# dimension of the v.e of the solutions of an homogeneous ODE

I was studying the book of ODE of Simmons, and he says that the Wronskian ($W$) is $0$ iff the solutions are LI. Even he present a proof, a counterexample that $W = 0$ does not implies LD is $$\left| x \right|, \ x$$ and he used that to prove the theorem, but since it is not well tested I can not follow, in addition to not really understand what he did.

Anyway, I want to know some proof of this fact, the fact that the solutions form a vector space (easy) of dimension $n$ (difficult to me T_T).

Thanks.

If someone knows some book to study ODE (for a beginner), I'll appreciate it.

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You have 8 more questions with no accepted answers. This fact may discourage people to answer you ( – Ilya Sep 25 '11 at 21:48
The pair $|x|,x$ is not a counterexample. Either your interval has $0$ in which case the wronskian doesn't exist there, or it doesn't include $0$ and then they are actually linearly dependent. However, the pair $x^2,x|x|$ both have continuous derivatives, their wronskian vanishes everywhere, and they are not linearly independent on intervals that include $0$. Generally, $W=0$ can only guarantee linear dependence if the functions are analytic. (I don't think the other direction requires analyticity, though.) – anon Sep 25 '11 at 23:01
What does "v.e" in the title mean? I don't see any pair of neighboring words in the question that even start with those letters. I suppose "LI" means linearly independent, but what is "LD"? – Henning Makholm Sep 25 '11 at 23:03
@Henning: Obviously LD means linearly *de*$\text{}$pendent. But I have no idea what v.e means either. It's possible OP mistyped "VS" for vector space (as 'e' and 's' are right next to each other on the keyboard). And I think OP's asking (1) how the mentioned proof could be valid given the stated 'counterexample' (which I addressed), (2) for help understanding the proof, and (3) for other ODE books for a beginner. – anon Sep 25 '11 at 23:15
@anon, of course "linearly dependent". It was not as easy to figure out because it is not clearly a predicate where it appears. – Henning Makholm Sep 25 '11 at 23:20