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Found in: Page 326

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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 $\left(0, \infty \right)$

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

See the step by step solution

## 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,

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

Solve the above equation,

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

## 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 $\left(\mathbf{0}\mathbf{,} \infty \right)$.