Given a system of nonlinear differential equation \begin{eqnarray}\frac{dx}{dt}=2x(3-y) \\ \frac{dy}{dt}=3y(4-x)\end{eqnarray} If $r(t)=$($x(t)$,$y(t)$) is a solution of the system with initial value $x(0)>0$ and $y(0)>0$, would you help me to prove that $x(t)>0$ and $y(t)>0$ for all real $t$.
Here is my argument: I prove it by contradiction. Let $r_1(t)=(x_1(t),x_2(t))$ be an orbit of the ODE with $x_1(0)=0$. Since, for $x=0$ we get $\frac{dx}{dt}=0$ then $x_1(t)=x_1(0)=0$. Thus, $x=0$ is an invariant manifold of the ODE. By the same argument, $y=0$ is also an invariant manifold. So the orbit that passing through a point in $x=0$ (resp $y=0$) always lies in $x=0$ (resp $y=0$). Assume there is a $t_1$ such that $x(t_1) \leq 0$ then $r(t)$ must intersect an orbit in $x=0 $( or $y=0$) hence $r(t)$ must lies on $x=0 $ (or $y=0$), contradicting $x(0)>0$ and $y(0)>0$.
Is there any a direct proof (or even elementary/simple proof)?