Need help finding Green's function for $x^2y''-2xy'+2y=x\ln x$

Question

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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.