5. First-order linear equations with a variable coefficient | 5. 一阶变系数线性微分方程

📚 5. First-order linear equations with a variable coefficient | 5. 一阶变系数线性微分方程

First-order linear differential equations form a cornerstone of the IB Mathematics analysis and approaches syllabus. When the coefficient of the dependent variable is a constant, the equation can often be solved by simple separation or by recognising a standard form. However, many real-world problems require us to handle equations where the coefficient itself depends on the independent variable. This chapter explores variable-coefficient first-order linear equations, derives the integrating factor method, and works through a range of examples to build confidence and fluency.

一阶线性微分方程是IB数学分析与方法课程的核心内容之一。当因变量的系数为常数时，通常可以通过简单的分离变量法或识别标准形式来求解。然而，许多实际问题要求我们处理系数本身依赖于自变量的方程。本章探讨变系数一阶线性方程，推导积分因子法，并通过一系列实例帮助你建立信心、熟练求解。

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1. Standard Form and Recognition | 标准形式与识别

rential equation can always be written in the standard form dy/dx + P(x)y = Q(x). The term "linear" ltiplied together. The coefficient P(x) is a function of x alone, which makes the equation "variable-coefficient" ith a coefficient in front of dy/dx, you must divide every term by that coefficient first.

一阶线性微分方程总是可以写成标准形式 dy/dx + P(x)y = Q(x)。所谓“线性”是指 y 及其导数 dy/dx 只以一次幂出现而且不相互乘在一起。系数 P(x) 仅是 x 的函数，当 P(x) 不是简单常数时方程就成为“变系数”方程。如果给出的方程在 dy/dx 前面有系数，必须先用这项系数除每一项。

ation contains terms like y² or y·dy/dx, it is not linear and a different technique is required.

识别方程是一阶线性是至关重要的第一步。例如 x·dy/dx + 2y = x² 可以重新整理成 dy/dx + (2/x)y = x，其中 P(x) = 2/x，Q(x) = x。如果方程含有类似 y² 或 y·dy/dx 的项，那么它就不是线性方程，需要采用不同的方法。

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2. The Integrating Factor Method | 积分因子法

chosen so that the left-hand side becomes the derivative of a product. We define μ(x) = e^∫P(x)dx. When we multiply the standard form by μ(x), the left side becomes d(μ·y)/dx exactly, which can then be integrated directly.

求解 dy/dx + P(x)y = Q(x) 的关键在于两边同时乘以一个恰当选择的积分因子 μ(x)，使得左边变成某个乘积的导数。我们定义 μ(x) = e^∫P(x)dx。当用 μ(x) 乘以标准形式后，左边恰好变成 d(μ·y)/dx，从而可以直接积分。

)y. Cancelling y gives dμ/dx = μ·P(x), a separable equation whose solution is indeed μ = e^∫P(x)dx. The arbitrary constant of integration in the exponent can be omitted because any constant factor will cancel out later.

这一巧妙的选择之所以有效，是因为乘法法则保证 d(μy)/dx = μ·dy/dx + y·dμ/dx，而我们要求 y·dμ/dx 等于 μ·P(x)y。消去 y 得到 dμ/dx = μ·P(x)，这是一个可分离变量的方程，其解正是 μ = e^∫P(x)dx。指数中积分产生的任意常数可以省略，因为任何常数因子之后都会被约掉。

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3. Deriving the General Solution Formula | 推导通解公式

to obtain μ(x)y = ∫ μ(x)Q(x) dx + C. Dividing through by μ(x) gives the explicit solution.

从 μ(x)·dy/dx + μ(x)P(x)y = μ(x)Q(x) 出发，并识别左边为 d(μy)/dx，对 x 两边积分得到 μ(x)y = ∫ μ(x)Q(x) dx + C。再除以 μ(x) 就给出显式解。

y = (1/μ(x)) ∫ μ(x)Q(x) dx + C/μ(x)

This formula is extremely powerful but you should understand the process rather than memorise it blindly. In practice, it is usually easier to find μ(x), multiply the original equation, and integrate step by step. The constant C is determined from an initial condition if one is provided.

这个公式非常强大，但你应该理解过程而不是盲目记忆。实际上，通常更容易的做法是求出 μ(x)，乘以原方程，然后逐步积分。如果给出了初始条件，常数 C 就由此确定。

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4. Step-by-Step Procedure | 逐步求解步骤

quation in standard form dy/dx + P(x)y = Q(x). (2) Compute the integrating factor μ(x) = e^∫P(x)dx, simplifying the exponent as much as possible. (3) Multiply the entire standard-form equation by μ(x). (4) Recognise the left side as d(μy)/dx and integrate both sides. (5) Solve for y and apply any given initial condition to find C.

要解任意一个一阶变系数线性方程，可遵循以下系统化步骤。(1) 将方程写成标准形式 dy/dx + P(x)y = Q(x)。(2) 计算积分因子 μ(x) = e^∫P(x)dx，尽可能简化指数部分。(3) 用 μ(x) 乘以标准形式的整个方程。(4) 识别左边为 d(μy)/dx，然后两边积分。(5) 解出 y，如果给出了初始条件就代入求出 C。

Many students find it helpful to box μ(x) and keep the integration constant inside the final integral until the very last step. Remember that the integral of P(x) does not need an added constant, but the integral of μQ must include a +C. Always check your final answer by differentiating and substituting back into the original equation.

许多学生发现把 μ(x) 框起来并把积分常数保留在最后的积分里直到最后一步是很有帮助的。记住 P(x) 的积分不需要加常数，但 μQ 的积分必须包含 +C。一定要通过求导并代回原方程来检验最后的答案。

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5. Worked Example 1: Basic Form | 实例1：基本形式

Solve the differential equation x·dy/dx + 2y = x² for x > 0.

解微分方程 x·dy/dx + 2y = x²，其中 x > 0。

ding by x gives dy/dx + (2/x)y = x. Here P(x) = 2/x and Q(x) = x. Step 2: Compute μ(x) = e^∫(2/x)dx = e^{2 ln x} s d(x²y)/dx, so integrate: x²y = ∫ x³ dx = x⁴/4 + C. Step 5: Solve for y: y = x²/4 + C/x².

步骤1：写成标准形式。除以 x 得到 dy/dx + (2/x)y = x。这里 P(x) = 2/x，Q(x) = x。步骤2：计算 μ(x) = e^∫(2/x)dx = e^{2 ln x} = x²。步骤3：用 x² 乘以标准形式：x²·dy/dx + 2x·y = x³。步骤4：左边是 d(x²y)/dx，因此积分：x²y = ∫ x³ dx = x⁴/4 + C。步骤5：解出 y：y = x²/4 + C/x²。

l condition is given. Always state the domain; here x > 0 avoids the singularity at x = 0.

请注意 x⁻² 项是自然地从积分过程中产生的。这是一族曲线，若给定初始条件就可以选出其中特定的一条。务必标明定义域；这里 x > 0 避开了 x = 0 处的奇点。

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6. Worked Example 2: Logarithmic Integrating Factor | 实例2：含对数的积分因子

Find the general solution of dy/dx + y/(x ln x) = 3x for x > 1.

求 dy/dx + y/(x ln x) = 3x 的通解，其中 x > 1。

dx/(x ln x). Using the substitution u = ln x, du = dx/x, the integral becomes ∫ du/u = ln|ln x|. For x > 1, ln x > 0, so we can write ln(ln x). Thus μ(x) = e^{ln(ln x)} dx + y/x = 3x ln x. The left side is d(y ln x)/dx. Integrating: y ln x = ∫ 3x ln x dx + C.

这里 P(x) = 1/(x ln x)，Q(x) = 3x。计算 ∫P(x)dx = ∫ dx/(x ln x)。使用代换 u = ln x，du = dx/x，积分变为 ∫ du/u = ln|ln x|。对于 x > 1，ln x > 0，故可写成 ln(ln x)。因此 μ(x) = e^{ln(ln x)} = ln x。用 ln x 乘方程两边：ln x·dy/dx + y/x = 3x ln x。左边是 d(y ln x)/dx。积分：y ln x = ∫ 3x ln x dx + C。

efore y ln x = (3x²/2) ln x − (3x²/4) + C. Finally, y = (3x²/2) − (3x²)/(4 ln x) + C/ln x.

剩下的积分用分部积分法：令 u = ln x，dv = 3x dx。则 du = dx/x，v = 3x²/2。所以 x = (3x²/2) ln x − ∫ (3x²/2)(1/x) dx = (3x²/2) ln x − (3/2)∫ x dx = (3x²/2) ln x − (3x²/4)。因此 y ln x = (3x²/2) ln x − (3x²/4) + C。最终 y = (3x²/2) − (3x²)/(4 ln x) + C/ln x。

This example highlights that P(x) does not have to be a simple power function. Recognising the integral of 1/(x ln x) is a standard technique and appears frequently in IB exam questions.

这个例子强调 P(x) 不一定是简单的幂函数。识别 1/(x ln x) 的积分是标准技巧，在 IB 考题中经常出现。

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7. Worked Example 3: Initial-Value Problem | 实例3：初值问题

Solve dy/dx − y·tan x = sec x, given that y(0) = 1, for −π/2 < x < π/2.

求解 dy/dx − y·tan x = sec x，已知 y(0) = 1，其中 −π/2 < x < π/2。

Standard form already: P(x) = −tan x, Q(x) = sec x. Compute ∫P(x)dx = ∫(−tan x)dx = ln|cos x|. On the given interval cos x > 0, so we can write ln(cos x). Thus μ(x) = e^{ln(cos x)} x·sec x = 1. Recognise left side as d(y cos x)/dx. Integrating: y cos x = ∫ 1 dx = x + C.

方程已为标准形式：P(x) = −tan x，Q(x) = sec x。计算 ∫P(x)dx = ∫(−tan x)dx = ln|cos x|。在给定区间上 cos x > 0，故可写成 ln(cos x)。因此 μ(x) = e^{ln(cos x)} = cos x。相乘得：cos x·dy/dx − y·sin x = cos x·sec x = 1。识别左边为 d(y cos x)/dx。积分：y cos x = ∫ 1 dx = x + C。

Apply the initial condition y(0) = 1: at x = 0, cos 0 = 1, so 1·1 = 0 + C ⇒ os x = (x + 1) sec x. Since sec x = 1/cos x, the solution is well defined on the interval.

应用初始条件 y(0) = 1：当 x = 0 时 cos 0 = 1，故 1·1 = 0 + C ⇒ C = 1。于是 y cos x = x + 1，y = (x + 1)/cos x = (x + 1) sec x。因为 sec x = 1/cos x，解在该区间上定义良好。

Initial-value problems are a common requirement. Once the general solution is found, substituting the condition pins down the constant and provides a unique solution curve.

初值问题是常见的要求。一旦找到通解，代入条件即可确定常数并给出唯一解曲线。

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8. Special Cases: Bernoulli Equation | 特殊情形：伯努利方程

A Bernoulli equation has the form dy/dx + P(x)y = Q(x)yⁿ a real number not equal to 0 or 1. Although it is non-linear, a clever substitution v = y¹⁻ⁿ transforms it into a first-order linear equation in v. This greatly extends the reach of the integrating factor technique.

伯努利方程具有形式 dy/dx + P(x)y = Q(x)yⁿ，其中 n 是不等于 0 或 1 的实数。虽然它是非线性的，但巧妙代换 v = y¹⁻ⁿ 可将它转化为关于 v 的一阶线性方程。这极大地扩展了积分因子法的应用范围。

Setting v = y¹⁻ⁿ implies dv/dx = (1−n)y⁻ⁿ·dy/dx. Multiply the original Bernoulli equation by (1−n)y⁻ⁿ a linear first-order equation in v with new coefficient (1−n)P(x). Solve for v, then back-substitute to find y.

令 v = y¹⁻ⁿ，则 dv/dx = (1−n)y⁻ⁿ·dy/dx。将原伯努利方程乘以 (1−n)y⁻ⁿ，并利用代换得到 dv/dx + (1−n)P(x)v = (1−n)Q(x)。这是一个关于 v 的、系数为 (1−n)P(x) 的一阶线性方程。解出 v，再回代求得 y。

For example, dy/dx + (1/x)y = x·y². Here n = 2, so v = y¹⁻² = y⁻¹. The transformed linear equation is dv/dx − (1/x)v = −x. The integrating factor is e^{∫(−1/x)dx} /3 + Cx. Then y = 1/v = 1/(−x²/3 + Cx). Such techniques are highly valued in IB HL papers.

例如 dy/dx + (1/x)y = x·y²。这里 n = 2，因此 v = y⁻¹。转化后的线性方程为 dv/dx − (1/x)v = −x。积分因子为 e^{∫(−1/x)dx} = 1/x，得到 v = −x²/3 + Cx。于是 y = 1/v = 1/(−x²/3 + Cx)。这类技巧在 IB HL 试卷中很有价值。

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9. Applications: A Variable-Coefficient Model | 应用：一个变系数模型

flows out at a rate proportional to the instantaneous volume V(t), given by outflow rate = 0.02V(t). The volume changes according to dV/dt = 3 − 0.02V.

考虑一个混合槽，其中流出速率与容积成正比，导致浓度方程出现变系数。假设一个槽最初装有 100 升纯水。浓度为 2 g/L 的盐溶液以 3 L/min 流入，搅拌充分的混合液以与瞬时体积 V(t) 成正比的速率流出，流出速率 = 0.02V(t)。体积变化满足 dV/dt = 3 − 0.02V。

ate in − rate out = 2·3 − (x/V)·(0.02V) = 6 − 0.02x. This is actually constant-coefficient! scribes a process where the volume dilutes the concentration with a time-dependent factor.

设 x(t) 为盐的质量（克）。盐量平衡给出 dx/dt = 流入速率 − 流出速率 = 2·3 − (x/V)·(0.02V) = 6 − 0.02x。结果实际上是常系数！更好的变系数模型出现在流入浓度是时间的函数时，例如 cᵢ(t) = 2 + sin t，或者流出速率显式依赖于时间。例如 dx/dt + (2/t)x = 3t 描述了一个体积以时间依赖因子稀释浓度的过程。

/4)(t² − 1/t²). Such models show how variable coefficients naturally emerge from changing physical conditions and why the integrating factor method is essential.

用 μ = t² 求解 dx/dt + (2/t)x = 3t 得到 x = (3/4)t² + C/t²。若 t=1 时初始质量为 0，则 C = −3/4，故 x = (3/4)(t² − 1/t²)。这类模型展示了变系数如何从变化的物理条件中自然产生，以及积分因子法为何必不可少。

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10. Common Pitfalls and Tips | 常见陷阱与提示

1. Forgetting to put the equation in exact standard form: always make sure the coefficient of dy/dx is 1 and that the y-term is on the left with its coefficient correctly identified as P(x). 2. Losing a minus sign: if P(x) is negative, the integrating factor becomes e^−∫…, and the integration must respect that sign. 3. Omitting the constant C in the final integration: the +C belongs after integrating μQ, not before.

1. 忘记将方程化为准确的标准形式：始终确保 dy/dx 的系数为 1，并且 y 项在左边，其系数被正确地识别为 P(x)。2. 遗漏负号：如果 P(x) 为负，积分因子变成 e^−∫…，积分时必须尊重这个符号。3. 在最终积分中遗漏常数 C：+C 应在对 μQ 积分之后，而不是之前。

4. Misapplying the product rule verification: always differentiate your solution and plug into the original equation to check. 5. Ignoring the domain: solutions may have restricted intervals because of logarithms or division by zero. 6. Confusing the roles of P(x) and Q(x): know which one goes into the integrating factor and which one is integrated later.

4. 错误应用乘积法则验证：始终对你的解求导并代回原方程进行检验。5. 忽略定义域：由于对数或除零问题，解可能存在受限区间。6. 混淆 P(x) 和 Q(x) 的角色：清楚哪一个进入积分因子，哪一个稍后积分。

Finally, practice is the key. Set yourself a routine: identify P and Q, calculate μ, rewrite the left side as a derivative, integrate, and apply initial conditions. With consistent practice, first-order linear equations with variable coefficients become a reliable source of marks.

最后，练习是关键。为自己设定一套常规：识别 P 和 Q，计算 μ，将左边改写为导数，积分，并应用初始条件。通过持续练习，一阶变系数线性方程会成为可靠的得分点。

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