Does the Wronskian of three or more linearly independent functions change its sign? If $y_1, y_2$ are two Linearly independent solutions of a differential equation of order $2$ then we know that if the Wronskian is not zero then it never changes its sign (using the Abel's identity).

How far is this true for a third order linear differential equation having $y_1, y_2, y_3$ as linearly independent solutions?

I tried finding a solution but i dont think i have an answer. How far is this true in a general case that if $W(y_1,\dots,y_n)$ if non zero then it never changes its sign?
 A: 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.
