## Calculus (3rd Edition)

$$y= \ \frac{1}{5} x^{2}+\frac{1}{3}+Cx^{-3} .$$
This is a linear equation and it has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{\int 3x^{-1} dx}=e^{3\ln x}=x^{3}$$ Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=x^{-3}\left( \int x^{4} +x^2dx +C\right)\\ &=x^{-3}\left( \frac{1}{5} x^{5}+\frac{1}{3}x^3+C\right) \end{align} Thus, the general solution is $$y= \ \frac{1}{5} x^{2}+\frac{1}{3}+Cx^{-3} .$$