Differential Equations Fundamentals and Solution Strategies | 微分方程基础与求解策略

📚 Differential Equations Fundamentals and Solution Strategies | 微分方程基础与求解策略

A differential equation is a mathematical equation that relates a function with its derivatives. In A-Level mathematics, differential equations are a core topic in both pure mathematics and applied contexts, forming the bridge between algebraic manipulation and real-world modeling.

微分方程是联系函数与其导数的数学方程。在A-Level数学中，微分方程是纯数学与应用情境中的核心考点，是连接代数运算与现实建模的桥梁。

---

1. What Is a Differential Equation? | 什么是微分方程？

A differential equation involves an unknown function and one or more of its derivatives. The order of a differential equation is determined by the highest derivative present. For example, dy/dx = 2x is a first-order equation, while d²y/dx² + y = 0 is second-order.

微分方程包含一个未知函数及其一个或多个导数。阶数由方程中最高阶导数决定。例如，dy/dx = 2x 是一阶方程，而 d²y/dx² + y = 0 是二阶方程。

The degree of a differential equation is the power of the highest-order derivative after the equation has been made rational and integral in all derivatives. Most A-Level questions involve first-degree (linear) differential equations.

次数是在方程化为有理整式后最高阶导数的幂次。大多数A-Level题目涉及一次（线性）微分方程。

A general solution contains arbitrary constants; a particular solution is obtained by substituting initial conditions or boundary conditions to find those constants.

通解含有任意常数；通过代入初始条件或边界条件求得具体常数后，得到特解。

---

2. Classifying Differential Equations | 微分方程的分类

In the A-Level syllabus, differential equations are classified primarily in two ways: by order (first-order or second-order) and by linearity (linear or non-linear). A linear differential equation has the unknown function and its derivatives appearing only to the first power, with no products of these terms.

在A-Level大纲中，微分方程主要从两个角度分类：按阶数（一阶或二阶）和按线性（线性或非线性）。线性微分方程中未知函数及其导数只以一次幂出现，且不包含它们的乘积项。

A first-order linear differential equation can be written in the general form:

dy/dx + P(x)y = Q(x)

where P(x) and Q(x) are functions of x only. Similarly, a second-order linear differential equation has the form a·d²y/dx² + b·dy/dx + c·y = f(x), where a, b, c are constants.

一阶线性微分方程可写成一般形式：

dy/dx + P(x)y = Q(x)

其中 P(x) 和 Q(x) 仅是 x 的函数。类似地，二阶线性微分方程具有形式 a·d²y/dx² + b·dy/dx + c·y = f(x)，其中 a、b、c 为常数。

---

3. Method of Separation of Variables | 分离变量法

The separation of variables method is applicable when a first-order differential equation can be rearranged so that all terms involving y (and dy) are on one side, and all terms involving x (and dx) are on the other side. This technique is a cornerstone of the A-Level 微分方程 unit.

分离变量法适用于可以整理为一边只含 y（和 dy）、另一边只含 x（和 dx）的一阶微分方程。这一技巧是A-Level微分方程单元的基石。

Step 1: Rearrange the equation into the form f(y) dy = g(x) dx.

步骤一：将方程整理为 f(y) dy = g(x) dx 的形式。

Step 2: Integrate both sides: ∫ f(y) dy = ∫ g(x) dx.

步骤二：两边同时积分：∫ f(y) dy = ∫ g(x) dx。

Step 3: Add the constant of integration C and solve for y explicitly if possible.

步骤三：加上积分常数 C，尽可能显式解出 y。

Consider the worked example: dy/dx = 3x²y. Separating variables gives (1/y) dy = 3x² dx. Integrating both sides yields ln|y| = x³ + C, hence y = Ae^x³, where A = ±e^C.

看一个经典例题：dy/dx = 3x²y。分离变量得到 (1/y) dy = 3x² dx。两边积分得 ln|y| = x³ + C，因此 y = Ae^x³，其中 A = ±e^C。

Key examination tip: always include the absolute value in ln|y|, and rewrite ±e^C as a single constant A before finalizing the answer.

关键考试提示：ln 中始终保留绝对值 |y|，并在最终答案前将 ±e^C 合并为单一常数 A。

---

4. Integrating Factor Method | 积分因子法

When a first-order differential equation of the form dy/dx + P(x)y = Q(x) cannot be separated, the integrating factor method is the standard approach. The integrating factor is defined as:

当形如 dy/dx + P(x)y = Q(x) 的一阶微分方程无法分离变量时，积分因子法成为标准解法。积分因子定义为：

I = e^(∫ P(x) dx)

Multiplying both sides of the differential equation by I converts the left-hand side into the derivative of I·y:

将微分方程两边同乘以 I，可将左边化为 I·y 的导数：

d/dx (I·y) = I·Q(x)

Then integrate both sides: I·y = ∫ I·Q(x) dx + C, and finally divide by I to obtain y.

然后两边积分：I·y = ∫ I·Q(x) dx + C，最后除以 I 得到 y。

Let us examine an example: dy/dx + 2y/x = x³. Here P(x) = 2/x. The integrating factor is I = e^{∫ 2/x dx} = e^{2 ln|x|} = x². Multiplying through gives x² · dy/dx + 2xy = x⁵, which simplifies to d/dx (x²y) = x⁵. Integrating: x²y = x⁶/6 + C, so y = x⁴/6 + C/x².

我们来看一个例子：dy/dx + 2y/x = x³。这里 P(x) = 2/x。积分因子为 I = e^{∫ 2/x dx} = e^{2 ln|x|} = x²。两边同乘得 x² · dy/dx + 2xy = x⁵，即 d/dx (x²y) = x⁵。积分：x²y = x⁶/6 + C，所以 y = x⁴/6 + C/x²。

---

5. Second-Order Homogeneous Equations | 二阶齐次方程

A second-order homogeneous linear differential equation with constant coefficients has the form a·d²y/dx² + b·dy/dx + c·y = 0. To solve it, we assume a solution of the form y = e^λx, which leads to the auxiliary equation:

常系数二阶齐次线性微分方程具有形式 a·d²y/dx² + b·dy/dx + c·y = 0。求解时假设解具有形式 y = e^λx，从而得到辅助方程：

aλ² + bλ + c = 0

There are three cases to consider based on the discriminant Δ = b² − 4ac:

根据判别式 Δ = b² − 4ac 分为三种情况：

• If Δ > 0: two distinct real roots λ₁, λ₂, giving the general solution y = Ae^λ₁x + Be^λ₂x.

  若 Δ > 0：两个不等实根 λ₁、λ₂，通解为 y = Ae^λ₁x + Be^λ₂x。

• If Δ = 0: one repeated real root λ, giving y = (A + Bx)e^λx.

  若 Δ = 0：一个二重实根 λ，通解为 y = (A + Bx)e^λx。

• If Δ < 0: complex conjugate roots λ = p ± qi, giving y = e^px(A cos(qx) + B sin(qx)).

  若 Δ < 0：共轭复根 λ = p ± qi，通解为 y = e^px(A cos(qx) + B sin(qx))。

For example, the equation d²y/dx² − 5·dy/dx + 6y = 0 has auxiliary equation λ² − 5λ + 6 = 0 = (λ − 2)(λ − 3). Thus the general solution is y = Ae^2x + Be^3x.

例如，方程 d²y/dx² − 5·dy/dx + 6y = 0 的辅助方程为 λ² − 5λ + 6 = 0 = (λ − 2)(λ − 3)。因此通解为 y = Ae^2x + Be^3x。

---

6. Second-Order Non-Homogeneous Equations | 二阶非齐次方程

A non-homogeneous equation has the form a·d²y/dx² + b·dy/dx + c·y = f(x). The general solution is the sum of the complementary function (CF, the solution of the corresponding homogeneous equation) and a particular integral (PI, any one solution of the full equation).

非齐次方程具有形式 a·d²y/dx² + b·dy/dx + c·y = f(x)。通解为互补函数（CF，对应齐次方程的解）与特解（PI，原方程的任意一个解）之和。

The table below summarises the standard trial functions for choosing a PI:

下表总结了选择特解时的标准试探函数：

f(x) 的形式	试探特解形式
多项式（如 x²）	同次数多项式（如 ax² + bx + c）
指数型（如 ekx）	Aekx（若 k 是辅助方程的根，则需乘 x）
三角函数（如 sin(kx) 或 cos(kx)）	A sin(kx) + B cos(kx)

Consider d²y/dx² + y = 3 sin(2x). The CF is y = A cos(x) + B sin(x). For the PI, try y = p sin(2x) + q cos(2x). Substitution gives (−4p sin(2x) − 4q cos(2x)) + (p sin(2x) + q cos(2x)) = 3 sin(2x). Therefore −3p = 3 and −3q = 0, so p = −1, q = 0. The PI is y = −sin(2x), and the general solution is y = A cos(x) + B sin(x) − sin(2x).

考虑 d²y/dx² + y = 3 sin(2x)。CF 为 y = A cos(x) + B sin(x)。对特解尝试 y = p sin(2x) + q cos(2x)。代入得 (−4p sin(2x) − 4q cos(2x)) + (p sin(2x) + q cos(2x)) = 3 sin(2x)。因此 −3p = 3、−3q = 0，所以 p = −1、q = 0。特解为 y = −sin(2x)，通解为 y = A cos(x) + B sin(x) − sin(2x)。

Special caution: if the trial function duplicates a term in the CF, multiply the trial by x (or x² in resonance cases) to avoid linear dependence.

特别注意：若试探解与CF中的某一项重复，需将试探解乘以 x（共振情形下乘 x²）以避免线性相关。

---

广告 / Sponsored

7. Initial Value Problems and Boundary Conditions | 初值问题与边界条件

An initial value problem specifies conditions at a single point, typically y(x₀) = y₀ and y′(x₀) = y₁. A boundary value problem specifies conditions at two different points. Both are resolved by substituting the conditions into the general solution to find the arbitrary constants.

初值问题在单个点处给出条件，通常为 y(x₀) = y₀ 和 y′(x₀) = y₁。边值问题在两个不同点处给出条件。两者的求解方法都是将条件代入通解以确定任意常数。

Worked example: Solve dy/dx = 6x² − 4x with y(1) = 5. Integrating gives y = 2x³ − 2x² + C. Substituting x = 1, y = 5 gives 5 = 2 − 2 + C, so C = 5. Therefore y = 2x³ − 2x² + 5.

例题：求解 dy/dx = 6x² − 4x，且 y(1) = 5。积分得 y = 2x³ − 2x² + C。代入 x = 1、y = 5 得 5 = 2 − 2 + C，即 C = 5。因此 y = 2x³ − 2x² + 5。

For second-order initial value problems, both y and dy/dx at x₀ are used to determine the two constants A and B simultaneously.

对于二阶初值问题，利用 x₀ 处的 y 和 dy/dx 两个条件联立求出常数 A 和 B。

---

8. Applications in Real-World Modelling | 微分方程的实际建模应用

Differential equations are powerful tools for modelling real-world phenomena. In A-Level examinations, common applications include population growth (unrestricted and restricted), radioactive decay, Newton's law of cooling, and simple harmonic motion.

微分方程是模拟现实世界现象的有力工具。在A-Level考试中，常见应用包括人口增长（无限制与有限制）、放射性衰变、牛顿冷却定律和简谐运动。

Exponential growth/decay: dN/dt = kN, where k is the growth (k > 0) or decay (k < 0) constant. The solution is N = N₀e^kt, where N₀ is the initial population.

指数增长/衰减：dN/dt = kN，其中 k 为增长（k > 0）或衰减（k < 0）常数。解为 N = N₀e^kt，其中 N₀ 为初始数量。

Restricted growth (logistic model): dN/dt = kN(1 − N/M), where M is the carrying capacity. This yields a sigmoidal growth curve that plateaus at N = M.

有限增长（逻辑斯蒂模型）：dN/dt = kN(1 − N/M)，其中 M 为环境容量。其解为一条在 N = M 处趋于平稳的S形增长曲线。

Newton's law of cooling: dθ/dt = −k(θ − θₑ), where θ is the object temperature and θₑ is the ambient temperature. The solution is θ = θₑ + (θ₀ − θₑ)e^−kt.

牛顿冷却定律：dθ/dt = −k(θ − θₑ)，其中 θ 为物体温度、θₑ 为环境温度。解为 θ = θₑ + (θ₀ − θₑ)e^−kt。

For second-order applications, simple harmonic motion is described by d²x/dt² = −ω²x, whose solution is x = A cos(ωt) + B sin(ωt), or equivalently x = R cos(ωt + φ).

在二阶应用中，简谐运动由 d²x/dt² = −ω²x 描述，其通解为 x = A cos(ωt) + B sin(ωt)，等价形式为 x = R cos(ωt + φ)。

---

9. Common Pitfalls and Exam Strategies | 常见错误与考试策略

Students frequently lose marks on differential equation questions due to avoidable errors. Being aware of these pitfalls before the exam is half the battle won.

学生常在微分方程题目中因可避免的错误而失分。考前了解这些常见陷阱，等于赢了一半。

• Forgetting the constant of integration: always include C after every integration in indefinite integrals.

  忘记积分常数：不定积分后务必加上常数 C。

• Incorrectly separating variables: if the equation cannot be written as f(y)dy = g(x)dx, do not force it; use the integrating factor instead.

  不当分离变量：若方程无法写成 f(y)dy = g(x)dx 的形式，不要强行分离，应改用积分因子法。

• Sign errors when substituting the integrating factor: double-check that P(x) is transcribed correctly and the exponent sign matches.

  代入积分因子时符号错误：仔细核对 P(x) 是否抄写正确，指数符号是否对应。

• Wrong trial function for PI: if the standard trial form is also a solution of the homogeneous equation, multiply by x (or x²) until it is independent of the CF.

  特解试探函数选择错误：若标准试探形式恰好是齐次方程的解，则需乘以 x（或 x²）直到与CF线性无关。

• Not applying both initial conditions for second-order equations: both y and y′ conditions must be used to determine A and B.

  二阶方程未同时利用两个初始条件：必须使用 y 和 y′ 两个条件联立求出 A 和 B。

Exam strategy: always state the auxiliary equation explicitly, show the general solution before applying conditions, and verify the particular solution by substituting back where feasible.

考试策略：始终明确写出辅助方程，先写出通解再应用条件，并在可行时将特解代回验证。

---

10. Worked Example Combining Multiple Skills | 综合技能例题

Let us now solve a comprehensive problem that integrates several techniques studied above.

下面我们求解一道综合运用以上多种技巧的完整题目。

Problem: Solve the differential equation d²y/dx² − 4y = 8x, subject to y(0) = 2 and y′(0) = 4.

题目：求解微分方程 d²y/dx² − 4y = 8x，满足条件 y(0) = 2 和 y′(0) = 4。

Solution:

解答：

Step 1 — complementary function. The auxiliary equation is λ² − 4 = 0, giving λ = ±2. Hence CF = Ae^2x + Be^−2x.

第一步——互补函数：辅助方程为 λ² − 4 = 0，得 λ = ±2。因此 CF = Ae^2x + Be^−2x。

Step 2 — particular integral. Since f(x) = 8x is linear, try PI = ax + b. Then d²(PI)/dx² = 0, and substituting gives −4(ax + b) = 8x, so −4a = 8 and −4b = 0, yielding a = −2, b = 0. PI = −2x.

第二步——特解：因 f(x) = 8x 为一次多项式，尝试 PI = ax + b。则 d²(PI)/dx² = 0，代入得 −4(ax + b) = 8x，即 −4a = 8、−4b = 0，解得 a = −2、b = 0。PI = −2x。

Step 3 — general solution: y = Ae^2x + Be^−2x − 2x.

第三步——通解：y = Ae^2x + Be^−2x − 2x。

Step 4 — apply conditions. y(0) = A + B = 2. Also y′ = 2Ae^2x − 2Be^−2x − 2, so y′(0) = 2A − 2B − 2 = 4, giving A − B = 3. Solving the system A + B = 2 and A − B = 3 yields A = 5/2, B = −1/2.

第四步——代入条件：y(0) = A + B = 2。又 y′ = 2Ae^2x − 2Be^−2x − 2，所以 y′(0) = 2A − 2B − 2 = 4，即 A − B = 3。联立 A + B = 2 与 A − B = 3，得 A = 5/2、B = −1/2。

Final answer:

最终答案：

y = (5/2)e^2x − (1/2)e^−2x − 2x

This example demonstrates the full workflow: finding the CF, constructing the PI, combining them, and solving for constants using boundary conditions.

该例题展示了完整流程：求互补函数、构造特解、合并为通解，并利用边界条件求解常数。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）