= Solution
{c}
The <power-law ansatz> $y=x^r$ gives
$$
\epsilon r(r-1)+r-1=(r-1)(\epsilon r+1)=0.
$$
Since $\epsilon>0$, the <equation roots> $1$ and $-1/\epsilon$ are distinct. The <Cauchy-Euler equation> therefore has <general solution>
$$
\boxed{y(x)=C_1x^{-1/\epsilon}+C_2x.}
$$
For $C_1=C_2=1$, the first <derivative> and second <derivative> are
$$
y'=1-\frac1\epsilon x^{-1/\epsilon-1},\qquad
y''=\frac{1+\epsilon}{\epsilon^2}x^{-1/\epsilon-2}>0.
$$
The <strictly convex function> diverges to $+\infty$ as $x\to0^+$, passes through $(1,2)$, and has a unique lowest point at
$$
\boxed{x_{\min}=\epsilon^{-\epsilon/(1+\epsilon)},\qquad
y_{\min}=(1+\epsilon)x_{\min}.}
$$
As $x\to\infty$, $y-x\to0$, so the line $y=x$ is an <asymptote>. When $0<\epsilon<1$, the lowest point is to the right of one; at $\epsilon=1$ it is at one, and for $\epsilon>1$ it is to the left of one.
The ratio of the two terms is
$$
\frac{x^{-1/\epsilon}}x=x^{-(1+1/\epsilon)}.
$$
Thus $x^{-1/\epsilon}$ 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 $\epsilon\to0^+$, $x^{-1/\epsilon}$ dominates at each fixed $x<1$, and $x$ dominates at each fixed $x>1$. Their crossover has $|\log x|=O(\epsilon)$, equivalently $x=1+O(\epsilon)$, as described by the <power-law crossover in a singular perturbation>. In that region $x=1+\epsilon X$ with bounded $X$ gives $y\sim1+e^{-X}$.
Substitute the proposed <regular perturbation> $y=y_0+\epsilon y_1+\cdots$ into the <differential equation>. The leading and subsequent equations are
$$
xy_0'-y_0=0,\qquad xy_j'-y_j=-x^2y_{j-1}''\quad(j\geq1).
$$
Consequently $y_0=Cx$, and in fact each successive coefficient has the form $y_j=C_jx$, because the preceding second <derivative> is zero. The leading approximation is therefore
$$
\boxed{y\sim Cx;\qquad y\sim x\text{ when matched to the specified solution at fixed }x>1.}
$$
It has only one adjustable constant and generally cannot impose two independent <boundary conditions>. Those assumed in the question require a nonzero $x^{-1/\epsilon}$ coefficient, which a <power series> in $\epsilon$ cannot recover: at fixed $x>1$ it is exponentially small, and at fixed $x<1$ it grows without bound as $\epsilon\to0^+$. 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 $(0,\infty)$.
\Image[/past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2021-power-law-modes.png]
{title=The exact solution and its two power-law contributions}
{description=Original sketch for $\epsilon=0.08$, showing the decreasing $x^{-1/\epsilon}$ contribution, the increasing $x$ contribution, their sum, and its lowest point. The inset resolves the region where both terms contribute.}
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