求下列微分方程的通解：(1) (dy)/(dx) + y = e^-x;(2) xy' + y = x^2 + 3x + 2;(3) y' + y cos x = e^-sin x;(4) y' + y tan x = sin 2x;

解 (1) $y = \mathrm{e}^{-\int dx} \left[ \int \mathrm{e}^{-x} \cdot \mathrm{e}^{\int dx} \, dx + C \right] = \mathrm{e}^{-x} \left( \int \mathrm{e}^{-x} \cdot \mathrm{e}^{x} \, dx + C \right)$ = $\mathrm{e}^{-x} (x + C)$. (2) 将方程改写成 $y' + \frac{1}{x}y = x + 3 + \frac{2}{x}$, 则 $y = \mathrm{e}^{-\int \frac{1}{x} dx} \left[ \int \left( x + 3 + \frac{2}{x} \right) \mathrm{e}^{\int \frac{1}{x} dx} \, dx + C \right] = \frac{1}{x} \left[ \int \left( x + 3 + \frac{2}{x} \right) x \, dx + C \right]$ = $\frac{1}{x} \left[ \int (x^2 + 3x + 2) \, dx + C \right] = \frac{1}{x} \left( \frac{x^3}{3} + \frac{3}{2}x^2 + 2x + C \right)$ = $\frac{x^2}{3} + \frac{3x}{2} + 2 + \frac{C}{x}$. (3) $y = \mathrm{e}^{-\int \cos x \, dx} \left( \int \mathrm{e}^{-\sin x} \cdot \mathrm{e}^{\int \cos x \, dx} \, dx + C \right) = \mathrm{e}^{-\sin x} \left( \int \mathrm{e}^{-\sin x} \cdot \mathrm{e}^{\sin x} \, dx + C \right)$ = $\mathrm{e}^{-\sin x} (x + C)$. (4) $y = \mathrm{e}^{-\int \tan x \, dx} \left( \int \sin 2x \cdot \mathrm{e}^{\int \tan x \, dx} \, dx + C \right)$ = $\cos x \left( \int \frac{\sin 2x}{\cos x} \, dx + C \right) = \cos x \left( \int 2 \sin x \, dx + C \right)$ = $C \cos x - 2 \cos^2 x$.