。在本文中，我们将探讨可分离方程背后的理论，逐步讲解解题示例，强调常见错误，并提供掌握练习26B的实用建议。

1. Introduction to Differential Equations | 微分方程简介

A differential equation is an equation that involves an unknown function and its derivatives. In IB Mathematics, we typically encounter first-order ordinary differential equations (ODEs) where the highest derivative is the first derivative. These equations describe how a quantity changes with respect to another, such as time. For example, dy/dx = 2x states that the rate of change of y with respect to x is 2x.

微分方程是包含未知函数及其导数的方程。在IB数学中，我们通常遇到一阶常微分方程，其中最高阶导数为一次导数。这些方程描述一个量如何随另一个量（例如时间）变化。例如，dy/dx = 2x 表示 y 关于 x 的变化率为 2x。

2. Separable Differential Equations | 可分离变量的微分方程

A differential equation is called separable if it can be written in the form dy/dx = g(x)h(y), where the right-hand side is a product of a function of x and a function of y. The key idea is to rearrange the equation so that all terms involving y appear on one side with dy, and all terms involving x appear on the other side with dx. Then both sides can be integrated.

若一个微分方程可写成 dy/dx = g(x)h(y) 的形式，即右侧是 x 的函数与 y 的函数的乘积，则称其为可分离变量的。关键思路是通过移项，使得所有包含 y 的项与 dy 在一侧，所有包含 x 的项与 dx 在另一侧。然后对两边积分。

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3. Step-by-Step Separation Method | 分离变量法的步骤分解

Step 1: Write the differential equation as dy/dx = g(x)h(y). Step 2: Separate the variables: (1/h(y)) dy = g(x) dx, provided h(y) ≠ 0. Step 3: Integrate both sides: ∫ (1/h(y)) dy = ∫ g(x) dx. Step 4: Solve for y if possible, to obtain an explicit solution; otherwise, leave the solution in implicit form. Step 5: If an initial condition is given, substitute to find the constant of integration.

步骤1：将微分方程写成 dy/dx = g(x)h(y)。步骤2：分离变量：(1/h(y)) dy = g(x) dx，前提是 h(y) ≠ 0。步骤3：两边积分：∫ (1/h(y)) dy = ∫ g(x) dx。步骤4：如有可能，解出 y 得显式解；否则保留隐式形式。步骤5：若给出初始条件，代入求积分常数。

4. Example 1: Exponential Growth and Decay | 例1：指数增长与衰减

Consider the differential equation dy/dt = k y, where k is a constant. This models unlimited population growth (k > 0) or radioactive decay (k < 0). Separate variables: (1/y) dy = k dt. Integrate: ∫ (1/y) dy = ∫ k dt → ln|y| = k t + C. Solve for y: |y| = e^{k t + C} = e^C e^{k t}. Let A = ± e^C, then y = A e^{k t}. If an initial condition y(0) = y₀ is given, we find A = y₀, so y = y₀ e^{k t}.

考虑微分方程 dy/dt = k y，其中 k 为常数。此模型描述无限制种群增长（k > 0）或放射性衰变（k < 0）。分离变量：(1/y) dy = k dt。积分：∫ (1/y) dy = ∫ k dt → ln|y| = k t + C。解出 y：|y| = e^{k t + C} = e^C e^{k t}。令 A = ± e^C，则 y = A e^{k t}。若给定初始条件 y(0) = y₀，可得 A = y₀，于是 y = y₀ e^{k t}。

5. Example 2: Logistic-Type Equation | 例2：逻辑型方程

Solve dy/dx = (y – 1)(x + 2). Separate: dy/(y – 1) = (x + 2) dx, provided y ≠ 1. Integrate: ∫ dy/(y – 1) = ∫ (x + 2) dx → ln|y Published by TutorHao | IB Mathematics Revision Series | aleveler.com

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