Solution: Before proceeding for the solution, let us recall some of the facts for this problem. Consider the second order ODE, $a_0(x) y^{\prime\prime} + a_1(x) y^\prime + a_2(x) y = 0$ for $x > x_0$. Also, at $x = x_0 ,\ a_0(x_0) = 0$. Then the point $x = x_0$ is said to be a regular singular point of the ODE if the following two limits exist: \begin{align*} \lim_{x \to x_0} \left( x - x_0 \right) \frac{a_1(x)}{a_0(x)} = p_0 (\text{ say } ) \\ \lim_{x \to x_0} \left( x - x_0 \right)^2 \frac{a_2(x)}{a_0(x)} = q_0 (\text{ say }). \end{align*} Then the indicial equation of the given ODE is the characteristic equation of the Cauchy-Euler differential equation \[ y^{\prime\prime} (x) + \frac{p_0}{x - x_0} y^\prime (x) + \frac{q_0}{(x - x_0)^2} y(x) = 0 \] which is \[ r(r - 1) + p_0 r + q_0 = 0. \]

We have the differential equation \begin{equation}\label{eq:03Nov2023-1} 4 \ln x y^{\prime\prime} + 3y^\prime + y = 0,\ x > 1. \end{equation} Given that $x = 1$ is the regular singular point of \eqref{eq:03Nov2023-1}. Comparing with the standard form discussed above, we have \[ a_2(x) = \ln x,\ a_1(x) = 3, \text{ and } a_2(x) = 1. \] Thus, \begin{align*} p_0 & = \lim_{x \to 1} (x - 1) \frac{a_1(x)}{a_0(x)} = \lim_{x \to 1} \frac{3(x - 1)}{4 \ln x} = \lim_{x \to 1} \frac{3}{4 / x} = \frac{3}{4} \\ q_0 & = \lim_{x \to 1} (x - 1)^2 \frac{a_2(x)}{a_0(x)} = \lim_{x \to 1} \frac{(x-1)^2}{4 \ln x} = \lim_{x \to 1} \frac{2(x-1)}{4 / x} = 0. \end{align*} Thus, the indicial equation at the regular singular point $x = 1$ will be \begin{align*} & r(r - 1) + p_0 r + q_0 = 0 \\ \implies & r^2 - r + \frac{3}{4} r + 0 = 0 \\ \implies & r^2 - \frac{1}{4} r = 0 \\ \implies & r = 0, \frac{1}{4}. \end{align*} Therefore, \[ \left\vert r_1 - r_2 \right\vert = \left\vert \frac{1}{4} - 0 \right\vert = \frac{1}{4}. \]