Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (b) i/Solution 1

Let's follow the hint and sketch a graph of the polynomial on the right hand side. Note that

$\frac{dy}{dt}=y^3-y^2-2y=y(y^2-y-2)=y(y-2)(y+1).$

The roots are at 0, 2 and -1.

A graph of $\frac{dy}{dt}$ vs. $y$

We are given that y(0)=1, which is in between 0 and 2. In this region, $y$ and $y+1$ are both positive and $y-2$ is negative. Therefore, $\frac{dy}{dt}<0$, which means that $y$ is decreasing (throughout the interval (0,2)). As $y$ approaches zero, we see that $\frac{dy}{dt}$ also approaches zero. (Note that $y$ could never cross zero, since $\frac{dy}{dt}=0$ at the point where $y=0$. In other words, if $y$ ever hits 0, it stays there!)

Therefore,

$\underset{t\to \mathrm{\infty }}{\lim }y(t)=0.$