Answer
$W(1,x,x^3)(t)=6x$
Work Step by Step
we verify the given functions are the solutions of the differential equation by plugging them into it:
$y=1\;\;\;\rightarrow \;\;\;x({1})'''-({1})''=\;0+0=0\\\\$
$y=x\;\;\;\rightarrow \;\;\;x({x})'''-({x})''=\;0+0=0\\\\$
$y=x^3\;\;\;\rightarrow \;\;\;x({x^3})'''-({x^3})''=\;6x-6x=0\\\\$
$W(1,x,x^3)(t)=\begin{vmatrix}
1 & x & x^3 \\
(1)' &(x)' & ({x^{3}})' \\
(1)'' & (x)'' &({x^{3}})''
\end{vmatrix}\;\;=\;\;\begin{vmatrix}
1 & x & x^3 \\
0 & 1 & 3x^2\\
0 & 0 & 6x
\end{vmatrix}\;\;=$
$1.\begin{vmatrix}
1 & 3x^2\\
0 & 6x
\end{vmatrix}\;\;=
1.(6x)\;=\;6x$
$W(1,x,x^3)(t)=6x$
