Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 2 / 1B / Solution

The logistic equation has the constant solutions $x=0$ and $x=1$. For a nonconstant solution, separation of variables givesSolving and imposing the initial value yieldsThis expression also includes the two equilibria when $x_0=0$ or $1$. For other initial values it is taken on the maximal interval containing $t=0$ on which its denominator does not vanish.

The phase portrait follows from the sign of $x(1-x)$: arrows point left for $x<0$, right for $0<x<1$, and left for $x>1$. Thus zero is an unstable equilibrium point and one is asymptotically stable. The requested sketch is shown below.

If $0<x_0<1$, the denominator is positive for every real $t$. The solution stays between the equilibria and increases monotonically, withMore precisely $x\sim [x_0/(1-x_0)]e^t$ in the remote past and $1-x\sim [(1-x_0)/x_0]e^{-t}$ in the remote future. Outside that interval there is a finite-time pole: for $x_0<0$ it occurs in the future, and for $x_0>1$ in the past, at $t_{\ast }=\log [(x_0-1)/x_0]$. This does not affect the globally defined trajectories requested between zero and one.