= Solution
Let $L_0u=(x+1)u''+xu'$. Matching powers in the <regular perturbation> expansion and in the <Taylor series> of the left endpoint value gives the complete recursive <boundary value problem>
$$
\boxed{L_0y_n=-y_{n-1},\qquad y_n(0)=\frac2{n!},\quad y_n(1)=0,\qquad n\geq1.}
$$
The higher-order left endpoint values are not zero. The <parameter-dependent boundary data in a regular perturbation hierarchy> rule applies here.
First prove uniqueness for any one of these forced problems. The difference $v$ of two solutions obeys $L_0v=0$ and homogeneous <Dirichlet boundary conditions>. The positive <integrating factor> $r_0=e^x/(x+1)^2$ gives
$$
r_0L_0v=(pv')',\qquad p=\frac{e^x}{x+1}>0.
$$
Multiplication by $\overline v$ and <integration by parts> yield
$$
0=\int_0^1\overline v\,(pv')'\,dx=-\int_0^1p|v'|^2\,dx.
$$
Hence $v'=0$, and the endpoint values force $v=0$. The <kernel> is trivial: zero is not an <eigenvalue> of this homogeneous <Dirichlet boundary value problem>.
Existence is also explicit. For arbitrary smooth forcing $g$ and endpoint values $u(0)=a$, $u(1)=b$, two integrations of $(pu')'=r_0g$ give
$$
u(x)=a+C H(x)+J_g(x),
$$
where
$$
H(x)=\int_0^x(s+1)e^{-s}\,ds=2-(x+2)e^{-x},
\qquad
J_g(x)=\int_0^x(s+1)e^{-s}\int_0^s\frac{e^tg(t)}{(t+1)^2}\,dt\,ds.
$$
Since $H(1)=2-3/e>0$, the unique choice
$$
\boxed{C=\frac{b-a-J_g(1)}{2-3/e}}
$$
always meets the second endpoint value. Starting from the smooth $y_0$, induction with $g=-y_{n-1}$, $a=2/n!$, $b=0$ proves \b[existence and uniqueness at every order].
Equivalently, the <Fredholm alternative> has no compatibility obstruction because the homogeneous <kernel> vanishes; the corresponding <adjoint operator> has no zero <eigenvalue> either. The direct construction above proves this without computing any particular $y_n$ beyond the leading term.
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