Science:Math Exam Resources/Courses/MATH215/December 2011/Question 03 (c)/Solution 1

From (a) we know that the fixed points are given by $y_0=-1,1,2$.

In part (b) we calculated that y is increases in $(-\mathrm{\infty },-1)\cup (1,\mathrm{\infty })$ and decreases in (-1,1).

This means, that starting at $y(0)=1.5$, the solution increases towards the critical point $y_0=2$. Hence $\underset{t\to \mathrm{\infty }}{\lim }y(t)=2$.

Starting at $y(0)=-3$, the solution increases towards the critical point $y_0=-1$. Hence $\underset{t\to \mathrm{\infty }}{\lim }y(t)=-1$.

Starting at $y(0)=4$, the solution increases. Since there is no larger critical point, the solution increases indefinitely. Hence $\underset{t\to \mathrm{\infty }}{\lim }y(t)=\mathrm{\infty }$.