Show that there exists a $k>o$ such that solutions of this system of differential equations never cross the line $y = kx$. For the system: 
$\frac{d x}{d t} = -y$
$\frac{d y}{d t} = x(1-x) - Ay$
Where $A \geq 2$. 
I want to show that there exists a $k > 0$ such that $(x(t), y(t))$ cannot cross through the line $\{ y = k x, x > 0 \}$ when we start below this line.
I started with showing that for the line $y = k x$ we have that $\frac{d y}{d t} = k \frac{d x}{ d t} = -k y$, and now I wanted to show that if we start below this line, we have a lower $\frac{dy}{dt}$ so we can never cross the line.
But I have been unsuccesful.
Any ideas? Thanks.
 A: Hint:
You need to prove that there exists $k$ such that on the line $y=kx$ the inequality $
\frac{y'(t)}{x'(t)}\le k$ holds.
A: I think the following consideration can answer your question. 
So, the $\lbrace y = kx \rbrace$ is a boundary for domain of interest. Let's define the normal field at $y = kx$ such that it points "in" $\lbrace y \leqslant kx \rbrace$; for $k > 0$ constant vector field $(k, -1)$ is such. Then the idea is following: if we compare the vector field on boundary with normal field on the same boundary, we can define where trajectories go in our out of domain. So, the sign of dotproduct between normal field and vector field shows the direction of crossing, "in" or "out: if it's positive, trajectory goes in, if it's negative it goes out. You're interested in situation such that trajectories don't go out of domain, so there's a positive sign of dotproduct along boundary. So, let's check that for our vector field:
$$ \vec{n} \equiv (k, -1), \; \vec{v}\vert_{\lbrace y = kx \rbrace} = (-kx, x(1-x) - kAx), $$
$$ (\vec{n}, \vec{v}) = x \lbrack x + Ak - k^2 -1 \rbrack $$
Here it goes: if you want trajectories not to leave through upper part of boundary, then $Ak - k^2 -1 \geqslant 0$ would be necessary and sufficient for that. In other cases you'll have a segment where trajectories can leave through boundary.  
