一阶线性微分方程公式推导

$$\begin{aligned} f'(x) + p(x)f(x) &= q(x) \\ {f'(x) \over f(x)}+ p(x) &\xlongequal{构造 {f'(x) \over f(x)}} {q(x) \over f(x)} \\ \ln[f(x)] + \int p(x) dx &\xlongequal{两端积分} \int {q(x) \over f(x)} dx \\ f(x) e^{\int p(x) dx} &\xlongequal{去对数} e^{\int {q(x) \over f(x)} dx} \\ f(x) &= e^{-\int p(x) dx} e^{\int {q(x) \over f(x)} dx} \\ f'(x) &\xlongequal{求导} e^{-\int p(x) dx}[-p(x)] e^{\int {q(x) \over f(x)} dx} + e^{-\int p(x) dx}\left(e^{\int {q(x) \over f(x)} dx}\right)' \\ f'(x) &\xlongequal{待定系数} q(x) - p(x)f(x)\\ &\Rightarrow e^{-\int p(x) dx}\left(e^{\int {q(x) \over f(x)} dx}\right)' = q(x) \\ &\Rightarrow \left(e^{\int {q(x) \over f(x)} dx}\right)' = q(x)e^{\int p(x) dx} \\ &\Rightarrow e^{\int {q(x) \over f(x)} dx} = \int q(x)e^{\int p(x) dx} dx + C \\ f(x) &= e^{-\int p(x) dx} \left[ \int q(x)e^{\int p(x) dx} dx + C\right] \\ \end{aligned}$$

$$\begin{aligned} u(x)y' =& f(x)g'(x) \\ u(x)p(x)y =& f'(x)g(x) \\ y =& g(x) \\ f'(x) =& u'(x) = u(x)p(x) \\ u(x) =& e^{\int p(x) dx} \\ [u(x)y]' =& u(x)[y' + p(x)y] = u(x)q(x) \\ u(x)y =& \int u(x)q(x) dx + C\\ y =& {1\over u(x)} \left[\int u(x)q(x) dx + C\right] \end{aligned}$$

$$\begin{gathered} v(x) = e^{\int p(x)dx} \\ y = \frac{1}{v(x)}\left[\int v(x) \cdot q(x) dx + C \right] \end{gathered}$$