Answer
$y=\frac{1}{5}x^3+Cx^{-2}$
Work Step by Step
This equation can be rewritten as
$$\frac{dy}{dx} = x^2-\frac{2y}{x}=x^2-\frac{2}{x}y$$
Therefore, $$\frac{dy}{dx}+\frac{2}{x}y=x^2$$
Multiplying both sides of the equation by $x^2$ gives
$$x^2\frac{dy}{dx}+2xy=x^4$$
By product rule, $(x^2y)'=2xy+x^2y'$, so $$\frac{d}{dx}(x^2y)=x^4$$ Integrating both sides with respect to $x$ gives $$x^2y=\frac{1}{5}x^5+C$$ Finally, dividing both sides by $x^2$ gives $$y=\frac{1}{5}x^3+Cx^{-2}$$
