I have a little problem here:
I need to find the 2 differential equations and build a state space representation of them.
First I get these 3 equations:
$$ u = u_C + u_R $$
$$ u_R = u_L + u_{r1} $$
$$ i_C=i_R+i_L $$
Then a little bit of thinking gives me my differential equations:
$$ i_L' = -\frac{u_C}{L} - \frac{R_1 i_L}{L} + \frac{u}{L}$$
$$ u_C' = -\frac{\alpha u_C}{C} + \frac{i_L}{C}+\frac{\beta}{C} + \frac{\alpha u}{C} $$
But here is my problem. The $\frac{\beta}{C}$ is without a state from the state vector and without an input. What am I supposed to do with that?
I just ignored it, then my state representation looks like that:
$$\frac{dx}{dt} = \begin{bmatrix}-\frac{\alpha}{C} & \frac{1}{C}\\-\frac{1}{L} & - \frac{R_1 }{L}\end{bmatrix}x+\begin{bmatrix}\frac{\alpha}{C} \\\frac{1}{L}\end{bmatrix}u$$
But I am not sure with this.
It shouldn't be a problem unless the question specifies that $f(x,u)$ must be a linear function of $x$ and $u$. Otherwise, the solution to the problem is simply
$$\frac{dx}{dt} = \begin{bmatrix}-\frac{\alpha}{C} & \frac{1}{C}\\-\frac{1}{L} & - \frac{R_1 }{L}\end{bmatrix}x+\begin{bmatrix}\frac{\alpha}{C} \\\frac{1}{L}\end{bmatrix}u+\begin{bmatrix}\frac{\beta}{C}\\0\end{bmatrix}=f(x,u).$$
In this case the dynamics of the circuit are not described by a linear system, but by a non-linear one (in particular, the system is said to be affine). This is a consequence of the resistor fact that the resistor $R_2$ has an affine current/voltage relationship instead of a linear one.
Note that state space description of a phenomenon is one which describes the phenomenon by a set of real variables (contained in the vector $x$) where the time-evolution of each is determined by a first order ODE. It has nothing to do with the ODEs being linear.