（部分内容参考了这篇博客)

定义：形如 $y’+P(x)y = Q(x)$ 的微分方程。

解法公式： \(y = e^{-\int P(x)dx}[\int Q(x)e^{\int P(x)dx}dx+C]\)

证明：

1. 变量代换

\[\begin{aligned} &\text{令 } y = u \cdot v\\ &\text{带入 }y'+P(x)y = Q(x)\text{ 可得 } u'v+uv'+P(x) \cdot (uv) = Q(x)\\ &\therefore u'v+u \cdot [v'+P(x) \cdot v] = Q(x)\\ &\text{为能够利用分离变量法来求 }u\text{ 对应于 }x\text{ 的函数关系不妨设 } v'+P(x) \cdot v = 0\\ &\text{易得 } v = C_1 \cdot e^{-\int P(x)dx} \\ &\text{此时 } u'v+u \cdot [v'+P(x)dx \cdot v] = u'v = u' \cdot C_1 \cdot e^{-\int P(x)dx} = Q(x)\\ &\therefore du = Q(x)dx \cdot \frac{1}{C_1} \cdot e^{\int P(x)dx}\\ &\therefore u = \frac{1}{C_1} \cdot e^{\int P(x)dx} \int Q(x)dx + C_2\\ &\therefore y = u \cdot v\\ &= [\frac{1}{C_1} \cdot e^{\int P(x)dx} \int Q(x)dx + C_2] \cdot [C_1 \cdot e^{-\int P(x)dx}]\\ &= \int Q(x)dx + C_1C_2e^{-\int P(x)dx}\\ &= [\int Q(x)e^{\int P(x)dx}dx+C]e^{-\int P(x)dx} &(C = C_1C_2) \end{aligned}\]

2. 常数变易法

\[\begin{aligned} &\text{注意到上文中有设 }v'+P(x) \cdot v = 0\text{ 的步骤，其形式与 }y'+P(x)y = Q(x)\text{ 很像}\\ &\text{可认为是 }Q(x) = 0\text{ 时的特殊情况}\\ &\text{我们先解出 }y'+P(x)y = 0\text{ 的解 }y = C \cdot e^{-\int P(x)dx}\\ &\text{比较上文 }y = u \cdot v = u \cdot e^{-\int P(x)dx}\text{ 与此处 }y = C \cdot e^{-\int P(x)dx}\\ &\text{可令 }u = C(x)\text{，此时 } y = C(x)e^{-\int P(x)dx}\\ &\text{代入 } \frac{dy}{dx}+P(x)y = Q(x)\\ &\text{得 } \frac{d}{dx}C(x)e^{-\int P(x)dx}+P(x)C(x)e^{-\int P(x)dx} = Q(x)\\ &\text{令 } u = -\int P(x)dx\\ &\therefore [e^{-\int P(x)dx}]' = u' \cdot (e^u)' = [-\int P(x)dx]' \cdot (e^u) = -P(x) \cdot e^{-\int P(x)dx}\\ &\therefore \frac{d}{dx}C(x)e^{-\int P(x)dx} = C'(x)e^{-\int P(x)dx}+C(x) \cdot [-P(x) \cdot e^{-\int P(x)dx}]\\ &= C'(x)e^{-\int P(x)dx}-C(x)P(x) \cdot e^{-\int P(x)dx}\\ &\therefore \text{原式化为 } C'(x)e^{-\int P(x)dx}-C(x)P(x)e^{-\int P(x)dx}+P(x)C(x)e^{-\int P(x)dx} = Q(x)\\ &\text{即 } C'(x)e^{-\int P(x)dx} = Q(x)\\ &\therefore C'(x) = Q(x)e^{\int P(x)dx}\\ &\therefore \int C'(x) = \int Q(x)e^{\int P(x)dx}dx+C\\ &\therefore C(x) = \int Q(x)e^{\int P(x)dx}dx+C\\ &\therefore y = C(x)e^{-\int P(x)dx} = [\int Q(x)e^{\int P(x)dx}dx+C]e^{-\int P(x)dx}\\ \end{aligned}\]