Solving First-Order Constant Coefficient Homogeneous Linear Differential Equations | 一阶常系数齐次线性微分方程的解法

📚 Solving First-Order Constant Coefficient Homogeneous Linear Differential Equations | 一阶常系数齐次线性微分方程的解法

Differential equations are a cornerstone of IB Mathematics HL, particularly in the Analysis and Approaches and Applications and Interpretation courses. Among the simplest yet most powerful types is the first-order linear differential equation with constant coefficients. This article provides a complete, exam-focused guide to solving the homogeneous form of such equations, with step-by-step methods, worked examples, and common pitfalls to avoid.

微分方程是 IB 数学 HL 的基石内容，尤其在分析与方法（AA）以及应用与解释（AI）课程中占有重要地位。在众多类型中，一阶常系数线性微分方程既简单又强大。本文将围绕齐次形式，提供一套完整且紧扣考试大纲的解法指南，包含逐步方法、实例演练与常见误区提醒。

1. What Is a First-Order Linear Differential Equation? | 什么是一阶线性微分方程？

A first-order linear differential equation is an equation that involves only the first derivative of an unknown function. It can be written in a standard form that is linear in both the dependent variable and its derivative, meaning neither is raised to a power, multiplied together, or nested inside another function.

一阶线性微分方程是仅涉及未知函数一阶导数的方程。它可以写成关于因变量及其导数的线性标准形式，即二者既不会被升幂，也不会相乘，更不会被嵌套在其他函数中。

In general, a first-order linear differential equation has the form:

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

When Q(x) = 0, the equation is called homogeneous. When Q(x) ≠ 0, it is non-homogeneous. This article focuses on the homogeneous case with constant coefficient P(x) = a, where a is a real constant.

当 Q(x) = 0 时，该方程被称为齐次方程；当 Q(x) ≠ 0 时，则为非齐次方程。本文聚焦于齐次情形，且系数 P(x) = a 为实常数。

2. The Standard Form | 标准形式

The homogeneous first-order constant coefficient linear differential equation is written as:

dy/dx + ay = 0, where a is a constant

Equivalently, it can be written as dy/dx = −ay. This form clearly shows that the rate of change of y is proportional to y itself, with proportionality constant −a. This property leads directly to exponential solutions.

等价地，它可以写成 dy/dx = −ay。这种形式清晰地表明：y 的变化率与 y 自身成正比，比例常数为 −a。这一性质直接导向指数形式的解。

Typical IB questions will give the equation in one of these forms and require you to find the general solution, then apply an initial condition to find the particular solution. Recognising the standard form quickly is essential for efficient problem solving.

典型的 IB 考题会以其中一种形式给出方程，要求你求出通解，再运用初值条件确定特解。快速识别标准形式是高效解题的关键。

3. The Separation of Variables Method | 变量分离法

Because the equation dy/dx = −ay can be rewritten to place all y-terms on one side and all x-terms on the other, the method of separation of variables is the most direct approach. This method works whenever the differential equation can be written in the form f(y) dy = g(x) dx.

由于 dy/dx = −ay 可以改写为所有含 y 的项在一边、所有含 x 的项在另一边，变量分离法是最直接的解法。只要微分方程能写成 f(y) dy = g(x) dx 的形式，该方法便适用。

Step 1: Separate the variables. Divide both sides by y and multiply by dx:

(1/y) dy = −a dx

第一步：分离变量。两边同时除以 y 并乘以 dx：

(1/y) dy = −a dx

Step 2: Integrate both sides.

∫(1/y) dy = ∫(−a) dx

第二步：两边积分。

∫(1/y) dy = ∫(−a) dx

Step 3: Evaluate the integrals. The left side gives ln|y|, and the right side gives −ax + C, where C is the constant of integration:

ln|y| = −ax + C

第三步：计算积分。左边得到 ln|y|，右边得到 −ax + C，其中 C 为积分常数：

ln|y| = −ax + C

Step 4: Solve for y. Exponentiate both sides:

|y| = e^(−ax + C) = e^C · e^(−ax)

y = ±e^C · e^(−ax)

第四步：解出 y。两边取指数：

|y| = e^(−ax + C) = e^C · e^(−ax)

y = ±e^C · e^(−ax)

Since ±e^C is just an arbitrary constant, we denote it by A (or C₁). The general solution is:

y = A·e^(−ax), where A is an arbitrary constant

由于 ±e^C 仍为任意常数，将其记作 A（或 C₁）。通解为：

y = A·e^(−ax)，其中 A 为任意常数

4. The Integrating Factor Method | 积分因子法

While separation of variables works perfectly for the homogeneous equation, the integrating factor method is a more general tool that also extends to non-homogeneous equations. IB examiners expect you to know both methods.

尽管变量分离法对于齐次方程非常有效，但积分因子法是一个更通用的工具，同样适用于非齐次方程。IB 考官期望你掌握这两种方法。

For the equation dy/dx + ay = 0, the integrating factor is defined as:

I(x) = e^(∫a dx) = e^(ax)

对于方程 dy/dx + ay = 0，积分因子定义为：

I(x) = e^(∫a dx) = e^(ax)

Multiply both sides of the original equation by I(x):

e^(ax) · dy/dx + a·e^(ax) · y = 0

两边同乘 I(x)：

e^(ax) · dy/dx + a·e^(ax) · y = 0

Notice that the left-hand side is exactly the derivative of e^(ax) · y by the product rule:

d/dx [e^(ax) · y] = 0

注意：根据乘积法则，左边恰好是 e^(ax) · y 的导数：

d/dx [e^(ax) · y] = 0

Integrating both sides with respect to x gives e^(ax) · y = C, and therefore:

y = C·e^(−ax)

两边对 x 积分得 e^(ax) · y = C，因此：

y = C·e^(−ax)

This method may seem more elaborate for the homogeneous case, but it is indispensable for solving non-homogeneous equations such as dy/dx + ay = f(x), which appear in later parts of the IB course.

对于齐次情形，此方法看似繁琐，但在求解非齐次方程（如 dy/dx + ay = f(x)）时不可或缺，后者在 IB 课程后续部分会出现。

5. Why the Constant Coefficient Matters | 常系数为何重要

The constancy of a greatly simplifies the solution process. The integrating factor e^(∫a dx) becomes a simple exponential e^(ax), and the general solution is always an exponential function. If a were a function of x, we would need more advanced techniques.

a 为常数这一性质极大简化了求解过程。积分因子 e^(∫a dx) 变成简单的指数函数 e^(ax)，通解也始终是指数函数。若 a 是 x 的函数，则需用到更高级的技巧。

Furthermore, the sign of a determines the qualitative behaviour of the solution:

此外，a 的符号决定了解的定性行为：

If a > 0, then y = A·e^(−ax) decays exponentially toward 0 as x → ∞. This models processes like radioactive decay and Newton’s law of cooling.
如果 a > 0，则 y = A·e^(−ax) 随 x → ∞ 指数衰减至 0。这可用于模拟放射性衰变和牛顿冷却定律等过程。
If a < 0, then y = A·e^(−ax) grows exponentially in magnitude as x increases. This models population growth in unrestricted environments.
如果 a < 0，则 y = A·e^(−ax) 随 x 增大而指数增长。这可用于模拟无约束环境下的种群增长。
If a = 0, the equation reduces to dy/dx = 0, giving the constant solution y = A.
如果 a = 0，方程退化为 dy/dx = 0，解为常数函数 y = A。

Understanding this qualitative behaviour helps you check whether your algebraic solution is sensible in a given real-world context.

理解这些定性行为有助于你在实际情境中检验代数解是否合理。

6. Interpretation of the Solution | 解的含义

The general solution y = A·e^(−ax) contains one arbitrary constant A. This is expected because a first-order differential equation requires one integration step, producing one constant of integration. Geometrically, this represents a family of curves—infinitely many solutions, each corresponding to a different value of A.

通解 y = A·e^(−ax) 含有一个任意常数 A。这并不意外，因为一阶微分方程只需一次积分，产生一个积分常数。从几何上看，它代表一族曲线——无穷多个解，每个解对应一个不同的 A 值。

To select a single curve from this family, we need one additional piece of information, typically an initial condition such as y(x₀) = y₀. This leads to what is called an initial value problem (IVP).

要从这族曲线中选出一条，需要额外一个条件，通常是初值条件，如 y(x₀) = y₀。这就引出了所谓的初值问题（IVP）。

In IB exams, the initial condition is almost always given at x = 0, so the solution simplifies nicely. Substituting x = 0 into y = A·e^(−ax) gives y(0) = A, meaning the constant A is simply the initial value of y.

在 IB 考试中，初值条件几乎总是在 x = 0 处给出，从而使解得以简化。将 x = 0 代入 y = A·e^(−ax)，得 y(0) = A，即常数 A 恰好是 y 的初始值。

7. Initial Value Problems | 初值问题

An initial value problem consists of the differential equation together with a condition specifying the value of y at a particular x. The procedure for solving an IVP is straightforward.

初值问题由微分方程和指定某点处 y 值的条件共同组成。求解 IVP 的步骤非常直接。

Worked Example: Solve the initial value problem

dy/dx + 3y = 0, y(0) = 5

示例：求解初值问题

dy/dx + 3y = 0, y(0) = 5

Solution: First, find the general solution. With a = 3, we have:

y = A·e^(−3x)

解答：首先求通解。由 a = 3，得：

y = A·e^(−3x)

Now apply the initial condition y(0) = 5:

5 = A·e⁰ = A

代入初始条件 y(0) = 5：

5 = A·e⁰ = A

Therefore, the particular solution is:

y = 5·e^(−3x)

因此，特解为：

y = 5·e^(−3x)

Always remember to write the final answer with the initial condition used. In IB mark schemes, finding the general solution typically earns the first set of marks, and substituting the initial condition earns the final mark.

务必记得写出已运用初值条件的最终答案。在 IB 评分标准中，求出通解通常可得前几项分，而代入初值条件则获得最后一分。

8. Applications: Exponential Growth and Decay | 应用：指数增长与衰减

The homogeneous first-order linear differential equation is the mathematical engine behind many natural phenomena. The most prominent applications in the IB syllabus are exponential growth and exponential decay.

一阶齐次线性微分方程是许多自然现象背后的数学引擎。IB 大纲中最突出的应用是指数增长与指数衰减。

Exponential decay: The rate at which a radioactive substance decays is proportional to the amount of substance present. If N(t) is the amount at time t, then:

dN/dt = −λN, N(0) = N₀

指数衰减：放射性物质的衰变速率与现存物质质量成正比。设 N(t) 为 t 时刻的质量，则：

dN/dt = −λN, N(0) = N₀

The solution is N(t) = N₀·e^(−λt), where λ is the decay constant. From this, IB students are often asked to derive the half-life using the relation N(t₁/₂) = N₀/2:

N₀/2 = N₀·e^(−λt₁/₂) → t₁/₂ = ln2/λ

其解为 N(t) = N₀·e^(−λt)，其中 λ 为衰变常数。IB 常要求学生由此推导半衰期，利用 N(t₁/₂) = N₀/2：

N₀/2 = N₀·e^(−λt₁/₂) → t₁/₂ = ln2/λ

Exponential growth: For an unrestricted population, the growth rate is proportional to the current population:

dP/dt = kP, P(0) = P₀

指数增长：对于无约束种群，增长率与当前种群数量成正比：

dP/dt = kP, P(0) = P₀

The solution P(t) = P₀·e^(kt) is used to model population growth, compound interest, and the spread of disease in early stages.

解 P(t) = P₀·e^(kt) 用于模拟种群增长、复利以及疾病传播的早期阶段。

Newton’s law of cooling also fits this pattern. The temperature T of a cooling body satisfies dT/dt = −k(T − Tₐ), which is a non-homogeneous equation. By defining u = T − Tₐ, it reduces to the homogeneous form du/dt = −ku.

牛顿冷却定律同样符合这一模式。冷却物体的温度 T 满足 dT/dt = −k(T − Tₐ)，这是非齐次方程。通过定义 u = T − Tₐ，可化为齐次形式 du/dt = −ku。

9. Common Mistakes in IB Exams | IB考试常见错误

Through years of marking IB scripts, examiners have identified several recurring errors. Being aware of these will help you avoid losing easy marks.

通过多年 IB 阅卷，考官发现了若干反复出现的错误。了解这些将帮助你避免不必要的失分。

Forgetting the absolute value: When integrating 1/y, write ln|y|, not ln y. While the absolute value often disappears in the final answer, omitting it in intermediate steps can be penalised.
忘记绝对值：对 1/y 积分时，应写 ln|y| 而不是 ln y。虽然绝对值在最终答案中通常会消失，但中间步骤缺失可能被扣分。
Missing the constant of integration: The general solution must include an arbitrary constant. A solution without +C (or the equivalent) is incomplete.
遗漏积分常数：通解必须包含任意常数。没有 +C（或等价形式）的解是不完整的。
Confusing a and −a: The sign in the exponent must match the equation. For dy/dx + ay = 0, the solution is A·e^(−ax), not A·e^(ax). Check the sign carefully.
混淆 a 与 −a：指数中的符号必须与方程一致。对于 dy/dx + ay = 0，解为 A·e^(−ax)，而非 A·e^(ax)。务必仔细检查符号。
Not applying the initial condition: Some students stop after finding the general solution. Always check whether the question asks for a particular solution.
未代入初值条件：部分学生在求出通解后便停止答题。务必检查题目是否要求特解。
Using the wrong variable: In applied problems, the independent variable is often t (time), not x. Write the solution in terms of the variable used in the question.
变量使用错误：在应用问题中，自变量通常是 t（时间），而非 x。请用题目中的变量写出解。

10. Practice Problems | 练习题

The following exercises cover the range of homogeneous first-order equations you can expect in the IB exam. Try them before checking the solutions provided.

以下练习覆盖了 IB 考试中可能出现的一阶齐次方程题型。请先自行尝试，再核对解答。

Problem 1: Find the general solution of dy/dx − 2y = 0.

题 1：求 dy/dx − 2y = 0 的通解。

Problem 2: Solve the initial value problem dy/dx + 4y = 0, y(2) = 3.

题 2：求解初值问题 dy/dx + 4y = 0, y(2) = 3。

Problem 3: A radioactive isotope decays according to dN/dt = −0.05N, where N is measured in grams and t in years. If N₀ = 100, find the amount remaining after 10 years, correct to 1 decimal place.

题 3：一种放射性同位素按 dN/dt = −0.05N 衰变，N 的单位为克，t 的单位为年。若 N₀ = 100，求 10 年后的剩余量，精确到 1 位小数。

Solutions:

解答：

Problem 1: Here a = −2, so y = A·e^(2x). The general solution is y = A·e^(2x).

题 1：这里 a = −2，故 y = A·e^(2x)。通解为 y = A·e^(2x)。

Problem 2: The general solution is y = A·e^(−4x). Using y(2) = 3:

3 = A·e^(−8) → A = 3e⁸

y = 3e⁸·e^(−4x) = 3e^(8−4x)

题 2：通解为 y = A·e^(−4x)。利用 y(2) = 3：

3 = A·e^(−8) → A = 3e⁸

y = 3e⁸·e^(−4x) = 3e^(8−4x)

Problem 3: The solution is N(t) = 100e^(−0.05t). At t = 10:

N(10) = 100e^(−0.5) ≈ 60.7 grams

题 3：解为 N(t) = 100e^(−0.05t)。当 t = 10 时：

N(10) = 100e^(−0.5) ≈ 60.7 克

11. Summary and Exam Strategy | 总结与应试策略

To master homogeneous constant coefficient first-order differential equations, remember the core result:

要掌握一阶常系数齐次线性微分方程，请牢记核心结论：

dy/dx + ay = 0 ⇔ y = A·e^(−ax)

In the exam, follow this efficient sequence:

考试中，按照以下高效步骤作答：

Identify the form: Confirm the equation is first-order, linear, homogeneous, with constant coefficient.
识别形式：确认方程是一阶、线性、齐次且常系数。
Write the general solution: State y = A·e^(−ax) directly if you have memorised the result, or derive it via separation of variables if the question requires working.
写出通解：若已牢记结论，可直接写出 y = A·e^(−ax)；若题目要求过程，则用变量分离法推导。
Apply the initial condition: Substitute the given pair (x₀, y₀) to find A.
代入初值条件：将给定的 (x₀, y₀) 代入以求出 A。
Write the particular solution: Present the final explicit form with the numerical or symbolic value of A.
写出特解：给出包含 A 的具体数值或符号值的最终显式形式。
Verify when time permits: Differentiate your answer and substitute back into the original equation to check correctness.
有时间则验算：将你的答案求导后代回原方程，检验是否正确。

With consistent practice, solving these equations becomes a straightforward, reliable source of marks in both Paper 2 and Paper 3 of IB Mathematics HL.

通过持续练习，求解这类方程将成为 IB 数学 HL 试卷 2 和试卷 3 中稳定、可靠的得分点。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving First-Order Constant Coefficient Homogeneous Linear Differential Equations | 一阶常系数齐次线性微分方程的解法