= Solution
Substituting $y=x^p$ in the homogeneous <ordinary differential equation> gives $p(p-1)-2=(p-2)(p+1)=0$. Thus
$$
\boxed{y_h=Cx^2+D/x}.
$$
For a <Green's function> decaying at the two ends, use $u(x)=x^2$ on the side of zero and $v(x)=x^{-1}$ on the side of infinity. Their <Wronskian> is $uv'-u'v=-3$. The <function> must be continuous at $x=\xi$, or its second <derivative> would contain a <derivative> of a <Dirac delta>. Its <derivative> jump must be one, to give the specified positive delta source. These two conditions yield the <boundary-decaying Green kernel for an inverse-square differential operator>:
$$
\boxed{G(x,\xi)=-\frac13\begin{cases}x^2/\xi,&x\leq\xi,\\\xi^2/x,&x\geq\xi.\end{cases}}
$$
Its <derivatives> at the join are $-2/3$ on the left and $1/3$ on the right, confirming the jump one. Away from the join each branch solves the homogeneous equation, and both endpoint conditions hold.
Integrate this <Green's function> against the forcing, whose support is between zero and one. For $0<x\leq1$,
$$
y(x)=-\frac13\left(\frac1x\int_0^x\xi^2\,d\xi+x^2\int_x^1\frac{d\xi}{\xi}\right)=\frac{x^2}{3}\log x-\frac{x^2}{9}.
$$
For $x\geq1$, only the right branch contributes. Therefore
$$
\boxed{y(x)=\begin{cases}\dfrac{x^2}{3}\log x-\dfrac{x^2}{9},&0<x\leq1,\\-\dfrac1{9x},&x\geq1.\end{cases}}
$$
On the first branch, $y'=(2x/3)\log x+x/9$ and $y''=(2/3)\log x+7/9$, so $y''-2y/x^2=1$. The second branch is a multiple of the homogeneous solution $x^{-1}$, so its forcing is zero. At $x=1$, both $y$ and $y'$ agree, with values $-1/9$ and $1/9$, so there is no additional delta source. Finally $x^2\log x\to0$ as $x\downarrow0$ and $x^{-1}\to0$ as $x\to\infty$. A homogeneous solution obeying both <boundary conditions> has $C=D=0$, establishing uniqueness.
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