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

MATH102 December 2011

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[hide] Question 02 (b) i
Consider the differential equation ${\frac {dy}{dt}}=y^{3}-y^{2}-2y$ Find $\lim _{t\to \infty }y(t)$ for y(0) = 1. Hint: A sketch of the polynomial may be helpful.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Is $y$ increasing or decreasing when $t=0$? Using the graph of $dy/dt$ vs. $y$ as the hint in the problem suggests, how can you predict the behaviour of $y$ as time passes?

[show] Hint 2
What do the zeros of ${\frac {dy}{dt}}=y^{3}-y^{2}-2y$ represent?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] Solution

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, $\lim _{t\to \infty }y(t)=0.$

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