I am trying to determine if $u_1(x)=x^3, \ u_2(x)=|x|^3$ is linearly independent/dependent on $\mathbb{R}$.
I first computed the Wronskian using two cases.
Case $1$ $(x\geq 0)$: $$W(x)=\begin{vmatrix} x^3 & x^3 \\ 3x^2 & 3x^2 \\ \notag \end{vmatrix}=0$$ Case $2$ $(x< 0)$: $$W(x)=\begin{vmatrix} x^3 & -x^3 \\ 3x^2 & -3x^2 \\ \notag \end{vmatrix}=0$$
I then used the definition of linear independence $(c_1,c_2\in\mathbb{R})$: $$(x\ge 0): \ \ c_1x^3+c_2x^3=0 \ \ \ \ \ (1) \\ (x<0): \ \ c_1x^3-c_2x^3=0 \ \ \ \ \ (2)$$ $$(1)+(2)=2c_1x^3=0\implies c_1=0 \\ \therefore c_2x^3=0\implies c_2=0$$ So since $c_1=c_2=0$, both functions should be linearly independent.
What I don't understand is why can both functions be linearly constucted from the other, doesn't this show linear dependence?
e.g. if $$x<0\implies |x|^3=-x^3 \\ \therefore -u_1(x)=u_2(x)$$ e.g. $$u_1(x)=\sin(2x), \ u_2(x)=\sin(x)\cos(x)$$ are linearly dependent as they can be written as a linear combination of each other. Linearly independent functions can not be written as a linear combination of said functions.
The Theorem about Wronskian and linear independent is a statement about functions which are solutions to linear differential equations not arbitrary functions such as $x^3$ and $|x|^3$
What you have discovered is that these two functions are not solutions to a lineare differential equation on the whole real line.