ODE existence of specific solutions I have difficulties with this question : 
Given the ODE named (1) : $$x'=y+\sin (x^2y)$$ $$y'=x+\sin(xy^2)$$
and the :
Definition. A Petal is a solution $(x(t),y(t))$ that verifies $\displaystyle \lim _{t \to \pm \infty} (x(t),y(t)) =(0,0)$.
How can I show that the exists at most two distinct petals ? 
First I remarked that the question has sense since : 
$$\|(x',y')\| \leqslant \|(x,y)\| + C$$ where $C$ is a wisely chosen constant. It follows that the solution $(x(t),y(t))$ is defined on $\mathbb{R}$.
Next, the linearization in $(0,0)$ shows that $(0,0)$ is an hyperbolic point. 
But how can I say more ? 
Thank you
 A: It looks to me that the non-linear perturbation is not strong enough to alter the behaviour of the linearized dynamical system:
$\hspace2in$
so there are no petals at all: solutions cannot cross each other, and if we rotate our reference system $45^\circ$ counter-clockwise, any solution stays in the same quadrant all the time.
A: What is important is that you've found that origin is a hyperbolic saddle point. There are exactly 2 trajectories in $\omega$-limit set (incoming separatrices) and 2 trajectories in $\alpha$-limit set (outgoing separatrices) (if we don't count the origin itself, but it's not needed here) — this is a consequence of Grobman-Hartman theorem. So this pretty answers your question: if there is a petal for this saddle, then it must contain one of the incoming and one of the outgoing separatrices. And since we have only 2 of each kind, we can't have more than 2 petals for hyperbolic planar saddle.
If I'm not confusing here, complex systems of form $\dot{z} = z^m, \; m>1,$ can be adjusted to have as many petals as needed. 
