= Solution
The homogeneous equation is a <Cauchy-Euler equation>. Trying $y=x^m$ gives the repeated <indicial root>
$$
m(m-1)+3m+1=(m+1)^2=0.
$$
Hence, for $x>0$,
$$
\boxed{y_h=\frac{C_1+C_2\log x}{x}.}
$$
Both independent solutions diverge at the origin, so no nonzero homogeneous combination satisfies the zero-origin conditions by continuous extension.
For a source point $\xi>0$, finiteness on $0<x<\xi$ forces the <Green function> to vanish on that side. On $x>\xi$, continuity and the unit first-derivative jump require
$$
G(\xi^+,\xi)=0,\qquad G_x(\xi^+,\xi)=1.
$$
Using the homogeneous solutions gives the <causal Green function for a repeated-root Euler operator>
$$
\boxed{G(x,\xi)=\begin{cases}0,&0\leq x<\xi,\\\displaystyle\frac{\xi^2}{x}\log\frac{x}{\xi},&x\geq\xi>0.\end{cases}}
$$
Indeed, the <Wronskian> of $x^{-1}$ and $x^{-1}\log x$ is $x^{-3}$, and direct <differentiation> gives $G_x(\xi^+,\xi)=1$. Since $G$ is continuous, its second derivative produces precisely a unit <Dirac delta>, with no derivative of a delta. For each positive source it is identically zero near the origin, so both required origin values vanish. The differential operator itself is interpreted on $x>0$; no value of its singular coefficients at zero is needed.
Integrating against the forcing yields
$$
y(x)=\int_0^\infty G(x,\xi)\xi\,d\xi=\frac1x\int_0^x\xi^3\log\frac{x}{\xi}\,d\xi=x^3\int_0^1s^3\log\frac1s\,ds.
$$
The last <integral> equals $1/16$, by <integration by parts> or differentiating $\int_0^1s^pds=1/(p+1)$. Consequently
$$
\boxed{y(x)=\frac{x^3}{16}.}
$$
Substitution gives $y''+3y'/x+y/x^2=(6+9+1)x/16=x$, while $y(0)=y'(0)=0$. The divergent homogeneous solutions prove uniqueness in the regular class. No boundary condition at infinity was imposed or required.
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