Solve the Following Questions
Question:

If $y^{14}+y^{-1 / 4}=2 x$, and $\left(x^{2}-1\right) \frac{d^{2} y}{d x^{2}}+\alpha x \frac{d y}{d x}+\beta y=0$, then $|\alpha-\beta|$ is equal to

Solution:

$y^{\frac{1}{4}}+\frac{1}{y^{\frac{1}{4}}}=2 x$

$\Rightarrow\left(\mathrm{y}^{\frac{1}{4}}\right)^{2}-2 \mathrm{xy}^{\left(\frac{1}{4}\right)}+1=0$

$\Rightarrow \mathrm{y}^{\frac{1}{4}}=\mathrm{x}+\sqrt{\mathrm{x}^{2}-1}$ or $\mathrm{x}-\sqrt{\mathrm{x}^{2}-1}$

So, $\frac{1}{4} \frac{1}{y^{\frac{3}{4}}} \frac{d y}{d x}=1+\frac{x}{\sqrt{x^{2}-1}}$

$\Rightarrow \frac{1}{4} \frac{1}{y^{3 / 4}} \frac{d y}{d x}=\frac{y^{\frac{1}{4}}}{\sqrt{x^{2}-1}}$

$\Rightarrow \frac{d y}{d x}=\frac{4 y}{\sqrt{x^{2}-1}} \ldots(1)$

Hence, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=4 \frac{\left(\sqrt{\mathrm{x}^{2}-1}\right) \mathrm{y}^{\prime}-\frac{\mathrm{yx}}{\sqrt{\mathrm{x}^{2}-1}}}{\mathrm{x}^{2}-1}$

$\Rightarrow\left(x^{2}-1\right) y^{\prime \prime}=4 \frac{\left(x^{2}-1\right) y^{\prime}-x y}{\sqrt{x^{2}-1}}$

$\Rightarrow\left(x^{2}-1\right) y^{\prime \prime}=4\left(\sqrt{x^{2}-1} y^{\prime}-\frac{x y}{\sqrt{x^{2}-1}}\right)$

$\Rightarrow\left(x^{2}-1\right) y^{\prime \prime}=4\left(4 y-\frac{x y^{\prime}}{4}\right)($ from I $)$

$\Rightarrow\left(x^{2}-1\right) y^{\prime \prime}+x y^{\prime}-16 y=0$

So, $|\alpha-\beta|=17$