First Order Differential Equations and Euler’s Method — IB数学：一阶微分方程与欧拉方法

引言 | Introduction

微分方程是IB数学AA HL和AI HL课程中最具挑战性也最令人着迷的主题之一。一阶微分方程不仅是微积分学习的制高点，更是连接纯数学与现实世界建模的桥梁。从放射性衰变到药物代谢，从牛顿冷却定律到种群增长模型，一阶微分方程帮助我们用量化方式理解变化的世界。而欧拉方法（Euler’s Method）则为那些无法求得解析解的复杂方程提供了一条数值逼近的道路。本文将深入探讨一阶微分方程的三种核心解析方法（可分离变量法、积分因子法、齐次方程法），并系统讲解欧拉方法的原理、应用与IB考试技巧。

Differential equations represent one of the most challenging yet fascinating topics in the IB Mathematics AA HL and AI HL syllabi. First-order differential equations are not only the pinnacle of calculus learning but also serve as a bridge connecting pure mathematics with real-world modelling. From radioactive decay to drug metabolism, from Newton’s Law of Cooling to population growth models, first-order differential equations help us understand the changing world in quantitative terms. And when analytical solutions are elusive, Euler’s Method provides a numerical pathway. This article explores in depth the three core analytical methods for first-order differential equations – separable equations, integrating factor method, and homogeneous equations – and systematically explains the principles, applications, and IB exam techniques of Euler’s Method.

一阶微分方程概述 | Overview of First-Order Differential Equations

一阶微分方程的一般形式为 dy/dx = f(x, y)，其中 x 为自变量，y = y(x) 为未知函数。这类方程描述了函数在某一点的变化率与该点坐标之间的关系。在物理上，这常常对应着”当前状态决定了下一步的变化”这一直觉 – 例如，当前温度决定了冷却的速率，当前人口数量决定了增长的速率。

The general form of a first-order differential equation is dy/dx = f(x, y), where x is the independent variable and y = y(x) is the unknown function. Such equations describe the relationship between the rate of change of a function at a point and the coordinates of that point. Physically, this often corresponds to the intuition that “the current state determines the next step of change” – for example, the current temperature determines the rate of cooling, and the current population determines the rate of growth.

在IB课程中，学生需要掌握三种解析求解方法以及一种数值方法。解析方法给出精确的函数表达式，而数值方法给出近似的数值序列。两者在考试和IA（内部评估）中都有重要的考察位置。

In the IB curriculum, students are expected to master three analytical solution methods and one numerical method. Analytical methods yield exact functional expressions, while the numerical method yields approximate numerical sequences. Both have important examination weighting in external assessments and the IA (Internal Assessment).

方法一：可分离变量方程 | Method 1: Separable Equations

理论基础 | Theoretical Foundation

当一阶微分方程可以写成 dy/dx = g(x)·h(y) 的形式时，我们就称之为可分离变量方程。这里的”可分离”是指包含 x 和 y 的因子可以分别归到等式的两侧。这是三种方法中最直观的 – 其核心思路是”把同类项放在一起”，然后分别积分。

When a first-order differential equation can be written as dy/dx = g(x)·h(y), we call it a separable equation. The term “separable” means that the factors involving x and y can be placed on separate sides of the equation. This is the most intuitive of the three methods – the core idea is to “group like terms together” and then integrate each side independently.

求解步骤十分清晰：首先将方程重写为 (1/h(y))·dy = g(x)·dx，然后对两边同时积分，得到隐式解。最后代入初始条件确定积分常数，得到特解。需要注意的是，当 h(y) = 0 时，y = 常数 也是方程的解（奇解），IB考试中偶尔会考察这一点。

The solution procedure is very clear: first rewrite the equation as (1/h(y))·dy = g(x)·dx, then integrate both sides to obtain the implicit solution. Finally, substitute the initial conditions to determine the constant of integration and obtain the particular solution. Note that when h(y) = 0, y = constant is also a solution (singular solution), which occasionally appears in IB examinations.

例题1：基本型 | Worked Example 1: Basic Type

题目：求解微分方程 dy/dx = 2xy，已知 y(0) = 3。

Problem: Solve the differential equation dy/dx = 2xy given that y(0) = 3.

解答：分离变量得 dy/y = 2x·dx。两边积分：∫ dy/y = ∫ 2x dx，得到 ln|y| = x² + C。因此 y = Ae^(x²)，其中 A = ±e^C。代入 y(0) = 3：3 = Ae^0 = A，所以 A = 3。特解为 y = 3e^(x²)。

Solution: Separating variables gives dy/y = 2x·dx. Integrating both sides: ∫ dy/y = ∫ 2x dx, yielding ln|y| = x² + C. Therefore y = Ae^(x²), where A = ±e^C. Substituting y(0) = 3: 3 = Ae^0 = A, so A = 3. The particular solution is y = 3e^(x²).

例题2：含三角函数的可分离方程 | Worked Example 2: Separable with Trigonometry

题目：求解 dy/dx = y·cos(x)，已知 y(0) = 1。

Problem: Solve dy/dx = y·cos(x) given that y(0) = 1.

解答：分离变量：dy/y = cos(x)·dx。积分：∫ dy/y = ∫ cos(x) dx，得到 ln|y| = sin(x) + C。即 y = Ae^(sin x)。代入 y(0) = 1：1 = Ae^0 = A。特解：y = e^(sin x)。这个解是一个周期函数，其值在 e^(-1) 和 e^1 之间振荡，体现了周期驱动力下的一阶衰减/增长行为。

Solution: Separating variables: dy/y = cos(x)·dx. Integrating: ∫ dy/y = ∫ cos(x) dx, yielding ln|y| = sin(x) + C. Thus y = Ae^(sin x). Substituting y(0) = 1: 1 = Ae^0 = A. Particular solution: y = e^(sin x). This solution is a periodic function oscillating between e^(-1) and e^1, reflecting first-order decay/growth behaviour under a periodic driving force.

方法二：一阶线性方程与积分因子 | Method 2: First-Order Linear Equations and Integrating Factors

标准形式与积分因子 | Standard Form and Integrating Factor

一阶线性微分方程的标准形式为 dy/dx + P(x)·y = Q(x)。这里的”线性”是指 y 及其导数都是一次的，不出现 y²、sin(y) 等非线性项。这类方程的求解核心是一个巧妙的技巧 – 积分因子法。积分因子 I(x) = e^(∫P(x)dx) 的设计非常精巧：乘以 I(x) 之后，方程左侧恰好变成了 y·I(x) 的导数。这本质上是一个乘积法则的逆运用。

The standard form of a first-order linear differential equation is dy/dx + P(x)·y = Q(x). Here “linear” means that y and its derivative appear only to the first power, with no nonlinear terms such as y², sin(y), etc. The core of solving such equations is an ingenious technique – the integrating factor method. The integrating factor I(x) = e^(∫P(x)dx) is elegantly designed: after multiplying by I(x), the left-hand side of the equation becomes exactly the derivative of y·I(x). This is essentially a reverse application of the product rule.

数学推导如下：d/dx[y·I(x)] = y’·I + y·I’ = y’·I + y·I·P(x) = I·[y’ + P(x)y] = I·Q(x)。因此，方程简化为一个可以直接积分的等式。然后两边积分，除以I(x)，就得到了通解。

The mathematical derivation is as follows: d/dx[y·I(x)] = y’·I + y·I’ = y’·I + y·I·P(x) = I·[y’ + P(x)y] = I·Q(x). Thus the equation reduces to one that can be directly integrated. Integrating both sides and dividing by I(x) gives the general solution.

例题3：经典积分因子 | Worked Example 3: Classic Integrating Factor

题目：求解 dy/dx + (2/x)·y = x³，已知 y(1) = 2（x > 0）。

Problem: Solve dy/dx + (2/x)·y = x³ given that y(1) = 2 (x > 0).

解答：这里 P(x) = 2/x，Q(x) = x³。积分因子 I(x) = e^(∫2/x dx) = e^(2ln x) = x²。乘以积分因子后，方程变为 d/dx[y·x²] = x³·x² = x⁵。积分得 y·x² = x⁶/6 + C。因此通解为 y = x⁴/6 + C/x²。代入 y(1) = 2：2 = 1/6 + C，所以 C = 11/6。特解：y = x⁴/6 + 11/(6x²)。

Solution: Here P(x) = 2/x, Q(x) = x³. Integrating factor I(x) = e^(∫2/x dx) = e^(2ln x) = x². After multiplying by the integrating factor, the equation becomes d/dx[y·x²] = x³·x² = x⁵. Integrating gives y·x² = x⁶/6 + C. Therefore the general solution is y = x⁴/6 + C/x². Substituting y(1) = 2: 2 = 1/6 + C, so C = 11/6. Particular solution: y = x⁴/6 + 11/(6x²).

例题4：常系数线性方程 | Worked Example 4: Constant-Coefficient Linear Equation

题目：求解 dy/dx + 3y = 6，已知 y(0) = 1。

Problem: Solve dy/dx + 3y = 6 given that y(0) = 1.

解答：P(x) = 3，Q(x) = 6。积分因子 I(x) = e^(∫3 dx) = e^(3x)。乘以积分因子：d/dx[y·e^(3x)] = 6e^(3x)。积分：y·e^(3x) = 2e^(3x) + C。即 y = 2 + Ce^(-3x)。代入 y(0) = 1：1 = 2 + C，所以 C = -1。特解：y = 2 – e^(-3x)。注意，当 x→∞ 时，y→2，这个稳态值恰好等于 Q/P = 6/3 = 2，富有物理意义。

Solution: P(x) = 3, Q(x) = 6. Integrating factor I(x) = e^(∫3 dx) = e^(3x). Multiplying by the integrating factor: d/dx[y·e^(3x)] = 6e^(3x). Integrating: y·e^(3x) = 2e^(3x) + C. Thus y = 2 + Ce^(-3x). Substituting y(0) = 1: 1 = 2 + C, so C = -1. Particular solution: y = 2 – e^(-3x). Note that as x→∞, y→2, and this steady-state value equals Q/P = 6/3 = 2, which is physically meaningful.

方法三：齐次方程 | Method 3: Homogeneous Equations

定义与代换技巧 | Definition and Substitution Technique

当一阶微分方程可以写成 dy/dx = F(y/x) 的形式时，我们称之为齐次方程。注意这里的”齐次”指的是一阶齐次（所有项关于x和y的次数相同），而非线性代数中的齐次概念。求解齐次方程的关键技巧是做代换 v = y/x（即 y = vx），利用乘积法则 dy/dx = v + x·dv/dx，将原方程转化为关于 v 和 x 的可分离变量方程。

When a first-order differential equation can be written as dy/dx = F(y/x), we call it homogeneous. Note that “homogeneous” here refers to first-order homogeneity (all terms have the same degree in x and y), not the concept of homogeneity from linear algebra. The key technique for solving homogeneous equations is the substitution v = y/x (i.e., y = vx). Using the product rule dy/dx = v + x·dv/dx, the original equation transforms into a separable equation in v and x.

例题5：典型齐次方程 | Worked Example 5: Typical Homogeneous Equation

题目：求解 dy/dx = (x² + y²)/(xy)，x > 0，y > 0。

Problem: Solve dy/dx = (x² + y²)/(xy), x > 0, y > 0.

解答：将右端改写：dy/dx = x/y + y/x = 1/(y/x) + y/x。令 v = y/x，则 y = vx，dy/dx = v + x·dv/dx。代入得：v + x·dv/dx = 1/v + v。消去 v 得 x·dv/dx = 1/v，即 v·dv = dx/x。积分：v²/2 = ln|x| + C。代回 y = vx：y²/(2x²) = ln|x| + C，即 y² = 2x²(ln|x| + C)。这是方程的通解。

Solution: Rewrite the right-hand side: dy/dx = x/y + y/x = 1/(y/x) + y/x. Let v = y/x, then y = vx, dy/dx = v + x·dv/dx. Substituting: v + x·dv/dx = 1/v + v. Cancelling v gives x·dv/dx = 1/v, i.e., v·dv = dx/x. Integrating: v²/2 = ln|x| + C. Substituting back y = vx: y²/(2x²) = ln|x| + C, i.e., y² = 2x²(ln|x| + C). This is the general solution of the equation.

欧拉方法 | Euler’s Method

为什么需要数值方法？ | Why Do We Need Numerical Methods?

不是所有微分方程都能求得解析解。事实上，绝大多数现实世界中的微分方程都无法用初等函数表示其解。比如 dy/dx = sin(xy)、dy/dx = e^(x²+y) 等看似简单的方程就没有封闭形式的解析解。在这种情况下，我们需要借助数值方法来近似求解。欧拉方法是最基本、最直观的一阶数值方法，也是理解所有更高级数值方法（如改进欧拉法、龙格-库塔法）的基石。

Not all differential equations can be solved analytically. In fact, the overwhelming majority of real-world differential equations cannot be expressed in terms of elementary functions. For instance, seemingly simple equations such as dy/dx = sin(xy) and dy/dx = e^(x²+y) have no closed-form analytical solutions. In such cases, we resort to numerical methods for approximation. Euler’s Method is the most fundamental and intuitive first-order numerical method, and it serves as the foundation for understanding all more advanced numerical methods such as the Improved Euler Method and Runge-Kutta methods.

递推公式与几何直观 | Recursion Formula and Geometric Intuition

给定初值问题 dy/dx = f(x, y)，y(x₀) = y₀，欧拉方法的递推公式为：x_{n+1} = x_n + h，y_{n+1} = y_n + h·f(x_n, y_n)，其中 h 为步长（step size）。

Given the initial value problem dy/dx = f(x, y), y(x₀) = y₀, the recursion formula for Euler’s Method is: x_{n+1} = x_n + h, y_{n+1} = y_n + h·f(x_n, y_n), where h is the step size.

几何上，我们从已知点 (x_n, y_n) 出发，沿着该点处的切线方向（斜率为 f(x_n, y_n)）前进一个步长 h。在新到达的点 (x_{n+1}, y_{n+1}) 处，我们重新计算斜率，再向前迈进。这一过程不断地用短线段来近似真实的解曲线。步长 h 越小，近似越精确，但计算量也越大。

Geometrically, we start from the known point (x_n, y_n) and advance one step h along the tangent direction at that point (with slope f(x_n, y_n)). At the newly reached point (x_{n+1}, y_{n+1}), we recompute the slope and take another step forward. This process continually approximates the true solution curve with short line segments. The smaller the step size h, the more accurate the approximation, but the greater the computational burden.

误差分析 | Error Analysis

欧拉方法的误差有两个层次。局部截断误差（local truncation error）衡量的是单步计算中的误差，量级为 O(h²)。这意味着，如果步长减半，单步误差将降至原来的四分之一。全局误差（global error）衡量的是从起点到终点的累计误差，量级为 O(h)。因此，步长减半使全局误差减半。这是欧拉方法作为”一阶方法”的本质特征。

The error of Euler’s Method operates on two levels. The local truncation error measures the error in a single step and is of order O(h²). This means that halving the step size reduces the single-step error to one quarter. The global error measures the accumulated error from the starting point to the endpoint and is of order O(h). Therefore, halving the step size halves the global error. This is the essential characteristic of Euler’s Method as a “first-order method.”

完整例题：欧拉方法逐步计算 | Full Worked Example: Step-by-Step Euler’s Method

题目：使用欧拉方法，取步长 h = 0.1，从 x = 0 到 x = 0.5，估计微分方程 dy/dx = x + y 满足 y(0) = 1 的解。同时给出该方程的精确解析解，并计算 y(0.5) 处的百分比误差。

Problem: Use Euler’s Method with step size h = 0.1 to estimate the solution of dy/dx = x + y with y(0) = 1 from x = 0 to x = 0.5. Also find the exact analytical solution and calculate the percentage error at y(0.5).

解答：这里 f(x, y) = x + y。欧拉递推：y_{n+1} = y_n + 0.1·(x_n + y_n)。

Solution: Here f(x, y) = x + y. Euler recursion: y_{n+1} = y_n + 0.1·(x_n + y_n).

逐步计算过程：

n=0: x₀=0, y₀=1, f(x₀,y₀)=0+1=1, y₁=1+0.1×1=1.1

n=1: x₁=0.1, y₁=1.1, f(x₁,y₁)=0.1+1.1=1.2, y₂=1.1+0.1×1.2=1.22

n=2: x₂=0.2, y₂=1.22, f(x₂,y₂)=0.2+1.22=1.42, y₃=1.22+0.1×1.42=1.362

n=3: x₃=0.3, y₃=1.362, f(x₃,y₃)=0.3+1.362=1.662, y₄=1.362+0.1×1.662=1.5282

n=4: x₄=0.4, y₄=1.5282, f(x₄,y₄)=0.4+1.5282=1.9282, y₅=1.5282+0.1×1.9282=1.72102

因此，y(0.5) ≈ 1.7210（保留四位小数）。

精确解：该方程为一阶线性方程：dy/dx – y = x。P(x) = -1，积分因子 I(x) = e^(-x)。通解：y = 2e^x – x – 1。代入 y(0) = 1 验证：2 – 0 – 1 = 1，正确。y(0.5) = 2e^0.5 – 0.5 – 1 = 2×1.64872 – 1.5 = 1.79744。

百分比误差：|1.79744 – 1.72102| / 1.79744 × 100% = 4.25%。

Step-by-step calculation:

Therefore, y(0.5) ≈ 1.7210 (to four decimal places).

Exact solution: This is a first-order linear equation: dy/dx – y = x. P(x) = -1, integrating factor I(x) = e^(-x). General solution: y = 2e^x – x – 1. Verify with y(0) = 1: 2 – 0 – 1 = 1, correct. y(0.5) = 2e^0.5 – 0.5 – 1 = 2×1.64872 – 1.5 = 1.79744.

Percentage error: |1.79744 – 1.72102| / 1.79744 × 100% = 4.25%.

步长的影响 | Effect of Step Size

为了展示步长对精度的影响，我们再用 h = 0.05 重新计算。这将需要10步而不是5步，但预期精度会显著提高。

To demonstrate the effect of step size on accuracy, we recalculate with h = 0.05. This requires 10 steps rather than 5, but the accuracy is expected to improve significantly.

h = 0.05 的计算结果（简略）：经过10步递推后，y(0.5) ≈ 1.7616。百分比误差为 |1.79744 – 1.7616| / 1.79744 × 100% = 1.99%。与 h = 0.1 时的 4.25% 相比，步长减半使误差约减半，符合欧拉方法全局误差为 O(h) 的预期。

Result with h = 0.05 (abbreviated): After 10 recursion steps, y(0.5) ≈ 1.7616. Percentage error = |1.79744 – 1.7616| / 1.79744 × 100% = 1.99%. Compared with 4.25% for h = 0.1, halving the step size roughly halves the error, consistent with the expectation that Euler’s Method has a global error of O(h).

实际应用场景 | Real-World Application Scenarios

1. 牛顿冷却定律 | Newton’s Law of Cooling

牛顿冷却定律描述了物体温度随时间趋近于环境温度的过程：dT/dt = -k(T – T_env)，其中 T 为物体温度，T_env 为环境温度，k > 0 为冷却常数。这是一阶线性的，可用积分因子法或分离变量法求解。通解为 T(t) = T_env + (T₀ – T_env)·e^(-kt)。在法医学中，这个模型被用来根据尸体温度推算死亡时间。在IB IA中，学生经常用温度传感器收集数据，然后拟合这个模型，是一个经典的实验选题。

Newton’s Law of Cooling describes how an object’s temperature approaches the ambient temperature over time: dT/dt = -k(T – T_env), where T is the object’s temperature, T_env is the ambient temperature, and k > 0 is the cooling constant. This is first-order linear and can be solved using either the integrating factor method or separation of variables. The general solution is T(t) = T_env + (T₀ – T_env)·e^(-kt). In forensic science, this model is used to estimate the time of death from a body’s temperature. In IB IAs, students frequently collect temperature data with sensors and fit this model, making it a classic experimental topic.

2. 放射性衰变 | Radioactive Decay

放射性衰变遵循一阶动力学：dN/dt = -λN，其中 N 为未衰变的原子核数量，λ 为衰变常数。这是最简单的可分离变量方程，通解为 N(t) = N₀·e^(-λt)。半衰期 T_{1/2} = ln(2)/λ。碳-14定年法就是基于这一原理，广泛应用于考古学中确定有机物的年代。IB物理和化学课程中都会涉及这一模型，而IB数学AA HL则将其作为微分方程的经典应用来考察。

Radioactive decay follows first-order kinetics: dN/dt = -λN, where N is the number of undecayed nuclei and λ is the decay constant. This is the simplest separable equation, with general solution N(t) = N₀·e^(-λt). The half-life T_{1/2} = ln(2)/λ. Carbon-14 dating is based on this principle and is widely used in archaeology to determine the age of organic materials. Both IB Physics and Chemistry courses cover this model, while IB Mathematics AA HL examines it as a classic application of differential equations.

3. Logistic 种群增长模型 | Logistic Population Growth Model

Logistic 方程 dP/dt = rP(1 – P/K) 是种群生态学中的经典模型，其中 r 为内禀增长率，K 为环境容纳量。该方程不再是线性的（包含 P² 项），但仍然是可分离变量的。通过分离变量和部分分式分解，可得通解 P(t) = K / (1 + ((K-P₀)/P₀)·e^(-rt))。该模型预测种群增长呈现S形曲线 – 先是指数增长，然后在接近K时趋于平稳。这正是大多数真实种群的增长模式。

The logistic equation dP/dt = rP(1 – P/K) is a classic model in population ecology, where r is the intrinsic growth rate and K is the carrying capacity. This equation is no longer linear (it contains a P² term) but is still separable. Through separation of variables and partial fraction decomposition, the general solution is P(t) = K / (1 + ((K-P₀)/P₀)·e^(-rt)). The model predicts an S-shaped growth curve – exponential growth initially, then levelling off as P approaches K. This matches the growth pattern of most real populations.

IB考试重点与答题策略 | IB Exam Focus and Answering Strategies

Paper 3 专项 | Paper 3 Special Focus

在IB数学AA HL的Paper 3中，欧拉方法经常作为12分大题的压轴考点出现。典型的题目结构包括：首先给出一个微分方程和初始条件，要求学生使用给定步长完成欧拉表格（约占4-6分）；然后要求学生求出解析解（约占3-4分）；最后要求学生比较数值解与解析解，讨论步长对精度的影响（约占2-3分）。

In IB Mathematics AA HL Paper 3, Euler’s Method frequently appears as the capstone 12-mark extended response question. The typical question structure includes: first, a differential equation and initial condition are given, and students are asked to complete an Euler table with a given step size (approximately 4-6 marks); then students are asked to find the analytical solution (approximately 3-4 marks); finally, students are asked to compare the numerical and analytical solutions and discuss the effect of step size on accuracy (approximately 2-3 marks).

常见失分点 | Common Pitfalls

1. 计算错误：欧拉方法需要多步迭代计算，任何一步的计算错误都会导致后续所有值偏差。建议在草稿纸上列出清晰的表格，每完成一步就进行粗略的合理性检查（如：y值是否在增大/减小，趋势是否合理）。

1. Calculation errors: Euler’s Method requires multi-step iterative calculations, and an error in any single step will propagate through all subsequent values. It is advisable to draw a clear table on scratch paper and perform a rough sanity check after each step (e.g., is y increasing or decreasing as expected, does the trend make sense?).

2. 指数函数积分错误：∫(1/y)dy = ln|y| + C，而不是 ln(y) + C（特别注意绝对值符号）。当初始条件暗示 y > 0 时，可以省略绝对值。

2. Exponential integration errors: ∫(1/y)dy = ln|y| + C, not ln(y) + C (pay special attention to the absolute value symbol). When the initial condition implies y > 0, the absolute value can be omitted.

3. 混淆通解与特解：很多学生求出通解后忘记代入初始条件。IB通常会明确要求”find the particular solution”，所以一定要把C求出来。

3. Confusing general and particular solutions: Many students forget to substitute the initial conditions after finding the general solution. IB typically explicitly asks to “find the particular solution,” so always determine C.

4. 积分因子符号错误：标准形式是 dy/dx + P(x)y = Q(x)。如果题目给的是 dy/dx – P(x)y = Q(x)，务必先将负号吸收进P(x)中，再计算积分因子。

4. Sign errors with integrating factors: The standard form is dy/dx + P(x)y = Q(x). If the question gives dy/dx – P(x)y = Q(x), be sure to absorb the negative sign into P(x) before computing the integrating factor.

三种方法的比较 | Comparison of the Three Methods

在选择求解方法时，可以遵循以下决策流程：首先检查方程是否可以写成 dy/dx = g(x)·h(y) 的形式 – 如果可以，使用可分离变量法。如果不行，检查是否能写成 dy/dx + P(x)y = Q(x) 的线性形式 – 如果可以，使用积分因子法。如果以上两种形式都不符合，检查是否能写成 dy/dx = F(y/x) 的齐次形式 – 如果可以，使用代换法。如果三种方法都不适用，那么该方程可能没有初等解析解，此时欧拉方法就成为了有力的替代方案。

When choosing a solution method, follow this decision flow: first check whether the equation can be written as dy/dx = g(x)·h(y) – if so, use the separable method. If not, check whether it can be written in the linear form dy/dx + P(x)y = Q(x) – if so, use the integrating factor method. If neither form fits, check whether it can be written in the homogeneous form dy/dx = F(y/x) – if so, use the substitution method. If none of the three methods apply, the equation likely has no elementary analytical solution, and Euler’s Method becomes a powerful alternative.

Summary | 总结

一阶微分方程与欧拉方法是IB数学分析方向的核心内容，也是衔接高中与大学数学的关键桥梁。本文系统介绍了三种解析方法 – 可分离变量法（适用于 dy/dx = g(x)h(y) 型方程）、积分因子法（适用于线性方程 dy/dx + P(x)y = Q(x)）和齐次方程代换法（适用于 dy/dx = F(y/x) 型方程） – 以及数值方法欧拉方法（适用于任意初值问题，全局误差为O(h)）。掌握这些方法不仅足以应对IB考试中的各类题型，更为大学学习微分方程、数值分析和数学建模奠定了坚实的基础。在复习过程中，建议多进行”解析求解+数值验证”的混合练习，这样既能加深对解析方法的理解，又能直观感受数值近似的威力与局限。

First-order differential equations and Euler’s Method constitute core content in the IB Mathematics analysis stream and serve as a crucial bridge between high school and university mathematics. This article has systematically introduced three analytical methods – the separable method (for equations of the form dy/dx = g(x)h(y)), the integrating factor method (for linear equations dy/dx + P(x)y = Q(x)), and the homogeneous substitution method (for equations of the form dy/dx = F(y/x)) – along with the numerical method Euler’s Method (applicable to any initial value problem, with global error O(h)). Mastering these methods not only adequately prepares you for all question types in IB examinations but also builds a solid foundation for university-level study of differential equations, numerical analysis, and mathematical modelling. During revision, it is recommended to undertake mixed practice combining “analytical solution + numerical verification,” which deepens understanding of analytical methods while providing an intuitive feel for both the power and the limitations of numerical approximation.

---

更多咨询请联系16621398022（同微信）