Answer
$W(x,x^2,\frac{1}{x})(t)=\frac{6}{x}$
Work Step by Step
we verify the given functions are the solutions of the differential equation by plugging them into it:
$y=x\;\;\;\rightarrow \;\;\;x^3({x})'''+x^2({x})''-2x(x)'+2(x)=\;0+0-2x+2x=0\\\\$
$y=x^2\;\;\;\rightarrow \;\;\;x^3({x^2})'''+x^2({x^2})''-2x(x^2)'+2(x^2)=\;0+2x^2-4x^2+2x^2=0\\\\$
$y=\frac{1}{x}\;\;\;\rightarrow \;\;\;x^3({\frac{1}{x}})'''+x^2({\frac{1}{x}})''-2x(\frac{1}{x})'+2(\frac{1}{x})=\;\frac{-6}{x}+\frac{2}{x}+\frac{2}{x}+\frac{2}{x}=0\\\\$
$W(x,x^2,\frac{1}{x})(t)=\begin{vmatrix}
x & x^2 & \frac{1}{x}\\
(x)' & ({x^{2}})' & (\frac{1}{x})'\\
(x)''&({e^{2}})'' & (\frac{1}{x})''
\end{vmatrix}\;\;=\;\;\begin{vmatrix}
x & x^2 & \frac{1}{x}\\
1 & 2x & \frac{-1}{x^2}\\
0 & 2 & \frac{2}{x^3}
\end{vmatrix}\;\;=$
$x.\begin{vmatrix}
2x & \frac{-1}{x^2}\\
2 & \frac{2}{x^3}
\end{vmatrix}\;\;-\;\begin{vmatrix}
x^2 & \frac{1}{x}\\
2 & \frac{2}{x^3}
\end{vmatrix}\;=\;$
$x.(\frac{4}{x^2}+\frac{2}{x^2})\;\;-\;(\frac{2}{x}-\frac{2}{x})\;=\;
\frac{6}{x}-0\;=\;\frac{6}{x}$
$W(x,x^2,\frac{1}{x})(t)=\frac{6}{x}$
