Proving Stability for Dynamical System (Delta-Epsilon)

We have to prove that $$ \forall \epsilon>0 \, \exists \delta>0:\; \|\bar x(0)\|<\delta\,\Rightarrow\,\forall t\ge 0\;\|\bar x(t)\|<\epsilon, $$ where $\bar x(t)$ is a vector of solution: $$ \bar x(t)=\left(\begin{array}{c}x_1(t)\\x_2(t)\end{array}\right). $$ For the considered system $$ x_1(t)= e^{-\zeta t}(C_1 + C_2t + C_1\zeta t) $$ $$ x_2(t)= - e^{-\zeta t}(C_1\zeta^2 t+ C_2\zeta t -C_2 ), $$ where $x_1(0)=C_1$, $x_2(0)=C_2$. Hence $$ \|\bar x(0)\|=\sqrt{C_1^2+C_2^2}, $$ (note that $|C_1|\le\|\bar x(0)\|$, $|C_2|\le\|\bar x(0)\|$), $$ \|\bar x(t)\|=\sqrt{x_1^2(t)+x_2^2(t)} =e^{-\zeta t} \sqrt{(C_1 + C_2t + C_1\zeta t)^2+ (C_1\zeta^2 t+ C_2\zeta t -C_2 )^2} $$ We can use the triangle inequality: $$ \sqrt{(a_1+b_1)^2+(a_2+b_2)^2}\le \sqrt{a_1^2+a_2^2}+\sqrt{b_1^2+b_2^2} $$ to obtain $$ \|\bar x(t)\|\le e^{-\zeta t} \left( \sqrt{C_1^2(1+\zeta t)^2+C_1^2\zeta^4t^2}+ \sqrt{C_2^2t^2+C_2^2(\zeta t-1)^2} \right) $$ $$ =e^{-\zeta t} \left( |C_1|\sqrt{(1+\zeta t)^2+\zeta^4t^2}+ |C_2|\sqrt{t^2+(\zeta t-1)^2} \right) $$ $$ \le \|\bar x(0)\|e^{-\zeta t} \left( \sqrt{(1+\zeta t)^2+\zeta^4t^2}+ \sqrt{t^2+(\zeta t-1)^2} \right) $$ Since $\zeta>0$, $e^{-\zeta t} \left( \sqrt{(1+\zeta t)^2+\zeta^4t^2}+ \sqrt{t^2+(\zeta t-1)^2} \right)$ tends to zero at $t\to\infty$. It implies that there exists some maximum value $$ M=\max_{t\ge 0} e^{-\zeta t} \left( \sqrt{(1+\zeta t)^2+\zeta^4t^2}+ \sqrt{t^2+(\zeta t-1)^2} \right) $$ and $$ \|\bar x(t)\|\le \|\bar x(0)\|M e^{-\zeta t} . $$ We have proved the exponential stability of the system. Usual (Lyapunov) stability is obvious if we take $$ \delta(\epsilon)=\frac{\epsilon}M $$ Update

Indeed, $\forall t\ge 0$ $$ \|\bar x(0)\|<\delta=\frac{\epsilon}{M}\quad\Rightarrow\quad \|\bar x(t)\|<\frac{\epsilon}{M}M e^{-\zeta t}=\epsilon e^{-\zeta t} \le\epsilon $$