Question

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

Answer 1

$y^{1/4}+y^{-1/4}=2x$
$\Rightarrow y^{1/2}-2x{y}^{1/4}+1=0$
$\Rightarrow y^{1/4}=\frac{2x\pm \sqrt{4x^2-4}}{2}$
$\Rightarrow y=(x\pm \sqrt{x^2-1}{)}^4$
Differentiate w.r.t. $x$, we get
$\frac{dy}{dx}=4(x\pm \sqrt{x^2-1}{)}^3(1\pm \frac{x}{\sqrt{x^2-1}})$
$\Rightarrow \frac{dy}{dx}=\frac{4(x\pm \sqrt{x^2-1}{)}^4}{\sqrt{x^2-1}}$
$\Rightarrow \frac{dy}{dx}=\frac{4y}{\sqrt{x^2-1}}$
Squaring on both the sides, we get
$(x^2-1){\left(\frac{dy}{dx}\right)}^2=16y^2$
Differentiate w.r.t. $x$, we get
$(x^2-1)\times 2\left(\frac{dy}{dx}\right)\left(\frac{d^{2}y}{d{x}^2}\right)+2x{\left(\frac{dy}{dx}\right)}^2=32y\frac{dy}{dx}$
Dividing by $2\frac{dy}{dx}$, we get
$(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=16y$
$\Rightarrow (x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-16y=0$
$\Rightarrow \alpha =1,\beta =-16$
$\therefore |\alpha -\beta |=17$