How is this equation solved for YSolutionYou have the right

How is this equation solved for Y?

Solution

You have the right equilibrium points. To determine whether or not they\'re stable, you should draw a vertical line(the y-axis), and label the two equilibrium points. Then your line is divided into 3 parts by these 2 points.
E<r, so E/r<1, so K(1-E/r)>0.
The derivative dy/dt has constant sign between equilibrium points because
f(y) = ry(1 - y/K) - Ey = -Kry^2 + (r-E)y is a polynomial in y, and polynomials have constant sign between it\'s zeros.
So you just need to pick one point from each of the 3 intervals that the equilibrium points break the line into.
Let\'s pick -1 from the lower interval, (k(1-E/r)/2) in the second interval,
and 2K(1-E/r) in the higher interval.
f(-1) = -Kr + E - r <0 because -Kr<0 and E-r<0
f(K(1-E/r)/2) >0
f(2K(1-E/r)) <0
So if y is below 0, the derivative is negative and y decreases, so it moves farther away from 0. If y is between 0 and K(1-E/r), then dy/dt is positive, so y increases away from 0 and towards K(1-E/r). If y is above K(1-E/r) then dy/dt is negative and so y decreases back down towards K(1-E/r).
So y tends away from 0 so 0 is unstable;
and y tends towards K(1-E/r) so K(1-E/r) is stable.