The problem: The equations of motion of a lumped mechanical system undergoing small motions can be expressed as $$ Mq'' + Dq' + Kq = f ~~~~ (1) $$ (using primes instead of dots on top of the $q$s) where $q(t) \in \mathbb{R}^k$ is the vector of deflections, $M$, $D$, and $K$ are the mass, damping, and stiffness matrices, respectively, and $f(t) \in \mathbb{R}^k$ is the vector of externally applied forces. Assuming $M$ is invertible, write linear system equations for the mechanical system, with the state $x = [q^T~~q'^T]^T$, input $u = f$, and output $y = q$.
(The above is a homework problem corresponding to a class I'm voluntarily taking on my own.)
What I've done so far: From (1), we can apply $M^{-1}$ to both sides to obtain $$ q'' = M^{-1}f - (M^{-1}Dq' + M^{-1}Kq). $$
What I'm struggling with is that the output is supposed to be $q$, when what I have can be written as a linear equation $y(t) = Ax(t) + Bu(t)$ with $$ A = -\begin{bmatrix} M^{-1}K \\ M^{-1}D \end{bmatrix}, ~~~ x = [q^T~~q'^T]^T, ~~~ B = M^{-1}, ~~~ u = f $$ but where $y = q''$, not $y = q$.
Question: Am I missing something here? Did the problem mean to say that the output $y$ should be $q''$? If I am missing something (and I probably am), any hints as to what that is would be much appreciated.
Calling $q_1 = q, q_2 = \dot q_1$ we have
$$ \cases{ \dot q_1 = q_2\\ \dot q_2 = -M^{-1}K q_1-M^{-1}D q_2+M^{-1}f } $$
and then
$$ \left(\begin{array}{c} \dot q_1\\ \dot q_2 \end{array}\right) = \left(\begin{array}{cc} 0 & I\\ -M^{-1}K &-M^{-1}D \end{array}\right) \left(\begin{array}{c} q_1\\ q_2 \end{array}\right)+\left(\begin{array}{c} 0\\ M^{-1} \end{array}\right)u $$
and
$$ y = \left(\begin{array}{cc} I & 0 \end{array}\right)\left(\begin{array}{c} q_1\\ q_2 \end{array}\right) $$