"Differential Equations" by Blanchard, Devaney, and Hall is what I use. It goes over this really well. So you figured out the equilibrium points of zero and

. That's when the derivative is zero and since this is an autonumous equation, the value of I completely determines the value of the derivative so once I reaches one of the equilibrium points, then the infections stay at that value (although below it's negative which is not applicable in real life). So we have:

and I want to study just one particular case:

That means of course the recovery rate is much higher than the infection rate. Suppose I then plot

. That's the first plot and it represents the derivative as a function of the number of infections (I).. Note the derivative is positive between -10 and zero. That means if we start I anywhere in that range (not realistic in actual practice), the derivative is positive and I will increase until it reaches the nearest equilibrium point of zero. If I<-10, then the derivative is negative and the value of I will continue to decrease. And if I is greater than zero, then again the derivative is negative and so I will decrease again towards it's nearest equilibrium point. I've shown all these cases in the second plot which is I(t) with the red lines as the equilibrium points. Note that if under these conditions, the initial value of I>0, since it's derivative is negative, the value of I will continue to drop until it reaches the nearest equilibrium point which is zero (the number of infections is then zero and stays there right?). So what happens if I then use different values of b and v (just leave N=100)? How will this change the first plot and by the shape of that plot, since it's the value of the derivative, how then will the solutions behave for various starting values of I based on the value of the derivative there? Just try a few different values and see what happens