Taylor of second order for System of Differential Equations I need to solve the next system
\begin{eqnarray}
x' &=& y+x(x^2 + y^2) \\
y' &=& -x + y(x^2 + y^2)
\end{eqnarray}
with $x(0) = 4$, $y(0) = 0$
I don't know how to start so I know use this method (Taylor of second order) with the case "$x' = f(x, t)$, $x(t_{0}) = x_{0}$" using the Taylor's expansion.
 A: You know that $x(0) = 4, y(0) = 0$.  Therefore $x'(0) = y + x(x^2 + y^2) = 0 + 4(16+0) = 64$  and $y'(0) = -x + y(x^2 + y^2) = -4 + 0(16 + 0) = -4$.
So $x(t) = 4 + 64t + ?$ and $y(t) = 0 - 4t + ?$.
Put that into your expressions for $x'(t)$ and $y'(t)$ and you will find first-order approximations for $x'(t)$ and $y'(t)$ near $t=0$.  Differentiate that and you get $x''(0)$ and $y''(0)$ exactly.  That gives you the next term in the Taylor series, which is all you are asked for.
If you wanted to continue, you would plug the second order approximations for $x(t)$ and $y(t)$ into the expressions for the derivative, differentiate twice, then get the third derivative.
This gets messy, fast.  But in theory you can keep going as long as you want.
A: The question asks how to solve the equation using Taylor series (which is well explained in the other answer). However the given equation can be solved analytically which is what I will show here even though this is not exactly what the question asks for. Having an analytical solution is anyway useful to check a numerical or series solution.

The two coupled ODEs can be combined to give us the two equations
$$xx' + yy' = (x^2+y^2)^2$$
$$y'x - yx' = -(x^2+y^2)$$
Since $2(xx'+yy') = (x^2+y^2)'$ and $(y/x)' = \frac{y'x-yx'}{x^2}$ we can define $R^2 = x^2+y^2$ and $\omega = \frac{y}{x}$ to get two simple uncoupled ODEs
$$R'= R^3~~~\text{and}~~~\omega' = -1-\omega^2$$
whose solution is
$$R(t) = \frac{1}{\sqrt{c-2t}}~~~\text{and}~~~\omega(t) = -\tan(t + d)$$
where $c,d$ are constants that can be fixed by imposing the initial conditions $R(0) = 4$ and $\omega(0) = 0$. The solution for $x(t)$ and $y(t)$ can now be found from the algebratic relationships
$$x^2+y^2 = R^2~~~\text{and}~~~\omega = \frac{y}{x}\implies x^2 = \frac{R^2}{1+\omega^2}~~~\text{and}~~~y^2 = \frac{R^2\omega^2}{1+\omega^2}$$
which gives us the solution
$$x(t) = \frac{4\cos(t)}{\sqrt{1-32t}}~~~\text{and}~~~y(t) = -\frac{4\sin(t)}{\sqrt{1-32t}}$$

We can Taylor expand the analytical solution to get a series solution
$$x(t) = 4+64 t+1534 t^2+40928 t^3+\frac{6876673 t^4}{6}+\mathcal{O}\left(t^5\right)$$
$$y(t) = -4 t-64 t^2-\frac{4606 t^3}{3}-\frac{122848 t^4}{3}+\mathcal{O}\left(t^5\right)$$
Notice the huge numbers that occur as the Taylor-coefficients above. This signifies that the solution is probably only valid in a short interval around $t=0$. This can be understood from the analytical solution as it blows up at $t = \frac{1}{32}$ which is also the convergence radius of the Taylor series.
