Different state space models derived from the same differential equation I hace come accross the book "Robust Adaptive Control" from the authors "Petros A. Ioannou" & "Jing Sun". There is a free PDF version of the book at the webpage of one of the authors if anyone is interested. At page 457 there is a transfer function of the form:
$$ Y(s) = \frac{b(s+1)}{(s+a)s}U(s) $$
which is converted to the differential equation:
$$ \ddot{y}+a\dot{y}=b\dot{u}+bu $$
I convert into a state space form by following the below classical procedure:
$x_1 = y \ \Rightarrow \ \dot{x_1} = x_2$
$x_2 = \dot{y} - bu \ \Rightarrow \ \dot{x_2} = -a(x_2+bu)+b\dot{u}+bu-b\dot{u} \ \Rightarrow \ \dot{x_2} = -ax_2+b(1-a)u$
The state-space matrix form is:
$$ \dot{x} = \begin{bmatrix}0 & 1\\0 & -a\end{bmatrix}x+\begin{bmatrix}b\\b(1-a)\end{bmatrix}u $$
$$ y = \begin{bmatrix}1 & 0\end{bmatrix}x $$ 
However, the author obtains another state-space form which also is right for this particular transfer function (I tested it with MATLAB) it is this one:
$$ \dot{x} = \begin{bmatrix}-a & 1\\0 & 0\end{bmatrix}x+\begin{bmatrix}b\\b\end{bmatrix}u $$
$$ y = \begin{bmatrix}1 & 0\end{bmatrix}x $$
I tried to "reverse engineer" to figure out which state variables he chose but was unable to do so. For example, I thought this:
$\begin{equation} x_2 = \dot{y}+ay-bu \ \Rightarrow \ \dot{x_2} = \ddot{y}+a\dot{y}-b\dot{u}  \end{equation} $ 
and by replacing $\ddot{y}$ from the differential equation we get $\dot{x_2}=bu$ which is the state space equation derived at the book. But this gets me nowhere regarding the choice of state variable $x_1$. Could use some help here. 
 A: Let us have a system defined by the general differential equation:
$$\frac{d^n y}{dt^n} + p_1 \frac{d^{n-1} y}{dt^{n-1}} + \cdots p_{n-1} \frac{d y}{dt} + p_n y = q_0 \frac{d^n u}{dt^n} + q_1 \frac{d^{n-1} u}{dt^{n-1}} + \cdots q_{n-1} \frac{d u}{dt} + q_n u$$
where $u$ is the input and $y$ is the output.
For the case that you are studying, $n = 2$, and the corresponding differential equation becomes:
$$\frac{d^2 y}{dt^2} + p_1 \frac{d y}{dt} + p_2 y = q_0 \frac{d^n u}{dt^n} + q_1 \frac{d^{n-1} u}{dt^{n-1}} + \cdots q_{n-1} \frac{du}{dt} + q_n u$$
where the coefficients are: $p_1 = a$, $p_2 = 0$, $q_0 = 0$, $q_1 = q_2 = b$.
Taking Laplace transform on both sides results in:
$$(s^2 + p_1 s + p_2) Y(s) = (q_0 s^2 + q_1 s + q_2) U(s)$$
Rearrange the equation by collating the terms containing $s^2$, $s$ and $s^0$ (or 1) such that:
$$s^2 (Y(s) - q_0 U(s)) + s(p_1 Y(s) - q_1 U(s)) + (p_2 Y(s) - q_2 U(s)) = 0$$
Dividing both sides by $s^2$ and transposing terms in $s$ to the right:
\begin{align} 
  Y(s) &=  q_0 U(s)) + \frac{1}{s}\left(q_1 U(s) - p_1 Y(s) \right) + \frac{1}{s^2}\left(q_2 U(s) - p_2 Y(s) \right) \\ 
\text{or, } Y(s) &=  q_0 U(s)) + \frac{1}{s}\left(q_1 U(s) - p_1 Y(s)+ \frac{1}{s}\left(q_2 U(s) - p_2 Y(s) \right) \right)     
\end{align}
Take, $Y(s) = q_0 U(s) + X_1(s)$, where
$$X_1(s) = \frac{1}{s}\left(q_1 U(s) - p_1 Y(s) + \frac{1}{s}\left(q_2 U(s) - p_2 Y(s) \right) \right) $$
Now, write $X_1(s) = \frac{1}{s}\left(q_1 U(s) - p_1 Y(s) + X_2(s)\right)$ such that
$$X_2(s) = \frac{1}{s}\left(q_2 U(s) - p_2 Y(s) \right)$$
Substituting $Y(s)$ in the expression for $X_1(s)$ and also multiply both sides by $s$, we get
$$sX_1(s) = q_1 U(s)- p_1 (q_0 U(s) + X_1(s))  +  X_2(s)$$
$$\therefore sX_1(s) = - p_1 X_1(s) + X_2(s) + (q_1- p_1 q_0) U(s)$$
Taking inverse Laplace transform (L.T.) of the above, we get
$$\dot{x_1} = - p_1 x_1 + x_2 + (q_1- p_1 q_0) u$$
Similarly, upon substituting $Y(s)$ in the expression for $X_2(s)$ and multiplying by $s$ on both sides, we get
$$sX_2(s) = q_2 U(s) - p_2 (q_0 U(s) + X_1(s))$$
$$\therefore sX_2(s) = - p_2 X_1(s) + (q_2 - p_2 q_0) U(s)$$
Taking inverse L.T. of the above, we obtain
$$\dot{x_2} = -p_2 x_1 + (q_2 - p_2 q_0) u$$
Also, take inverse L.T. of the expression for $Y(s)$ to get: 
$$y = q_0 u + x_1 $$
Now, substitute the values for $p_1 = a,~p_2 = 0,~q_0 = 0,~q_1 = b$ and $q_2 = b$ to get the following state equations:
\begin{align}
  \dot{x_1} &= -ax_1 + x_2 + bu \\
  \dot{x_2} &= bu \\
  y &= x_1
\end{align}
In state-space form, 
\begin{gather}
\begin{bmatrix}
x_1 \\
x_2 
\end{bmatrix}
= 
\begin{bmatrix} 
-a & 1 \\ 
0 & 0 
\end{bmatrix}
+
\begin{bmatrix}
b\\
b
\end{bmatrix}
u \\
y =
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
\end{gather}
This form of the state-space equation is also known as observable canonical form. You could read more about it in Chapter 9 of the book: Modern Control
Engineering (Fifth edition) by Katsuhiko Ogata.
