Differential Equation and Picard Iteration Write out $x_n(t)$ for the following differential equation:
$$\frac{dx}{dt} = f(t,x) = x^2, x(0)=1$$
So using the Picard Iteration : $x_n(t) = x_0 + \int_{0}^t f(s,x_{n-1}(s)) ds$,
we have $x_0(t) = 1 \quad \forall t$,
$x_1= 1+ \int_0^t (1)^2 ds = 1+t$
$x_2 = 1+ \int_0^t (1+s)^2 ds = 1+t+t^2+ \frac{t^3}{3}$
$x_3 = 1+ \int_0^t (1+s+s^2+ \frac{s^3}{3})^2 ds = 1+t+t^2+ \frac{t^3}{3}+ \frac{2}{3}t^4+ \frac{1}{3}t^5+ \frac{1}{9}t^6+ \frac{1}{56}t^7$
I am stuck in giving a general $x_n(t)$ since I could not figure out the pattern.
EDIT:
Yes I do know the solution is $\frac{1}{1-t}$, which is why I wonder there should be a recognizable pattern $1+t+t^2+t^3+...$ or something but yet my calculation seems to lead me to nowhere.
 A: I don't see any simple pattern in the coefficients of $t^k$ in polynomial $x_n$ for $k>n$. But there is such a pattern when $k\le n$: the coefficients are all $1$. That is, 
$$x_n=1+t+t^2+\dots+t^n +O(t^{n+1})\tag1$$
where $O(t^{n+1})$ stands for powers $n+1$ and higher. 
It's not hard to prove (1) by induction. The base case is $x_0=1$. If (1) holds for $n$, then squaring (1) yields $$x_n^2=1+2t+3t^2+\dots+(n+1)t^n+O(t^{n+1}) \tag2$$ which integrates to 
$$x_{n+1}=1+t+t^2+\dots+t^{n+1} +O(t^{n+2})\tag3$$
At least in the sense of convergence of coefficients, (1) is enough to tell you what happens as $n\to\infty$.
A: Well, the solution is $x(t)=1/(1-t)$, so you should generate the geometric series.  Also, note that coefficients can change from iteration to iteration.  So, if you have a $t^3/3$ at one step, that doesn't guarantee a $t^3/3$ at the next.
Incidentally, this procedure can be automated very easily with a computer algebra system.  Here's how to do so with Mathematica:
F[t_, x_] := x^2;
T[x_] := 1 + Integrate[F[s, x /. t -> s], {s, 0, t}];
NestList[T, 1, 3]

\begin{align}
  x_0(t)&=1 \\
  x_1(t)&=t+1 \\
  x_2(t)&=\frac{t^3}{3}+t^2+t+1 \\
  x_3(t)&=\frac{t^7}{63}+\frac{t^6}{9}+\frac{t^5}{3}+\frac{2 t^4}{3}+t^3+t^2+t+1 \\
\end{align}
Note that the coefficient of $t^3$ indeed changes from $1/3$ to $1$ as we move from the second to the third iterate. It stabilizes after that, though.
