Explicit solution to ODE $rg''(r)= \gamma g(r)^2$. While working on a project, I have come across a function $g\in C^\infty((0,\infty),\mathbb{R})$ satisfying the ODE
$$ r g''(r) = \gamma g(r)^2,\quad r > 0,$$
where $\gamma$ is some positive constant. I know also that $g > 0$, $g$ must be decreasing, $\lim_{r\to 0} g(r) < \infty$ is a constant I know, and $g(r) = o(r^{-1/2})$ for large $r$.
I believe these constraints should determine $g$ uniquely, and the ODE looks (to my untrained eye) simple enough that I'm hoping there is an explicit solution, but I couldn't find one (other than $\frac{2}{\gamma r}$, which doesn't fit my constraints). As I don't have a particularly rich background in differential equations, I would be grateful for an insight of any kind.

After a bit of research, I found that this equation falls into the class of Emden-Fowler equations ($n=-1,m=2$ in the notation of the link), but I couldn't find explicit solutions for this particular case.

Another thing I noted was that defining $f\colon \mathbb{R}^3\setminus\{0\}\to \mathbb{R}$ by $f(x) = \frac{g(|x|)}{|x|}$ for $x\in \mathbb{R}^3\setminus\{0\}$ yields a positive function $f\in C^\infty(\mathbb{R}^3\setminus \{0\})$ with
$$\Delta f = \gamma f^2,$$
since the Laplacian in three dimensions, when applied to spherically symmetric functions, can be written as $\Delta_\text{radial} = \frac 1r \frac{\partial^2}{\partial r^2} (r \, \cdot)$.

Edit: Thanks to @Jacob Manaker and @Eli, whose answers are below, I am now quite convinced that no explicit general solution to the above equation exists (what a shame).
 A: I have come to the sad conclusion that there probably are no explicit solutions to your equation.
The original solution I posted was based on a calculation mistake that I do not see how to fix.  I've tried a number of substitutions since then, and gotten nowhere.  The best I can do is restate the discussion of Emden-Fowler equations in your link.
I can give a sociological argument for why this should be true: simple differential equations need not have simple solutions, but they will get studied before more complicated ones.  Your link is an excerpt from a handbook of exact solutions for differential equations.  Those sorts of books scour the literature.  If this equation had a solution, it would have been published (since this sort of equation has certainly been studied), and would appear in that handbook.
Anyways, on to what your link says ("On the shoulders of giants…"):
Let $w(r)=rg'(r)/g(r)$.  Then \begin{align*}
w'(r)&=\frac{rg''(r)}{g(r)}+\frac{g'(r)}{g(r)}-\frac{rg'(r)^2}{g(r)^2} \\
&=\gamma g(r)+\frac{w(r)}{r}-\frac{w(r)^2}{r}
\end{align*}  I have no intuition for this change of variables, but apparently it reduces the order of your equation — except, of course, that we still have $g(r)$.  But that's actually a fixable problem!
First, multiply by $r$: $$rw'(r)=\gamma rg(r)+w(r)(1-w(r))$$  Now write $w(r)=f(rg(r))$.  Then $$rw'(r)=rf'(rg(r))(g(r)+rg'(r))=rg(r)f'(rg(r))(1+w(r))$$  We've removed all the $r$-dependence, so let's change variables: call $z=rg(r)$.  Then $$zf'(z)(1+f(z))=\gamma z+f(z)(1-f(z))$$
Rearranging, $$\frac{f'(z)(1+f(z))-\gamma}{f(z)(1-f(z))}=z$$  That looks awful, but it shows that our equation is separable.  In fact, if $\gamma=0$, then we would be able to solve this equation with quadratures directly.  Since $\gamma\neq0$, the best we can do is numerically solve each side as an autonomous ODE.
Now suppose $F$ does solve that ODE; i.e. $F(f(z))=z^2/2$.  Then $f(z)=F^{-1}(z^2/2)$; that is, $$w(r)=F^{-1}(r^2g(r)^2/2)$$  Substituting our definition of $w$, we have $$g'(r)=\frac{g(r)}{r}\cdot F^{-1}(r^2g(r)^2/2)$$  In principle, this should be numerically solvable too.
A: Let me put the final nail in the coffin, if you had any doubts left. Starting from about half-way into where he was, he had
\begin{align}
z(1+f)\frac{\mathrm df}{\mathrm dz}=\gamma z+f(1-f).
\end{align}
This is an Abel equation of the second kind, it just doesn't look like it yet.
Let $1+f(z)=u(z)$ and $\gamma z=\xi$, so that
\begin{align}
u\frac{\mathrm du}{\mathrm d\xi}&=-\frac{u^2}{\xi}+3\frac{u}{\xi}-\frac{2}{\xi}+1.
\end{align}
Then, substituting $u(\xi)=v(\xi)/\xi$ and $3\xi=\zeta$ eliminates the square term, bringing the ODE into the canonical form
\begin{align}
v\frac{\mathrm dv}{\mathrm d\zeta}-v=-\frac{2}{9}\zeta+\frac{1}{27}\zeta^2.
\end{align}
Unless I made a mistake with my constants, and that -2/9 is actually supposed to be a $\pm$6/25 this does not have a solution.
You may learn that given a particular solution to the Abel equation you can construct a general solution around it. . . and you might recall that we have a particular solution for the equation we started with, so all you would have to do is transform it right? When you go through all the transformations you find that $v(\zeta)\Big|_{\zeta=6}=0$, which could be found out simply by setting $v(\zeta_0)=0$. Darkness triumphs over light once again.
Hopefully that gives some closure to the problem.
