## Calculus (3rd Edition)

$$y=\frac{1}{3x} +\frac{1}{9} +Ce^{3 x^{-1}}.$$
Rewrite the equation as $$y'+3x^{-2}y=x^{-3}$$ This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{3\int x^{-2} dx}=e^{-3 x^{-1}}$$ Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=e^{3 x^{-1}}\left( \int e^{-3 x^{-1}} x^{-3}dx +C\right)\\ &= e^{3 x^{-1}}\left( \frac{1}{3}x^{-1}e^{-3 x^{-1}} +\frac{1}{9} e^{-3 x^{-1}}+C\right) \end{align} where we did the integration by parts. The general solution is $$y=\frac{1}{3x} +\frac{1}{9} +Ce^{3 x^{-1}}.$$