where did I go wrong in solving this sytem of nonlinear first-order ODEs? To communicate my experience level and intent: I'm an undergraduate, this is not homework, I'm trying to write a physical simulation for fun and xp and am stuck just before (what looks to me like) the end of the tunnel. 
I am attempting to solve a system of ordinary nonlinear differential equations and have either made a mistake that I can't see or have run out of tools to hit this thing with.
$v'(t) = \alpha r^2 v^2$
$r'(t) = \beta v^2$
In my attempt to solve it, I assume $v(t) = c_v e^{st}$, which implies:
$v'(t) = sce^{st} = \alpha r^2 c_v^2 e^{2st}$ 
$r'(t) = \beta c_v^2 e^{2st}$.
I solve $v'(t)$ for $c_v$ and make the substitution $c_v = \dfrac{s}{\alpha r^2}e^{-st}$ to eliminate $c_v$ from $r'(t)$:
$r'(t) = \dfrac{\beta s^2}{\alpha^2r^4}$
Now I solve for $r(t)$ by the separation of variables $\int{r^{-4}dr} = \int{\dfrac{\beta s^2}{\alpha^2}dt}$:
$r(t) = \bigg(\dfrac{1}{-3\beta(\frac{s}{a})^2t-3c_r)}\bigg)^{1/3}$
Setting $r(0) = r_0$, I solve the IVP for $c_r = \dfrac{-1}{3r_0^3}$,
and in the interest of carpal health I make the substitutions:
$-3\beta\big(\dfrac{s}{a}\big)^2 = \gamma$
$-3c_r = \dfrac{1}{r_0^3} = \delta$
which rewrites $r(t)$ into
$r(t) = (\gamma t + \delta)^{-1/3}$
Okay, one down. Now back to $v(t)$... I expand $r(t)$ into its new form to give
$v'(t) = \alpha v^2 (\gamma t + \delta)^{-2/3}$
Now I use separation of variables again, though because $r'(t)$ is a function of $v'(t)$ and I already assumed $v = c_v e^{st}$ I'm not totally sure this is ok, but...
$\int{\dfrac{dv}{v^2}} = \int{\alpha (\gamma t + \delta)^{-2/3}dt}$
which gives
$v(t) = \dfrac{-\gamma}{\gamma c_v + 3(\gamma t + \delta)(\gamma t + \delta)^{-2/3}}$
Now I get stuck. I isolate $c_v$ by setting $v(0) = v_0$ and substitute back in all of my $\gamma$s and $\delta$s to get the full picture:
$c_v = -\dfrac{1}{v_0} + \dfrac{1}{r_0\beta(\frac{s}{\alpha})^2}$
I've still got that pesky $s$ in there from when I assumed $v(t) = c_v e^{st}$ and I'm out of equations to extract it with. I can't isolate $s$ in my equation for $r(t)$, I don't have the skill to begin to unravel that mess and I don't even know if its possible. Wolfram Alpha reports no solutions, for what its worth.
So, how do I get at that $s$, and if I can't, where did I go wrong?
 A: Well, the Mathematica solution is horrifying, involving inverse arbitrary functions.  I can at least verify that your initial ansatz is definitely wrong; $v$ is not an exponential function of $t$.  The inverse functions will require numerical solutions, so you may be just as well off numerically solving your differential equations.
A: Starting with
$$
v'(t) = \alpha r^2 v^2,\\
r'(t) = \beta v^2.
$$
we can write 
$$
v' = \dfrac{\alpha}{\beta}r^2r' = \dfrac{\alpha}{\beta}\frac{d}{dt}\dfrac{r^{3}}{3}
$$
spo yeilding
$$
v(t) = \dfrac{\alpha}{3\beta}r^3 + C_{1},\\
r(t) = \left(\dfrac{3\beta}{\alpha}\right)^{1/3}\left(v(t)-C_1\right)^{1/3}
$$
now we have 
$$
v' = \alpha\left(\dfrac{3\beta}{\alpha}\right)^{2/3}\left(v(t)-C_1\right)^{2/3}v^2 = \alpha\left(\dfrac{3\beta}{\alpha}\right)^{2/3}\left[v\left(v-C_1\right)^{1/3}\right]^2,\\
r'=\beta \left[\dfrac{\alpha}{3\beta}r^3 + C_{1}\right]^2
$$
Looking at $v'$ first
$$
\int\frac{dv}{\left[v\left(v-C_1\right)^{1/3}\right]^2} = \lambda_0 t + C_2
$$
where $\lambda_0 = \alpha\left(\dfrac{3\beta}{\alpha}\right)^{2/3}$
using the change of variable $X = (v-C_{1})^{1/3} \rightarrow dX = \frac{1}{3}(v-C_{1})^{-2/3}dv$
we obtain
$$
\int \dfrac{dX}{\left(X^3+C_1\right)^2} = \lambda_1 t + C_2
$$
and $\lambda_1 = 3\lambda_0$
focus on $r'$
$$
\int \frac{dr}{\left(r^3 + \lambda_2\right)^2} = \lambda_3 t + C_3 
$$
where 
$$
\lambda_2 = \dfrac{3\beta}{\alpha}C_1,\\
\lambda_3 = \dfrac{9\beta}{\alpha^2}
$$
So basically the analytic solution will be found $\textbf{if}$ one can compute the integral
$$
\int \frac{dy}{\left(y^3+b\right)^2}?
$$
Though this rather extended comment might all be for nought :).
$\textbf{update:}$
I will not claim to solve this integral myself (wolfram ;)) but apparently the integral above equates to
$$
\frac{1}{9b^{5/3}}\left[\frac{3b^{2/3}y}{b+y^3} - \log\left(b^{2/3}-b^{1/3}y+y^2\right) + 2\log\left(b^{1/3} + y\right) - 2\sqrt{3}\tan^{-1}\left(\frac{1-\frac{2y}{b^{1/3}}}{\sqrt{3}}\right)\right]
$$
..apparently. Hopefully someone can verify this in the meantime.
