Science:Math Exam Resources/Courses/MATH215/December 2013/Question 07 (a)

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MATH215 December 2013

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• April 2013 • December 2011 • December 2013 •

Question 07 (a)
Consider the following initial value problem for a first-order autonomous ODE: ${\frac {dy}{dt}}=-y(y-1)^{2},\quad y(0)=y_{0}$ (a) Find the equilibrium solutions (critical points), determine the stability (stable or unstable) of each, and find $\lim _{t\to \infty }y(t)$ for all possible values of the initial condition $\displaystyle y_{0}$ .

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
The equilibrium solutions are the constant solutions where $dy/dt=0$ . To assess the stability, consider what the sign of the derivative $dy/dt$ tells you about solutions near the equilibrium points.

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Solution
Setting $dy/dt=0$ , we have $-y(y-1)^{2}=0$ so $y=0,1$ . Looking at the sign of $dy/dt$ , we have ${\begin{cases}y'>0,\quad y<0\\y'<0,\quad 0<y<1\\y'<0,\quad y>1\end{cases}}$ . Thus, if we start below $y=0$ , $y(t)$ increases; if we start above $y=0$ but below $y=1$ , $y(t)$ decreases. Either way, we approach $y=0$ so $y=0$ is a stable equilibrium solution. If we start below $y=1$ but above $y=0$ , $y(t)$ decreases and moves away from $y=1$ ( $\displaystyle y'<0$ ) and if we start above $y=1$ , $y(t)$ decreases and approaches $y=1$ ( $\displaystyle y'<0$ ). However, because the solution moves away from $y=1$ locally on at least one side, $y=1$ is an unstable equilibrium solution. All possible values for $\lim _{t\rightarrow \infty }y(t)$ include 0 and 1. If $y(0)<0$ , $y\rightarrow 0$ as $t\rightarrow \infty$ . If $0<y(0)<1$ , $y\rightarrow 0$ as $t\rightarrow \infty$ . Finally, if $y(0)>1$ then $y\rightarrow 1$ as $t\rightarrow \infty$ .