Find the Green's function for an ODE $$ Lu= (x-2)u''+(1-x)u'+u , \, u'(0)=(1)=0$$
It can be shown that ${x-1,e^{x}}$ is a fundamental set for $L$ on this interval.
$$ g(x,y)= c_1(x-1)+c_2e^{x}, 0\leq x<y ,  c_3(x-1)+c_4e^{x}, y<x\leq 1 $$
So we see
$$ g_x(x,y)= c_1+c_2e^{x}, 0\leq x<y ,  c_3+c_4e^{x}, y<x\leq 1 $$
Applying the boundary conditions
$$ g_x(0,y)= c_1+c_2 = 0 \implies c_1=-c_2 $$
$$ g(1,y)= c_4e^1 = 0 \implies c_4=0 $$
So 
$$ g(x,y)= c_1(x-1-e^{x}), 0\leq x<y ,  c_3(x-1), y<x\leq 1 $$
$$ g_x(x,y)= c_1(1-e^{x}), 0\leq x<y ,  c_3, y<x\leq 1 $$
Considering the continuity at $x=y$,
$$ c_1(y-1-e^{y})-c_3(y-1)=0\implies c_1=c_3(y-1-e^{y})^{-1}(y-1)$$
Considering the jump condition at $x=y$
$$ c_3-c_1(1-e^{y})=(y-2)^{-1} \implies c_3=c_1(1-e^{y})+(y-2)^{-1}$$
So we have 
$$ \left( \begin{array}{cc} y-1-e^{y} & 1-y \\ e^{y}-1 & 1\\ \end{array}\right)
 \left( \begin{array}{c} c_1  \\c_3 \\ \end{array}\right)
=  \left( \begin{array}{c}0 \\ (y-2)^{-1} \end{array}\right) $$
or
$$ 
 \left( \begin{array}{c} c_1  \\c_3 \\ \end{array}\right)
=\left(y-1-e^{y}- (1-y)(e^{y}-1 ) \right)^{-1} \left( \begin{array}{cc}1 & y-1 \\ 1-e^{y} &  y-1-e^{y}\\ \end{array}\right)
 \left( \begin{array}{c}0 \\ (y-2)^{-1} \end{array}\right) $$
$$ 
 \left( \begin{array}{c} c_1  \\c_3 \\ \end{array}\right)
=e^{-y} \left(y-2 \right)^{-1} \left( \begin{array}{c}y-1 \\ y-1-e^{y}\\ \end{array}\right)(y-2)^{-1} $$
$$ 
 \left( \begin{array}{c} c_1  \\c_3 \\ \end{array}\right)
=e^{-y} \left( \begin{array}{c}y-1 \\ y-1-e^{y}\\ \end{array}\right)(y-2)^{-2} $$
 A: Let $\phi$, $\psi$ be non-zero solutions of $L\phi = 0$, $L\psi=0$, chosen to satisfy $\phi'(0)=0$, $\psi(1)=0$. Let $w$ be the Wronskian $w=W(\phi,\psi)=\phi\psi'-\psi\phi'$.
I believe the solution $u$ of $Lu=f$ with $u'(0)=u(1)=0$ is
$$
        u = \psi(x)\int_{0}^{x}\frac{\phi(t)f(t)}{w(t)(t-2)}\,dt
            +\phi(x)\int_{x}^{1}\frac{\psi(t)f(t)}{w(t)(t-2)}\,dt.
$$
The $t-2$ comes from the coefficient of the highest-order derivative term in your equation. Try forming $u'$, $u''$, and you'll see if everything works; and the arrangement of integral limits makes the endpoint conditions work out, too. What remains is to find $\psi$ and $\phi$, where $\phi$ satisfies the left endpoint condition and $\psi$ satisfies the right endpoint condition. In this case, two such solutions are $\phi(x) = e^{x}+(1-x)$ and $\psi(x)=x-1$. And their Wronskian is
$$
\begin{align}
           w(x) & =W(\phi,\psi)=W(e^{x}+1-x,x-1)=W(e^{x},x-1) \\
                & = e^{x}\frac{d}{dx}(x-1)-\frac{de^{x}}{dx}(x-1) = -e^{x}(x-2).
\end{align}
$$
So now you can read off the Green function from
$$
   u = -\int_{0}^{x}\frac{(x-1)[e^{t}+1-t]}{e^{t}(t-2)^{2}}f(t)\,dt
        -\int_{x}^{1}\frac{(t-1)[e^{x}+1-x]}{e^{t}(t-2)^{2}}f(t)\,dt.
$$
(Note: It's easy to start with $u=\psi\int_{0}^{x}\phi f\,dt+\phi\int_{x}^{1}\psi f\,dt$, plug it in, find that $Lu = hf$ for some function $h$, and then adjust by dividing the $f$ inside the integrals by $h$ in order to get $Lu_{new}=f$.)
