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Expert-verified Found in: Page 343 ### 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 interval ${\mathbf{\left(}}{\mathbf{0}}{\mathbf{,}}{\mathbf{\infty }}{\mathbf{\right)}}$ .(a) $\left\{{e}^{2x},{x}^{2}{e}^{2x},{e}^{-x}\right\}$(b) $\left\{{e}^{x}sin2x,x{e}^{x}sin2x,{e}^{x},x{e}^{x}\right\}$(c) $\left\{2{e}^{2x}-{e}^{x},{e}^{2x}+1,{e}^{2x}-3,{e}^{x}+1\right\}$

1. Given set of functions are Linearly Independent (LI).
2. Given set of functions are Linearly Independent (LI).
3. Given set of functions are Linearly Independent (LI).
See the step by step solution

## Step 1: Determine the given functions are linearly dependent or linearly independent

$<{e}^{2x},{x}^{2}{e}^{2x},{e}^{-x}>$

Assuming scalars ${c}_{1},{c}_{2},{c}_{3}$ such that

${c}_{1}{e}^{2x}+{c}_{2}{x}^{2}{e}^{2x}+{c}_{3}{e}^{-x}=0$

At

$x=0\phantom{\rule{0ex}{0ex}}{c}_{4}=0={c}_{2}\phantom{\rule{0ex}{0ex}}{c}_{1}+{c}_{3}=0\phantom{\rule{0ex}{0ex}}$

At $x\to \infty$

${c}_{1}\left(\infty \right)+{c}_{2}\left(\infty \right)=0\phantom{\rule{0ex}{0ex}}{c}_{1}={c}_{2}=0={c}_{3}\phantom{\rule{0ex}{0ex}}$

Hence

Given set of functions are Linearly Independent (LI).

## Step 2: Determine the given functions are linearly dependent or linearly independent

$<{e}^{x}\mathrm{sin}2x,x{e}^{x}\mathrm{sin}2x,{e}^{x},x{e}^{x}>$

Assuming scalars ${c}_{1},{c}_{2},{c}_{3},{c}_{4}$ such that

${c}_{1}{e}^{x}\mathrm{sin}2x+{c}_{2}x{e}^{x}\mathrm{sin}2x+{c}_{3}{e}^{x}+{c}_{4}x{e}^{x}=0$

At $x\to 0$

${c}_{4}=0={c}_{2}$

At x = 0

${c}_{3}=0\phantom{\rule{0ex}{0ex}}{c}_{1}={c}_{2}={c}_{3}={c}_{4}=0$

Hence

Given set of functions are Linearly Independent (LI).

## Step 3: Determine the given functions are linearly dependent or linearly independent

$<2{e}^{2x}-{e}^{x},{e}^{2x}+1,{e}^{2x}-3,{e}^{x}+1>$

Assuming scalars ${c}_{1},{c}_{2},{c}_{3},{c}_{4}$ such that

${c}_{1}\left(2{e}^{2x}-{e}^{x}\right)+{c}_{2}\left({e}^{2x}+1\right)+{c}_{3}\left({e}^{2x}-3\right)+{c}_{4}\left({e}^{x}+1\right)=0$

At $x\to -\infty$

${c}_{2}-2{c}_{3}+{c}_{4}=0$

At x = 0

${c}_{1}+2{c}_{2}-2{c}_{3}+2{c}_{4}=0$

At x = 1

${e}^{2}\left(2{c}_{1}+{c}_{2}+{c}_{3}\right)={c}_{1}-{c}_{1}$

At x = -1

$2{c}_{1}+{c}_{2}+{c}_{3}+{c}_{\left({c}_{1}}-{c}_{1}\right)=0$

From equation third and fourth we get

${c}_{2}=-{c}_{1}\phantom{\rule{0ex}{0ex}}{c}_{4}={c}_{1}\phantom{\rule{0ex}{0ex}}{c}_{3}=-{c}_{1}$

From equation second, we can get

${c}_{1}=0\phantom{\rule{0ex}{0ex}}{c}_{1}={c}_{2}={c}_{3}={c}_{4}=0$

Hence

Given set of functions are Linearly Independent (LI). ### Want to see more solutions like these? 