## Calculus (3rd Edition)

$$y= x\left( -\ln x+C\right) .$$
Rewrite the equation as $$y'-x^{-1}y=-1$$ This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{-\int x^{-1} dx}=e^{-\ln x}=\frac{1}{x}$$ Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=x \left( \int -\frac{1}{x}dx +C\right)\\ &= x\left( -\ln x+C\right) \end{align} Thus, the general solution is $$y= x\left( -\ln x+C\right) .$$