• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q Review Problems-8E

Expert-verified Found in: Page 344 ### 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 # Use the annihilator method to determine the form of a particular solution for the given equation.(a) ${{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{6}}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{+}}{\mathbf{5}}{\mathbit{y}}{\mathbf{=}}{{\mathbit{e}}}^{\mathbf{-}\mathbf{x}}{\mathbf{+}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{1}}$(b) ${{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{19}}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{-}}{\mathbf{20}}{\mathbit{y}}{\mathbf{=}}{\mathbit{x}}{{\mathbit{e}}}^{\mathbf{-}\mathbf{x}}$(c) ${{\mathbit{y}}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}{\mathbf{+}}{\mathbf{6}}{{\mathbit{y}}}^{\mathbf{\text{'}}\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{9}}{\mathbit{y}}{\mathbf{=}}{{\mathbit{x}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{3}}{\mathbit{x}}$(d) ${\mathbit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathbit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathbit{y}\mathbf{=}\mathbit{x}\mathbit{s}\mathbit{i}\mathbit{n}\mathbit{x}$

1. ${y}_{p}={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+{C}_{3}x{e}^{x}$
2. ${y}_{p}={C}_{4}x{e}^{-x}$
3. $y={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+{C}_{3}\mathrm{cos}\sqrt{3}x+{C}_{4}\mathrm{sin}\sqrt{3}x+{C}_{5}x\mathrm{cos}\sqrt{3}x+{C}_{6}x\mathrm{sin}\sqrt{3}x+{C}_{7}\mathrm{cos}3x+{C}_{8}\mathrm{sin}3x$
4. ${y}_{p}={C}_{3}\mathrm{cos}x+{C}_{4}\mathrm{sin}x+{C}_{5}x\mathrm{cos}x+{C}_{6}x\mathrm{sin}x$
See the step by step solution

## Step 1: Determining the form of a particular solution for the given equation

${y}^{\text{'}\text{'}}+6{y}^{\text{'}}+5y={e}^{-x}+{x}^{2}-1\phantom{\rule{0ex}{0ex}}{e}^{-x}\to D+1\phantom{\rule{0ex}{0ex}}{x}^{2}-1\to {D}^{3}\phantom{\rule{0ex}{0ex}}\left({D}^{2}+6D+5\right)\left(D+1\right){D}^{3}=0\phantom{\rule{0ex}{0ex}}\left(D+1{\right)}^{2}\left(D+5\right){D}^{3}=0\phantom{\rule{0ex}{0ex}}D=0,0,0,-5,-1,-1\phantom{\rule{0ex}{0ex}}y={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+{C}_{3}{e}^{-5x}+{C}_{4}{e}^{-x}+{C}_{5}x{e}^{-x}\phantom{\rule{0ex}{0ex}}$

Since - 1 and - 5 are homogeneous solutions

${y}_{p}={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+{C}_{3}x{e}^{x}$

Hence,

${y}_{p}={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+{C}_{3}x{e}^{x}$

## Step 2: Determining the form of a particular solution for the given equation

${y}^{\text{'}\text{'}\text{'}}+2{y}^{\text{'}\text{'}}-19{y}^{\text{'}}-20y=x{e}^{-x}\phantom{\rule{0ex}{0ex}}{e}^{-x}\to D+1\phantom{\rule{0ex}{0ex}}x{e}^{-x}\to \left(D+1{\right)}^{2}\phantom{\rule{0ex}{0ex}}\left({D}^{3}+2{D}^{2}-19D-20\right)\left(D+1{\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\left(D+1\right)\left({D}^{2}+D-20\right)\left(D+1{\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\left(D-4\right)\left(D+5\right)\left(D+1{\right)}^{3}=0\phantom{\rule{0ex}{0ex}}D=4,-5,-1,-1\phantom{\rule{0ex}{0ex}}y={C}_{1}{e}^{4x}+{C}_{2}{e}^{5x}+{C}_{3}{e}^{-x}+{C}_{4}x{e}^{-x}\phantom{\rule{0ex}{0ex}}{y}_{p}={C}_{4}x{e}^{-x}\phantom{\rule{0ex}{0ex}}$

Hence,

${y}_{p}={C}_{4}x{e}^{-x}$

## Step 3: Determining the form of a particular solution for the given equation

${y}^{4}+6{y}^{\text{'}\text{'}}+9y={x}^{2}-\mathrm{sin}3x\phantom{\rule{0ex}{0ex}}{x}^{2}\to {D}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{sin}3x\to {D}^{2}+9\phantom{\rule{0ex}{0ex}}\left({D}^{4}+6{D}^{2}+9\right){D}^{3}\left({D}^{2}+9\right)=0\phantom{\rule{0ex}{0ex}}{D}^{3}{\left({D}^{2}+3\right)}^{2}\left({D}^{2}+9\right)=0\phantom{\rule{0ex}{0ex}}\to D=0,0,0,±\sqrt{3i},±\sqrt{3i},±3i\phantom{\rule{0ex}{0ex}}y={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+{C}_{3}\mathrm{cos}\sqrt{3}x+{C}_{4}\mathrm{sin}\sqrt{3}x\phantom{\rule{0ex}{0ex}}+{C}_{5}x\mathrm{cos}\sqrt{3}x+{C}_{6}x\mathrm{sin}\sqrt{3}x+{C}_{7}\mathrm{cos}3x+{C}_{8}\mathrm{sin}3x\phantom{\rule{0ex}{0ex}}$

Hence,

$y={C}_{0}+{C}_{1}x+{C}_{2}{x}^{2}+{C}_{3}\mathrm{cos}\sqrt{3}x+{C}_{4}\mathrm{sin}\sqrt{3}x+{C}_{5}x\mathrm{cos}\sqrt{3}x+{C}_{6}x\mathrm{sin}\sqrt{3}x+{C}_{7}\mathrm{cos}3x+{C}_{8}\mathrm{sin}3x$

## Step 4: Determining the form of a particular solution for the given equation

${y}^{\text{'}\text{'}\text{'}}-{y}^{\text{'}\text{'}}+2y=x\mathrm{sin}x$

the corresponding differential equation

role="math" localid="1663978603548" $\mathrm{sin}x\to {D}^{2}+1\phantom{\rule{0ex}{0ex}}x\mathrm{sin}x\to {\left({D}^{2}+1\right)}^{2}\phantom{\rule{0ex}{0ex}}$

$\left({D}^{3}-{D}^{2}+2\right){\left({D}^{2}+1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\left(D+1\right)\left({D}^{2}-2D+2\right){\left({D}^{2}+1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\left[{\left(D-1\right)}^{2}+1\right]\left(D+1\right){\left({D}^{2}+1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\to D=1±i,-1,±i,±i\phantom{\rule{0ex}{0ex}}y=\left[{C}_{0}\mathrm{cos}x+{C}_{1}\mathrm{sin}x\right]{e}^{x}+{C}_{2}{e}^{-x}+{C}_{3}\mathrm{cos}x+{C}_{4}\mathrm{sin}x\phantom{\rule{0ex}{0ex}}+{C}_{5}x\mathrm{cos}x+{C}_{6}x\mathrm{sin}x\phantom{\rule{0ex}{0ex}}{y}_{p}={C}_{3}\mathrm{cos}x+{C}_{4}\mathrm{sin}x+{C}_{5}x\mathrm{cos}x+{C}_{6}x\mathrm{sin}x\phantom{\rule{0ex}{0ex}}$

Hence,

${y}_{p}={C}_{3}\mathrm{cos}x+{C}_{4}\mathrm{sin}x+{C}_{5}x\mathrm{cos}x+{C}_{6}x\mathrm{sin}x$ ### Want to see more solutions like these? 