Explicit solutions for differential system $x'=y^2,y'=x^2$ Could the system
$$ \begin{cases}
x'=y^2 \\
y'=x^2 
\end{cases}
$$
be explicitly solved?
I should determine if for every initial condition the system has a solution defined in all $\mathbb{R}$.

You can eliminate $x$ by differentiating
$$y''=2xx'=2\sqrt{y'}y^2.$$
Then
$$\sqrt{y'}y''=2y^2y',$$
$$\frac23y'^{3/2}=\frac23y^3+K$$
and
$$(y^3+K)^{-2/3}y'=1.$$
Unfortunately, the antiderivative of the LHS is far from simple and cannot be inverted analytically.
In the special case $K=0$,
$$-\frac1y=t+C$$ and $$x=\sqrt{y'}=y.$$

In the case $K\ne0$, by a suitable rescaling of the coordinates, the relation can be normalized to 
$$(y^3+1)^{-2/3}y'=1,$$
giving
$$\int_{y_0}^y(y^3+1)^{-2/3}dy=F(y)-F(y_0)=t-t_0,$$ and
$$y=F^{-1}(t-t_0+F(y_0)).$$
The functions $F$ and $F^{-1}$ have no adjustable parameter and can be tabulated once for all, for instance by Runge-Kutta integration.
