Consider $p(x)y'' + q(x)y' + r(x)y = 0$ with a regular singular point at $x = x_0$. Let $\sigma_1$ and $\sigma_2$ (with $\sigma_2 \geq \sigma_1$) be the roots of the indicial equation. **Case 1:** If $\sigma_2 - \sigma_1 \notin \mathbb{Z}$, there are two linearly independent Frobenius solutions: \begin{align*} y_1 &= (x - x_0)^{\sigma_1} \sum_{n=0}^{\infty} a_n (x - x_0)^n, \\ y_2 &= (x - x_0)^{\sigma_2} \sum_{n=0}^{\infty} b_n (x - x_0)^n. \end{align*} **Case 2:** If $\sigma_2 - \sigma_1 \in \mathbb{N}$ (positive integer), the larger root $\sigma_2$ yields a Frobenius solution $y_1$. The second solution has the form \begin{align*} y_2 = (x - x_0)^{\sigma_1} \sum_{n=0}^{\infty} b_n (x - x_0)^n + A \, y_1(x) \ln(x - x_0), \end{align*} where the constant $A$ may or may not be zero. **Case 3:** If $\sigma_1 = \sigma_2 = \sigma$, the logarithmic term is always required ($A \neq 0$): \begin{align*} y_1 &= (x - x_0)^{\sigma} \sum_{n=0}^{\infty} a_n (x - x_0)^n, \\ y_2 &= (x - x_0)^{\sigma} \sum_{n=0}^{\infty} b_n (x - x_0)^n + A \, y_1(x) \ln(x - x_0). \end{align*}