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 $W[y_1, \ldots, y_n](t) = 0$.
Background :
- $f, g $ are differentiable functions of an open interval $I$ and $f, g$ both are real valued.
Let $W(f,g)(x) =\begin{vmatrix}f(x) &g(x) \\f'(x)&g'(x)\end{vmatrix}$ denote the Wronskian of $f, g$ at $x\in I$
- If two functions are solutions of a differential equation $y''+p(x) y'+q(x) y=0\tag 1$ on $I$ where $p, q\in C(I) $ then by Abel's identity we have
$$W(f, g) (x) =W(f, g) (x_o) e^{-\int_{x_0}^{x} p(t) dt}$$
Then $W(f, g) (x_0) \neq 0$ for some $x_0\in I$ implies $W(f, g) \neq 0$ on $I$
Moreover $W(f,g)$ different from zero with the same sign at every point ${\displaystyle x} \in {\displaystyle I}$
Back to the main problem:
$f, g$ are solution of $y''+p(x) y'+q(x) y=0$ on $I$ where $p, q\in C(I) $ then $f, g\in C^1(I) $ i.e $f, g$ both are continuously differentiable.
Then $W(f,g)(x) =\begin{vmatrix}f(x) &g(x) \\f'(x)&g'(x)\end{vmatrix}=f(x)g'(x)-f'(x)g(x) $
is continuous forall $x\in I$.
Hence $W(f, g) $ has Darboux property (or I.V.P) . Hence if $W(f, g) $ changes it's sign then $W(f, g)(x_0)=0$ for some $x_0\in I$.
Then by Abel's identity $W(f, g) =0$ on $I$.
Conclusion: If $f, g$ are two linearly independent solutions of $(1) $ then $W(f, g) $ can't change the sign on $I$.
Abel's identity can be generalized for $n$ solutions of a $n$ the order homogeneous linear equation. (See here).
Again $n$ solutions are $n$-th times differentiable.Hence $f_1, f_2, \ldots , f_n\in C^{n-1}(I) $.
Since $W(f_1, f_2, \ldots, f_n) $ are function of $f_1, f_2, \ldots, f_n$ and their derivatives of order upto $n-1$ , so $W(f_1, f_2, \ldots, f_n) $ is continuous on $I$ .
Now again by Darboux property we can conclude that if Wronskian changes sign then it must attains $0$.
Then by Generalized Abel's identity, it follows that $W(f_1, f_2, \ldots, f_n) =0$ on $I$.
Conclusion: If Wronskian of $n$ solutions of a $n$th order homogenous ode doesn't attin $0$ , it can't change the sign.
But more stronger result is true.
See
here
,here and
here.
Best Answer
This is a particular case of Theorem 1.50 in my Collected trivialities on algebra derivations (as of 18 October 2020), which says the following (up to different notations):
To recover your claim from this theorem, let $A$ be the ring of smooth functions, and set $\delta = \dfrac{d}{dx}$ and $a_i = y_i$ and $a = y$.
The proof of the theorem follows the same plan as my comment: The trick is to write the matrix \begin{align} \begin{pmatrix} \delta^0\left(aa_1\right) & \delta^1\left(aa_1\right) & \cdots & \delta^{n-1}\left(aa_1\right) \\ \delta^0\left(aa_2\right) & \delta^1\left(aa_2\right) & \cdots & \delta^{n-1}\left(aa_2\right) \\ \vdots & \vdots & \ddots & \vdots \\ \delta^0\left(aa_n\right) & \delta^1\left(aa_n\right) & \cdots & \delta^{n-1}\left(aa_n\right) \end{pmatrix} \end{align} (whose determinant is $W_\delta\left(aa_1, aa_2, \ldots, aa_n\right)$) as a matrix product $BC$, where $B$ is the matrix \begin{align} \begin{pmatrix} \delta^0\left(a_1\right) & \delta^1\left(a_1\right) & \cdots & \delta^{n-1}\left(a_1\right) \\ \delta^0\left(a_2\right) & \delta^1\left(a_2\right) & \cdots & \delta^{n-1}\left(a_2\right) \\ \vdots & \vdots & \ddots & \vdots \\ \delta^0\left(a_n\right) & \delta^1\left(a_n\right) & \cdots & \delta^{n-1}\left(a_n\right) \end{pmatrix} \end{align} (whose determinant is $W_\delta\left(a_1, a_2, \ldots, a_n\right)$) and where $C$ is a certain upper-triangular matrix whose diagonal entries are $a, a, \ldots, a$. More precisely, $C$ is the upper-triangular matrix whose $\left(i,j\right)$-th entry is $\dbinom{j-1}{i-1} \delta^{i-j}\left(a\right)$ whenever $j \geq i$.
The same argument applies in the more general case when the functions $y_i$ are merely $n-1$-times differentiable (as opposed to smooth), even though the derivation $\delta = \dfrac{d}{dx}$ no longer literally exists as a derivation on a ring.