If $y = e^{a \cos^{-1} x}$,$-1 \le x \le 1$,show that $(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - a^{2} y = 0$.

(N/A) Given,$y = e^{a \cos^{-1} x}$.
Taking the natural logarithm on both sides,we get $\ln y = a \cos^{-1} x$.
Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = a \left( -\frac{1}{\sqrt{1-x^2}} \right)$.
This simplifies to $\frac{dy}{dx} = -\frac{ay}{\sqrt{1-x^2}}$.
Squaring both sides,we get $\left( \frac{dy}{dx} \right)^2 = \frac{a^2 y^2}{1-x^2}$.
Rearranging,we have $(1-x^2) \left( \frac{dy}{dx} \right)^2 = a^2 y^2$.
Differentiating both sides with respect to $x$ using the product rule and chain rule:
$(1-x^2) \cdot 2 \frac{dy}{dx} \cdot \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \cdot (-2x) = a^2 \cdot 2y \cdot \frac{dy}{dx}$.
Dividing throughout by $2 \frac{dy}{dx}$ (assuming $\frac{dy}{dx} \neq 0$),we get $(1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = a^2 y$.
Thus,$(1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0$. Hence proved.