= Logarithmic matching in a small-exponent boundary layer
{title2=$y(0)=1+\epsilon\log\epsilon+O(\epsilon^2\log^2\epsilon)$}
In a singular first-order equation with coefficient $x+\epsilon$, an outer correction proportional to $\epsilon\log x$ cannot be evaluated at the endpoint. The inner scale $x=\epsilon X$ changes it to $\epsilon\log\epsilon+\epsilon\log(1+X)$. Matching retains the logarithmic constant, which is larger than a plain order-epsilon correction. For $(x+\epsilon)y'=\epsilon y$, $y(1)=1$, the exact solution $[(x+\epsilon)/(1+\epsilon)]^\epsilon$ verifies the displayed endpoint asymptotic and its remainder.
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