Finding the equilibrium solutions of a logistic equation Given a logistic equation 
$$dy/dt = r(1 − y/K)y − Ey$$
(a) Show that if $E < r$, then there are two equilibrium points, $y_{1} = 0$ and $y_{2} = K(1 −E/r) > 0$.
(b) Show that $y = y_{1}$ is unstable and $y = y_{2}$ is asymptotically stable.
What i tried, Equating $dy/dt=0$, the eqution becomes  $$r(1 − y/K)y − Ey=0$$.  Taking the common facter $y$ out, i got $y=0$ which corresponds to $y_{1}$ while the second portion becomes   $$r(1 − y/K) − E=0$$, then solving for $y$, i got $$y=1-(Ek/r)=0$$. But that dosent seem to be simillar to $y_{2}$. I might have missed out something. Could anyone please explain. Thanks

For part $a.$, you just have a slight algebra issue.
$$r(1 − y/K) − E=0 \implies (1-y/K) = E/r \implies y/K = (1-E/r) \implies y = K(1-E/r)$$
Also note, since $E \lt r$ (and $K \gt 0$), $y_2 \gt 0$.
For part $b.$, you can draw the direction field plot or phase line using the slope field.
Update
When drawing the direction field, you should end up with something that looks like:

You could look at the slope (positive or negative) for various values of $y$ and determine the directions.
In this case, you see the $y_1$ is unstable and $y_2$ is asymptotically stable.
Recommend you also do this for the phase line as it is quite instructive.
