Science:Math Exam Resources/Courses/MATH100 B/December 2023/Question 20

MATH100 B December 2023

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Question 20
Sketch the phase line for the differential equation $$y'=-y(y-2)(y-10).$$ Clearly indicate the steady states (also called equilibrium points), and indicate with arrows where ${\textstyle y}$ is increasing or decreasing on either side of each steady state.

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
Recall: if ${\textstyle y}$ is at an equilibrium point, then the function ${\textstyle y}$ is not changing with time. Given that the derivative is the rate of change, what is happening to the derivative at an equilibrium point?

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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. We know that if the function ${\textstyle y}$ is at an equilibrium point, then it is not changing, which means the derivative of the function is zero. Since we are given the differential equation in factored form, we can read off the zeros easily: ${\textstyle y'=0}$ when ${\textstyle y=0,2,10}$; these are the equilibrium points. To determine if the function is increasing or decreasing on either side of the equilibrium points, we look at whether ${\textstyle y'}$ is positive or negative. When ${\textstyle y<0}$, ${\textstyle y'}$ is positive. When ${\textstyle 0<y<2}$, ${\textstyle y'}$ is negative. When ${\textstyle 2<y<10}$, ${\textstyle y'}$ is positive. And when ${\textstyle y>10}$, ${\textstyle y'}$ is negative. So, we have that the function is increasing for ${\textstyle y<0}$, decreasing for ${\textstyle 0<y<2}$, increasing for ${\textstyle 2<y<10}$, and decreasing for ${\textstyle y>10}$, which means ${\textstyle y=0}$ is an unstable equilibrium, ${\textstyle y=2}$ is a stable equilibrium, and ${\textstyle y=10}$ is an unstable equilibrium. The phase line representing all of this is given below. math100 B 2023 Q20 solution: phase diagram