# Wronskian proof and linear dependency.

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 $$~I~$$.

• Do you know what it would mean for $x^3$ and $|x|^3$ to be linearly dependent? Do you know the definition of the Wronskian? What have you tried, and what's giving you trouble?
– user61527
Commented Oct 20, 2013 at 1:16
• ummm...well when the Wronskian is 0, it just vanishes and the results become inconclusive. Also I have taken the first and second derivatives of x^3 and the absolute value of x^3 and had a determinant of 0. Commented Oct 20, 2013 at 1:22
• Seriously guys...this question is insanely old and I'm getting downvoted. lol! hysterical rolls eyes Commented Feb 8, 2014 at 9:57

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?
• See, you want to consider each part of the function separately. How would you write (or graph) the function on the interval $[-1,0]$? Here's a hint: for the interval $[0,1]$, $|x|^3 \equiv x^3.$ So consider each sub-interval separately and find your constants (there should be two). Are they the same over both sub-intervals?