Given differential equation

$(4+x^2)\,dy-2x(x^2+3y+4)\,dx=0$

Rearranging in derivative form:

$(4+x^2)\frac{dy}{dx}=2x(x^2+3y+4)$

so

$\frac{dy}{dx}=\frac{2x}{x^2+4}(x^2+3y+4).$

Write in linear form

Expand the right side:

$\frac{dy}{dx}=\frac{2x}{x^2+4}(x^2+4)+\frac{6xy}{x^2+4}$

since $x^2+3y+4=(x^2+4)+3y.$

Therefore,

$\frac{dy}{dx}=2x+\frac{6x}{x^2+4}y.$

Bring the $y$-term to the left:

$\frac{dy}{dx}-\frac{6x}{x^2+4}y=2x.$

This is a linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x),$ with $P(x)=-\frac{6x}{x^2+4}, \qquad Q(x)=2x.$

Find the integrating factor

$\text{I.F.}=e^{\int P(x)dx}=e^{\int -\frac{6x}{x^2+4}dx}.$

Let $u=x^2+4$, so $du=2x\,dx$. Then

$\int -\frac{6x}{x^2+4}dx=-3\int \frac{2x}{x^2+4}dx=-3\ln(x^2+4).$

Hence,

$\text{I.F.}=e^{-3\ln(x^2+4)}=(x^2+4)^{-3}.$

Multiply the equation by the integrating factor

$\frac{1}{(x^2+4)^3}\frac{dy}{dx}-\frac{6x}{(x^2+4)^4}y=\frac{2x}{(x^2+4)^3}.$

Left-hand side becomes:

$\frac{d}{dx}\left(\frac{y}{(x^2+4)^3}\right)=\frac{2x}{(x^2+4)^3}.$

Integrate both sides

$\frac{y}{(x^2+4)^3}=\int \frac{2x}{(x^2+4)^3}dx + C.$

Again let $u=x^2+4$, $du=2x\,dx$:

$\int \frac{2x}{(x^2+4)^3}dx=\int u^{-3}du=\frac{u^{-2}}{-2}=-\frac{1}{2u^2}=-\frac{1}{2(x^2+4)^2}.$

So,

$\frac{y}{(x^2+4)^3}=-\frac{1}{2(x^2+4)^2}+C.$

Multiply through by $(x^2+4)^3$:

$y=-\frac{x^2+4}{2}+C(x^2+4)^3.$

Use the condition that the curve passes through the origin

Since it passes through $(0,0)$,

$0=-\frac{0+4}{2}+C(4)^3$ $0=-2+64C$ $C=\frac{1}{32}.$

Therefore,

$y=-\frac{x^2+4}{2}+\frac{(x^2+4)^3}{32}.$

Find $y(2)$

For $x=2$,

$x^2+4=8.$

Thus,

$y(2)=-\frac{8}{2}+\frac{8^3}{32}=-4+\frac{512}{32}=-4+16=12.$

Conclusion

$\boxed{12}$