# Maximal interval of definition as the initial data vary

I've been given the following problem, namely I have to study whether the solution of this system of differential equations $$\begin{cases}\dot x=\cos(xy)x^3\\ \dot y=\cos(xy)y^3\end{cases}$$ is defined on the whole of $\mathbb R$ as $(x(0),y(0))$ vary in $\mathbb R^2$.

I've begun my analysis considering the constant solutions. Due to the symmetry I may WLOG assume that $\dot x=0$ because if it were $\dot y=0$, then we would get the same conclusions. So if $\dot x=0$ then we may have $\cos (xy)=0$ which means $xy=(2k+1)\pi/2$, but, plugging this into the second, even $\dot y=0$, hence $y$ constant and then the only requirement needed would be $x(0)y(0)=(2k+1)\pi/2,\;k\in\mathbb Z$. If instead $x=0$, we would get $\dot y=y^3$, and then, eventually $$\frac{1}{y(t)^4}=\frac{1}{y(0)^4}-4t.$$ In this case $y$ cannot be defined on the whole of $\mathbb R$. This conclude my first part of the solution.

Then i passed to consider non constant solutions, and clearly I divided out the first equation by the second and I got $\dot x/x^3=\dot y/y^3$, which led me to $$\frac{1}{x(t)^4}=\frac{1}{y(t)^4}-\left(\frac{1}{y(0)^4}-\frac{1}{x(0)^4}\right),$$ but I couldn't say anything more from here, and thus I would like to see what else can be inferred from this discussion or to see different approaches to the problem in order to finish it clearly. Thank you very much for your kindness.

Regards.

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By symmetry it is enough to consider the case $x(0)\ge y(0)>0$ (for $y(0)=0$ you have found the solution). Suppose first that $x(0)>y(0)$. Let $$\frac{1}{C^4}=\frac{1}{y(0)^4}-\frac{1}{x(0)^4},\quad C>0.$$ Then the trajectory corresponding to the initial values $(x(0),y(0))$ is contained in the graph of the function $$y(x)=\frac{C\,x}{\bigl(C^4+x^4\bigr)^{1/4}}.$$ The function $y(x)$ is increasing and $\lim_{x\to\infty}y(x)=C$. It's graph cuts all the hyperbolae $$x\,y=\frac{\pi}{2}+k\,\pi,$$ whose points correspond all to constant solutions. Since by uniqueness two trajectories do not cut, the trajectory of the solution is an arc of the graph of $y(x)$ between two consecutive hyperbolae. In particular $x(t)$ and $y(t)$ remain bounded, and the solution is global.
When $x(0)=y(0)$ the trajectory is contained in the diagonal of the first quadrant, and the same conclussion holds.