Answer
$\rightarrow y=\frac{1}{\sqrt x^3+Cx}$
Work Step by Step
Given:
$$2x(y'+y^3x^2)+y=0$$
$$2x\frac{dy}{dx}+y=-2y^3x^3$$
$$\frac{dy}{dx}+\frac{y}{2x}=-y^3x^2$$
This equation is in form of Bernoulli equation, where $p(x)=\frac{1}{2x}, q(x)=-x^2$ and $n=3$
Dividing both sides by $y^3$
$$\frac{1}{y^3}\frac{dy}{dx}+\frac{1}{2xy^2}=-x^2$$
Let $u=y^{-2} \rightarrow \frac{du}{dx}=-\frac{2}{y^3}\frac{dy}{dx}$
Then it becomes:
$$-\frac{du}{2dx}+\frac{u}{2x}=-x^2$$
$$\frac{du}{dx}-\frac{u}{x}=2x^2$$
The integrating factor is:
$$I(x)=e^{\int -\frac{1}{x}dx}=e^{-\ln (x)}=x^{-1}$$
The equation becomes:
$$\frac{d}{dx}(ux^{-1})=x^{-1}\frac{du}{dx}-x^{-2}u=2x$$
Integrating both sides:
$$ux^{-1}=x^2+C$$
$$u=x^3+Cx$$
where $C$ is a constant of integration
Subtitute when $u=y^{-2}$ we have:
$$\rightarrow y^{-2}=x^3+Cx$$
$$\rightarrow y=\frac{1}{\sqrt x^3+Cx}$$
