Why a n-th order linear homogeneous ODE has exactly n linearly independent solutions? After some research I cannot find an answer that satisfies me for the question above. I know about the Cauchy criterion but I don't think that it proves the fact that there are exactly n linearly independent solutions to an n-th order linear homogeneous ODE.
Is it a dimension problem ? (like maybe we could construct an isomorphism from the ODE to Rn prove that ?) 
 A: Consider the n-th order linear ODE for $t \in \mathbb{R}$
$$y^{(n)}(t) + a_{n-1}(t) y^{(n-1)}(t)+ \ldots +a_1(t) y'(t) + a_0(t)y(t) =0.$$
Introducing the function $\mathbf{y} : \mathbb{R} \to \mathbb{R}^n$, 
$$\mathbf{y} = (y,y',y'', \ldots, y^{(n-2)}, y^{(n-1)}) = (y_1,y_2,y_3, \ldots , y_{n-1}, y_{n})$$ the ODE is equivalent to a system $\mathbf{y}' = A \, \mathbf{y}$ where component by component we have
$$y_1'(t) = y_2(t) \\ y_2'(t) = y_3'(t) \\ \cdots \\ y_n'(t) = - a_{n-1}(t)y_{n}(t) + \ldots +a_0(t) y_1(t)$$
Given the standard basis vectors $\mathbf{e}_1 = (1,0, \ldots, 0), \ldots , \mathbf{e}_n = (0,0, \ldots ,1)$ for $\mathbb{R}^n$, the existence-uniqueness theorem for homogeneous linear ODEs guarantees the existence of $n$ solutions $\mathbf{y}_{1}, \ldots , \mathbf{y}_{n}$ satisfying the initial conditions
$$\mathbf{y}_{k}(0) = \mathbf{e}_k, \quad \text{for } k = 1,2,\ldots ,n.$$
Forming a matrix using the $n$ solutions as columns $[\mathbf{y}_1^T(t), \ldots, \mathbf{y}_n^T(t)]$ we can show that the determinant (called the Wronskian) $W(t) = \det\, [\mathbf{y}_1^T(t), \ldots, \mathbf{y}_n^T(t)]$ satisfies the differential equation 
$$W'(t) = -a_{n-1}(t) W(t)$$
and, obviously, the initial condition 
$$W(0) = \det\, [\mathbf{e}_1^T, \ldots, \mathbf{e}_n^T] = \det\, I =1.$$
The solution is $W(t) = \exp\left(- \int_0^t a_{n-1}(s) \, ds \right) \neq 0$.  Since the determinant never vanishes, there exists a set of $n$ linearly independent solutions $\{\mathbf{y}_1(t), \ldots, \mathbf{y}_n(t)\}$. Solution of the ODE for any set of initial conditions is always some linear combination of this set. 
