Q11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

$\left\{x^{-1}, x^{\frac{1}{2}}, x\right\}$ on $\left(0, \infty \right)$

Step-by-Step Solution

Answer

Therefore, $\left\{{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{,} {\mathbf{x}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,} \mathbf{x}\right\}$ are linearly independent on $\left(0,\infty \right)$.

Step 1: Using the concept of Wronskian

The given function is $\left\{{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{,} {\mathbf{x}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,} \mathbf{x}\right\}.$

Apply the concept of Wronskian,

 

$W\left[f_1,f_2,\dots ,f_n\right]=\left|\begin{array}{cccc}{f}_1\left(x\right)& f_2\left(x\right)& \dots & f_n\left(x\right)\\ f_1\text{'}\left(x\right)& f_2\text{'}\left(x\right)& \dots & f_n\text{'}\left(x\right)\\ ⋮& ⋮& & ⋮\\ {f_1}^{n-1}\left(x\right)& {f_2}^{n-1}\left(x\right)& \cdots & {f_n}^{n-1}\left(x\right)\end{array}\right|$

Therefore,

$W\left[x^{-1},x^{\frac{1}{2}},x\right]=\left|\begin{array}{ccc}{x}^{-1}& x^{\frac{1}{2}}& x\\ -x^{-2}& \frac{1}{2}{x}^{-\frac{1}{2}}& 1\\ 2x^{-3}& -\frac{1}{4}{x}^{-\frac{3}{2}}& 0\end{array}\right|$

 

Solve the above equation,

$$\begin{gathered} \mathbf{W}\left[{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{,} {\mathbf{x}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,} \mathbf{x}\right]\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{0}\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}{\mathbf{x}}^{\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}}\right)\mathbf{-}{\mathbf{x}}^{\frac{\mathbf{1}}{\mathbf{2}}}\left(\mathbf{0}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{-}\mathbf{3}}\right)\mathbf{+}\mathbf{x}\left(\frac{\mathbf{1}}{\mathbf{4}}{\mathbf{x}}^{\mathbf{-}\mathbf{2}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}}\mathbf{-}{\mathbf{x}}^{\mathbf{-}\mathbf{3}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\right) \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\left({\mathbf{x}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}}\right)\mathbf{+}\left(\mathbf{2}{\mathbf{x}}^{\frac{\mathbf{-}\mathbf{5}}{\mathbf{2}}}\right)\mathbf{+}\left(\frac{\mathbf{1}}{\mathbf{4}}{\mathbf{x}}^{\mathbf{1}\mathbf{-}\frac{\mathbf{7}}{\mathbf{2}}}\mathbf{-}{\mathbf{x}}^{\mathbf{1}\mathbf{-}\frac{\mathbf{7}}{\mathbf{8}}}\right) \\ \mathbf{=}\frac{\mathbf{9}}{\mathbf{4}}\left({\mathbf{x}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}}\right)\mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}{\mathbf{x}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}} \\ \mathbf{=}\frac{\mathbf{6}}{\mathbf{4}}\left({\mathbf{x}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}}\right) \\ \mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\left({\mathbf{x}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}}\right) \end{gathered}$$

Step 2: Check the linearly independent or dependent

The above function is not equal to zero $\left(\forall \mathbf{x}\right)$

 

Therefore, $\left\{{\mathbf{x}}^{\mathbf{-}\mathbf{1}}\mathbf{,} {\mathbf{x}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{,} \mathbf{x}\right\}$ are linearly independent on $\left(\mathbf{0}\mathbf{,} \infty \right)$.