= Solution
Work first on $0<x<1$. Divide by $x-1$ to put the equation in leading-coefficient-one form. Its first-derivative coefficient is $-x/(x-1)$, so the <integrating factor> for the <self-adjoint differential operator> is
$$
p(x)=\exp\left(\int\frac{-x}{x-1}\,dx\right)=\frac{e^{-x}}{1-x}.
$$
The resulting coefficients are
$$
\boxed{p(x)=\frac{e^{-x}}{1-x},\qquad q(x)=-\frac{e^{-x}}{(1-x)^2},\qquad
F(x)=-\frac{e^{-x}}{(1-x)^2}.}
$$
Indeed $p'(x)=x e^{-x}/(1-x)^2$, and multiplying the original equation by $-e^{-x}/(1-x)^2$ gives $(py')'+qy=F$. These transformed coefficients are singular at the right endpoint; the boundary value there is understood as the endpoint limit.
Substituting $y=x$ and $y=e^x$ in the homogeneous original equation gives zero in both cases. Their <Wronskian> is
$$
W(x,e^x)=xe^x-e^x=(x-1)e^x,
$$
which is nonzero for $0<x<1$. Thus they are <linearly independent> on the solution interval.
To construct a <Dirichlet Green function>, choose homogeneous solutions adapted to the two endpoints:
$$
u(x)=x,\qquad v(x)=e^x-ex,\qquad u(0)=0,\quad v(1)=0.
$$
Their <Wronskian> is again $(x-1)e^x$, and $pW=-1$. Therefore the Green function for $L=(p\,d/dx)'+q$ with zero endpoint values is
$$
\boxed{G(x,s)=-u(\min(x,s))v(\max(x,s))
=\begin{cases}-x(e^s-es),&x\leq s,\\-s(e^x-ex),&s\leq x.\end{cases}}
$$
For fixed $s\in(0,1)$ it is continuous in $x$, vanishes at both endpoints, and satisfies the unit-flux jump condition
$$
p(s)\bigl[G_x(s^+,s)-G_x(s^-,s)\bigr]=1.
$$
Consequently $L_xG(x,s)=\delta(x-s)$ in the sense of <distributions>. This is a <singular-endpoint Green function with a double-zero right solution>.
The endpoint data are nonhomogeneous, so use the homogeneous lifting $2x$ and write
$$
y(x)=2x+\int_0^1G(x,s)F(s)\,ds.
$$
Although $F(s)$ is singular at one, the integral exists for each interior $x$: the right solution satisfies $v(s)\sim e(1-s)^2/2$, cancelling the $(1-s)^{-2}$ forcing when $s>x$.
Splitting the integral at $s=x$ gives
$$
y(x)=2x+v(x)I_1(x)+xI_2(x),
$$
where elementary antiderivatives yield
$$
I_1(x)=\int_0^x\frac{s e^{-s}}{(1-s)^2}\,ds=\frac{e^{-x}}{1-x}-1,
$$
$$
I_2(x)=\int_x^1\frac{(e^s-es)e^{-s}}{(1-s)^2}\,ds
=\frac{e^{1-x}-1}{1-x}-1.
$$
For the second equality, an antiderivative is $(1-e^{1-s})/(1-s)$, whose endpoint limit at one is $-1$. Simplifying the Green representation gives
$$
\boxed{y(x)=1+(1+e)x-e^x.}
$$
It has the prescribed endpoint values. Direct substitution gives $(x-1)y''-xy'+y=1$, and at the degenerate endpoint it also obeys $-y'(1)+y(1)=1$. The only homogeneous solution satisfying both zero endpoint conditions is zero, so this solution is unique in the corresponding classical class.
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