# Showing asymptotic stability

How can I show that $$\tilde x=0$$ is asymptotically stable for

$$x'=-t \cos x, t \in \mathbb R$$

I guess I need to find a Lyapunov function but I'm not sure

• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. – dantopa Jun 5 at 18:12
• – dantopa Jun 5 at 18:16
• Or this: [Using a Lyapunov function to classify stability and sketching a phase portrait ](math.stackexchange.com/questions/3204990/…) – dantopa Jun 5 at 18:16
• One problem with this question is that $0$ is not an equilibrium for $\dot x = -t\cos x$ for all $t$; on the other hand, if we consider $\dot x = -t \sin x$ . . . or look at the equilibria $x = (2n + 1)\pi / 2$ – Robert Lewis Jun 5 at 18:25
• @RobertLewis How would it work for $x'=-t \sin x$? – user679927 Jun 5 at 18:41