For $|x| < 1$, if $x = \cos\left(\frac{1}{a}\log y\right)$, then

The correct option is:

$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$

Step-by-Step Solution:

1. Rearrange the given expression:

$x = \cos\left(\frac{1}{a} \log y\right)$

$\cos^{-1}x = \frac{1}{a} \log y$

$a \cos^{-1}x = \log y$

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

2. Differentiate with respect to $x$:

$\frac{dy}{dx} = e^{a \cos^{-1}x} \cdot \frac{-a}{\sqrt{1-x^2}}$

$\frac{dy}{dx} = \frac{-ay}{\sqrt{1-x^2}}$

3. Square both sides to clear the square root:

$\left(\frac{dy}{dx}\right)^2 = \frac{a^2y^2}{1-x^2}$

$(1-x^2)\left(\frac{dy}{dx}\right)^2 = a^2y^2$

4. Differentiate again with respect to $x$ using the product rule:

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

5. Divide the entire equation by $2\frac{dy}{dx}$:

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

$(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} - a^2y = 0$