 Solution

The Taylor series of the logarithm givesTherefore the closed form of the larger-root Frobenius solution isThe series derivation holds initially for $0<x<1$; the closed form extends as a real solution to all $x>0$. To check it without series, put $y=\sqrt{x}v(x)$ in the original equation. All undifferentiated $v$ terms cancel and the remaining expression is $4x^{5/2}[(1+x)v^{\prime \prime }+v^{\prime }]$. Hence $[(1+x)v^{\prime }{]}^{\prime }=0$, and $v=\log (1+x)$ indeed solves it. Its logarithm is analytic near zero, so this is not a forced $\log x$ companion term at the Frobenius point.