Does this nasty non linear ODE have any "nice" solutions?

Question

The ODE is the following:

$$y' = y^4 + \dfrac{(y')^2}{1-y^2}$$

I've been trying to solve it but obviously had no luck. I'm not sure whether it can even be solved analitically, so any help even on seeing the behavior of $y$ might be helpful.

Answer 1

Solve it via quadratic equation:

$$y' = \frac{1\pm \sqrt{1+4y^4/(1-y^2)}}{2}.$$

Grouping the $y$'s together, you can evaluate the integral in $y$, in terms of elliptic functions. The solution will not be pretty.

Answer 2

Would you consider $y=0$ to be a nice solution?

Answer 3

$$y' = y^4 + \dfrac{(y')^2}{1-y^2}$$ This equation is of the form $y'=f(y)$ $$y'(1-y^2) = y^4-y^6 + y'^2$$ $$ y'^2-y'(1-y^2)= y^6-y^4 $$ $$ (y'-\frac 12(1-y^2))^2= y^6-y^4 + \frac 14(1-y^2)^2$$ $$ \frac {dy}{dx}=\frac 12(1-y^2)\pm \sqrt{ y^4(y^2-1) + \frac 14(1-y^2)^2}$$ $$ \frac {dy}{dx}=\frac 12(1-y^2) \pm \sqrt{ (y^2-1)(y^4 + \frac 14(y^2-1))}$$ The integral cant be expressed with elementary functions