Good references for analytic solutions to nonlinear ordinary differential equations? I am faced with a non-autonomous initial value problem for a function $x:[0,\infty) \to \mathbb{R}^2$ of the form
$$ x'(t) = f(t,x(t)) $$
for $f: [0,\infty) \times \mathbb{R}^2 \to \mathbb{R}^2$ with initial condition $x(0)$.  Now, the function $f$ is such that
$$ f(t,x) = t^{-1} f_{-1}(x) + f_0(x) + t f_1(x) $$
where the functions $f_n: \mathbb{R}^2 \to \mathbb{R}^2$ are analytic (in fact, polynomial).  Furthermore, the initial condition is such that
$$\lim_{t\to 0^+} f(t,x(t))$$
exists.  This means that there is a formal power series for $x(t)$ around $t=0$ which solves the initial value problem; although it depends on a parameter (related to $x'(0)$) which cannot be fixed.  In other words, I get a formal power series in $ct$, for some real number $c$ which cannot be determined.  My problem is to determine the radius of convergence of this power series in $ct$.
Alas, my expertise with analytic solutions of ODEs stops with the standard undergraduate fare of the Frobenius method,... but only for linear equations.  Hence I am asking the MO community for some readable reference(s) for the nonlinear case.
Thanks in advance.
Added (in response to comments below and particularly KConrad's answer)
What I am actually interested is in whether the solution will blow up in finite time.  (Actually, in the problem $t$ is not really time, but inverse distance from a black-hole-like singularity and by blowing up in finite time, what I am after is whether the solution is indeed a black hole; i.e., whether there is an event horizon.)  I cannot prove that this is the same as the formal power series solution having a finite radius of convergence, but this is precisely what happens in the black hole solutions I know: Schwarzschild and Reissner-Nordström, for instance.
 A: Since the $f_n$'s are polynomials, does that tell us that the radius of convergence should equal a blow-up time (where the solution first diverges)?  If the $f_n$'s were merely smooth I would be reluctant to suggest that.  All I can pass along is a suggestion of how to get a bound on the blow-up time, which hopefully is equal to the radius of convergence: use differential inequalities.  
An expert in non-linear ODEs told me once that in practice nobody tries to prove very sharp approximations for blow-up times; mere existence of a blow-up time usually suffices.  
Here is an example.  Consider $y'(t) = y(t)^2 - t$, with $y(0) = 1$.  It has a solution which blows up in finite time. To estimate the blow-up time, let $Y(t) = 1/y(t)$ and see where $Y(t) = 0$.  From $Y'(t) = tY(t)^2 - 1$ and $Y(0) = 1$, a computer algebra package 
has $Y(t) = 0$ at $t \approx 1.125$.  We'd like to prove a theorem related to this numerical observation.
Claim: The solution to $y'(t) = y(t)^2 - t$ satisfying $y(0) = 1$ is undefined somewhere before $t = 1.221$.
Remark: This is weaker than what numerics suggest (i.e., the blow-up time is 
around 1.125), but 
proving something sharper requires a more careful analysis than 
I wish to develop.  
Proof:
We know $y(t)$ is defined for small $t > 0$. 
Assume $y(t)$ is defined for $0 \leq t < c$.  We will show 
for an explicit number $c$ that 
$y(t) \geq c/(c-t)$ for $0 \leq t < c$, so $y(t) \rightarrow \infty$ as $t \rightarrow c^{+}$. 
Therefore $y(t)$ has to be undefined for some $t \leq c$. 
Set $z(t) = c/(c-t)$, with $c$ still to be determined, so 
$$
\frac{\rm d}{{\rm d}t}(y - z)  =  y^2 - t - \frac{{\rm d}z}{{\rm d}t}.
$$
After some algebra on the right, this becomes
$$
\frac{\rm d}{{\rm d}t}(y - z)  =  (y - z)(y + z) + \frac{(c-1)c}{(c-t)^2} - t.
$$
By calculus, $(c-1)c/(c-t)^2 - t \geq 0$ for $0 \leq t < c$ 
as long as $c - 1 \geq (4/27)c^2$, which happens for 
$c$ between the two roots of $x - 1 = (4/27)x^2$.  The roots are approximately 
$1.2207$ and $5.5292$.  So taking $c = 1.221$, we have $$(y(t) - z(t))' \geq (y(t)-z(t))(y(t)+z(t))$$ for $0 \leq t < c$.  Using an integrating factor, this differential inequality is the same as 
$$
\frac{{\rm d}}{{\rm d}t}\left(e^{-\int_0^t(y(s)+z(s)){\rm d}s}(y(t) - z(t))\right) \geq 0. 
$$
Since $e^{-\int_0^t(y(s)+z(s)){\rm d}s}(y(t)-z(t))|_{t = 0} = 0$, 
$e^{-\int_0^t(y(s)+z(s)){\rm d}s}(y(t)-z(t)) \geq 0$ for $t \geq 0$, so 
$y(t) - z(t) \geq 0$ because the exponential factor is positive.  Thus $y(t) \geq z(t) = c/(c-t)$.  QED
A: If what you're after is really time of existence, I think your best bet is to study the equivalent integral equation
$$ x(t) = \Phi(x)(t), $$
where
$$ \Phi(x)(t) = x(0) + \int_0^t f(t, x(t))\, dt $$
and try to find a boundedness or growth condition on $x$ which implies the the same condition for $\Phi(x)$.
