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

📚 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。

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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。

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

You cannot simply integrate both sides of dy/dx + P(x)y = Q(x) with respect 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’ = P R, then the left side of the differential 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。

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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。

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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。

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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

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7. Using Initial Conditions | 使用初始条件

When an initial condition is given, you must find the particular solution by substituting the condition into 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)

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

If Q(x) = 0, the equation becomes dy/dx + 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. If P(x) = 1/x, the integrating 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。

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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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