原方程化为标准一阶线性微分方程形式：$$\frac{dy}{dx} - y = e^x.$$其中$P(x) = -1$，$Q(x) = e^x$，积分因子为$$\mu(x) = e^{\int -1\,dx} = e^{-x}.$$两边同乘积分因子得$$e^{-x}\frac{dy}{dx} - e^{-x}y = 1,$$即$$\frac{d}{dx}(e^{-x}y) = 1.$$积分得$$e^{-x}y = x + C,$$故通解为$$y = e^x(x + C).$$