Show that $~x^3~$ and $~|x|^3~$ are not linearly dependent on $~[-1,1]~$, but that Wronskian,$$W(x^3, |x|^3) = 0~.$$ This shows that the converse of Theorem $5$ is false.
Are $~x^3~$ and $~|x|^3~$ solutions on $~[-1,1]~$ on any second order homogeneous linear equation?
Any third order homogeneous linear equation?
Theorem 5 : If $~u_1,\cdots,u_k~$ are any $~(k-1)~$ times differentiable functions which are linearly dependent on $~I~$, then Wronskian, $~W(u_1(x),\cdots,u_k(x) = 0~$ on $~I~$.
Consider what $|x|^3$ really is: a piecewise function. You can split it up around zero. Are there a single set of scalars you can multiply by these two functions over $[-1,1]$ that shows they are linearly dependent?
Recall what it means to be linearly (in)dependent. I gave you a hint above.