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

We start from the relation $$y^{1/4}+y^{-1/4}=2x.$$

Put $$t=y^{1/4}\;(\Rightarrow y=t^{4}).$$ The given relation becomes

$$t+\frac1t=2x\qquad\Longrightarrow\qquad x=\frac{t+\dfrac1t}{2}.$$

First we differentiate this relation to connect the derivatives of $$t$$ and $$x$$. Using the chain rule,

$$\frac{d}{dx}\left(t+\frac1t\right)=2\;\Longrightarrow\;\left(1-\frac1{t^{2}}\right)\frac{dt}{dx}=2,$$

so

$$\frac{dt}{dx}=\frac{2t^{2}}{t^{2}-1}.$$

Because $$y=t^{4},$$ we have by the chain rule

$$\frac{dy}{dx}=4t^{3}\frac{dt}{dx}=4t^{3}\left(\frac{2t^{2}}{t^{2}-1}\right)=\frac{8t^{5}}{t^{2}-1}.$$

Again differentiating, and writing $$y'=\frac{dy}{dx},\;y''=\frac{d^{2}y}{dx^{2}},$$ we find

$$y'=\frac{8t^{5}}{t^{2}-1}\;\Longrightarrow\;\frac{dy'}{dt}= \frac{d}{dt}\left(\frac{8t^{5}}{t^{2}-1}\right) =\frac{40t^{4}(t^{2}-1)-8t^{5}\cdot2t}{(t^{2}-1)^{2}} =\frac{24t^{6}-40t^{4}}{(t^{2}-1)^{2}} =\frac{8(3t^{6}-5t^{4})}{(t^{2}-1)^{2}}.$$

Hence

$$y''=\frac{dy'}{dx}=\frac{dy'}{dt}\frac{dt}{dx} =\frac{8(3t^{6}-5t^{4})}{(t^{2}-1)^{2}}\cdot\frac{2t^{2}}{t^{2}-1} =\frac{16t^{6}(3t^{2}-5)}{(t^{2}-1)^{3}}.$$

Next we need $$x^{2}-1$$ in terms of $$t$$. From $$x=\dfrac{t+\dfrac1t}{2},$$

$$x^{2}=\frac{(t+\frac1t)^{2}}{4} =\frac{t^{2}+2+\dfrac1{t^{2}}}{4},$$

so

$$x^{2}-1=\frac{t^{2}+\dfrac1{t^{2}}-2}{4} =\frac{(t-\frac1t)^{2}}{4} =\frac{(t^{2}-1)^{2}}{4t^{2}}.$$

We now substitute $$y,\;y',\;y''$$ and $$x^{2}-1$$ into the differential equation

$$(x^{2}-1)y''+\alpha xy'+\beta y=0.$$

The first term is

$$(x^{2}-1)y'' =\frac{(t^{2}-1)^{2}}{4t^{2}}\cdot\frac{16t^{6}(3t^{2}-5)}{(t^{2}-1)^{3}} =\frac{4t^{4}(3t^{2}-5)}{t^{2}-1}.$$

The second term uses $$xy'=\frac{t+\dfrac1t}{2}\cdot\frac{8t^{5}}{t^{2}-1} =\frac{4(t^{6}+t^{4})}{t^{2}-1},$$ so

$$\alpha xy'=\alpha\cdot\frac{4(t^{6}+t^{4})}{t^{2}-1}.$$

The third term is simply $$\beta y=\beta t^{4}.$$

Multiplying the whole equation by $$t^{2}-1$$ to clear denominators gives

$$4t^{4}(3t^{2}-5)+4\alpha(t^{6}+t^{4})+\beta t^{4}(t^{2}-1)=0.$$

Collecting like powers of $$t$$:

$$\bigl[12+4\alpha+\beta\bigr]t^{6}+\bigl[-20+4\alpha-\beta\bigr]t^{4}=0.$$

This must hold for all (non-zero) $$t$$, so the coefficients of both $$t^{6}$$ and $$t^{4}$$ must vanish:

$$\begin{cases} 12+4\alpha+\beta=0,\\ -20+4\alpha-\beta=0. \end{cases}$$

Adding the two equations gives $$8\alpha-8=0\;\Longrightarrow\;\alpha=1.$$

Substituting $$\alpha=1$$ into the first equation yields $$\beta=-12-4(1)=-16.$$

Therefore

$$|\alpha-\beta|=\bigl|1-(-16)\bigr|=17.$$

Hence, the correct answer is Option 17.