Need help finding Green's function for $x^2y''-2xy'+2y=x\ln x$ The problem is as follows:

Use the given solutions of the homogeneous equation to find a particular solution of the given equation. Do this by the Green's Function Method.
$x^2y''-2xy'+2y=x\ln x;\quad x,x^2$
Use initial conditions $y(1)=y'(1)=0$.

I keep getting 
$$G(x,x')= \begin{cases} 
      0 & 1\leq x<x' \\
      -x+\dfrac{x^2}{x'} & x'<x<\infty
   \end{cases}$$
but when I evaluate
$$y(x)=\int_1^x\left(-x+\dfrac{x^2}{x'}\right)\cdot x'\ln x' dx'$$
I am left with the result $$y(x)=\dfrac{1}{4}x\left[-1+4x+x^2(-3+2\ln x)\right].$$
This does not agree with the solution to the differential equation that Mathematica gives:
$$y(x)=\frac{1}{2} x \left(2 x-\log ^2(x)-2 \log (x)-2\right).$$
Could someone help me figure out what I'm doing wrong here?
Edit: I figured out that the Green function I found corresponds to a $f(x)$ of $\frac{\ln x}{x}$ (obtained when the original equation is divided on both sides by $x^2$), not $x\ln x$. I'm not exactly sure why this is though.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
The Green function derivative 'jump' at $\ds{x = x'}$ is
$\ds{\color{#f00}{1 \over x'^{2}}}$ $\pars{~\mbox{instead of}\ \color{#f00}{1}}$.
That yields
$\ds{-\,{x \over x'^{2}} + {x^{2} \over x'^{3}}}$ when $\ds{x > x'}$. The result is given by
$$
\int_{1}^{x}\pars{-\,{x \over x'^{2}} + {x^{2} \over x'^{3}}}x'\ln\pars{x'}
\,\dd x' =
\bbox[#ffe,15px,border:1px dotted navy]{\ds{%
-x + x^{2} - x\ln\pars{x} - {1 \over 2}\,x\ln^{2}\pars{x}}}
$$

Note that
  $\ds{\int_{x'^{-}}^{x'^{+}}x^{2}\,\partiald[2]{\mrm{G}''\pars{x,x'}}{x}\,\dd x =
\color{#f00}{x'^{2}}\bracks{\left.\partiald{\mrm{G}\pars{x,x'}}{x}
\right\vert_{\ x\ =\ x'^{+}} - \left.\partiald{\mrm{G}\pars{x,x'}}{x}
\right\vert_{\ x\ =\ x'^{-}}} = 1}$.

Moreover, with your 'original Green Function', the 'right integration' is given by
$$
\int_{1}^{x}\pars{-x + {x^{2} \over x'}}\,{\ln\pars{x'} \over x'}\,\dd x'
\,\dd x' =
\bbox[#ffe,15px,border:1px dotted navy]{\ds{%
-x + x^{2} - x\ln\pars{x} - {1 \over 2}\,x\ln^{2}\pars{x}}}
$$
which is the right result.
