First Order differential equation (Schaefer model) I have the equation:
$$\frac{dy}{dt} = ry (1-\frac{y}{K})-Ey$$
For part of my homework which is modelling the population of fish where $Ey$ is the rate of the which the fish are caught 
"Show that if $E < r$ there are two equilibrium points $Y = 0$ and $Y = K(1 − E/r)$
and that the first of these is unstable and the second stable.
From this solution find the yield (ie: $EY$ ) that will occur after a long time (this is called the sustainable yield).
Find the value of $E$ which gives the maximum sustainable yield (and hence the level of harvesting that will result in the maximum number of whales being caught on a sustainable basis).
Comment on what might occur if we take $E > r$?"
So I know how to find the two equilibrium points, but I am not sure how to show whether they are stable or not, I am also having trouble finding sustainable yield and the maximum  sustainable yield
Thanks for any help
 A: $Ey$ is the rate at which the fish are caught.  Thus, at a fixed point $Y$, the rate at which the fish are being caught (the yield) is $EY$.  Having computed the fixed points, the yield at $Y=0$ is $E \cdot 0 = 0$.  If there are no fish, the yield is 0.  For the nontrivial fixed point $Y=K\left( 1- \frac{E}{r} \right)$ we get that $EY = EK\left( 1- \frac{E}{r} \right) = -\frac{KE^2}{r} + Ek$.  This is the "sustainable" yield, since the population persists.
We now want to find the value of $E$ that maximizes this value.  The sustainable yield is a quadratic with a negative leading coefficient (upside down parabola), and thus the maximizing value is found by computing the vertex.  Recall that given a quadratic $ax^2+bx+c$ the vertex ix at $x = -\frac{b}{2a}$.  Thus the value of $E$ that maximizes the yield is $-\frac{K}{\frac{2k}{r}} = \frac{r}{2}$.  Thus the maximum sustainable yield occurs when $E = \frac{r}{2}$ and is plugging this into the formula for the sustainable yield we get the maximum sustainable yield of $\frac{rK}{4}$.
The question of stability depends largely on whgat you have learned already.  You can do it graphically, drawing out the phase line. For positive parameter values the graph of the RHS of the ODE is an upside down parabola with a positive vertex and the intersects at the fixed points.  The picture looks like this.  
Another method of establishing stability is to compute the derivative of the RHS with respect to $y$ and evaluate it at the fixed points.  When the derivative is positive, the fixed point is unstable.  When the derivative is negative the fixed point is stable.
