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