Consider the problem $Lu:=u''(x)+{1\over 4x^2}u(x)=0,\ u(1)=u(2)=0$
Green's function associated with $L$ is defined as $\Gamma(x,\xi)=\begin{cases}{1\over c}u_1(\xi)u_2(x),\ x\ge\xi,\ x\in[1,2]\\ {1\over c}u_2(\xi)u_1(x),\ x<\xi,\ x\in[1,2]\end{cases}$
where $c=p(x)(u_1u_2'-u_2u_1') \ne 0$ and $u_1,u_2$ is a fundamental system for $L$ i.e. linearly independent solutions of $Lu=0$
Find Green's function to the problem
The first step is finding linearly independent solutions. For that we do the following change of variables: $e^t=x$ and realize that $$\begin{align}u''(e^t)+{1\over 4x^2}u(e^t)&=0\\{d^2\over dt^2} [u(e^t)]-{d\over dt}[u(e^t)]+{1\over 4}u(e^t)&=0\end{align}$$
And by defining $g(t)=u(e^t)$ that equation is $g''-g'+{1\over4}g=0$
And has a solution of the form $c_1e^{{1\over2}t}+c_2e^{{1\over2}t}t$
The initial conditions $u(1)=u(2)$ are equivalent to $g(\log(1))=g(\log(2))=0$ and imply $c_1=c_2=0$
This is where I feel like I have made a mistake, I was looking for two non trivial independent solutions...
Once you arrive at
$$ g(t) = u(e^t) $$ you should return to
$$ u(\tau) = g(\ln \tau) = (c_1 + \ln\tau c_2)\sqrt{\tau} $$
and then apply the BC's on $c_1,c_2$ and the solution is $u(\tau) = 0$