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The power-law ansatz $y=x^r$ givesSince $ϵ>0$, the equation roots $1$ and $-1/ϵ$ are distinct. The Cauchy-Euler equation therefore has general solutionFor $C_1=C_2=1$, the first derivative and second derivative areThe strictly convex function diverges to $+\infty$ as $x\to 0^{+}$, passes through $(1,2)$, and has a unique lowest point atAs $x\to \infty$, $y-x\to 0$, so the line $y=x$ is an asymptote. When $0<ϵ<1$, the lowest point is to the right of one; at $ϵ=1$ it is at one, and for $ϵ>1$ it is to the left of one.

The ratio of the two terms isThus $x^{-1/ϵ}$ is larger for $0<x<1$, the terms are equal at $x=1$, and $x$ is larger for $x>1$. In the stronger asymptotic sense as $ϵ\to 0^{+}$, $x^{-1/ϵ}$ dominates at each fixed $x<1$, and $x$ dominates at each fixed $x>1$. Their crossover has $|\log x|=O(ϵ)$, equivalently $x=1+O(ϵ)$, as described by the power-law crossover in a singular perturbation. In that region $x=1+ϵX$ with bounded $X$ gives $y\sim 1+e^{-X}$.

Substitute the proposed regular perturbation $y=y_0+ϵy_1+\cdots$ into the differential equation. The leading and subsequent equations areConsequently $y_0=Cx$, and in fact each successive coefficient has the form $y_j=C_{j}x$, because the preceding second derivative is zero. The leading approximation is thereforeIt has only one adjustable constant and generally cannot impose two independent boundary conditions. Those assumed in the question require a nonzero $x^{-1/ϵ}$ coefficient, which a power series in $ϵ$ cannot recover: at fixed $x>1$ it is exponentially small, and at fixed $x<1$ it grows without bound as $ϵ\to 0^{+}$. Thus the regular perturbation describes the specified solution only in the region to the right of the crossover, away from $x=1$; it is not a uniform approximation on $(0,\infty )$.