= Solution
Seek an outer <asymptotic expansion> $y=y_0+\epsilon y_1+\cdots$ for fixed $x>0$. The leading equation is $xy_0'=0$, and the endpoint condition gives $y_0=1$. At first order, $xy_1'=y_0=1$ and $y_1(1)=0$, so $y_1=\log x$. This outer correction cannot be evaluated at $x=0$.
The coefficient $x+\epsilon$ selects the <boundary layer> coordinate $X=x/\epsilon$. With $Y(X)=y(\epsilon X)$, the equation becomes $(1+X)Y_X=\epsilon Y$. Its inner leading term is constant; matching it to the outer solution gives $Y_0=1$. At first order, $Y_{1X}=1/(1+X)$, hence $Y_1=\log(1+X)+C$. In the overlap $1\ll X\ll\epsilon^{-1}$, the outer solution is $1+\epsilon\log\epsilon+\epsilon\log X$. Matching the constants therefore gives
$$
Y(X)=1+\epsilon\log\epsilon+\epsilon\log(1+X)+o(\epsilon)
$$
for fixed $X$. The <logarithmic matching in a small-exponent boundary layer> must retain $\epsilon\log\epsilon$, which is larger in magnitude than a plain order-$\epsilon$ term. At the endpoint,
$$
\boxed{y(0)=1+\epsilon\log\epsilon+o(\epsilon).}
$$
The coefficient of a separate plain $\epsilon$ term is zero.
Direct integration provides the exact check:
$$
\boxed{y(x;\epsilon)=\left(\frac{x+\epsilon}{1+\epsilon}\right)^\epsilon.}
$$
At zero its logarithm is $\epsilon\log\epsilon-\epsilon\log(1+\epsilon)=\epsilon\log\epsilon-\epsilon^2+O(\epsilon^3)$. Exponentiating gives $y(0)=1+\epsilon\log\epsilon+O(\epsilon^2\log^2\epsilon)$, whose remainder is indeed $o(\epsilon)$. The first-order uniform composite $1+\epsilon\log[(x+\epsilon)/(1+\epsilon)]$ also satisfies the <boundary condition> exactly and reproduces both expansions to the requested order.
Back to article page
