Differential Equation and Picard Iteration

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 $n\to\infty$.

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.