一阶线性微分方程：积分因子法

积分因子 $I(x)=e^{\int (1/x)dx}=x$。

乘上 $x$ 得 $(xy)'=1/x$，积分：

$$xy=\ln x+C\Rightarrow y=\frac{\ln x+C}{x}.$$

用 $y(1)=1$ 得 $C=1$，所以

$$y(x)=\frac{\ln x+1}{x}.$$

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Original Explanation

• We identify $$P(x)=\frac{1}{x},\; Q(x)=\frac{1}{x^2}.$$

• The integrating factor is $$I(x) = \exp\!\Bigl(\int \tfrac{1}{x}\,dx\Bigr) = \exp(\ln x) = x.$$

• Then $$x\,\frac{dy}{dx} + y \;=\; \frac{1}{x}. \quad\Longrightarrow\quad \bigl(x\,y\bigr)' \;=\; 1.$$

• Integrate: $$x\,y \;=\; x + C \;\Longrightarrow\; y \;=\; 1 + \frac{C}{x}.$$

  However, from the direct formula: $$y = \frac{1}{I(x)} \int I(x)\,Q(x)\,dx = \frac{1}{x} \int x \,\frac{1}{x^2}\,dx = \frac{1}{x} \int \frac{1}{x}\,dx = \frac{1}{x} \Bigl(\ln x + C\Bigr).$$ Using $$y(1)=1$$ implies $$1 = 1\cdot(\ln 1 + C)=C.$$ So $$y = \frac{\ln x + 1}{x}.$$

Hence the solution: $$y(x) = \frac{\ln x + 1}{x}.$$