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

MATH215 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  

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  •  April 2013  •  December 2011  •  December 2013  •  

Question 03 (c)
For the following ODE ${\frac {dy}{dt}}=(y^{2}-1)(y-2)^{2}$ (c) Determine the $\lim _{t\rightarrow \infty }y(t)$ and sketch roughly the graph of the solutions for each of the following initial conditions: y(0) = 1.5, y(0) = -3, and y(0) = 4.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Consider the sign of ${\frac {dy}{dt}}$ between the fixed points.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $(-\infty ,-1)\cup (1,\infty )$ 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 $\ \lim _{t\rightarrow \infty }y(t)=2$. Starting at $\ y(0)=-3$, the solution increases towards the critical point $\ y_{0}=-1$. Hence $\ \lim _{t\rightarrow \infty }y(t)=-1$. Starting at $\ y(0)=4$, the solution increases. Since there is no larger critical point, the solution increases indefinitely. Hence $\ \lim _{t\rightarrow \infty }y(t)=\infty$.