= Solution
The three <distinguished limits> are the <outer expansion> at $x=O(1)$, an <intermediate asymptotic region> at $x=O(\epsilon)$, and an <inner expansion> at $x=O(\epsilon^2)$. The middle scale resolves the coefficient change from $x$ to $x+\epsilon$; the narrower scale resolves the unmet left <boundary condition>. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second <derivative> at $x=0$ would be too strong for this <degenerate ordinary differential equation>.
In the <outer expansion>, set $y=y_0+\epsilon y_1+\cdots$. The leading equation is $x(x+1)y_0'-xy_0=0$. Enforcing the right <boundary condition> gives $y_0=x+1$. Because $y_0''=0$, the next equation is particularly simple:
$$
x(x+1)y_1'-xy_1=-(x+1),\qquad y_1(1)=0.
$$
An <integrating factor>, or the substitution $y_1=(x+1)u$, gives $u'=-1/[x(x+1)]$. Thus
$$
\boxed{y_{\rm out}(x)=(1+x)\left[1+\epsilon\log\frac{1+x}{2x}\right]+O(\epsilon^2),\qquad x=O(1).}
$$
The error statement is for $x$ bounded away from zero. The left limiting form is $1+x-\epsilon\log(2x)$ through the terms needed for matching. A <logarithm> in the correction makes a pure integer-power inner ansatz inadequate.
For the <intermediate asymptotic region>, put $x=\epsilon s$ and $y=V(s)$. The transformed equation is
$$
\sqrt\epsilon\sqrt s(s+1)\frac{\sin[\epsilon(s+1)]}{\epsilon(s+1)}V''+(s+1)(1+\epsilon s)V'-\epsilon(s+\epsilon)V=0.
$$
The leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt\epsilon$ forcing, so that homogeneous constant is zero after matching. At order $\epsilon$, $(s+1)V_1'=s$, giving $V_1=s-\log(1+s)+C(\epsilon)$. Matching for $1\ll s\ll1/\epsilon$ with the <outer expansion> fixes $C=-\log(2\epsilon)$. Hence
$$
\boxed{y_{\rm mid}(\epsilon s)=1+\epsilon\left[s-\log(1+s)-\log(2\epsilon)\right]+o(\epsilon),\qquad s=O(1),\ s>0.}
$$
The <switchback term> $-\epsilon\log\epsilon$ is larger than an ordinary $O(\epsilon)$ constant correction and must be retained. The highest <derivative> term first changes this middle solution at order $\epsilon^{3/2}$, so it does not alter the displayed result.
To locate the <inner expansion>, compare the <derivative> terms for $x\ll\epsilon$. Their ratio at a layer width $\delta$ is $\epsilon/\sqrt\delta$, so the <square-root-degenerate endpoint layer> has $\delta=\epsilon^2$. Put $x=\epsilon^2X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $\epsilon$ is
$$
\sqrt X(1+\epsilon X)U''+(1+\epsilon X)U'+O(\epsilon^2)=0.
$$
Both orders therefore have the same homogeneous operator $\sqrt X\,d^2/dX^2+d/dX$. Its <derivative> mode is proportional to $e^{-2\sqrt X}$. Using $U(0)=0$ and the supplied elementary <integral>, define
$$
H(X)=2\int_0^X e^{-2\sqrt q}\,dq=1-(1+2\sqrt X)e^{-2\sqrt X}.
$$
The <intermediate asymptotic region> gives the overlap value $1-\epsilon\log(2\epsilon)$ as $s\to0$. Therefore
$$
\boxed{y_{\rm in}(\epsilon^2X)=\left[1-\epsilon\log(2\epsilon)\right]\left[1-(1+2\sqrt X)e^{-2\sqrt X}\right]+o(\epsilon),\qquad X=O(1).}
$$
It satisfies the left <boundary condition> exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll1/\epsilon$. The outer and middle limits match with $\epsilon\ll x\ll1$. These three formulae include all terms through $O(\epsilon)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O(\epsilon^2)$.
For a check on that last point, continue the middle equation by one order. Its $\epsilon^{3/2}$ term obeys
$$
V_{3/2}'=-\frac{\sqrt s}{(1+s)^2},\qquad
V_{3/2}(s)=\int_s^\infty\frac{\sqrt r}{(1+r)^2}\,dr=\frac\pi2-\arctan\sqrt s+\frac{\sqrt s}{1+s}.
$$
The integration constant is selected by the outer limit. Its value at $s=0$ is $\pi/2$, so the inner matching amplitude has a next correction $\pi\epsilon^{3/2}/2$. This independently explains why an $o(\epsilon)$ remainder is appropriate.
If a single <additive composite expansion> is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-\epsilon\log(2\epsilon)$ and interpreting $x\log x=0$ at zero gives
$$
y_{\rm comp}=1+x+\epsilon\left[(1+x)\log\frac{1+x}{2}-x\log x-\log(x+\epsilon)\right]-A\left(1+\frac{2\sqrt x}{\epsilon}\right)e^{-2\sqrt x/\epsilon}.
$$
It is zero at $x=0$, gives $2+O(\epsilon^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O(\epsilon^2)$ error uniformly through the nested regions.
Finally, near zero $H(X)=2X-\tfrac83X^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first <derivative> but generally a second <derivative> diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y''$ multiplies that singular <derivative> to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y'(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the <initial value problem> with $y(0)=y'(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y')$ with an integrable coefficient matrix; its integral equation and the <Gronwall inequality> force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.
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