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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


    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]


    [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


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    Homework Helper

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

  4. Jul 1, 2005 #3
    Absolutely! I see it now. Thanks
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