Answer
$y=\frac{1}{2x^5-Cx^2}$
Work Step by Step
Given:
$$y'+2x^{-1}y=6y^2x^4$$
This equation is in form of Bernoulli equation, where $p(x)=2x^{-1}, q(x)=6y^2x^4$ and $n=2$
$$\frac{dy}{dx}+2x^{-1}y=6y^2x^4$$
Dividing both sides by $y^2$
$$\frac{1}{y^2}\frac{du}{dx}+\frac{2}{xy}=6x^4$$
Let $u=y^{-1} \rightarrow \frac{du}{dx}=-\frac{1}{2y^2}\frac{dy}{dx}$
Then it becomes:
$$-\frac{du}{dx}+\frac{2u}{x}=6x^4$$
$$\frac{du}{dx}-\frac{2u}{x}=-6x^4$$
The integrating factor is:
$$I(x)=e^{\int -\frac{2}{x}dx}=e^{-2\ln (x)}=x^{-2}$$
The equation becomes:
$$\frac{d}{dx}(ux^{-2})=x^{-2}\frac{du}{dx}-2x^{-3}u=-6x^4x^{-2}=-6x^2$$
Integrating both sides:
$$ux^{-2}=-6\frac{x^3}{3}+C$$
$$u=-2x^5+Cx^2$$
where $C$ is a constant of integration
Subtitute when $u=y^{-1}$ we have:
$$\rightarrow y^{-1}=-2x^5+Cx^2$$
$$\rightarrow y=\frac{1}{2x^5-Cx^2}$$
