= Solution
The <Taylor series> of the logarithm gives
$$
\log(1+x)=\sum_{j=1}^\infty\frac{(-1)^{j+1}}j x^j=x\sum_{k=0}^\infty\frac{(-x)^k}{k+1}.
$$
Therefore the closed form of the larger-root <Frobenius solution> is
$$
\boxed{y(x)=a_0\sqrt{x}\log(1+x).}
$$
The 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 $4x^{5/2}[(1+x)v''+v']$. Hence $[(1+x)v']'=0$, 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.
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