Show that if \(e \neq 0\) and \(\alpha>1, \beta>0\), then no solution of $$ \ddot{\varphi}(t)+\beta \dot{\varphi}(t)+(\alpha-1) \varphi(t)=e $$

Find numbers \(\alpha, \beta, \mu\) with the following property: For each $e \in \mathbb{R}$, all of the solutions of the system of equations $$ \begin{aligned} \dot{x}_{0}(t) &=x_{1}(t) \\ \dot{x}_{1}(t) &=x_{2}(t) \\ \dot{x}_{2}(t) &=-\mu x_{0}(t)+(1-\alpha) x_{1}(t)-\beta x_{2}(t)+e \end{aligned} $$ converge to \((e / \mu, 0,0)\). (Hint: First solve this problem for the homogeneous system that results if \(e=0\) by finding parameters that make the matrix associated to the equations have all eigenvalues with negative real parts. Then show that the same parameters work in the general case.)

Consider again the pendulum linearized about its unstable upper position, given by the equation \(\ddot{\varphi}(t)-\varphi(t)=u(t)\), and assume that we use the PD control law \(u(t)=-\alpha \varphi(t)-\beta \dot{\varphi}(t)\) to obtain an asymptotically stable closed-loop system $\ddot{\varphi}(t)+b \dot{\varphi}(t)+a \varphi(t)=0\( (with \)a=\alpha-1>0\( and \)b=\beta>0$ ). Introduce the natural frequency \(\omega:=\sqrt{a}\) and the damping factor \(\zeta:=b /(2 \sqrt{a})\), so that the equation now reads $$ \ddot{\varphi}(t)+2 \zeta \omega \dot{\varphi}(t)+\omega^{2} \varphi(t)=0 . $$ (A) Prove the following facts: 1\. If \(\zeta<1\) (the "underdamped" case), all solutions are decaying oscillations. 2\. If \(\zeta=1\) (the "critically damped" case) or if \(\zeta>1\) ("overdamped"), then all solutions for which \(\varphi(0) \neq 0\) are such that \(\varphi(t)=0\) for at most one \(t>0\). 3\. If \(\zeta \geq 1\), then every solution that starts from rest at a displaced position, that is, \(\varphi(0) \neq 0\) and \(\dot{\varphi}(0)=0\), approaches zero monotonically. (In this case, we say that there is no "overshoot.") 4\. Show rough plots of typical solutions under the three cases $\zeta<1, \zeta=1\(, and \)\zeta>1$.