Exceptional solutions to ODE

Question

I'm trying to find the general solution for $y'=2(1-y^2)$ such that it also contains the solutions $y=\pm1$. I managed to arrive at $$\frac{\ln|y+1|-\ln|y-1|}{2}=2x+C$$ which gives $$\frac{y+1}{y-1}=Ae^{4x}$$ which further gives $$\frac{2}{y-1}=Ae^{4x}-1$$ so $$y=\frac{2}{Ae^{4x}-1}+1$$. Now, what I'm unsure about is how to check if this solution satisfies the exceptional solutions since the derivative is an expression containing only the variable $x$ and I'm trying to check if $y=\pm1$ satisfies the ODE. I'm also not sure how I would modify the solution so that it also contains the exceptional solutions.

Answer 1

Hint: What happens in the general solution if $A=0$ or $A\to \infty$?