I'm concerned with the system $\frac{dy}{dt} = \frac{1}{yt}$, of course restricting so that $y$ and $t$ are real and cannot be $0$. I've been able to come up with a form for the unique solution to this equation at any point $(t_0, y_0)$. I want to know if all of these unique solutions are stable/asymptotically stable.
The unqiue particular solution at $(t_0, y_0)$ I've come up with is $\varphi(t, t_0, y_0) = \sqrt{y_0^2 + t_0^{-2}-t^{-2}}$. Based on the definition then, wouldn't $\phi(t)$ and $\varphi(t,t_0,y_0)$ have the exact same form, since this solution is exactly that solution where plugging in $t_0$ yields $y_0$? Based on this I believe that all of these solution are in fact stable, and they must be asymptotically stable as well, since in the limit these two solutions will eventually equal. Am I thinking about this wrong?
Consider the case when the solution is positive. The case of a negative solution can be considered similarly. We have to estimate $$ |\phi(t)-\varphi(t, t_0, y_0)|=\left|\sqrt{\phi(t_0)^2 + t_0^{-2}-t^{-2}}-\sqrt{y_0^2 + t_0^{-2}-t^{-2}}\right| $$ using the fact that $|\phi(t_0) - y_0| < \delta$. First notice that $$ |\phi(t)^2-\varphi(t, t_0, y_0)^2|=\left|\phi(t_0)^2 + t_0^{-2}-t^{-2}-y_0^2 - t_0^{-2}+t^{-2}\right|=|\phi(t_0)^2-y_0^2| $$ $$ =|\phi(t_0)-y_0|\cdot|\phi(t_0)+y_0|<\delta |\phi(t_0)+y_0|. $$ Since the solution $\phi(t)$ is given, we know the value of $\phi(t_0)$; however, we don't know what is $y_0$ exactly. But $|\phi(t_0) - y_0| < \delta$, thus, $$ |\phi(t_0)+y_0|=|y_0-\phi(t_0)+2\phi(t_0)|\le2|\phi(t_0)|+\delta=2 \phi(t_0)+\delta. $$ Hence $$ |\phi(t)^2-\varphi(t, t_0, y_0)^2|<2\delta \phi(t_0)+\delta^2. $$ Then notice that $$ \phi(t)+\varphi(t, t_0, y_0)= \sqrt{\phi(t_0)^2 + t_0^{-2}-t^{-2}}+\sqrt{y_0^2 + t_0^{-2}-t^{-2}} $$ $$ >\sqrt{\phi(t_0)^2 }+\sqrt{y_0^2}>|\phi(t_0)|=\phi(t_0), $$ which implies $$ |\phi(t)-\varphi(t, t_0, y_0)|=\frac{|\phi(t)^2-\varphi(t, t_0, y_0)^2|} {\phi(t)+\varphi(t, t_0, y_0)}<\frac{2\delta \phi(t_0)+\delta^2}{\phi(t_0)} =2\delta +\frac{\delta^2}{\phi(t_0)}. $$ Now let $$ \epsilon=2\delta +\frac{\delta^2}{\phi(t_0)}. $$ Expressing $$ \delta = \phi(t_0)\left(\sqrt{1+\frac{\epsilon}{\phi(t_0)}}-1\right) $$ we obtain the $\delta$ from the definition. Hence, stability is proven.
The solution $\phi(t)$ is not asymptotically stable since $$ \lim_{t\to+\infty} \phi(t)= \lim_{t\to+\infty} \sqrt{\phi(t_0)^2 + t_0^{-2}-t^{-2}}= \sqrt{\phi(t_0)^2 + t_0^{-2}}, $$ but $$ \lim_{t\to+\infty} \varphi(t, t_0, y_0)= \lim_{t\to+\infty} \sqrt{y_0^2 + t_0^{-2}-t^{-2}}= \sqrt{y_0^2 + t_0^{-2}}, $$ i.e. $$ \lim_{t\to+\infty} \varphi(t, t_0, y_0)\ne \lim_{t\to+\infty} \phi(t). $$