Question

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

Answer 1

Answer: 17

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