= Solution
The <logistic equation> has the constant solutions $x=0$ and $x=1$. For a nonconstant solution, <separation of variables> gives
$$
\int\left(\frac1x+\frac1{1-x}\right)dx=t+C,\qquad
\log\left|\frac{x}{1-x}\right|=t+C.
$$
Solving and imposing the initial value yields
$$
\boxed{x(t)=\frac{x_0e^t}{1-x_0+x_0e^t}.}
$$
This 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ia/paper-2-logistic.png]
{title=Logistic phase portrait and representative trajectories between the equilibria}
If $0<x_0<1$, the denominator is positive for every real $t$. The solution stays between the equilibria and increases monotonically, with
$$
\boxed{\lim_{t\to-\infty}x(t)=0,\qquad\lim_{t\to+\infty}x(t)=1.}
$$
More 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 $t_* =\log[(x_0-1)/x_0]$. This does not affect the globally defined trajectories requested between zero and one.
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