How to use Picard-Lindelöf Fixpoint Iteration on this differential equation system I have this system of differential equations
$$ x'=yz $$
$$y'=-xz $$
$$z'=2 $$
with $ x(0)=1, y(0)=1, z(0)=0 $
I do know how to do the iteration with one equation.
My question is, how do I do it with the system of the 3 equations?
Any help, to start is very appreciated because I don't know how to tackle those.

We have the system
$$\begin{align} x'&=yz, \\ y'&=-xz \\ z'&=2\end{align} $$
The IC is given as $x(0)=1, y(0)=1, z(0)=0$.
We can solve this by finding $z$ and then substituting into $y'$, solving for $y$ and then for $x$ as
$$\begin{align} x(t) &= \sin \left(t^2\right)+\cos \left(t^2\right)\\ y(t) &= \cos \left(t^2\right)-\sin \left(t^2\right)\\ z(t) &= 2 t \end{align}$$
Using eighteen terms, we can write this as a series as
$x(t) = (t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\frac{t^{14}}{5040}+\frac{t^{18}}{362880}+O\left(t^{19}\right)) + (1-\frac{t^4}{2}+\frac{t^8}{24}-\frac{t^{12}}{720}+\frac{t^{16}}{40320}+O\left(t^{19}\right))$
$y(t) = (1-\frac{t^4}{2}+\frac{t^8}{24}-\frac{t^{12}}{720}+\frac{t^{16}}{40320}+O\left(t^{19}\right)) -(t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\frac{t^{14}}{5040}+\frac{t^{18}}{362880}+O\left(t^{19}\right))$
$z(t) = 2t$
The Picard-Lindelöf iteration  is given by
$$
f\left(\begin{pmatrix} x \\ y \\ z\end{pmatrix}\right) = \begin{pmatrix} y z \\ - x z \\ 2\end{pmatrix}, ~~~~\phi_0 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$$
Then
$$
\phi_1(t) = \phi_0 + \int_0^t f(\phi_0(s)) \,ds \\
 = \phi_0 + \int_0^t f\left(\begin{pmatrix} 1\\1 \\ 0\end{pmatrix}\right) \,ds \\
 = \phi_0 + \int_0^t \begin{pmatrix} 0 \\ 0 \\ 2\end{pmatrix} \,ds \\
= \begin{pmatrix}1 \\  1\\ 2t\end{pmatrix} 
$$
$$
\phi_2(t) = \phi_0 + \int_0^t f(\phi_1(s)) \,ds \\
 = \phi_0 + \int_0^t f\left(\begin{pmatrix} 1\\1 \\ 2t\end{pmatrix}\right) \,ds \\
 = \phi_0 + \int_0^t \begin{pmatrix} 2s \\ -2s \\ 2\end{pmatrix} \,ds \\
= \begin{pmatrix} 1+t^2 \\ 1-t^2 \\ 2t\end{pmatrix} 
$$
You can compare that to the series written above and see that we are converging to that solution.
