How can I solve the equation $x'(t)=y(t)^2, y'(t)=x(t)^2$

Question

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.

Answer 1

$$y=\pm \sqrt{x'}$$

Leaving out the $\pm$ ambiguity for now:

$$y'=\frac{x''}{2\sqrt{x'}}$$

$$\frac{x''}{2\sqrt{x'}}=x^2$$

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$$x'=u(x)$$

$$x''=u u'$$

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$$\sqrt{u} ~u'=2 x^2$$

$$\frac{2}{3} u^{3/2}=\frac{2}{3} x^3+c_1$$

$$u=(x^3+c_1)^{2/3}$$

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$$x'=(x^3+c_1)^{2/3}$$

$$dt=\frac{dx}{(x^3+c_1)^{2/3}}$$

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)

Answer 2

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

Answer 3

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 $$C= {x^3}- {y^3}={x^3_0}- {y^3_0}$$