How can I solve the equation $x'(t)=y(t)^2, y'(t)=x(t)^2$ How can I solve the differential equation
$$
\begin{cases}
x'(t)=y(t)^2 \\
y'(t)=x(t)^2
\end{cases}
$$
with the initial conditions $x(0)=x_0, y(0)=y_0$?
A little bit code in Mathematica can give the result, but just a pile of stuff as InverseFunction and Hypergeometric2F1 etc, which is useless.
 A: From
$$
\dot x = y^2\\
\dot y = x^2 
$$
we have
$$
\frac{dy}{dx} = \frac{x^2}{y^2}\Rightarrow y^3=x^3+C_0
$$
and then
$$
\dot x = \left(x^3+C_0\right)^{2/3}\\
\dot y = \left(y^3-C_0\right)^{2/3}
$$
etc
A: $$y=\pm \sqrt{x'}$$
Leaving out the $\pm$ ambiguity for now:
$$y'=\frac{x''}{2\sqrt{x'}}$$
$$\frac{x''}{2\sqrt{x'}}=x^2$$

$$x'=u(x)$$
$$x''=u u'$$

$$\sqrt{u} ~u'=2 x^2$$
$$\frac{2}{3} u^{3/2}=\frac{2}{3} x^3+c_1$$
$$u=(x^3+c_1)^{2/3}$$

$$x'=(x^3+c_1)^{2/3}$$
$$dt=\frac{dx}{(x^3+c_1)^{2/3}}$$
Integrating we have the function $t(x)$, which we need to invert to find $x(t)$, if it's possible to do explicitly.
The integral on the right can be found in terms of hypergeometric function (hence, Mathematica output), using the Euler integral (see wikipedia and other sources)
A: Note that
$$\begin{cases}
x'(t)=y(t)^2 \\
y'(t)=x(t)^2
\end{cases} \implies \frac {dx}{dy}=\frac {y^2}{x^2}$$
$$\implies \frac {x^3}3=\frac {y^3}3+K \implies x^3=y^3+C$$
With initial condition 
$$C= {x^3}- {y^3}={x^3_0}- {y^3_0}$$
