A-Level数学 一阶微分方程 分离变量 积分因子 应用
什么是微分方程？What is a Differential Equation?
A differential equation is an equation that relates an unknown function to one or more of its derivatives. Unlike algebraic equations that give you a number as the answer, differential equations give you a function. They are the language of change: whenever a quantity changes in response to something, a differential equation describes that relationship. In A-Level Mathematics, we focus on ordinary differential equations (ODEs) ： equations involving a function of a single variable and its derivatives.
微分方程是联系未知函数及其导数的方程。与代数方程给出一个数字作为答案不同，微分方程给出的答案是一个函数。它们是描述变化的语言：每当一个量因某些因素而变化时，微分方程就描述了这种关系。在A-Level数学中，我们关注常微分方程（ODE）：涉及单变量函数及其导数的方程。
is the highest derivative that appears. A first-order ODE contains only the first derivative dy/dx (or y&# x, y). The degree is the power to which the highest-order derivative is raised, after clearing radicals. First-order equations appear in population growth, cooling, electrical circuits, and mixing problems ： making them one of the most practically useful topics in A-Level Mathematics.
微分方程的阶是其最高导数的阶数。一阶常微分方程仅包含一阶导数dy/dx（或y’）。其一般形式为dy/dx = f(x, y)。方程的次数是最高阶导数在消去根式后的幂次。一阶方程出现在人口增长、冷却、电路和混合问题中：使其成为A-Level数学中最实用的主题之一。
一阶微分方程的分类 Classification of First-Order ODEs
duct of a function of x alone and a function of y alone. Linear first-order ODEs have the form dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x only.
A-Level考试局（Edexcel、AQA、OCR、CAIE）要求掌握两种主要的一阶常微分方程求解方法：可分离方程和积分因子法。可分离方程的形式为dy/dx = g(x)·h(y)，其右侧可分解为仅含x的函数与仅含y的函数的乘积。一阶线性常微分方程的形式为dy/dx + P(x)y = Q(x)，其中P(x)和Q(x)仅为x的函数。
method to a separable equation, or vice versa. Check: can you write the RHS as g(x) × h(y)? f yes, separate. Can you write it as dy/dx + P(x)y = Q(x) with P and Q depending only on x? If yes, use an integrating factor. Some equations ： like dy/dx = xy + x ： can be solved by both methods, giving you a built-in verification.
识别方程类型是第一步也是最关键的一步。考试中常见的陷阱是将积分因子法应用于可分离方程，反之亦然。检查方法：能否将右侧写为g(x) × h(y)？如果可以，用分离变量法。能否将其写为dy/dx + P(x)y = Q(x)，且P和Q仅依赖于x？如果可以，用积分因子法。某些方程：如dy/dx = xy + x：两种方法均可解，为你提供了内在验证。
分离变量法：理论 Separation of Variables: Theory
When a first-order ODE can be written as dy/dx = g(x)·h(y), we “separate&# ntegrate both sides: ∫(1/h(y)) dy = ∫g(x) dx + C. The constant of integration C is crucial ： ution expresses y implicitly or explicitly as a function of x with one arbitrary constant.
当一阶常微分方程可写为dy/dx = g(x)·h(y)时，我们通过将所有含y的项（包括dy）移到一边，所有含x的项（包括dx）移到另一边来”分离”变量：(1/h(y)) dy = g(x) dx。然后两边积分：∫(1/h(y)) dy = ∫g(x) dx + C。积分常数C至关重要：没有它，你只得到特解而非通解。通解将y隐式或显式地表示为含有一个任意常数的x的函数。
/[y(1−y)] = 1/y + 1/(1−y). If partial fractions are a weak point, review them before practising differential equations ： they appear in roughly 40% of A-Level separable-equation problems.
关键的代数技能是部分分式分解。许多A-Level可分离方程导致需要对有理函数进行积分的部分分式分解。例如，像dy/dx = y(1 − y)这样的方程分离后得到∫1/[y(1−y)] dy = ∫dx，左侧积分需要1/[y(1−y)] = 1/y + 1/(1−y)。如果部分分式是你的薄弱点，在练习微分方程之前请复习它们：它们出现在大约40%的A-Level可分离方程问题中。
分离变量法：实例 Worked Examples of Separation
Example 1: Solve dy/dx = 2xy, given y(0) = 3. First, separate: (1/y) dy = 2x dx. Integrate: ln|y| = x² + C. Exponentiate: |y| = ex²+C = eC · ex². Let A = ±eC, so y = A·ex². Apply the initial condition y(0) = 3: 3 = A·e0, so A = 3. The particular solution is y = 3ex².
例1：求解dy/dx = 2xy，已知y(0) = 3。首先分离变量：(1/y) dy = 2x dx。积分：ln|y| = x² + C。取指数：|y| = ex²+C = eC · ex²。令A = ±eC，则y = A·ex²。应用初始条件y(0) = 3：3 = A·e0，故A = 3。特解为y = 3ex²。
Example 2: rate: 1/[y(1−y)] dy = dx. Use partial fractions: 1/[y(1−y)] = 1/y + 1/(1−y). Integrate: ln|y| − ln|1−y| = x + C. Combine logs: ln|y/(1−y)| = x + C. Exponentiate: y/(1−y) = Kex where K = eC. Solve for y: y = Kex/(1+Kex). Apply y(0) = 0.5: 0.5 = K/(1+K), so K = 1. Thus y = ex/(1+ex). This is the logistic function ： fundamental to population modelling.
例2：求解dy/dx = y(1 − y)，y(0) = 0.5。分离变量：1/[y(1−y)] dy = dx。使用部分分式：1/[y(1−y)] = 1/y + 1/(1−y)。积分：ln|y| − ln|1−y| = x + C。合并对数：ln|y/(1−y)| = x + C。取指数：y/(1−y) = Kex，其中K = eC。解出y：y = Kex/(1+Kex)。应用y(0) = 0.5：0.5 = K/(1+K)，故K = 1。因此y = ex/(1+ex)。这是逻辑斯谛函数：种群建模的基础。
积分因子法：理论 Integrating Factor Method: Theory
C, and solve for y. The cleverness of this method is the observation that multiplying by e∫P dx always collapses the left side into a product-rule derivative.
对于标准形式的一阶线性常微分方程dy/dx + P(x)y = Q(x)，积分因子为I(x) = e^(∫P(x) dx)。将整个方程乘以I(x)使左侧成为I(x)·y的精确导数：d/dx[I(x)·y] = I(x)·Q(x)。然后两边积分：I(x)·y = ∫I(x)·Q(x) dx + C，解出y。此方法的巧妙之处在于观察到乘以e∫P dx总是将左侧压缩为乘积规则导数。
n variant: the equation is given as f(x) dy/dx + g(x)y = h(x). You must first divide through by f(x) to obtain the standard form before identifying P(x) and Q(x). Forgetting this step is the number-one error.
积分因子法在所有主要考试局中都涉及。Edexcel通常将其作为纯数学试卷1中的独立题目呈现，而CAIE常将其嵌入结合微分方程与分部积分或代换的多部分题目中。一个常见变体：方程以f(x) dy/dx + g(x)y = h(x)的形式给出。你必须先除以f(x)得到标准形式，然后再识别P(x)和Q(x)。忘记这一步是头号错误。
积分因子法：实例 Worked Examples of Integrating Factor
Example 3: lve dy/dx + 2xy = x, with y(0) = 1. Here P(x) = 2x, Q(x) = x. Integrating factor: I(x) = e∫2x dx = ex². Multiply through: ex² dy/dx + 2x ex² y = x ex². The left side is d/dx[y·ex²] = x ex². Integrate: y·ex² = ∫x ex² dx = (1/2)ex² + C. Solve: y = 1/2 + C·e−x². Apply y(0) = 1: 1 = 1/2 + C, so C = 1/2. Final answer: y = 1/2 + (1/2)e−x² = (1/2)(1 + e−x²).
例3：求解dy/dx + 2xy = x，y(0) = 1。此处P(x) = 2x，Q(x) = x。积分因子：I(x) = e∫2x dx = ex²。乘以积分因子：ex² dy/dx + 2x ex² y = x ex²。左侧为d/dx[y·ex²] = x ex²。积分：y·ex² = ∫x ex² dx = (1/2)ex² + C。求解：y = 1/2 + C·e−x²。应用y(0) = 1：1 = 1/2 + C，故C = 1/2。最终答案：y = 1/2 + (1/2)e−x² = (1/2)(1 + e−x²)。
Example 4: y/dx + (2/x)y = 4x. So P(x) = 2/x, Q(x) = 4x. Integrating factor: I(x) = e^(∫(2/x) dx) = e2 ln|x| y = x⁴ + C. Thus y = x² + C/x². Apply y(1) = 2: 2 = 1 + C, so C = 1. Final: y = x² + 1/x².
例4：求解x·dy/dx + 2y = 4x²，y(1) = 2。首先化为标准形式：除以x得dy/dx + (2/x)y = 4x。故P(x) = 2/x，Q(x) = 4x。积分因子：I(x) = e^(∫(2/x) dx) = e2 ln|x| = x²。乘以积分因子：x² dy/dx + 2x y = 4x³。左侧为d/dx[x²·y] = 4x³。积分：x²·y = x⁴ + C。因此y = x² + C/x²。应用y(1) = 2：2 = 1 + C，故C = 1。最终：y = x² + 1/x²。
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