$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$ how should I procced from here having the equation $$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$$
I am getting to the following $$\frac{1}{2^{\frac{3}{2}}}\ln \left(y+\sqrt{2}\right)-\frac{1}{2^{\frac{3}{2}}}\ln \left(y-\sqrt{2}\right)=\frac{x^3}{3}+4x+c$$ from here I can do $\frac{1}{2^{\frac{3}{2}}}(ln\frac{y+\sqrt(2)}{y-\sqrt(2)})=\frac{x^3}{3}+4x+c$
how should I get none implicit function of $y$?
 A: $$\frac{y+\sqrt{2}}{y-\sqrt{2}}=\exp\left[2^{3/2}\left(\frac{x^3}{3}+4x+c\right)\right]$$
to make a few steps easier I will call the RHS $f(x)$:
$$y+\sqrt{2}=(y-\sqrt{2})f(x)$$
$$y(1-f(x))=-\sqrt{2}(1+f(x))$$
and so:
$$y=-\sqrt{2}\frac{1+f(x)}{1-f(x)}$$
A: $$\frac{dy}{dx} = 8+2x^2-4y^2-y^2x^2 = (4+x^2)(2-y^2)$$
$$\frac{dy}{2-y^2}=(4+x^2)dx \iff 2^{-3/2}\ln \Big(\Big \vert\frac{y+\sqrt{2}}{y-\sqrt{2}}\Big \vert\Big) = 4x+\frac{1}{3}x^3+C$$
And this is where you got to. Now we can multiply by $2^{3/2}$ and take $\exp$ of both sides to get
$$\Big \vert \frac{y+\sqrt{2}}{y-\sqrt{2}}\Big \vert= \exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)$$
(considering the positive case of the absolute value)
$$y+\sqrt{2} = y\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big) - \sqrt{2}\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)$$
$$y\Big(1-\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)\Big) = -\sqrt{2}\Big( 1+\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)\Big)$$
$$y=-\sqrt{2} \frac{1+\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}{1-\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}$$
if $|y| > \sqrt{2}$, and (by the same method and considering the negative case)
$$y=-\sqrt{2} \frac{1-\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}{1+\exp \Big(2^{7/3}x+\frac{2^{3/2}}{3}x^3+C\Big)}$$
if $|y| < \sqrt{2}$
