Discussion problem, wronskian matrix, linear independence of solution

1. Jul 24, 2012

Edit: I think I may have posted this in the wrong section, sorry about that. Note that this isn't a homework problem though, I"m not enrolled in this class, I was just reading over some of this stuff and trying some problems since I"m majoring in physics.

I have a text book "discussion" problem that's stumping me, I'm given that these are two linearly independent solutions:

$$y_1=x^3$$ and $$y_2=\mid x\mid ^3$$

to the homogeneous differential equation $$x^2y''-4xy'+6y=0$$

on $$(-\infty,\infty)$$

I'm asked to show that the wronskian matrix for the solutions is equal to zero for every real number x. This is easy enough to do, I won't show that work here... but then I"m asked whether this violates the theorem for the wronskian matrix test, which is stated in my text:

If $$y_1,y_2...y_n$$ are n solutions of a homogenous linear nth-order differential equation on an interval I. Then the set of solutions is linearly independent on I if and if $$w(y_1,y_2....y_n)≠0$$ for every x in the interval.

Naturally I would say yes, it is a violation, but they seem to be implying that it isn't a violation (i can just tell by the way they're asking...). They're messing with me. Why is it not a violation?

2. Jul 24, 2012

LCKurtz

If you look at the theorem where it is proven that the Wronskian is either never zero or else identically zero, you will find that one of the hypotheses is that the leading coefficient of the DE is nonzero. If you look at the proof, you will see why. This equation fails at $x=0$ and the result may fail on any interval containing zero. Notice, however, on any interval not containing zero it all works.

3. Jul 25, 2012

thanks, wouldn't that usually be have stated in the theorem though? usually with theorems in textbooks I don't find there is anything crucial missing. You think the theorem would state that it should be a linear homogeneous equation with $$a_n(x)≠0$$ for x in I....or something like that, if I'm understanding correctly. My text doesn't provide a proof so I'll check that out online somewhere.
I think you will find that $a_n(x)\ne 0$ is explicitly stated or the leading coefficient is $1$. In that case the hypothesis is inherent. You can always divide both sides of the DE by $a_n(x)$ if it is nonzero to get a leading coefficient of $1$.