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Expert-verified 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.$\left\{\mathrm{cos}2x, {\mathrm{cos}}^{2}x, {\mathrm{sin}}^{2}x\right\}$ on $\left(-\infty , \infty \right)$

Therefore, the function $\left\{\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{,} {\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{,} {\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right\}$ is linearly dependent on $\left(\mathbf{-}\infty \mathbf{,} \infty \right)$.

See the step by step solution

## Step 1: Using the concept of Wronskian

The given function is $\left\{\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{,} {\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{,} {\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right\}.$

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,

$W\left[\mathrm{cos}2x,{\mathrm{cos}}^{2}x,{\mathrm{sin}}^{2}x\right]=W\left[\mathrm{cos}2x,\frac{1+\mathrm{cos}2x}{2},\frac{1-\mathrm{cos}2x}{2}\right]\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{ccc}\mathrm{cos}2x& \frac{1+\mathrm{cos}2x}{2}& \frac{1-\mathrm{cos}2x}{2}\\ -2\mathrm{sin}2x& -\mathrm{sin}2x& \mathrm{sin}2x\\ -4\mathrm{cos}2x& -2\mathrm{cos}2x& 2\mathrm{cos}2x\end{array}\right|$

Solve the above equation,

$\mathbf{W}\left[\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{,} {\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{,} {\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right]\mathbf{=}\mathbf{cos}\mathbf{2}\mathbf{x}\left(\mathbf{-}\mathbf{2}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\right)\phantom{\rule{0ex}{0ex}}\mathbf{-}\frac{\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{-}\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\right)\phantom{\rule{0ex}{0ex}}\mathbf{+}\frac{\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{4}\mathbf{sin}\mathbf{2}\mathbf{xcos}\mathbf{2}\mathbf{x}\right)\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{cos}\mathbf{2}\mathbf{x}\left(\mathbf{0}\right)\mathbf{-}\frac{\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{0}\right)\mathbf{+}\frac{\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{x}}{\mathbf{2}}\left(\mathbf{0}\right)\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{0}$

## Step 2:Check the linearly independent or dependent

The above function is equal to zero $\left(\forall \mathbf{x}\right).$

Therefore, $\left\{\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{,} {\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{,} {\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\right\}$ is linearly dependent on $\left(\mathbf{-}\infty \mathbf{,} \infty \right)$. ### Want to see more solutions like these? 