I want to obtain the necessary condition(s) such that the equilibrium (i.e., $x=-\frac{b}{a}$) of the following nonlinear (single and not a system of) ODE is asymptotically stable.

$\frac{dx}{dt}=e^t (ax+b); a,b \in ℝ$

Can I use the Lyapunov stability theorems? If so, how? I know that I can solve this specific ODE analytically and then discuss its asymptotic behavior, but suppose that I do not want to solve it.

  • $\begingroup$ Note that $e^t$ does not influence any stability, technically a time change can reduce your equation to $\dot y=ay+b$. Here the full analysis is trivial. $\endgroup$
    – Artem
    Commented Jul 19, 2022 at 20:07
  • $\begingroup$ Note also that you can get an explicit expression of $x(t)$ by separating the variables and integrating. $\endgroup$
    – pioneer
    Commented Jul 19, 2022 at 20:16
  • $\begingroup$ @pioneer Thanks. As I said, suppose that I cannot solve this ODE. I want to use some theorems regarding convergence or asymptotical stability. $\endgroup$ Commented Jul 19, 2022 at 20:18
  • 1
    $\begingroup$ Sorry, I missed it. what you can do is the following: 1) Translate your equilibrium point to the origin 2) Use the Lyapunov function $V(x)=x^2$ $\endgroup$
    – pioneer
    Commented Jul 19, 2022 at 20:31

1 Answer 1


Intuitively, it seems obvious that the 1st-order system is asymptotically stable when

$$a < 0,$$

because $b$ can be treated as an external input.

Let $a = - \alpha$, where $\alpha > 0$, then the solution is given by

$$x(t) = \frac{\alpha x_{0} - b}{\alpha} e^{\alpha - \alpha e^{t}} + \frac{b}{\alpha}.$$

where $x_{0}$ is the initial value. If we set $\alpha = 1$, then the solution becomes simply

$$x(t) = \left(x_{0} - b\right) e^{1 - e^{t}} + b.$$


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