Wronskian of $x^3$ and $x^2 |x|$ The question is fairly straight forward, given $2$ functions $y_1=x^3$ and $y_2=x^2|x|$ determine whether the functions are linearly dependent or independent.
I am conflicted because in my differential equations textbook it states that two functions are linearly independent if their quotient is a non-constant function for all $x$. Here it is evident that for $x<0$ the quotient $\frac{y1}{y2} = 1$ while for $x<0$, $\frac{y1}{y2}=-1$. This would lead to a conclusion that the functions are linearly independent.
However, in the textbook it is also stated that two linearly dependent functions will have a Wronskian of zero on the real number line. 
So the Wronskian $W(y_1(x), y_2(x)) = \det \begin{vmatrix}y_1(x)& y_2(x)\\ y_1'(x)& y_2'(x) \end{vmatrix}$ and $y_1'(x) = 3x^2$
the derivative of $y_2'(x)$ is slightly more tricky but checking with wolfram alpha
$y_2'(x)=\frac{3x^3}{|x|}$
so the Wronskian is given by:
$$W=y_1(x)y_2'(x)-y_2(x)y_1'(x)=x^3\times\frac{3x^3}{|x|}-x^2|x|\times3x^2$$
Now multiplying the second term by $\frac{|x|}{|x|}$ in order to add the terms yields
$$W=\frac{3x^6}{|x|}-3x^4\times \frac{|x|^2}{|x|}$$
At this point I assume $|x|^2=x^2$
which yields:
$$W=\frac{3x^6}{|x|}-\frac{3x^6}{|x|}=0 $$
This would imply the functions are linearly dependent....
The answer provided by the textbook states that the functions are linearly independent. Does anyone have some words of wisdom for me :)
Cheers
 A: The theorem says if the functions are linearly dependent, then the Wronskian is $0$.  It does not state that if the Wronskian is $0$, the functions are linearly dependent.  These functions  are an example 
that shows this.  Their Wronskian is $0$, and they would be linearly dependent if you just looked at the interval $(-\infty, 0)$ or $(0, \infty)$, but they are not linearly dependent on the whole real line
because neither is a constant multiple of the other.
The point (which will probably be made later) is that solutions of homogeneous second-order linear differential equations are rather special: if they are linearly dependent in some interval, this same linear dependence holds over the whole interval where the differential equation is defined; moreover, the value of the Wronskian at one point tells the whole story.
A: Since |x| is not a differentiable function, you get a weird result. If we consider the definition of linear dependency:
$$c_1x^3+c_2 x^2|x|=0$$
It is clear that no $c_1,c_2$ satisfies this expression so the functions are linearly independent.
A: If I consider two functions $x^3$ and $x^2|x|$ in the interval $[-1,1]$, then their wronskian vanishes identically, but they are linearly independent in that interval. What is the proper reason behind this? I saw one said that two functions are linearly dependent implies their wronskian vanishes identically, but the converse is not true...but I want to ask the reference book of this statement because it violates all the standard books like Ross etc..
A: If the Wronskian is not equal to zero for some x in I, then the functions are linearly independent on the interval. Also if the functions are linearly dependent on I, then the  Wronskian is equal to zero for all x in the interval I. But W=0 does not mean the functions are dependent.
