Suppose that $y_1(t), \ldots, y_n(t)$ are solutions of $\frac{d^n y}{dt} + p_{n-1}(t) \frac{d^{n-1} y}{dt} + \cdots + p_1(t) \frac{dy}{dt} + p_0(t) y = 0$, and suppose that their Wronskian is zero for $t = t_0$, i.e.
\begin{equation*}
\left|
\begin{array}{cccc}
y_1(t_0) & y_2(t_0) & \cdots & y_n(t_0) \\
y_1'(t_0) & y_2'(t_0) & \cdots & y_n'(t_0) \\
\vdots & \vdots & \ddots & \vdots \\
y_1^{(n-1)}(t_0) & y_2^{(n-1)}(t_0) & \cdots & y_n^{(n-1)}(t_0)
\end{array}
\right| = 0.
\end{equation*}
Then the corresponding matrix is not invertible, and the system of equations
\begin{array}{c}
c_1 y_1(t_0) &+& c_2 y_2(t_0) &+& \cdots &+& c_n y_n(t_0) &=& 0 \\
c_1 y_1'(t_0) &+& c_2 y_2'(t_0) &+& \cdots &+& c_n y_n'(t_0) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t_0) &+& c_2 y_2^{(n-1)}(t_0) &+& \cdots &+& c_n y_n^{(n-1)}(t_0) &=& 0 \\
\end{array}
has a nontrivial solution for $c_1, c_2, \ldots, c_n$ not all zero.
Let $y(t) = c_1 y_1(t) + \cdots + c_n y_n(t)$. Because $y(t)$ is a linear combination of solutions of the differential equation, $y(t)$ is also a solution of the differential equation. Additionally, because the weights satisfy the above system of equations, we have $y(t_0) = y'(t_0) = \cdots = y^{(n-1)}(t_0) = 0$.
These initial conditions and the original differential equation define an initial-value problem, of which $y(t)$ is a solution. If $p_0(t), p_1(t), \ldots, p_{n-1}(t)$ are continuous, then any initial-value problem associated with the differential equation has a unique solution. Obviously $y^*(t) = 0$ is a solution of the initial-value problem; since we know that $y(t)$ is also a solution of the same initial-value problem, it follows that $y(t) = 0$ for all $t$, not just $t = t_0$.
We now have $c_1 y_1(t) + \cdots + c_n y_n(t) = 0$ for all $t$, where $c_1, \ldots, c_n$ are not all zero. Thus the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent.
Conversely, if the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent, then the system of equations
\begin{array}{c}
c_1 y_1(t) &+& c_2 y_2(t) &+& \cdots &+& c_n y_n(t) &=& 0 \\
c_1 y_1'(t) &+& c_2 y_2'(t) &+& \cdots &+& c_n y_n'(t) &=& 0 \\
\vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\
c_1 y_1^{(n-1)}(t) &+& c_2 y_2^{(n-1)}(t) &+& \cdots &+& c_n y_n^{(n-1)}(t) &=& 0 \\
\end{array}
has a nontrivial solution for every $t$, the corresponding matrix is not invertible for any $t$, and $W[y_1, \ldots, y_n](t) = 0$.
Note: I wonder what DEQ this would be a solution for? Regardless, lets proceed.
We know that if $f(x)$ and $g(x)$ are linearly dependent on $I$ then $W(f,g)(x) = 0$ for all $x$ in the interval $I$.
Is this telling you anything about the linear dependence of the functions themselves? It does not imply that if $W(f,g)(x) = 0$ then $f(x)$ and $g(x)$ are linearly dependent. It is possible for two linearly independent functions to have a zero Wronskian!
We would analyze
$$ W(f,g) = \det\begin{bmatrix}f & g\\f' & g'\end{bmatrix} = fg' - gf'$$
- $x < 0 \rightarrow |x| = -x \rightarrow \text{Wronskian} = 0$ since $fg' - gf' = x(-1) - (-x)(1) = 0$
- $x = 0 \rightarrow |x| = 0 \rightarrow \text{Wronskian} = 0$ since $fg' - gf' = 0-0 = 0$
- $x > 0 \rightarrow |x| = x \rightarrow \text{Wronskian} = 0$ since $fg' - gf' = x(1) - (x)(1) = 0$
Since the Wronskian is zero, no conclusion can be drawn about linear independence!
For linear independence, we want to go back to the basic definitions again. We have:
- $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x \lt 0$. Thus, our equations to check for linear independence of these functions become:
$$c_1 x + c_2 x = 0~~~~ \text{for}~ x \ge 0 \\ c_1x - c_2 x = 0~~~~\text{for}~ x \lt 0$$
The only solution to this system is $c_1 = c_2 = 0 \rightarrow$ linear independence. Note that at the single point $x = 0$ does not matter.
You can also see the same argument for your second example Calculate the Wronskian of $f(t)=t|t|$ and $g(t)=t^2$ on the following intervals: $(0,+\infty)$, $(-\infty, 0)$ and $0$?
Best Answer
No. Linear dependence of a collection of functions $f_i$ on an interval $I\subset\mathbb{R}$ means that there exists a nontrivial linear combination of these functions that is identically equal to zero on $I$ — at all points, not just at some point. This is the only right way to interpret the definition: the result of a linear combination to establish linear dependence must be the zero element of the same vector space. The vector space here is a set of (differentiable) functions on $I$, whose zero element is the zero function — the function $O(x)$ such that $O(x)=0$ for all $x\in I$.
So for linearly dependent functions the Wronskian vanishes at all points of $I$. The negation of this statement is that for linearly independent functions the Wronskian is non-zero at least for one point $x_0\in I$.