How to show that an equation is exponentially stable

Question

I need to show that $\ddot{x}$ + $2\dot{x}$ + $2{x}$ = $0$ is exponentially stable (or the contrary).
The discriminant of the characteristic equation is $-4$, $x_1 = -1 +i$, $x_2 = -1 - i$, so the solution is $$y(x) = e^{-x}(C_{1} cos(x) + C_2sin(x))$$ What are my next steps here? Any help like books or articles with similar examples is appreciated.

Answer 1

There is a theorem that states that if an LTI system $$\tag{1} \dot {\bar x}=A{\bar x}, \quad {\bar x}\in\mathbb R^n $$ is asymptotically stable, then it is exponentially stable. Your equation is exactly the system (1) with $$ A=\left(\begin{array}{rr} 0&1\\-2&-2 \end{array}\right),\quad {\bar x}= \left(\begin{array}{c}x\\ \dot x \end{array}\right). $$ Indeed, $$ \ddot{x} + 2\dot{x} + 2x=0 $$ or $$ \ddot{x}=-2x - 2\dot{x} $$ can be rewritten as $$ \frac{d}{dt}x=\dot x= 0\cdot x + 1\cdot \dot x, $$ $$ \frac{d}{dt}\dot x=\ddot x= -2\cdot x - 2\cdot \dot{x}. $$ The eigenvalues of $A$ are $-1\pm i$ and since they have negative real parts, the system is asymptotically stable, thus it is exponentially stable.