Solving a differential equation: $y^4 \left(\frac{dy}{dx}\right)^4 = (y^2-1)^2$ How would one go ahead solving the following for these conditions: (i) passes through $(0,\frac{\sqrt{3}}{2})$ and (ii) $(0,\frac{\sqrt{5}}{2})$ :
$$y^4 \left(\frac{dy}{dx}\right)^4 = (y^2-1)^2$$
There is something going on with the signs here when we take the LHS and RHS to the power of $1/4$ here. I can't quite figure out what.

$$(yy')^4=(y^2-1)^2$$
$$(yy')^2=\pm(y^2-1)$$
$$yy'=\pm\sqrt{\pm(y^2-1)}$$
$Y=y^2$
$$\frac{1}{2}Y'=\pm\sqrt{\pm(Y-1)}$$
$$\frac{dY}{\pm 2\sqrt{\pm(Y-1)}}=dx$$
$$\pm\sqrt{\pm(Y-1)}=x+c$$
$$Y=1\pm(x+c)^2$$
$$y=\pm\sqrt{1\pm(x+c)^2}$$
Condition $y(0)=\frac{\sqrt{3}}{2}=+\sqrt{1- c^2} \qquad\to\qquad c=\pm\frac{1}{2}$
Condition $y(0)=\frac{\sqrt{5}}{2}=+\sqrt{1+ c^2} \qquad\to\qquad c=\pm\frac{1}{2}$
Solution :
$$y(x)=\sqrt{1\pm\left(x\pm\frac{1}{2}\right)^2}$$
