Picard Iteration/ index i have the following System of Differential Equations 
$$ \begin{pmatrix}
x'(t) \\
y'(t)
\end{pmatrix} =  \begin{pmatrix}
0 && 1\\
-1 && 0
\end{pmatrix} \begin{pmatrix}
x(t) \\
y(t)
\end{pmatrix} \ and \ \begin{pmatrix}
x(0) \\
y(0)
\end{pmatrix}= \begin{pmatrix}
2 \\
0
\end{pmatrix}  $$
When i use the Picard-Iteration, i get ($ s:= \begin{pmatrix}
x(t) \\
y(t)\end{pmatrix} $ )
$$ s_1 =  \begin{pmatrix}
2 \\
-2t\end{pmatrix} $$
$$ s_2 =  \begin{pmatrix}
2-t^2 \\
-2t\end{pmatrix} $$
$$ s_3 =  \begin{pmatrix}
2-t^2 \\
\frac{1}{3} t^3-2t\end{pmatrix} $$
I assume that $ s_{\infty} =   \begin{pmatrix}
2cos t \\
-2sint \end{pmatrix} $
But when i consider the series for sin and cos, for example for the index 1, $ s_1$ will be $ s_1 =  \begin{pmatrix}
2-t^2 \\
\frac{1}{3} t^3-2t\end{pmatrix}  $
How do i get this right?

There is no guarantee that taking finitely many steps of Picard iteration will result in successively matched Taylor polynomials between the approximant and the solution. This does happen to occur for $y'=y,y(0)=1$ but this is in some sense a coincidence.
