let $$y''+ p(x)y'+ q(x)y = 0$$
Given two different set initial conditions such that:
$ 1. ~~y'(x_0)= y_0~$ and $~y(x_0) = y_1$
$2.~~ y'(x_0)= y_2~$ and $~y(x_0) =y_3$
Let $y_1$ and $y_2$ corresponding solutions.
what Is wronskian of two solution, is $W(y_1,y_2)=0$?
If not then what is wronskian?
My guess is that since both have same solution set, they must be linearly dependent which implies $W=0$
By definition, $W(y_1,y_2)$ is the function $$W(y_1,y_2):x\mapsto \begin{vmatrix} y_1(x) & y_2(x) \\ y_1'(x) & y_2'(x) \end{vmatrix} = y_1(x)y_2'(x)-y_2(x)y_1'(x)$$ It is a well known result that $W(y_1,y_2)=0$ if and only if $y_1$ and $y_2$ are linearly dependent.
To see this, you compute $W'(y_1,y_2)(x)$, and find out that : $$W'(y_1,y_2)(x) = p(x)W(y_1,y_2)(x)$$ which means $W(y_1,y_2)$ is a solution of the first order equation $$u'=pu$$ and it is well known that a non null solution of an equation of this kind is never null.
So, to answer your question, saying that $W(y_1,y_2)=0$ is equivalent to saying $W(y_1,y_2)(x_0)=0$, which means $y_1$ and $y_2$ are two collinear solutions ($y_2=ky_1$ for a certain constant $k$). Another way of putting it is $(y_1,y_2)$ is a basis of the vectorial space of the solutions of the equation iff their wronskian is not $0$.