📚 Integrating Factors for First Order Differential Equations | 一阶微分方程的积分因子法
First order differential equations appear throughout A-level Mathematics, from population models to mechanical systems. When an equation is linear but not separable, the integrating factor method provides a systematic way to find an analytical solution. This article explains the standard form, derives the integrating factor, and works through examples and exam techniques.
一阶微分方程在 A-level 数学中广泛出现，从人口模型到力学系统。当方程是线性但不可分离变量时，积分因子法提供了一种系统的求解方法。本文将解释标准形式，推导积分因子，并通过例题和考试技巧加以说明。
1. Recognising the Linear First Order Form | 识别一阶线性微分方程标准形式
A first order linear differential equation can always be written in the standard form dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x only. The coefficient of dy/dx must be 1. If an equation appears as a(x)dy/dx + b(x)y = c(x), divide every term by a(x) first.
一阶线性微分方程总可以写成标准形式 dy/dx + P(x)y = Q(x)，其中 P(x) 和 Q(x) 只是 x 的函数。dy/dx 的系数必须为 1。如果方程以 a(x)dy/dx + b(x)y = c(x) 的形式出现，要先对每一项除以 a(x)。
dy/dx + P(x)y = Q(x)
This form is called linear because y and dy/dx appear only to the first power, with no products such as y dy/dx or sine of y. Equations that cannot be rearranged into this form are not suitable for the integrating factor method.
这种形式之所以称为线性，是因为 y 和 dy/dx 只以一次幂出现，没有 y dy/dx 或 sin y 等乘积或函数。不能整理成这种形式的方程不适合使用积分因子法。
2. The Concept of an Integrating Factor | 积分因子的概念
The integrating factor is defined as I(x) = e^(∫ P(x) dx). Multiplying the standard form by I(x) turns the left-hand side into the derivative of a product, so the equation can then be integrated directly. The arbitrary constant in the integrating factor can be omitted because it cancels after multiplication.
积分因子定义为 I(x) = e^(∫ P(x) dx)。将标准形式乘以 I(x) 后，左边就变成乘积的导数，因此方程可以直接积分。积分因子中的任意常数可以省略，因为乘法后会相互抵消。
I(x) = e^(∫ P(x) dx)
After multiplying, the equation becomes I(x)dy/dx + I(x)P(x)y = I(x)Q(x). The left-hand side is exactly d/dx[I(x)y], which allows the variables to be integrated without separating them.
相乘后，方程变为 I(x)dy/dx + I(x)P(x)y = I(x)Q(x)。左边恰好是 d/dx[I(x)y]，这样就可以直接积分，而无需分离变量。
3. Deriving the Product Rule Connection | 导数乘积法则的推导联系
We require I(x) to satisfy I(x)dy/dx + I(x)P(x)y = d/dx[I(x)y]. By the product rule, d/dx[I(x)y] = I(x)dy/dx + I'(x)y. Comparing the coefficients of y gives I'(x) = I(x)P(x).
我们需要 I(x) 满足 I(x)dy/dx + I(x)P(x)y = d/dx[I(x)y]。根据乘积法则，d/dx[I(x)y] = I(x)dy/dx + I'(x)y。比较 y 的系数得到 I'(x) = I(x)P(x)。
d/dx[I(x)y] = I(x)dy/dx + I'(x)y
I'(x) = I(x)P(x)
This is a separable equation in I. Rearranging gives ∫(1/I)dI = ∫P(x) dx, so ln|I| = ∫P(x) dx. Therefore I(x) = e^(∫ P(x) dx), which is the integrating factor.
这是关于 I 的可分离变量方程。整理得 ∫(1/I)dI = ∫P(x) dx，所以 ln|I| = ∫P(x) dx。因此 I(x) = e^(∫ P(x) dx)，这就是积分因子。
4. Step-by-Step Algorithm | 分步求解算法
Use the following sequence whenever a question asks for an analytical solution of a linear first order differential equation.
当题目要求一阶线性微分方程的解析解时，请按照以下步骤进行。
Step 1: Write the equation in the standard form dy/dx + P(x)y = Q(x). Step 2: Identify P(x) and compute the integrating factor I(x) = e^(∫ P(x) dx). Step 3: Multiply the entire standard form by I(x). Step 4: Rewrite the left-hand side as d/dx[I(x)y]. Step 5: Integrate both sides with respect to x. Step 6: Solve for y and include the constant of integration. Step 7: Use any initial condition to find the constant.
第一步：将方程写成标准形式 dy/dx + P(x)y = Q(x)。第二步：确定 P(x) 并计算积分因子 I(x) = e^(∫ P(x) dx)。第三步：将整个标准形式乘以 I(x)。第四步：把左边改写为 d/dx[I(x)y]。第五步：对两边关于 x 积分。第六步：解出 y，并保留积分常数。第七步：利用初始条件求出常数。
This algorithm works for any linear first order equation, including those with constant coefficients and those where the integrating factor is a power of x or an exponential function.
该算法适用于任何一阶线性方程，包括常系数方程以及积分因子为 x 的幂或指数函数的方程。
5. Worked Example with Constant Coefficient | 常系数例题
Solve dy/dx + 2y = 6. Here P(x) = 2, so the integrating factor is I(x) = e^(∫ 2 dx) = e^(2x).
求解 dy/dx + 2y = 6。这里 P(x) = 2，所以积分因子为 I(x) = e^(∫ 2 dx) = e^(2x)。
I(x) = e^(2x)
Multiplying both sides by e^(2x) gives e^(2x)dy/dx + 2e^(2x)y = 6e^(2x). The left-hand side is d/dx[e^(2x)y].
两边乘以 e^(2x) 得 e^(2x)dy/dx + 2e^(2x)y = 6e^(2x)。左边为 d/dx[e^(2x)y]。
d/dx[e^(2x)y] = 6e^(2x)
Integrating both sides with respect to x gives e^(2x)y = 3e^(2x) + C. Dividing by e^(2x) produces the general solution y = 3 + Ce^(-2x).
对两边关于 x 积分得 e^(2x)y = 3e^(2x) + C。两边除以 e^(2x) 得通解 y = 3 + Ce^(-2x)。
y = 3 + Ce^(-2x)
If the initial condition is y(0) = 4, then 4 = 3 + C, so C = 1. The particular solution is y = 3 + e^(-2x).
如果初始条件为 y(0) = 4，则 4 = 3 + C，所以 C = 1。特解为 y = 3 + e^(-2x)。
6. Worked Example with Non-Constant P(x) | 非常系数例题
Solve dy/dx + (2/x)y = x for x > 0. Here P(x) = 2/x, so the integrating factor is I(x) = e^(∫ 2/x dx) = e^(2 ln x) = x².
在 x > 0 的条件下求解 dy/dx + (2/x)y = x。这里 P(x) = 2/x，所以积分因子为 I(x) = e^(∫ 2/x dx) = e^(2 ln x) = x²。
I(x) = x²
Multiplying the standard form by x² gives x²dy/dx + 2xy = x³. The left-hand side is d/dx[x²y].
将标准形式乘以 x² 得 x²dy/dx + 2xy = x³。左边为 d/dx[x²y]。
d/dx[x²y] = x³
Integrating both sides gives x²y = x⁴/4 + C. The general solution is therefore y = x²/4 + C/x².
两边积分得 x²y = x⁴/4 + C。因此通解为 y = x²/4 + C/x²。
y = x²/4 + C/x²
Notice that the domain x > 0 allows ln x to be used without absolute value signs. This restriction is often stated in AQA questions when logarithmic functions appear in the integrating factor.
注意，定义域 x > 0 使得 ln x 可以直接使用而无需加绝对值符号。当积分因子中出现对数函数时，AQA 题目通常会给出这样的限制条件。
7. Dealing with Initial Conditions | 处理初始条件
Initial conditions are used after the general solution has been found. Solve dy/dx + y = 2e^x with y(0) = 1. Here P(x) = 1, so I(x) = e^x.
初始条件在求出通解后使用。求解 dy/dx + y = 2e^x，其中 y(0) = 1。这里 P(x) = 1，所以 I(x) = e^x。
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