If $y^{1/4}+y^{-1/4}=2x,$ and $\left(x^2-1\right)\frac{{\mathrm{d}}^{2}y}{d{x}^2}+\alpha x\frac{\mathrm{d}y}{ \mathrm{d}x}+\beta y=0,$ then $|\alpha -\beta |$ is equal to _______.

$y^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+\frac{1}{y^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=2x$

${\left(y^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}^2-2x{y}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+1=0$

$y^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=x+\sqrt{x^2-1}\& x-\sqrt{x^2-1}$

So, $\frac{1}{4}\frac{1}{y^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}} \frac{dy}{dx}=1+\frac{x}{\sqrt{x^2-1}}$

$\frac{1}{4}\frac{1}{y^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}\frac{dy}{dx}=\frac{y^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}{\sqrt{x^2-1}}$

$\frac{dy}{dx}=\frac{4y}{\sqrt{x^2-1}} ...\left(1\right)$

Hence 

$\frac{d^{2}y}{d{x}^2}=4\frac{\left(\sqrt{x^2-1}\right)y^{\text{'}}-{\displaystyle \frac{xy}{\sqrt{x^2-1}}}}{x^2-1}$

$\left(x^2-1\right)y^{"}=\frac{4\left(x^2-1\right)y^{\text{'}}-xy}{\sqrt{x^2-1}}$

$\left(x^2-1\right)y{\text{'}}^{}=4\left[\sqrt{x^2-1}{y}^{\text{'}}-\frac{xy}{\sqrt{x^2-1}}\right]$

$\left(x^2-1\right)y^{"}=4\left[4y-\frac{x{y}^{\text{'}}}{4}\right] \left(1\right)$

$\left(x^2-1\right)y^{"}+xy\text{'}-16y=0$

So  $\left|\alpha -\beta \right|=17$