Intuition of the Wronskian I've got a question regarding the intuition of a Wronskian, in the following sense:  
The intuition for the determinant of a square $n \times n $-matrix is that it represents the area/(hyper-)volume between vectors. But what is the intuition behind the Wronskian of let's say two linear functions $f_1 = 0.5x+4$ and $f_2 = -2x+4$ which is $-10$? What is the (geometric/graphical? )intuition of the value $-10$? Is there any intuition in such a sense possible?
Thank you in advance for hints.
Best regards
 A: This is best seen for dynamical systems, that is, systems of first order autonomous differential equations: 
$$
\dot{\mathbf x}(t)= F(\mathbf x(t)),$$ 
where $\mathbf x\in \mathbb R^n$ is considered as a column vector. In this case, the Wronskian of $n$ solutions $\mathbf x_1 \ldots \mathbf x_n$ is
$$
W(t)=\det \begin{bmatrix} \mathbf x_1(t)\ldots \mathbf x_n(t)\end{bmatrix}.$$
In this case it is clear that $W(t)$ is the signed volume of the parallelepiped spanned by $\mathbf x_1(t),\ldots, \mathbf x_n(t)$. 
For the higher-order equation 
$$\frac{d^{n} x}{dt^{n}} = G\left( \frac{d^{n-1} x}{dt^{n-1}} ,\ldots, \frac{dx}{dt}, x\right)$$
we define the Wronskian of $n$ solutions $x_1\ldots x_n$ as follows:
$$
W(t)=\det\begin{bmatrix} x_1 & \ldots & x_n \\ \frac{dx_1}{dt} & \ldots & \frac{dx_n}{dt} \\ 
\ldots & \ldots & \ldots \\ \frac{d^{n-1}x_1}{dt^{n-1}} & \ldots & \frac{d^{n-1}x_n}{dt^{n-1}}\end{bmatrix}.$$
This is the same as the Wronskian of the dynamical system obtained with the substitutions 
$$ \mathbf x(t) =\begin{bmatrix} x \\ \frac{dx}{dt} \\ \vdots \\ \frac{d^{n-1}x}{dt^{n-1}}\end{bmatrix}, \qquad F(\mathbf x)=\begin{bmatrix} x_2 \\ x_3 \\\vdots \\ G(x_n\ldots x_1)\end{bmatrix}. $$
This makes it apparent that the Wronskian is a signed volume in the phase space of $n$-uples $(x, \dot x\ldots \ x^{(n-1)})$.
