How many initial conditions are required? Consider the following system of ODE:
$$\begin{array}{ll}\ddot y + y + \ddot x + x = 0 \\
y+\dot x - x = 0 \end{array}$$
Question: How many initial conditions are required to determine a unique solution?
A naive reasoning leads to four: $y(0),\dot y(0), x(0)$ and $\dot x(0)$. However, if we write the system in a first-order form:
$$\begin{bmatrix} 1  & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \dot x \\ \ddot x \\ \dot y \\ \ddot y\end{bmatrix} = 
\begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & -1 & 0 \\ 0 & 0  & 0 & 1\\ 1 & 0 & -1 & 0 \end{bmatrix}\begin{bmatrix} x  \\ \dot x \\ y \\ \dot y\end{bmatrix}$$
the left matrix is not of full rank, which means the equations are not all independent. 
Indeed, by differentiating the second equation: $\dot y + \ddot x -\dot x=0$ which leads to $\ddot y + y + \dot x - \dot y + x =0$, or, in the first-order form:
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}
\begin{bmatrix} \dot x \\ \dot y \\ \ddot y \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ \dot y \end{bmatrix}$$
With the above form, it shows only three initial conditions are required: $x(0),y(0),\dot y(0)$.
And, what if I had:
$$\begin{array}{ll}\ddot y + y + x^{(n)} + x = 0 \\
\dot x - x = 0 \end{array}$$
for some $n$. Then, I could solve the second equation for some $x(0)$, then differentiate $x$ $n$ times and inject in the first equation, so only $x(0), y(0), \dot y(0)$ are needed. Can this be seen directly, in a robust manner?
 A: This problem depends on 3 arbitrary constants. Here is why.
Let $$\tag{1}\begin{cases}u&=&x+y\\v&=&x-y\end{cases}$$
The given differential system, written under the form:
$$\tag{2}\left\{\begin{array}{rclr}\ddot{(x+y)}& = & - (x+y) \ \ \ \ \ & (a) \\
 \dot x& = & x-y   \ \ \ \ \ & (b)\end{array}\right.$$
is equivalent to:
$$\tag{3}\begin{cases}\ddot u &=& -u \ \ \ \ \ &(a)\\\dot u +\dot v&=&2v \ \ \ \ \ &(b)\end{cases}$$
(equation (3b) comes from the addition of the two equations of (1), $2x=u+v$,  then the differentiation of this relationship, and, at last, the use of (2b)).
The solution of equation (3a), the harmonic oscillator, depends on 2 arbitrary constants, for example under the form 
$$u=A \cos(t)+B\sin(t)$$
This solution, plugged into equation (b) gives the first order linear differential equation:
$$-\dot v + 2v=-A \sin(t)+B \cos(t)$$
whose general solution depends on a supplementary constant $C$.
Knowing $u$ and $v$, it is immediate to deduce $x$ and $y$.
A: I guess that the first answer is $4$, even if the dynamical matrix is rank deficient. Consider this example:
$$\left[\begin{array}{c}\dot{x}\\\dot{y}
\end{array}\right] = \left[\begin{array}{cc}1 & 0\\ 1 & 0
\end{array}\right] \cdot \left[\begin{array}{c}x\\y
\end{array}\right].$$
Following you thoughts, even in this case we need just $1$ initial condition instead of $2$. Anyway:
$$x(t) = A e^{-t}, y(t) = B + Ce^{-t}.$$
Then:
$$\dot{y}(t) = -Ce^{-t} = x(t) \Rightarrow A=-C,$$ and hence:
$$x(t) = Ae^{-t}, y(t) = B - Ae^{-t}.$$
This means that you still need $2$ initial conditions!!!
In this case, you have that $A = x(0)$ and $B = x(0)+y(0)$.
For the second case, you need $n+2$ initial conditions.

Addition
If the dynamical matrix is rank deficient, then this means that it has at least one null eigenvalue. The presence of the null eigenvalue implies the presence of a "constant" term in the solution. This constant term still represents a degree of freedom of the system, even if the dynamical matrix lost a degree of freedom.
