# Autonomous differential equations

Consider the autonomous differential equation that follows

$$\frac{dy}{dt} = e^y - 1, \qquad -\infty < y_0 < \infty \mbox{.}$$

I'm supposed to plot $$f(y)$$ versus $$y$$, 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

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

and

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

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

Any help is highly appreciated.

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ehild
Homework Helper
thiago_j said:
Consider the autonomous differential equation that follows

$$\frac{dy}{dt} = e^y - 1, \qquad -\infty < y_0 < \infty \mbox{.}$$

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

and

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

However, I picture the critical point $$y = \phi (t) = 0$$ to be semistable. The book says it is unstable, but I really can't find my mistake.
When a negative number decreases its absolute value increases so it goes away further from the origin.

ehild

Absolutely! I see it now. Thanks