Green function of a boundary value problem 
Find the Green's function for the corresponding boundary value problem
  : $$x^2(\log_e x-1)y''-xy'+y=0$$ with the conditions $y(0)$ is finite
  and $y(1)=0$.

As a matter of fact, we need to solve the above equation in its general form then use the properties of Green's functions, i.e. its continuity, jump discontinuity of its first order derivative etc. to calculate the Green's function. But I cannot solve this ODE in general methods. I have tried reducing the equation to normal form and changing the independent variable but to no avail. Can someone help in this regard? It will be enough to tell just how to solve this ODE in general. Thanks in advance.
 A: By visual inspection, esp. of the last two terms, $y(x)=x$ is a solution. Insert $y(x)=xu(x)$ to get
$$
0=x^2(\ln x-1)(xu''+2u')-x(xu'+u)+xu=x^3(\ln x-1)u''+x^2(2\ln x-3)u'
$$
which now is first order and separable in $u'$ giving 
$$
\frac{u''}{u'}=-\frac1x\frac{2\ln x-3}{\ln x-1}\implies u'(x)=Cx^{-2}(\ln x-1)
\\~\\
\implies u(x)=D-Cx^{-1}\ln x
$$
so that $y(x)=\ln x$ is the second basis solution.
Check that $y(x)=x$ satisfies the left boundary condition and $y(x)=\ln x$ the right one. Now compute the Wronskian and compose the Green function.
A: Whenever you see monomial terms attached only to derivatives of the same order as the degree, think of the substitution $t = \log x$. Then, calculating the derivatives:
$$\begin{align}
& \frac{dy}{dx} = \frac{dt}{dx}\frac{dy}{dt} = \frac{1}{x}\frac{dy}{dt} \\
& \frac{d^2 y}{dx^2} = \frac{1}{x^2}\frac{d^2 y}{dt^2} - \frac{1}{x^2}\frac{dy}{dt} \\
\end{align} $$
by the chain rule. Then, plugging in,
$$(t-1)\left( \frac{d^2 y}{dt^2} - \frac{dy}{dt} \right) - \frac{dy}{dt} + y = 0$$
$$ \implies (t-1)\ddot{y} - t\dot{y} + y = 0$$
Then what do we do? Plug in our favorite, $e^{mt}$, of course. 
$$((t-1)m^2-tm+1)e^{mt} = 0 \implies m(m-1)t + 1 - m^2 = 0$$
Taking both coefficients of the polynomial to be zero, we only get $m=1$. Next we plug in $te^{mt}$:
$$((t-1)(2m+m^2t)-t(1+mt)+t)e^{mt} = 0 \implies m(m-1)t^2 - m(m-2)t - 2m = 0$$
We get that $m=0$ is the only solution here. From our guesses, we have two linear independent solutions for a second order linear ODE, so our work is done.
$$y(t) = C_1 e^t + C_2 t \implies y(x) = C_1 x + C_2 \log x$$
The two initial conditions correspond to $C_2 = 0$ and $C_1 = 0$, respectively.
