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Putting $\epsilon =0$ 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 $\epsilon y^{\prime }$ order one at $x=O({\epsilon }^{-1})$.

The useful exact change of variables isDirect substitution turns the equation into two autonomous differential equations,The initial values here are $z(0)=w(0)=\epsilon$. Near the origin, linearizing this system gives eigenvalues $1$ and $-1-\epsilon$. Equivalently, neglecting only the derivative-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\epsilon }$. Their matching coefficients areThus 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 $R\to -\infty$ selectsThen $z_R+z=w$, and the growing-mode matching $z\sim X/2$ givesThis proves the logistic crossover in a weakly nonlinear Euler equation. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-2}H(\epsilon x)$ with the constants taken at their lowest order.

A composite asymptotic expansion retaining the small-distance stable mode and satisfying the initial value exactly isSince $H(X)=X^2/2+O(X^3)$, this matches $A{x}^2+B{x}^{-\epsilon }$ when $Cx\ll 1$ and hence $(x^2+1)/2$ at fixed $x$ as $\epsilon \to 0$. Its derivative at one differs from the prescribed value only by $O(\epsilon )$, within leading-order accuracy. For $Cx\gg 1$ it gives $y\sim x/\epsilon$. 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 $[0,1+\epsilon {]}^2$, because its vector field points inward at every boundary. Its positive equilibrium point is $z=w=1+\epsilon$, with Jacobian matrix having trace $-2-3\epsilon$ and determinant $1+\epsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\epsilon -2w<0$ excludes cycles by the Bendixson-Dulac theorem. The Poincaré-Bendixson theorem therefore gives convergence to the positive stable equilibrium. At fixed small positive $\epsilon$, the exact far-field slope is $y/x\to (1+\epsilon )/\epsilon$, whose difference from $1/\epsilon$ is only relative $O(\epsilon )$, consistent with the global leading composite.

Here the zero-parameter solution is $y_0=1$, but stopping there erases a weakly excited growing mode. Expand $y=1+\epsilon y_1+\cdots$ at fixed $x$. The first correction satisfiesand thereforeThe $\epsilon x^2/2$ term eventually dominates the constant, so this is another singular perturbation on the unbounded domain.

Using the exact logarithmic-variable system from part (a), the initial values are now $z(0)=\epsilon$ and $w(0)=0$. Its linear growing/stable modes have matching coefficientsThus $w\sim Cx$ is order one only at $x=O({\epsilon }^{-2})$, rather than $O({\epsilon }^{-1})$. This is the nonlinear crossover delayed by a small unstable-mode coefficient. The first change in the size of $y$ occurs much earlier, at $x=O({\epsilon }^{-1/2})$, when its weak growing mode becomes comparable to the initial constant; that is not yet the derivative-nonlinearity crossover.

At $X=Cx=O(1)$ the same logistic crossover in a weakly nonlinear Euler equation gives $w=X/(1+X)$ and $H(X)=X-\log (1+X)$. The nonlinear-region solution is now $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-3}H({\epsilon }^{2}x)$ at the lowest order. A composite asymptotic expansion covering all these regions isIt gives $y(1)=1$ exactly, and its initial derivative is $O({\epsilon }^3)$ rather than zero, within leading accuracy. At fixed $x$ its expansion is precisely $1+\epsilon [(x^2-1)/2-\log x]+O({\epsilon }^2)$; for $Cx\ll 1$ it retains the weak $A{x}^2$ growth, and for $Cx\gg 1$ it gives $y\sim x/\epsilon$. The positive equilibrium and boundedness argument from part (a) applies after the initially zero derivative immediately becomes positive. Thus the same eventual slope saturation occurs, but much farther downstream. Both the seeding calculation and the changed scale are essential to a leading approximation on the entire half-line.