The following question is from a System Theory test without answers or solutions.
Let a continuous-time LTI system be given by the following differential equations:
$\frac{d^2}{dt^2}y_1(t)+4\frac{d}{dt}y_1(t)-\frac{d}{dt}y_2(t)+4y_1(t)-2y_2(t)=0\quad (1)$
$\frac{d}{dt}y_2(t)+y_2(t)-\frac{d}{dt}u_1(t)+u_2(t)=0\qquad \qquad \quad \qquad (2)$
Here, $u_i (t) \in R$ are inputs and $y_i(t) \in R$ are outputs, $i = 1,2$ at time $t \in R^+$. a minimal state-space representation of this system is given by $\frac{d}{dt}x(t)=Ax(t)+Bu(t), \quad y(t)=Cx(t)+Du(t), \quad t \in R^+$
$\qquad a) \quad$ What is the order of this system?
$\qquad b) \quad$ What can be said about the stability of this system?
$\qquad \qquad b_A)$ Only Lyapunov stable.
$\qquad \qquad b_B)$ Both BIBO stable and Lyapunov stable but not asymptotically stable.
$\qquad \qquad b_C)$ BIBO stable, Lyapunov stable and asymptotically stable.
$\qquad \qquad b_D)$ It is BIBO stable, but there is not enough information to know wether it is Lyapunov $\qquad$ $\qquad$ $\quad$ or asymptotically stable.
$\qquad \qquad b_E)$ Not Lyapunov, not BIBO and not asymptotically stable.
My approach:
First take the Laplace transforms of $(1)$ and $(2)$. This gives:
$(3)\quad s^2y_1+4sy_1-sy_2+4y_1-2y_2=0$
$(4)\quad sy_2+y_2-su_1+u_2=0$
Rewriting $(2)$ gives:
$y_2=\frac{su_1-u_2}{s+1}$
Substituting this in $(3)$ we get, after simplifying:
$y_1=\frac{u_1s-u_2}{(s+1)(s+2)}$
The order of this system is 2 and looking at the pole locations we can conclude that this system is BIBO, Lyapunov and asymptotically stable.
Is this approach correct?