= Solution
Use the <Frobenius method> and write $y=x^r\sum_{n\geq0}a_nx^n$, $a_0\ne0$, with coefficients at negative indices defined as zero. The lowest power gives the <indicial equation> $r(2r-1)=0$, so $r=0$ or $r=1/2$. Equating the coefficient of $x^{n+r-1}$ gives, for $n\geq1$,
$$
\boxed{a_n=\frac{[4(n+r-1)^2+2]a_{n-1}-12(n+r-2)a_{n-2}+8a_{n-3}}{(n+r)[2(n+r)-1]}.}
$$
For $r=0$ and $a_0=1$, the first coefficients are $1,2,2,4/3,\ldots$. For $r=1/2$ they are $1,1,1/2,1/6,\ldots$, multiplying $x^{1/2}$. The distinct nonintegrally separated indicial roots give two independent <Frobenius series>. The half-power solution is a generalized power series, rather than an ordinary Taylor series at zero.
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