1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Autonomous differential equations

  1. Jul 1, 2005 #1
    Consider the autonomous differential equation that follows

    [tex]\frac{dy}{dt} = e^y - 1, \qquad -\infty < y_0 < \infty \mbox{.}[/tex]

    I'm supposed to plot [tex]f(y)[/tex] versus [tex]y[/tex], and determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. Anyhow, you can find my plot at

    http://mygraph.cjb.net/

    which gives

    [tex]y_0 < 0 \Longrightarrow \left\{ \begin{array}{ll} y^{\prime} < 0 & \mbox{(y is decreasing),} \\ y^{\prime \prime} < 0 & \mbox{(y is concave down),} \end{array} \right.[/tex]

    and

    [tex]y_0 > 0 \Longrightarrow \left\{ \begin{array}{ll} y^{\prime} > 0 & \mbox{(y is increasing),} \\ y^{\prime \prime} > 0 & \mbox{(y is concave up).} \end{array} \right.[/tex]

    However, I picture the critical point [tex]y = \phi (t) = 0[/tex] to be semistable. The book says it is unstable, but I really can't find my mistake.

    Any help is highly appreciated.
     
    Last edited: Jul 1, 2005
  2. jcsd
  3. Jul 1, 2005 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    When a negative number decreases its absolute value increases so it goes away further from the origin.

    ehild
     
  4. Jul 1, 2005 #3
    Absolutely! I see it now. Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Autonomous differential equations
  1. Differential equations (Replies: 5)

  2. Differential Equations (Replies: 1)

  3. Differential Equations (Replies: 4)

Loading...