Question

If $y=e^{m.{\cos }^{-1}x}$, prove that :
$(1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=m^{2}y$

Answer 1

Given : $y=e^{{m\cos }^{-1}x}$ ...$\left(i\right)$

Differentiating $\left(i\right)$, w.r.t. $x$, we get
$\frac{dy}{dx}=e^{{m\cos }^{-1}x}.m\left(\begin{array}{c}\frac{-1}{\sqrt{1-x^2}}\end{array}\right)$
$\Rightarrow \sqrt{1-x^2}\frac{dy}{dx}=-my$ [Using $\left(i\right)$]
$\Rightarrow (1-x^2){\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)}^2=m^2{y}^2$

Differentiating again w.r.t. $x$, we get
$(1-x^2)2.\frac{dy}{dx}\frac{d^{2}y}{d{x}^2}+{\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)}^2.(-2x)=m^{2}2y\frac{dy}{dx}$

$\Rightarrow (1-x^2)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}=m^{2}y$