On Lyapunov stability of a solution I have recently encountered the following exercise: prove or disprove that the solution of the ODE $\dot x = -\frac{2x}{t} + t^2$ that satisfies $x(1)=1$ is Lyapunov stable. I have proved that the solution is stable, however, the correct answer (as per my teacher) is the opposite.
My proof is as follows:
$$\begin{aligned}
\dot x &= -\frac{2x}{t} + t^2 \\
t^2\dot x + 2xt &= t^4 \\
t^2\dot x + x(t^2)' &= t^4 \\
(t^2 x)' &= t^4 \\
t^2 x &= \frac{t^5}{5} + C \\
x &= \frac{t^3}{5} + \frac{C}{t^2} \\
\end{aligned}$$
The solution that satisfies $x(1)=1$ is $x=\varphi(t)=\cfrac{t^3}{5} + \cfrac{4}{5t^2}$.
The definition of Lyapunov stability:
$$\forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ |x(t_0) - \varphi(t_0)|< \delta \implies \forall t > t_0 \ \ |x(t) - \varphi(t)| < \varepsilon$$
$$|x(t_0) - \varphi(t_0)|< \delta \implies \left|\frac{1}{5} + \frac{C}{t^2} - \frac{1}{5} - \frac{4}{5}\right| < \delta \implies \left|C - \frac{4}{5}\right| < \delta$$
$$|x(t) - \varphi(t)| < \varepsilon \implies \left|\frac{t^3}{5} + \frac{C}{t^2} - \frac{t^3}{5} - \frac{4}{5t^2}\right| < \varepsilon \implies \left|\frac{1}{t^2}\left(C-\frac{4}{5}\right)\right| < \varepsilon$$
Let $\delta=\varepsilon$. Then
$$\left|\frac{1}{t^2}\left(C-\frac{4}{5}\right)\right| < \varepsilon \implies \varepsilon\left|\frac{1}{t^2}\right| < \varepsilon$$
, which holds.
 A: You obtained the following equation
$$\frac{d}{dt}(t^2 x)  = t^4 \quad (1) $$
then you found the following general solution:
$$t^2 x(t) = \frac{t^5}{5} + C $$
which is also true, but you can not treat $C$ as an initial condition of the original differential equation. To use the defintion of stability, you sould integrate both sides of Equaiton (1) form $t_0$ to $t\geq t_0$. Integrating (1) gives:
$$t^2x(t)-t_0^2x(t_0)=\frac{1}{5}t^5-\frac{1}{5}t_0^5$$
or equivilanetly:
$$x(t)=\frac{t_0^2}{t^2}x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5$$
To prove the stability, the following should hold for any   $\epsilon$ .....
$$\|x(t)-x(t_0)||=\bigg\|(\frac{t_0^2}{t^2}-1)x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5\bigg\|\leq\epsilon \qquad (2)$$
But if there exists at least one $\epsilon$ such that Equaiton (2) is not satisfied for all initial conditions then the system is not stable. Slecet $\epsilon =1$, then one can see that $\bigg\|(\frac{t_0^2}{t^2}-1)x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5\bigg\|$ is not bounded because of $t^3$.
