A-Level数学 一阶微分方程 积分因子 应用

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。

The order of a differential equation is the highest derivative that appears. A first-order ODE contains only the first derivative dy/dx (or y’). The general form is dy/dx = f(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

A-Level exam boards (Edexcel, AQA, OCR, CAIE) expect you to solve two main types of first-order ODEs analytically: separable equations and linear equations solvable by an integrating factor. Separable equations have the form dy/dx = g(x)·h(y), where the right-hand side factorises into a product 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。

Recognising the type is the first and most critical step. A common exam pitfall is applying the integrating factor method to a separable equation, or vice versa. Check: can you write the RHS as g(x) × h(y)? If 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” the variables by moving all y-terms (including dy) to one side and all x-terms (including dx) to the other: (1/h(y)) dy = g(x) dx. Then integrate both sides: ∫(1/h(y)) dy = ∫g(x) dx + C. The constant of integration C is crucial ： without it, you only have one particular solution rather than the general solution. The general solution 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。

The key algebraic skill is partial fraction decomposition. Many A-Level separable equations lead to integrals of rational functions that require splitting into partial fractions. For example, an equation like dy/dx = y(1 − y) separates to ∫1/[y(1−y)] dy = ∫dx, and the left integral requires 1/[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| = e^(x²+C) = e^C · e^(x²). Let A = ±e^C, so y = A·e^(x²). Apply the initial condition y(0) = 3: 3 = A·e^0, so A = 3. The particular solution is y = 3e^(x²).

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| = e^(x²+C) = e^C · e^(x²)。令A = ±e^C，则y = A·e^(x²)。应用初始条件y(0) = 3：3 = A·e^0，故A = 3。特解为y = 3e^(x²)。

Example 2: Solve dy/dx = y(1 − y), with y(0) = 0.5. Separate: 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) = Ke^x where K = e^C. Solve for y: y = Ke^x/(1+Ke^x). Apply y(0) = 0.5: 0.5 = K/(1+K), so K = 1. Thus y = e^x/(1+e^x). 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) = Ke^x，其中K = e^C。解出y：y = Ke^x/(1+Ke^x)。应用y(0) = 0.5：0.5 = K/(1+K)，故K = 1。因此y = e^x/(1+e^x)。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

For a linear first-order ODE in standard form, dy/dx + P(x)y = Q(x), the integrating factor is I(x) = e^(∫P(x) dx). Multiplying the entire equation by I(x) makes the left side the exact derivative of I(x)·y: d/dx[I(x)·y] = I(x)·Q(x). Then integrate both sides: I(x)·y = ∫I(x)·Q(x) dx + 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。

The integrating factor method is tested across all major exam boards. Edexcel typically presents it as a standalone question in Paper 1 (Pure Mathematics), while CAIE often embeds it in a multi-part question combining differential equations with integration by parts or substitution. A common 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: Solve dy/dx + 2xy = x, with y(0) = 1. Here P(x) = 2x, Q(x) = x. Integrating factor: I(x) = e^(∫2x dx) = e^(x²). Multiply through: e^(x²) dy/dx + 2x e^(x²) y = x e^(x²). The left side is d/dx[y·e^(x²)] = x e^(x²). Integrate: y·e^(x²) = ∫x e^(x²) dx = (1/2)e^(x²) + 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) = e^(x²)。乘以积分因子：e^(x²) dy/dx + 2x e^(x²) y = x e^(x²)。左侧为d/dx[y·e^(x²)] = x e^(x²)。积分：y·e^(x²) = ∫x e^(x²) dx = (1/2)e^(x²) + 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: Solve x·dy/dx + 2y = 4x², y(1) = 2. First, put into standard form: divide by x to get dy/dx + (2/x)y = 4x. So P(x) = 2/x, Q(x) = 4x. Integrating factor: I(x) = e^(∫(2/x) dx) = e^(2 ln|x|) = x². Multiply: x² dy/dx + 2x y = 4x³. Left side is d/dx[x²·y] = 4x³. Integrate: x²·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) = e^(2 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。