In Problems$\mathbf{8}\mathbf{}\mathbf{to}\mathbf{}\mathbf{15}$,use$\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{)}$to show that the given functions are linearly independent.

$\mathbf{x}\mathbf{,}{\mathbf{e}}^{\mathbf{x}}\mathbf{,}{\mathbf{xe}}^{\mathbf{x}}$

The functions ${\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{f}}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{f}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}$are linearly independent if the determinant

$\mathbf{W}\mathbf{=}\mathbf{\left|}\begin{array}{cccc}{\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& {\mathbf{f}}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{\cdots }& {\mathbf{f}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\\ {\mathbf{f}}_{\mathbf{1}}^{\mathbf{\prime }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& {\mathbf{f}}_{\mathbf{2}}^{\mathbf{\prime }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{\cdots }& {\mathbf{f}}_{\mathbf{n}}^{\mathbf{\prime }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\\ \mathbf{⋮}& \mathbf{⋮}& \mathbf{\ddots }& \mathbf{⋮}\\ {\mathbf{f}}_{\mathbf{1}}^{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& {\mathbf{f}}_{\mathbf{2}}^{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{\cdots }& {\mathbf{f}}_{\mathbf{n}}^{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\end{array}\mathbf{\right|}\mathbf{\ne }\mathbf{0}$

Here, W is called the Wronksian of function.

Find the derivatives of the function ${\mathrm{f}}_1\left(\mathrm{x}\right)=\mathrm{x}$of order $\mathrm{n}-1.$

$$\begin{gathered} {\mathrm{f}}_1\left(\mathrm{x}\right)=\mathrm{x} \\ {{\mathrm{f}}_2}^{\text{'}}\left(\mathrm{x}\right)=1 \\ {{\mathrm{f}}_3}^{"}\left(\mathrm{x}\right)=0 \end{gathered}$$

Find the derivatives of the function ${\mathrm{f}}_2\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$of order $\mathrm{n}-1.$

${\mathrm{f}}_2\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$

${\mathrm{f}}_2\text{'}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$

${\mathrm{f}}_2"\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}$

Find the derivatives of the function ${\mathrm{f}}_3\left(\mathrm{x}\right)={\mathrm{xe}}^{\mathrm{x}}$of order $\mathrm{n}-1.$

${\mathrm{f}}_3\left(\mathrm{x}\right)={\mathrm{xe}}^{\mathrm{x}}$

${{\mathrm{f}}_3}^{\text{'}}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})$

${{\mathrm{f}}_3}^{\text{'}}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})+{\mathrm{e}}^{\mathrm{x}}$

$={\mathrm{e}}^{\mathrm{x}}(2+\mathrm{x})$

Substitute the derivatives in the Wronksian formula and simplify as follows:

$\mathrm{W}=\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{e}}^{\mathrm{x}}& {\mathrm{xe}}^{\mathrm{x}}\\ 1& {\mathrm{e}}^{\mathrm{x}}& {\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})\\ 0& {\mathrm{e}}^{\mathrm{x}}& {\mathrm{e}}^{\mathrm{x}}(2+\mathrm{x})\end{array}\right|$

$=\mathrm{x}\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{x}}& {\mathrm{e}}^{\mathrm{x}}(1+\mathrm{x})\\ {\mathrm{e}}^{\mathrm{x}}& {\mathrm{e}}^{\mathrm{x}}(2+\mathrm{x})\end{array}\right|-\left|\begin{array}{cc}{\mathrm{e}}^{\mathrm{x}}& {\mathrm{xe}}^{\mathrm{x}}\\ {\mathrm{e}}^{\mathrm{x}}& {\mathrm{e}}^{\mathrm{x}}(2+\mathrm{x})\end{array}\right|$

$=\mathrm{x}\left[{\mathrm{e}}^{2\mathrm{x}}\right(2+\mathrm{x})-{\mathrm{e}}^{2\mathrm{x}}(1+\mathrm{x}\left)\right]-\left[{\mathrm{e}}^{2\mathrm{x}}\right(2+\mathrm{x})-{\mathrm{xe}}^{2\mathrm{x}}]$

$={\mathrm{xe}}^{2\mathrm{x}}-2{\mathrm{e}}^{2\mathrm{x}}$

Simplify the equation further.

$\mathrm{W}={\mathrm{e}}^{2\mathrm{x}}(\mathrm{x}-2)$

Here, $\mathrm{W}\ne 0$, therefore,$\mathrm{x},{\mathrm{e}}^{\mathrm{x}}\mathrm{and}{\mathrm{xe}}^{\mathrm{x}}$are three linearly independent functions.