Frobenius & Special Functions · Challenge problem

Yes, because any time two indicial roots differ by a positive integer, the smaller-root solution always needs a logarithm.

No, because this is an exact Cauchy–Euler equation (no extra $x$-dependent terms), so both Frobenius solutions are exactly $x$ and $x^2$ with no logarithm.

Yes, because a repeated indicial root always forces a logarithm, and this equation's roots are close together.

No, because logarithms only ever appear in equations with a singular point at $x=1$, not at $x=0$.