I have the following vector field \begin{align} X(x,y)=(y^2,x^2) \end{align}
and I need to prove it's not complete. I've seen several answer to this question, as this, and this, but I'd like to understand a bit better it from a theroretical point of view and have a check on my approach
My attempt
As noticed in the other answers, the integral curves are $y^3=x^3+c$, and plugging this into the equation for $\dot x$ leads to
\begin{align} \dot x = (x^3-c)^{2/3}, \quad x(0)=x_0 \end{align}
It follows immediately that \begin{align} t(x)= \int_{x_0}^x \frac{ds}{(s^3-c)^{2/3}}\end{align}
As pointed out in the second of the linked answers, in order to have that the integral curve is defined for all times, I need this integral to be divergent, i.e.:
$\lim_{x \rightarrow +\infty} t(x)=\int_{x_0}^\infty \frac{ds}{(s^3-c)^{2/3}}=+\infty$
At infinity, it is asymptotic to $\frac{1}{s^2}$, therefore I have no problem.
There could ie a problem if there exists $\bar{s}$ such that the denominator vanishes, but I can assume that such an $\bar{s}$ is outside the interval of integration, indeed:
I know that the stationary solution is $x(t)=c^{1/3}$, provided $x_0=c^{1/3}$ and by existence and uniqueness thm. I can therefore consider $x_0$ to be grater or lesser than $c^{1/3}$.
Therefore the integral is converging, and then the integral curve is not defined for all times, therefore $X$ is not complete over $\mathbb{R}^2$
If you write $y^3 = x^3 + c$ then it should be $\dot{x} = (x^3 + c)^{2/3}$, and
$$ t = \int_{x_0}^x \dfrac{ds}{(s^3+c)^{2/3}} $$
Although the integrand blows up when $s^3+c = 0$, that's not really a problem because (if $c \ne 0$) the singularity is integrable.
The only stationary solution is $x=0, y=0$.