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

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

The correct answer is Option (3) → $(1-x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx}-a^2y=0$

Given $x=\cos\!\big(\frac{1}{a}\log y\big)$ with $|x|<1$. Put $u=\arccos x=\frac{1}{a}\log y$, so $y=e^{a u}$.

$\frac{du}{dx}=-\frac{1}{\sqrt{1-x^2}}$, hence

$\frac{dy}{dx}=a e^{a u}\frac{du}{dx}=-a y\frac{1}{\sqrt{1-x^2}}$

Differentiate once more:

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

Use $\frac{d}{dx}\big((1-x^2)^{-1/2}\big)=x(1-x^2)^{-3/2}$ and substitute $\frac{dy}{dx}=-a y(1-x^2)^{-1/2}$ to get

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

Multiply by $(1-x^2)$ and use $\frac{dy}{dx}=-a y(1-x^2)^{-1/2}$ to eliminate radicals. This yields

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