Let's assume the following set of differential equations describes some biological system:
$$\begin{align} \dot{x}_1&=u-\alpha x_1\\ \dot{x}_2&=\beta (1 - x_1)-\gamma x_3\\ \dot{x}_3&=\delta(x_2-x_3)\\ y &= x_2 \end{align}$$
where I defined $x_2$ to be the output, and $\alpha$, $\beta$, $\gamma$ and $\delta$ are constants.
How can I now transform this system into state-space representation? I.e.
$$\begin{align} \boldsymbol{\dot{x}} &= \boldsymbol{Ax}+\boldsymbol{b}u\\ y &= \boldsymbol{c}^T\boldsymbol{x} + du \end{align}$$
I don't know what to do with the constant term in the equation for $x_2$.
Thanks a lot!
Update:
As I recently learned,a standard way to do this is to shift the state variables by their equilibrium $\bar{\boldsymbol{x}}$. Thus,
$$ \bar{x}_1=\frac{\bar{u}}{\alpha},\quad\bar{x}_2=\bar{x}_3=\frac{(\alpha-\bar{u})\beta}{\alpha\gamma} $$
with the new variables $\boldsymbol{z}=\boldsymbol{x}-\bar{\boldsymbol{x}}$ and $v=u-\bar{u}$, we end up with
$$ \boldsymbol{\dot{z}} = \begin{bmatrix}-\alpha & 0 & 0\\-\beta & 0 &-\gamma\\0&\delta&-\delta\end{bmatrix}\boldsymbol{z}+\begin{bmatrix}1\\0\\0\end{bmatrix}v\\ $$ This raises a couple of new questions, though. How do I handle the output $y$ which would certainly not fit into the state-space representation after that transformation since it would read $y_\text{new}=z_2+\bar{x}_2$. Furthermore, did I treat the input $u$ in the correct manner or would such a transformation be carried out on the unforced system, i.e. with $u=0$?