= Solution
Putting $\varepsilon=0$ gives the local <regular perturbation> result $y_0=(x^2+1)/2$. It cannot describe the whole half-line: its <derivative> grows like $x$, making $\varepsilon y'$ order one at $x=O(\varepsilon^{-1})$.
The useful exact <change of variables> is
$$
r=\log x,\qquad z(r)=\frac{\varepsilon y(x)}x,\qquad
w(r)=\varepsilon y'(x).
$$
Direct substitution turns the equation into two <autonomous differential equations>,
$$
\boxed{z_r=w-z,\qquad w_r=(1-\varepsilon-w)w+2\varepsilon z}.
$$
The initial values here are $z(0)=w(0)=\varepsilon$. Near the origin, linearizing this system gives <eigenvalues> $1$ and $-1-\varepsilon$. Equivalently, neglecting only the <derivative>-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\varepsilon}$. Their matching coefficients are
$$
A=\frac{1+\varepsilon}{2+\varepsilon},\qquad
B=\frac1{2+\varepsilon},\qquad
C=2\varepsilon A=\frac{2\varepsilon(1+\varepsilon)}{2+\varepsilon}.
$$
Thus the growing <derivative> has $w\sim Cx$. Keeping these unexpanded linear-mode coefficients is convenient for matching; it does not assert that the linearized solution is the exact nonlinear one.
At the <distinguished limit> $X=Cx=O(1)$, set $R=\log X$. To leading order the <derivative> equation becomes $w_R=w(1-w)$, the <logistic differential equation>. Matching $w\sim e^R$ as $R\to-\infty$ selects
$$
w=\frac{X}{1+X}.
$$
Then $z_R+z=w$, and the growing-mode matching $z\sim X/2$ gives
$$
z=\frac1X\int_0^X\frac{s}{1+s}\,ds
=\frac{H(X)}X,\qquad H(X)=X-\log(1+X).
$$
This proves the <logistic crossover in a weakly nonlinear Euler equation>. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\varepsilon C)$, or $\varepsilon^{-2}H(\varepsilon x)$ with the constants taken at their lowest order.
A <composite asymptotic expansion> retaining the small-distance stable mode and satisfying the initial value exactly is
$$
\boxed{y_{\rm comp}(x)=A+B x^{-\varepsilon}
+\frac{H(Cx)-H(C)}{\varepsilon C},\qquad
A=\frac{1+\varepsilon}{2+\varepsilon},\ B=\frac1{2+\varepsilon},\ C=2\varepsilon A}.
$$
Since $H(X)=X^2/2+O(X^3)$, this matches $Ax^2+Bx^{-\varepsilon}$ when $Cx\ll1$ and hence $(x^2+1)/2$ at fixed $x$ as $\varepsilon\to0$. Its <derivative> at one differs from the prescribed value only by $O(\varepsilon)$, within leading-order accuracy. For $Cx\gg1$ it gives $y\sim x/\varepsilon$. Thus the composite covers the initial region, the nonlinear transition and the arbitrarily large-$x$ regime; uniform leading accuracy is understood relative to the growing solution, not as an absolute error independent of unbounded $x$.
There is no later hidden growth regime. The exact positive system keeps $z,w$ in $[0,1+\varepsilon]^2$, because its vector field points inward at every boundary. Its positive <equilibrium point> is $z=w=1+\varepsilon$, with <Jacobian matrix> having trace $-2-3\varepsilon$ and determinant $1+\varepsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\varepsilon-2w<0$ excludes cycles by the <Bendixson-Dulac theorem>. The <Poincaré-Bendixson theorem> therefore gives convergence to the positive <stable equilibrium>. At fixed small positive $\varepsilon$, the exact far-field slope is $y/x\to(1+\varepsilon)/\varepsilon$, whose difference from $1/\varepsilon$ is only relative $O(\varepsilon)$, consistent with the global leading composite.
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