In this text https://www.springer.com/gp/book/9781461454762 on p. 95, it has the following
The text does not explain how to get to this solution. $B, C, D$ are arbitrary constants. With $A>0$, I came to something like this $$Y_0=B\tan\left(\frac{B\bar{x}}{2}+C\right)$$ Is that and the positive case solution the same? Also how did the negative case solution come about?
You can look at this like a Riccati equation and set $Y=-2\frac{u'}{u}$ so that $$ Y'=-2\frac{u''}{u}+2\frac{u'^2}{u^2}=2\frac{u'^2}{u^2}+A\iff u''+\frac{A}2u=0 $$ This now is a standard linear DE with constant coefficients and you can easily treat the 3 cases $A=2B^2$, $A=-2B^2$ and $A=0$.
\begin{align} u''+B^2u&=0\implies& u&=\cos(Bx+C),&~~Y&=2B\tan(Bx+C)\\ u''&=0&\implies u&=Dx+C,&~~Y&=-\frac{2D}{Dx+C}\\ u''-B^2u&=0\implies & u&=Ce^{Bx}+De^{-Bx},&~~ Y&= -2B\frac{Ce^{Bx}-De^{-Bx}}{Ce^{Bx}+De^{-Bx}} \end{align}