If y ^ 1 / 4 + y ^ - 1 / 4 = 2x, and ( x ^ 2 - 1) d ^ 2 y d x ^ 2 +...

y\dfrac{1}{4} + \dfrac{1}{y\dfrac{1}{4}} = 2x$</p><p>${\left(y\dfrac{1}{4}\right)}^{2} - 2xy\dfrac{1}{4} + 1 = 0$</p><p>$y\dfrac{1}{4} = x + \sqrt{{x}^{2} - 1} \text{$and$} x - \sqrt{{x}^{2} - 1}$</p><p>So,$\dfrac{1}{4}\dfrac{1}{{y}^{\dfrac{3}{4}}}\dfrac{dy}{dx} = 1 + \dfrac{x}{\sqrt{{x}^{2} - 1}}$</p><p>$\dfrac{1}{4}\dfrac{1}{{y}^{\dfrac{3}{4}}}\dfrac{dy}{dx} = \dfrac{{y}^{\dfrac{1}{4}}}{\sqrt{{x}^{2} - 1}}$</p><p>$\dfrac{dy}{dx} = \dfrac{4y}{\sqrt{{x}^{2} - 1}} . . . \left(1\right)$</p><p>Hence </p><p>$\dfrac{{d}^{2}y}{d{x}^{2}} = 4\dfrac{\left(\sqrt{{x}^{2} - 1}\right){y}^{'} - \dfrac{xy}{\sqrt{{x}^{2} - 1}}}{{x}^{2} - 1}$</p><p>\left({x}^{2} - 1\right){y}^{"} = \dfrac{4\left({x}^{2} - 1\right){y}^{'} - xy}{\sqrt{{x}^{2} - 1}}$

\left({x}^{2} - 1\right)y{'}^{} = 4\left[\sqrt{{x}^{2} - 1}{y}^{'} - \dfrac{xy}{\sqrt{{x}^{2} - 1}}\right]$</p><p>\left({x}^{2} - 1\right){y}^{"} = 4\left[4y - \dfrac{x{y}^{'}}{4}\right]\left(1\right)$

\left({x}^{2} - 1\right){y}^{"} + xy ' - 16y = 0$</p><p>So \left|\alpha - \beta\right| = 17$