= Solution
In the <outer solution>, derivative terms are small and $\sqrt y\sim\varepsilon k$, so $y_{\rm out}\sim\varepsilon^2k^2$. To resolve its mismatch with the zero left boundary, set $x=\varepsilon\xi$, $y=\varepsilon^2Y$. The original equation becomes exactly $Y''-\sqrt Y+k=0$. The left <inner variable> is therefore $\xi=x/\varepsilon$, with $Y(0)=0$ and $Y\to k^2$ as $\xi\to+\infty$.
Multiplying by $Y'$ gives the <first integral of a square-root reaction layer>. Matching $Y=k^2$, $Y'=0$ fixes its constant:
$$
\frac12(Y')^2-\frac23Y^{3/2}+kY=\frac{k^3}{3},\qquad \boxed{(Y')^2=\frac23(\sqrt Y-k)^2(2\sqrt Y+k).}
$$
The positive derivative branch below $k^2$ begins with $Y'(0)=\sqrt{2k^3/3}$ and approaches the plateau exponentially. Thus the left layer has thickness $O(\varepsilon)$, not $O(\sqrt\varepsilon)$ despite the coefficient on the original second <derivative>. Its small amplitude is essential in the <dominant balance>.
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