Asymptotic Evaluation of Differential equation: $a\frac{d y}{dx} = -\frac{1}{y(x)} e^{-\frac{1}{y(x)}}$ I'm interested in solving the differential equation
$$
a\frac{d y}{dx} = -\frac{1}{y(x)} e^{-\frac{1}{y(x)}}
$$
where $a>0$. 
$Mathematica$ can solve this but it gives the answer in terms of InverseFunctions of exponential integrals. I'm only interested in the asymptotic behavior of $y(x)$ as $x \to \infty$ but I'm not sure how to proceed with this sort of a problem. 
Edit: The suggestion was to transform to $z(x) = \frac{1}{y(x)}$. In terms of this new variable, the problem becomes
$$
a \frac{dz}{dx} = z(x)^3 e^{-z(x)}
$$
As far as initial conditions are concerned, let's say $z(0) = 1$. Even in these new variables, I don't really know how to proceed.
 A: Motivated by Felix Marin's response yesterday, I was able to figure this out for a more general class of problems. I am interested in solving
$$
a \frac{dy}{dt} = - y(t)^\alpha e^{-1/y(t)}
$$
for $\alpha \in \mathbb{Z}$, $a>0$, and in the limit $t\to \infty$. Let us also consider the initial condition $y(0) = y_0$. The solution proceeds as follows
$$
a \frac{d y}{d t} = - y^\alpha e^{-1/y} \implies \int_{y(0)}^{y(t)} dz\,\frac{e^{1/z}}{z^\alpha} = -\frac{t}{a}
$$
Change variables, $z = -\frac{1}{x}$
$$
(-1)^\alpha \int_{-\frac{1}{y_0}}^{-\frac{1}{y(t)}} dx\, x^{\alpha - 2} e^{-x} = - \frac{t}{a}
\implies \int_{-\frac{1}{y_0}}^{\infty} dx\, x^{\alpha - 2} e^{-x} - \int_{-\frac{1}{y(t)}}^{\infty} dx\, x^{\alpha - 2} e^{-x} = (-1)^{\alpha + 1}\frac{t}{a}
\implies \Gamma\left(\alpha - 1,-\frac{1}{y_0}\right) -  \Gamma\left(\alpha - 1,-\frac{1}{y(t)}\right) = (-1)^{\alpha + 1}\frac{t}{a}
$$
Now, we are interested in the behavior of $y(t)$ as $t\to \infty$. We note that $\forall n \in \mathbb{Z}$, 
$$
\Gamma(n,z) \sim \frac{e^{-z}}{z^{1-n}} \quad z \to -\infty
$$
Since we want the LHS to blow up as well, we can then see that the appropriate limit is $y(t) \to 0^+$ so that
$$
\Gamma\left(\alpha - 1, - \frac{-1}{y(t)} \right) \sim (-1)^{\alpha - 2} y(t)^{\alpha - 2} e^{1/y(t)}
$$
So, asymptotically, 
$$
y(t)^{\alpha - 2} e^{1/y(t)} \sim \frac{t}{a}
$$
We can invert this to find
$$
y(t) \sim \frac{-1}{W\left(- \frac{\left(\frac{t}{a} \right)^{\frac{1}{2 - \alpha}}}{\alpha - 2}\right)}
$$
where $W(z)$ is the Lambert-W function. We can push this even further by considering that
$$
W(x) \sim \log(x) - \log(\log(x)) \quad x\to \infty
$$
So,
$$
y(t) \sim \frac{1}{\log(\frac{t}{a}) + (\alpha - 2)\log(\log(\frac{t}{a}))}
$$
A: Here's a possible start,
where I wander around for a while
and don't come to
any satisfactory conclusion.
From
$a\frac{d y}{dx} 
= -\frac{1}{y(x)} e^{-\frac{1}{y(x)}}
$
we get
$yy'
=-e^{-1/y}
$
or
$-yy'
=e^{-1/y}
$.
Let $z = -1/y$,
or
$y = -1/z$
so
$y' = z'/z^2$
and
$yy'
=(-1/z)(z'/z^2)
=-z'/z^3
$.
This becomes
$z'/z^3
=e^z
$,
so
$z' = z^3 e^z
$.
Differentiating,
$\begin{array}\\
z''
&=3z^2z'e^z + z^3z'e^z\\
&=z'z^2e^z(3+z)\\
&=z'z^2(z'z^{-3})(3+z)\\
&=z'^2z^{-1}(3+z)\\
\text{or}\\
zz''
&=z'^2(3+z)\\
\end{array}
$
At this point,
we could try a
power series for $z$
and get a recurrence for
the coefficients.
I'm tired,
so I'll try something else.
Let
$z = x^aw$
where
$w(0) \ne 0$.
$z' = x^aw'+ax^{a-1}w$
and
$\begin{array}\\
z''
&=x^aw''+ax^{a-1}w'+a((a-1)x^{a-2}w+x^{a-1}w')\\
&=x^{a-2}(x^2w''+axw'+a((a-1)w+xw')\\
&=x^{a-2}(x^2w''+2axw'+a(a-1)w)\\
\end{array}
$
From
$zz''
=z'^2(3+z)
$
we get
$(x^aw)x^{a-2}(x^2w''+2axw'+a(a-1)w)
=(x^aw'+ax^{a-1}w)^2(3+x^aw)
$
or
$x^{2a-2}w(x^2w''+2axw'+a(a-1)w)
=x^{2a-2}(xw'+aw)^2(3+x^aw)
$
or
$w(x^2w''+2axw'+a(a-1)w)
=(xw'+aw)^2(3+x^aw)
$.
I'm not sure what 
to do from here,
and everything looks messy,
so I'll stop.
