📚 Common Methods for Solving Ordinary Differential Equations | 常微分方程的常见解法梳理
An ordinary differential equation (ODE) is an equation that relates a function to its derivatives. In the IB Mathematics Analysis and Approaches HL course, solving ODEs is a key skill that connects calculus with real-world modeling. This article reviews the most common solution methods you need for your exams, with clear steps and worked examples.
常微分方程（ODE）是联系函数与其导数的方程。在 IB 数学分析与方法 HL 课程中，解常微分方程是连接微积分与现实建模的关键技能。本文梳理考试中最常用的解法，帮助你理清步骤并掌握题型。
1. Classifying ODEs: First Steps | 微分方程的分类：解题前要做什么
Before you choose a solution method, identify the order of the equation (the highest derivative present) and whether it is linear. A linear ODE has the dependent variable and all its derivatives occurring to the first power only, with no products of the dependent variable with its derivatives.
在选择解法之前，需要先判断方程的阶数（最高导数的导数阶数）以及是否为线性方程。线性常微分方程中，因变量及其各阶导数只能以一次幂出现，且不能彼此相乘。
First-order ODE: dy/dx = f(x, y)
一阶常微分方程：dy/dx = f(x, y)
Second-order linear ODE: a d²y/dx² + b dy/dx + cy = 0
二阶线性常微分方程：a d²y/dx² + b dy/dx + cy = 0
Nonlinear example: (dy/dx)² + y = sin x
非线性例子：(dy/dx)² + y = sin x
2. Separation of Variables | 分离变量法
If a first-order equation can be written in the form dy/dx = g(x)h(y), we can separate the variables and integrate both sides.
如果一个一阶方程可以写成 dy/dx = g(x)h(y) 的形式，就可以分离变量并对两边积分。
∫ 1/h(y) dy = ∫ g(x) dx + C
For example, solve dy/dx = x/y. Rewrite as y dy = x dx, integrate both sides: y²/2 = x²/2 + C, so y² = x² + 2C. The arbitrary constant remains until initial conditions are given.
例如，求解 dy/dx = x/y。改写为 y dy = x dx，两边积分得 y²/2 = x²/2 + C，即 y² = x² + 2C。任意常数在给定初始条件之前保持不变。
3. Integrating Factor for First-Order Linear ODEs | 一阶线性微分方程的积分因子法
For an equation of the form dy/dx + P(x)y g factor μ(x) = e^(∫P(x) dx). This converts the left-hand side into the derivative of μ(x)y.
对于形如 dy/dx + P(x)y = Q(x) 的方程，将两边乘以积分因子 μ(x) = e^(∫P(x) dx)，左边就能变成 μ(x)y 的导数。
d/dx [ μ(x) y ] = μ(x) Q(x)
Then integrate both sides and solve for y:
然后两边积分并解出 y：
y = ( ∫ μ(x) Q(x) dx + C ) / μ(x)
Example: dy/dx + 2y/x = x². Here P(x) = 2/x, so μ = e∫2/x dx = x². Multiplying through gives x² y' + 2x y = x⁴, which is (x² y)' = x⁴. Therefore x² y = x⁵/5 + C, and y = x³/5 + C/x².
例：dy/dx + 2y/x = x²。这里 P(x) = 2/x，所以 μ = e∫2/x dx = x²。两边乘以 μ 得 x² y' + 2x y = x⁴，即 (x² y)' = x⁴。因此 x² y = x⁵/5 + C，所以 y = x³/5 + C/x²。
4. Homogeneous Second-Order Linear ODEs with Constant Coefficients | 常系数齐次二阶线性微分方程
For ay'' + by' + cy = 0, assume a solution of the form y = er x tic equation ar² + br + c = 0. The roots of this quadratic determine the general solution.
对于 ay'' + by' + cy = 0，设解为 y = er x，代入得到特征方程 ar² + br + c = 0。这个二次方程的根决定了通解的形式。
| Discriminant Δ = b² − 4ac | Roots | General solution |
| Δ > 0 | r₁, r₂ real and distinct | y = C₁er₁x + C₂er₂x |
| Δ = 0 | r repeated | y = (C₁ + C₂x)erx |
| Δ < 0 | r = α ± βi | y = eαx[C₁ cos(βx) + C₂ sin(βx)] |
Remember that C₁ and C₂ are arbitrary constants. If initial conditions are given, you will use them after obtaining the general solution.
注意 C₁ 和 C₂ 为任意常数。如果题目给出了初始条件，需要在得到通解之后再利用它们确定常数。
5. Non-Homogeneous Second-Order Linear ODEs: Method of Undetermined Coefficients | 非齐次二阶线性微分方程：待定系数法
For ay'' + by' + cy m of the complementary function (from the homogeneous equation) and a particular integral. A common technique is the method of undetermined coefficients, where we guess a trial solution based on the form of f(x).
对于 ay'' + by' + cy = f(x)，通解等于齐次方程的通解（互补函数）加一个特解。常用方法为待定系数法，即根据 f(x) 的形式猜测特解的形式。
| f(x) | Trial particular integral |
| Polynomial of degree n | Aₙxⁿ + Aₙ₋₁xⁿ⁻¹ + ... + A₀ |
| k emx | A emx |
| k sin(px) or k cos(px) | A cos(px) + B sin(px) |
| emx sin(px) or emx cos(px) | emx[A cos(px) + B sin(px)] |
If any term in your trial solution already appears in the complementary function, multiply the entire trial solution by x (or x² if the root repeats). This adjustment is essential for the method to work.
如果试解中的某一项已经出现在互补函数中，则需要将整个试解乘以 x（若特征根是重根则乘以 x²）。这一调整必不可少。
6. The Structure of the General Solution | 通解的结构
For any linear non-homogeneous ODE, the general solution is written as:
对于任何线性非齐次常微分方程，通解都可以写成：
y = yc + yp
where y_c is the complementary function (the general solution of the homogeneous equation) and y lar integral. This structure holds for both first-order and higher-order linear equations.
其中 y_c 是互补函数（齐次方程的通解），y_p 是一个特解。这一结构对一阶和高阶线性方程都成立。
7. Initial Value Problems and Boundary Conditions | 初值问题与边界条件
After solving an ODE, you will often be given conditions to determine the arbitrary constants. For a first-order equation, one condition is enough; for a second-order equation, you need two conditions.
解出微分方程后，题目通常会给出条件来确定任意常数。一阶方程需要一个条件，二阶方程需要两个条件。
Initial value problem: conditions at the same point, e.g. y(0) = 2 and y'(0) = −1.
初值问题：在同一点给出条件，例如 y(0) = 2 和 y'(0) = −1。
Boundary value problem: conditions at different points, e.g. y(0) = 1 and y(π) = 0.
边值问题：在不同点给出条件，例如 y(0) = 1 和 y(π) = 0。
For second-order equations, first find the general solution, then compute its derivative if y'(x₀) is specified, and solve the resulting simultaneous equations for the constants.
对于二阶方程，先求出通解，如果给出 y'(x₀)，则需先求导，再联立方程组确定常数。
8. Numerical Methods: Euler's Method | 数值方法：欧拉法
When an ODE cannot be solved analytically, Euler's method provides a simple numerical approximation. For the initial value problem y' = f(x, y), y(x₀) = y₀, divide the interval into steps of size h.
当常微分方程无法解析求解时，欧拉法提供了一种简单的数值逼近。对于初值问题 y' = f(x, y)，y(x₀) = y₀，将区间分成步长为 h 的小段。
yn+1 = yn + h f(xn, yn)
Here xn+1 = xn + h. Each iteration uses the previous point to estimate the next value. The accuracy improves as h becomes smaller, but more steps are then required.
其中 xn+1 = xn + h。每次迭代都利用前一点的值来估计下一点。步长 h 越小，精度越高，但所需步数也会增加。
9. Real-World Modeling with ODEs | 常微分方程与真实世界建模
Many IB questions ask you to set up an ODE from a verbal description and then solve it. In each case, the key is to identify the rate of change and the proportionality relationship.
许多 IB 题目要求你根据文字描述建立微分方程并求解。关键是要识别变化率和比例关系。
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Exponential growth or decay: dP/dt = kP, solution P = P₀ekt.
指数增长或衰减：dP/dt = kP，解为 P = P₀ekt。
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Newton's law of cooling: dT/dt = −k(T − Tenv), solution T = Tenv + (T₀ − Tenv)e−kt.
牛顿冷却定律：dT/dt = −k(T − Tenv)，解为 T = Tenv + (T₀ − Tenv)e−kt。
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/dt = rP(1 − P/K), solved by separation of variables; the solution is a logistic function.
逻辑斯谛增长：dP/dt = rP(1 − P/K)，可用分离变量法求解，解为逻辑斯谛函数。
s P = 0 and P = K are often worth noting. The constant K represents the carrying capacity.
对于逻辑斯谛增长，注意平衡解 P = 0 和 P = K。常数 K 代表环境容纳量。
10. Exam Tips and Common Mistakes | 考试技巧与常见错误
Use this checklist to avoid losing marks in the exam:
使用以下检查清单，避免在考试中丢分：
Check whether the equation is separable before trying the integrating factor.
先检查方程是否可分离变量，再考虑积分因子法。
Always include the constant of integration when integrating.
积分时不要忘记加积分常数 C。
For second-order equations, always solve the homogeneous equation first.
对于二阶方程，必须先解齐次方程。
If your trial particular solution overlaps with the complementary function, multiply by x (or x²).
如果待定特解与互补函数有重叠，需要乘以 x（或 x²）。
Apply initial or boundary conditions only after the general solution is complete.
在通解完整之后才能代入初值或边界条件。
When using Euler's method, check whether the step size h is given in the question.
使用欧拉法时，注意题目是否给出了步长 h。
By practising these methods with past paper questions, you will quickly recognise which technique to apply to each ODE.
通过用真题练习这些方法，你会快速识别每道常微分方程题该用哪种技巧。
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