Question

If $y^{1/4}+y^{-1/4}=2 x$, 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 ___

Answer 1

$y ^{\frac{1}{4}}+\frac{1}{ y ^{\frac{1}{4}}}=2 x$
$\Rightarrow \left(y^{\frac{1}{4}}\right)^{2}-2 x y^{\left(\frac{1}{4}\right)}+1=0$
$\Rightarrow y^{\frac{1}{4}}=x+\sqrt{x^{2}-1}$
or $x-\sqrt{x^{2}-1}$
So, $\frac{1}{4} \frac{1}{ y ^{\frac{3}{4}}} \frac{ dy }{ dx }=1+\frac{ x }{\sqrt{ x ^{2}-1}}$
$\Rightarrow \frac{1}{4} \frac{1}{y^{3 / 4}} \frac{d y}{d x}=\frac{y^{\frac{1}{4}}}{\sqrt{x^{2}-1}}$
$\Rightarrow \frac{d y}{d x}=\frac{4 y}{\sqrt{x^{2}-1}} \ldots(1)$
Hence, $\frac{ d ^{2} y }{ dx ^{2}}=4 \frac{\left(\sqrt{ x ^{2}-1}\right) y'-\frac{ yx }{\sqrt{ x ^{2}-1}}}{ x ^{2}-1}$
$\Rightarrow \left(x^{2}-1\right) y''=4 \frac{\left(x^{2}-1\right) y'-x y}{\sqrt{x^{2}-1}}$
$\Rightarrow \left(x^{2}-1\right) y''=4\left(\sqrt{x^{2}-1} y'-\frac{x y}{\sqrt{x^{2}-1}}\right)$
$\Rightarrow \left(x^{2}-1\right) y''=4\left(4 y-\frac{x y'}{4}\right)($ from $I )$
$\Rightarrow \left(x^{2}-1\right) y''+x y'-16 y=0$
So, $|\alpha-\beta|=17$