A phase diagram outlining I'm trying to solve this differential equation
$$x^{ \prime}=f(x)-nx-y$$ $$y^{\prime}=\frac{(f^{\prime}(x)-r)y}{\alpha}$$ 
where
$f:[0,+\infty[\rightarrow \mathbb{R}_{+}$ is an increasing and concave function such that $f(0)=0$, $\displaystyle\lim _{x \rightarrow  0}f^{\prime}(x)=+\infty$, $\displaystyle\lim _{x \rightarrow  +\infty}f^{\prime}(x)=0$ and $\{ n,r,\alpha\} \subset ]0,1[$. 


*

*Determine the existence and number of equilibria

*Draw the phase diagram for the region $x\geq 0, y\geq 0$. 


Note: $(0,0)$ is equilibrium, but there is another part of this? 
grateful if you give me any suggestions how to solve this problem
 A: This is not a complete solution.  I just give you some hints that I will be using if I have to solve this problem.
The equations to determine the equilibrium points are given by:
$$0=x^{ \prime}=f(x)-nx-y......(1)$$
$$0=y^{\prime}=\frac{(f^{\prime}(x)-r)y}{\alpha}......(2)$$
Because $f(0)=0$, we can verify that $x=y=0$ satisfies (1) and (2).
The other solutions of the equilibrium points are given by:
$$0=f(x)-nx-y......(3)$$
$$0=f^{\prime}(x)-r......(4)$$
These equilibrium points are then independent of $\alpha$.
Suppose that the solution to (4) is 
$$x=x_1(r)$$
from (3) we may then get
$$y_1(n,r)=f(x_1(r))+n x_1(r)$$.
To draw phase diagram, we now pick $n=r=\alpha=1/2$ and $f(x)=x^{1/3}$ so that 
$f'(x)=(1/3)x^{-2/3}$.
From (4) we obtain
$$x_1=(2/3)^{-1/2}=0.544$$
and thus from (3) we obtain
$$y_1=f(x_1)+(1/2) x_1=(2/3)^{-1/2}=x_1=0.544$$.
Now you can pick a point $(x(0),y(0))$ with $0 < x(0) <x_1$ and $0 < y(0) <y_1$ as the initial point and use the equations (1) and (2) to calculate the initial velocity $x'(0),y'(0)$ to deduce how the initial point should move.
