#### Answer

$y=x^3+Cx^{-2}$

#### Work Step by Step

Use an integration factor of $$\mu(x)=e^{\int P(x) dx}$$ where $P(x)=\frac{2}{x}$. $$\mu(x)=e^{\int \frac{2}{x}dx}=e^{2lnx}=e^{lnx^2}=x^2$$ Multiply the entire equation by this factor. $$x^2(\frac{dy}{dx}+\frac{2}{x}y=5x^2)$$ $$x^2\frac{dy}{dx}+2xy=5x^4$$ Integrate each side. Note how the left side becomes $y \times \mu(x)$. $$\int {x^2\frac{dy}{dx}+2xy}=\int 5x^4 dx$$ $$yx^2=x^5+C$$ Solve for $y$. $$y=x^3+Cx^{-2}$$