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Virtual Math Learning Center

Virtual Math Learning Center Texas A&M University Virtual Math Learning Center

Section 2.5 – Autonomous Equations and Population Dynamics


Exercises

Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.

1. Given the differential equation \[y'=y^3-4y\]

  1. Find the equilibrium solutions.
  2. Graph the phase line. Classify each equilibrium solution as either stable, unstable, semistable.
  3. Graph some solutions.
  4. If \(y(t)\) is the solution of the equation satisfying the initial condition \(y(0)=y_0\), where \(-\infty< y_0<\infty\), find the limit of \(y(t)\) when \(t\) increases.

  1. Equilibrium Solutions: \(y=0,\) \(y=2,\) \(y=-2\)
  2. Stable: \(y=0,\) Unstable: \(y=2,\) \(y=-2.\) See the video for the graph of the phase line
  3. See the video
  4. \(\begin{cases}  y_0<-2 & \Rightarrow & y\to -\infty \\ y_0=-2 & \Rightarrow & y=-2 \\ -2< y_0<0 & \Rightarrow & y\to 0 \\ y_0=0 & \Rightarrow & y= 0 \\ 0< y_0<2 & \Rightarrow & y\to 0 \\y_0=2 & \Rightarrow & y=2 \\ y_0>2 & \Rightarrow & y\to \infty  \end{cases}\)

If you would like to see more videos on this topic, click the following link and check the related videos.

2. Given the differential equation \[y'(t)=y^3-2y^2+y\]

  1. Find the equilibrium solutions.
  2. Graph the phase line. Classify each equilibrium solution as either stable, unstable, or semistable.
  3. Sketch the graph of some solutions. 
  4. If \(y(t)\) is the solution of the equation satisfying the initial condition \(y(0)=y_0\), where \(-\infty< y_0<\infty\), determine the behavior of \(y(t)\) as \(t\) increases.
  5. Do any solutions 'blow up in finite time,' namely, do they admit a vertical asymptote?

  1. Equilibrium Solutions: \(y=0,\) \(y=1,\) 
  2. Semistable: \(y=1,\) Unstable: \(y=0,\) See the video for the graph of the phase line
  3. See the video
  4. \(\begin{cases}  y_0<0 & \Rightarrow & y\to -\infty \\ y_0=0 & \Rightarrow & y=0 \\ 0< y_0<1 & \Rightarrow & y\to 1 \\ y_0=1 & \Rightarrow & y= 1 \\ y_0>1 & \Rightarrow & y\to \infty\end{cases}\)
  5. Yes, \(y(t)\) blows up in finite time. See the video for an explanation.

If you would like to see more videos on this topic, click the following link and check the related videos.