Analytical Solution of First-Order Linear Differential Equations: Integrating Factors | 一阶线性微分方程的解析解法：积分因子法

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📚 Analytical Solution of First-Order Linear Differential Equations: Integrating Factors | 一阶线性微分方程的解析解法：积分因子法

First-order linear differential equations appear throughout mathematics, physics and engineering. In AQA A-level Mathematics, the analytical method of integrating factors provides a systematic routine for solving equations of the form dy/dx + P(x)y = Q(x). This article explains the method step by step and shows how to apply it in examinations.

一阶线性微分方程在数学、物理和工程中广泛出现。在 AQA A-level 数学中，积分因子法为形如 dy/dx + P(x)y = Q(x) 的方程提供了一套系统的解析求解步骤。本文将逐步讲解此方法，并说明如何在考试中应用。

1. Recognising the Standard Form | 识别标准形式

The first step is to identify the standard form. A first-order linear differential equation is linear in y and dy/dx. It can be written as

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

where P and Q are functions of x only. If the coefficient of dy/dx is not 1, divide the whole equation by that coefficient before comparing it with the standard form.

第一步是识别标准形式。一阶线性微分方程关于 y 和 dy/dx 是线性的。它可以写成

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

其中 P 和 Q 仅为 x 的函数。如果 dy/dx 的系数不是 1，应先将整个方程除以该系数，再与标准形式对照。

2. Why Do We Need an Integrating Factor? | 为什么需要积分因子？

to x, because the left-hand side contains terms involving both y and dy/dx. The product rule suggests that a combination such as d/dx(yR) might simplify the equation, since

d/dx(yR) = R dy/dx + R' y

If we can find a function R such that R' ntial equation becomes an exact derivative. This is exactly the idea behind the integrating factor.

不能直接对 dy/dx + P(x)y = Q(x) 两边关于 x 积分，因为左边同时含有 y 和 dy/dx 项。乘积法则提示我们，像 d/dx(yR) 这样的组合可能化简方程：

d/dx(yR) = R dy/dx + R' y

如果能找到函数 R 使 R' = P R，那么微分方程左边就变成一个完全导数。这正是积分因子的核心思想。

3. Deriving the Integrating Factor | 积分因子的推导

Let the integrating factor be I(x). Multiply the standard equation by I:

I dy/dx + I P y = I Q

We want the left-hand side to equal d/dx(I y). Expanding this derivative gives I dy/dx + I' y. Therefore we require I' = I P.

设积分因子为 I(x)。将标准方程乘以 I：

I dy/dx + I P y = I Q

我们希望左边等于 d/dx(I y)。展开该导数得到 I dy/dx + I' y。因此需要 I' = I P。

This is a separable equation: dI/dx = P I. Rearranging and integrating gives

∫(1/I) dI = ∫P dx

so ln|I| = ∫P dx. Hence

I(x) = e^{∫P(x) dx}

No constant of integration is needed here, because multiplying I by a scalar does not change the solution process.

这是一个可分离变量的方程：dI/dx = P I。整理并积分得

∫(1/I) dI = ∫P dx

所以 ln|I| = ∫P dx。因此

I(x) = e^{∫P(x) dx}

这里不需要加积分常数，因为给 I 乘以一个常数不会改变求解过程。

4. The General Solution Formula | 通解公式

After multiplying by the integrating factor, the equation becomes

d/dx(y I(x)) = I(x)Q(x)

Integrating both sides with respect to x gives

y I(x) = ∫I(x)Q(x) dx + C

Finally, divide by I(x) to obtain the general solution:

y = (1/I(x))(∫I(x)Q(x) dx + C)

This is the master formula for solving first-order linear differential equations using an integrating factor.

乘以积分因子后，方程变为

d/dx(y I(x)) = I(x)Q(x)

两边关于 x 积分得

y I(x) = ∫I(x)Q(x) dx + C

最后除以 I(x) 得到通解：

y = (1/I(x))(∫I(x)Q(x) dx + C)

这就是用积分因子求解一阶线性微分方程的主公式。

5. Worked Example 1: Constant Coefficient | 例 1：常系数方程

Solve the differential equation

dy/dx + 2y = e^3x

Here P(x) = 2 and Q(x) = e^3x. The integrating factor is

I(x) = e^{∫2 dx} = e^2x

Multiplying through by e^2x gives

e^2x dy/dx + 2e^2x y = e^5x

The left-hand side is exactly d/dx(y e^2x). Therefore

y e^2x = ∫e^5x dx = (1/5)e^5x + C

Dividing by e^2x, the general solution is

y = (1/5)e^3x + C e^-2x

解微分方程

dy/dx + 2y = e^3x

这里 P(x) = 2，Q(x) = e^3x。积分因子为

I(x) = e^{∫2 dx} = e^2x

两边乘以 e^2x 得

e^2x dy/dx + 2e^2x y = e^5x

左边恰好是 d/dx(y e^2x)。因此

y e^2x = ∫e^5x dx = (1/5)e^5x + C

除以 e^2x，得通解

y = (1/5)e^3x + C e^-2x

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6. Worked Example 2: Variable Coefficient | 例 2：变系数方程

Solve, for x > 0,

dy/dx + (1/x)y = x²

Here P(x) = 1/x and Q(x) = x². The integrating factor is

I(x) = e^{∫(1/x) dx} = e^{ln x} = x

Multiplying by x gives

x dy/dx + y = x³

The left-hand side is d/dx(xy). Integrating both sides:

xy = ∫x³ dx = (1/4)x⁴ + C

Thus the general solution is

y = (1/4)x³ + C/x

求解以下方程（x > 0）：

dy/dx + (1/x)y = x²

这里 P(x) = 1/x，Q(x) = x²。积分因子为

I(x) = e^{∫(1/x) dx} = e^{ln x} = x

乘以 x 得

x dy/dx + y = x³

左边是 d/dx(xy)。两边积分：

xy = ∫x³ dx = (1/4)x⁴ + C

因此通解为

y = (1/4)x³ + C/x

7. Using Initial Conditions | 使用初始条件

to the general solution. For example, suppose that in Example 2 we are told that y(1) = 5.

当给出初始条件时，必须将该条件代入通解，以求出特解。例如，在例 2 中假设 y(1) = 5。

Substituting x = 1 and y = 5 into y = (1/4)x³ + C/x gives

5 = (1/4) + C

so C = 5 - 1/4 = 19/4. The particular solution is therefore

y = (1/4)x³ + 19/(4x)

把 x = 1 和 y = 5 代入 y = (1/4)x³ + C/x 得

5 = (1/4) + C

所以 C = 5 - 1/4 = 19/4。因此特解为

y = (1/4)x³ + 19/(4x)

8. Special Cases and Quick Checks | 特殊情况与快速检验

+ P(x)y = 0, which is also separable. The integrating factor method still works and gives

y = C e^{-∫P(x) dx}

If P(x) is a constant k, the integrating factor is simply e^kx grating factor simplifies to x (assuming x > 0). These shortcuts can save time in an exam.

如果 Q(x) = 0，方程变为 dy/dx + P(x)y = 0，这也是可分离变量的。积分因子法仍然适用，并给出

y = C e^{-∫P(x) dx}

如果 P(x) 是常数 k，则积分因子就是 e^kx。如果 P(x) = 1/x，则积分因子简化为 x（假设 x > 0）。这些快捷结论在考试中可以节省时间。

9. Common Pitfalls | 常见错误

Forgetting to divide by the coefficient of dy/dx. The standard form must always have coefficient 1 for dy/dx.
忘记除以 dy/dx 的系数。标准形式中 dy/dx 的系数必须始终为 1。
Adding a constant inside the integral for I(x). The constant is unnecessary and, if included, will cancel anyway.
在积分因子 I(x) 的指数积分中加常数。该常数并不需要，即使加上也会抵消。
Forgetting to multiply Q(x) by the integrating factor. Every term on both sides must be multiplied.
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