## Calculus (3rd Edition)

$$y=x^{-1}\left( \frac{x^2}{2} +C\right).$$
Rewriting the equation as $$y'+x^{-1}y=1$$ This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{\int x^{-1} dx}=e^{\ln x}=x$$ Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int \alpha(x)Q(x)dx +C\right)\\ &=x^{-1}\left( \int x dx +C\right)\\ &=x^{-1}\left( \frac{x^2}{2} +C\right) \end{align} The general soluiton is $$y=x^{-1}\left( \frac{x^2}{2} +C\right).$$