Q:

Solving an ODE using Picard's Iteration Method

Find the exact solution of the IVP
$y'=y^2$, $y(0)=1$
Starting with $y_0(x)=1$, apply Picard's method to calculate $y_1(x),y_2(x),y_3(x)$, and compare these results with the exact solution.

Solving this IVP with separation of variables, I get that $y(x)=\frac{1}{1-x}$. 

Now using Picard's method ($y_n(x)=y_0+\int_0^xf[t,y_{n-1}(t)]dt$), i get
\begin{align}
y_0(x) & =1 \\
y_1(x) & =1+x \\
y_2(x) & =1+x+x^2+\frac{x^3}{3} \\
y_3(x) & =1+x+x^2+x^3+\frac{2x^4}{3}+\frac{x^5}{3}+\frac{x^6}{9}+\frac{x^7}{63}
\end{align}
By now, I would have expected to see some pattern, but I don't really see anything, and continuing to iterate I don't think will get me anywhere. 

A:

Maybe this will help you to better understand what is going on:
http://mathworld.wolfram.com/GeometricSeries.html
The two results are actually equivalent for $|x|<1$ since the Picard iteration is converging to a geometric series, which is $\frac{1}{1-x}$ for $|x|<1$.

