Asymptotic for $y'' + \frac{\epsilon y'}{y^2} - y' = 0$, $y(-\infty) = 1$, $y(+\infty) = \epsilon$. 
Asymptotic for $y'' + \frac{\epsilon y'}{y^2} - y' = 0$, $y(-\infty) = 1$, $y(+\infty) = \epsilon$.

I started with a regular expansion for 
$$y^2y'' + \epsilon y' - y^2 y' = 0$$
and 
$$y = y_0 + \epsilon y_1 + O(\epsilon^2)$$ 
with $y_0(-\infty) = 1$, $y_0(+\infty) = 0$ and $y_1(-\infty) = 0$, $y_1 (+\infty) = 1$.
The zero order ODE is 
$$ y_0'' -  y_0' = 0, $$
this gives $y_0 = Ae^{x} + B$, but I can this can not satisfy the boundary condition. How should I approach differently? So this problem is a singular perturbation problem. We can not use WKB because this is not linear, and I am not familar with explicit boundary layer calculation on an unbounded domain.

The differential equation has an exact solution in implicit form
$$ x - x_0 =  \frac{\ln  \left( cy +  y 
  ^{2}+\epsilon \right)}{2} +{\frac {c}{\sqrt {{c}^{2}-4\, \epsilon}}{\rm arctanh}
 \left({\frac {2\,y  +c}{\sqrt {{c}^{2}-4\, \epsilon}}}\right)
}
$$
Actually it's better (changing the constant $x_0$) to write this as
$$ 
x - x_0 =  \frac{\ln  \left(- cy -  y 
  ^{2}-\epsilon \right)}{2} +{\frac {c}{\sqrt {{c}^{2}-4\, \epsilon}}{\rm arctanh}
 \left({\frac {2\,y  +c}{\sqrt {{c}^{2}-4\, \epsilon}}}\right)
}$$
If $c = -1-\epsilon$, it turns out that the right side will go to $-\infty$ as $y \to \epsilon+$ and $+\infty$ as $y \to 1-$.
