Question please show your work (1) Obtain state space representations for the following differential equations for a dynamic system output (a) y - 2y + y = 2u. (b) Y - 3y + 2y - y - y = 3u.

(a) \dot{y}-2 \dot{y}+y=2 u\omega . K . T\begin{array}{l}x^{0}=A x+B U \\y=C x+D U\end{array}D= feed forward component, leads to instability, usually 'Zero'foluStept: Transfer lower order terms to RHS\dot{y}=-y+2 \dot{y}+2 uStep-2: State variables: State variables must be linearly independent of other state variables\Rightarrow we can use derivatives of state variable as thry are not linearly dependent on each other\therefore \quad y=x_{1}then \quad \dot{y}=x_{1}^{0}=x_{2}y=x_{2}Replacing terms in eq (1) with state variables ' x '\begin{array}{l}\Rightarrow \quad x_{1}^{0}=x_{2} \\y=x \\x_{2}^{0}=-x_{1}+2 x_{2}+2 u \\\end{array}In matsix form\begin{aligned}{\left[\begin{array}{l}x_{1}^{0} \\\dot{x}_{2}\end{array}\right] } & =\left[\begin{array}{ll}0 & 1 \\-1 & 2\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]+\left[\begin{array}{l}0 \\2\end{array}\right] U \\y & =\left[\begin{array}{ll}1 & 0\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]+[0] U\end{aligned}\xi(b)\text { b) } \begin{array}{r}\quad y-3 y^{0 .}+2 y^{0}-\dot{y}-y=3 u \\w \cdot k-7 \quad x=A x+B u \\y=c x+D u \\\Rightarrow \quad y=y+\dot{y}-2 \ddot{y}+3 \ddot{y}+3 u\end{array}(tolloning previous steps)\begin{aligned}y & =x_{1} \\\dot{y} & =\dot{x}_{1}=x_{2} \\\ddot{y}^{\prime} & =\dot{x}_{2}=x_{3} \\\dot{y}^{0} & =\dot{x}_{3}=x_{4} \\y_{0} & =\dot{x}_{4} \\\dot{y}_{1} & =x_{2} \\\dot{x}_{2} & =x_{3} \\\dot{x}_{3} & =x_{4} \\\dot{x}_{4} & =x_{1}+x_{2}-2 x_{3}+3 x_{4}+3 u\end{aligned}y=x_{1}In Mataix form\begin{array}{l}\text { In Matsix form } \\{\left[\begin{array}{l}0 \\x_{1} \\\dot{x}_{2} \\\dot{x}_{3} \\\dot{x}_{4}\end{array}\right]=\left[\begin{array}{llll}0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3} \\x_{4}\end{array}\right]+\left[\begin{array}{l}0 \\0 \\0 \\3\end{array}\right] \cup\left[\begin{array}{l}x_{1} \\x_{2} \\x_{3} \\x_{4}\end{array}\right]+[0] U}\end{array}Care should be taken while representing in final matrix form.  The value of D - feed forward component is not always zero, it's value depends on  ...