Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 04 (b)/Solution 1

The below is the phase line for the given equation.

As you see, based on the graph of $f(x)=rx(1-x^{2})$, we can easily get $x'=f(x)>0$ for $x<-1$ and $0<x<1$, while $x'=f(x)<0$ for $-1<x<0$ and $x>1$. Therefore, on $(-\infty ,-1)\cup (0,1)$, $x$ is increasing (i.e, move to the right on the graph), while on $(-1,0)\cup (1,\infty )$, $x$ is decreasing (i.e., move to the left). This gives the phase line as you see in the picture above.

In particular, if we start at the point $x=3$, since it lies on $(1,\infty )$ it moves to the left and approaches to $x=1$ as $t\to \infty$.

Answer: $\color {blue}\lim _{t\to \infty }x(t)=1$