If the Wronskian of two arbitrary functions is zero, are they linearly dependent only if they are not zero at some point?

Question

Given that the functions are not the solutions of the same linear differential equation, I know that $W = 0$ doesn't mean they are linearly dependent, for example $x^3$ and $ |x|^3$ are linearly independent but their Wronskian is zero.

On the other hand I saw this proof:

$$ W(y_1, y_2) = y_1 y_2' - y_1'y_2 = 0$$

$$ \frac{y_1'}{y_1} = \frac{y_2'}{y_2} $$

$$\int\frac{y_1'}{y_1} = \int\frac{y_2'}{y_2} $$

$$ ln(y_1) = ln(y_2) + C $$

$$ y_1 = Ay_2 $$

Is this proof valid and $W=0$ means LD only if you can divide by the functions?

Answer 1

For the question in the title, the general answer is $\color{black}{NO}$.
In the field $\mathbb Z_2$, the polynomials $p_1(x)=x^4+x^2+1$ and $p_2(x)=1$ do not have any zero and the Wronskian determinant
$W(p_1(x),p_2(x))=\begin{vmatrix}x^4+x^2+1&1\\4x^3+2x&0 \end{vmatrix}=\begin{vmatrix}x^4+x^2+1&1\\0&0 \end{vmatrix}=0$
even then the polynomials $p_1(x)$ and $p_2(x)$ are linearly independent.