Frobenius method solution

Question

Suppose I have a second order linear differential equation. I have a solution about a regular singular point say $x=0$. Suppose the indicial equation has a repeated root for the indicial equation. I get the first non trivial solution say $y_1$. Now in order to get the second non trivial linearly independent solution can I just write $y_2= y_1 \log(x)$. Hence the complete solution can be $y= (a+ b\log(x))y_1$. Can this be the final answer?

Answer 1

Let $$y_1 = x^\sigma \sum_{n=0}^\infty a_n x^n$$

be the first solution. To obtain the second solution it is always necessary to add a log term if the roots of the indicial equation are repeated (in general, there are two linearly independent series solutions if the roots to the indicial equation $\sigma_1 $ and $\sigma_2$ obey $\sigma_1 - \sigma_2 \not\in \mathbb{Z}$, and one otherwise). The form of the second solution is

$$y_2 = x^\sigma \sum_{n=0}^\infty b_n x^n + cy_1\log{x}$$ where $c$ is some nonzero constant. $c$ is not arbitrary, it is fixed by $a_0$ and $b_0$.

The form of $y_2$ can be derived using reduction of order, or you can assume it as the trial solution and substitute into the ODE.