If $y = e^{a\cos^{-1}x},-1 < x < 1$, then $(1 - x^2)\frac{d^2y}{dx^2} -x\frac{dy}{dx}$ is equal to

$a^2y$

The correct answer is Option (1) → $a^2y$

$y=e^{a\cos^{-1}x},\ \theta=\cos^{-1}x\Rightarrow y=e^{a\theta}$

$\frac{d\theta}{dx}=-\frac{1}{\sqrt{1-x^2}}\Rightarrow \frac{dy}{dx}=a e^{a\theta}\frac{d\theta}{dx}=-\frac{a y}{\sqrt{1-x^2}}$

$\frac{d^2y}{dx^2}=-a\left[\frac{dy}{dx}(1-x^2)^{-1/2}+y\cdot \frac{d}{dx}(1-x^2)^{-1/2}\right]$

$\frac{d}{dx}(1-x^2)^{-1/2}=x(1-x^2)^{-3/2}$

$\frac{d^2y}{dx^2}=-a\left[\frac{dy}{dx}(1-x^2)^{-1/2}+x y(1-x^2)^{-3/2}\right]$

Multiply by $(1-x^2)$ and subtract $x\frac{dy}{dx}$:

$\begin{aligned} (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx} &=-a\Big[\frac{dy}{dx}(1-x^2)^{1/2}+x y(1-x^2)^{-1/2}\Big]-x\frac{dy}{dx}\\[6pt] &=\frac{dy}{dx}\big(-a(1-x^2)^{1/2}-x\big)-a x y(1-x^2)^{-1/2} \end{aligned}$

Substitute $\frac{dy}{dx}=-a y(1-x^2)^{-1/2}$:

$\begin{aligned} (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx} &=-a y(1-x^2)^{-1/2}\big(-a(1-x^2)^{1/2}-x\big)-a x y(1-x^2)^{-1/2}\\[6pt] &=a^2 y + a x y(1-x^2)^{-1/2}-a x y(1-x^2)^{-1/2}=a^2 y \end{aligned}$