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

MATH100 B December 2023

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[hide] 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.

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[show] 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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[show] Solution

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