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Expert-verified Found in: Page 391 ### 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 # solve the given initial value problem using the method of Laplace transforms.${\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbit{y}}{\mathbf{=}}{\mathbit{u}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{-}}{\mathbf{2}}{\mathbf{\pi }}{\mathbf{\right)}}{\mathbf{-}}{\mathbf{-}}{\mathbit{u}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{-}}{\mathbf{4}}{\mathbf{\pi }}{\mathbf{\right)}}{\mathbf{;}}{\mathbit{y}}\left(0\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{\mathbit{y}}{\mathbf{\text{'}}}\left(0\right){\mathbf{=}}{\mathbf{1}}$

On solving the given initial value problem using the method of Laplace transforms,the solution is

$y\left(t\right)={e}^{-t}\mathrm{cos}t+2{e}^{-t}\mathrm{sin}t+\frac{1}{2}\left[1-{e}^{-t+2\pi }\left[\mathrm{cos}t+\mathrm{sin}t\right]\right]u\left(t-2\pi \right)-\frac{1}{2}\left[1-{e}^{-t+4\pi }\left[\mathrm{cos}t+\mathrm{sin}t\right]\right]u\left(t-4\pi \right)$

See the step by step solution

## Step 1: Definition

The Laplace transform, is an integral transform that converts a function of a real variable usually t, the time domain to a function of a complex variable s.

## Step 2: Laplace transform function

Given initial value problem,

$y\text{'}\text{'}+2y\text{'}+2y=u\left(t-2\pi \right)-u\left(t-4\pi \right)$

Where $y\left(0\right)=1$ and $y\text{'}\left(0\right)=1$

Taking Laplace transform for the initial value problem

$\mathcal{L}y\text{'}\text{'}\left(s\right)+2\mathcal{L}y\text{'}\left(s\right)+2\mathcal{L}y\left(s\right)=\mathcal{L}\left[u\left(t-2\pi \right)-u\left(t-4\pi \right)\right]\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+2s\mathcal{L}y\left(s\right)-2y\left(0\right)+2\mathcal{L}y\left(s\right)=\frac{{e}^{-2\pi s}}{s}-\frac{{e}^{-4\pi s}}{s}\phantom{\rule{0ex}{0ex}}$

${s}^{2}\mathcal{L}y\left(s\right)-s-1+2s\mathcal{L}y\left(s\right)-2+2\mathcal{L}y\left(s\right)=\frac{{e}^{-2\pi s}}{s}-\frac{{e}^{-4\pi s}}{s}\phantom{\rule{0ex}{0ex}}\left({s}^{2}+2s+2\right)\mathcal{L}y\left(s\right)-\left(s+3\right)=\frac{{e}^{-2\pi s}}{s}-\frac{{e}^{-4\pi s}}{s}$

$\mathcal{L}y\left(s\right)=\frac{s+3}{\left({s}^{2}+2s+2\right)}+\frac{{e}^{-2\pi s}}{s\left({s}^{2}+2s+2\right)}-\frac{{e}^{-4\pi s}}{s\left({s}^{2}+2s+2\right)}\phantom{\rule{0ex}{0ex}}=\frac{s+1+2}{{\left(s+1\right)}^{2}+1}+\frac{{e}^{-2\pi s}}{s\left({s}^{2}+2s+2\right)}-\frac{{e}^{-4\pi s}}{s\left({s}^{2}+2s+2\right)}$

## Step 3: By partial function method

$\frac{1}{s\left({s}^{2}+2s+2\right)}=\frac{\frac{1}{2}}{s}-\frac{\frac{1}{2}s+1}{{s}^{2}+2s+2}$

$\mathcal{L}y\left(s\right)=\frac{s+1}{{\left(s+1\right)}^{2}+1}+\frac{2}{{\left(s+1\right)}^{2}+1}+\frac{\frac{{e}^{-2\pi s}}{2}}{s}-\frac{\frac{{e}^{-2\pi s}}{2}s+{e}^{-2\pi s}}{{s}^{2}+2s+2}-\frac{\frac{{e}^{-4\pi s}}{2}}{s}+\frac{\frac{{e}^{-4\pi s}}{2}s+{e}^{-4\pi s}}{{s}^{2}+2s+2}$

## Step 4: Taking inverse Laplace transform

$y\left(t\right)={e}^{-t}\mathrm{cos}t+2{e}^{-t}\mathrm{sin}t+\frac{1}{2}u\left(t-2\pi \right)-\frac{1}{2}{e}^{-t+2\pi }\mathrm{cos}\left(t-2\pi \right)u\left(t-2\pi \right)\phantom{\rule{0ex}{0ex}}+\frac{1}{2}{e}^{-t+2\pi }\mathrm{sin}\left(t-2\pi \right)u\left(t-2\pi \right)-{e}^{-t+2\pi }\mathrm{sin}\left(t-2\pi \right)u\left(t-2\pi \right)\phantom{\rule{0ex}{0ex}}-\frac{1}{2}u\left(t-4\pi \right)+\frac{1}{2}{e}^{-t+4\pi }\mathrm{cos}\left(t-4\pi \right)u\left(t-4\pi \right)\phantom{\rule{0ex}{0ex}}-\frac{1}{2}{e}^{-t+4\pi }\mathrm{sin}\left(t-4\pi \right)u\left(t-4\pi \right)+{e}^{-t+4\pi }\mathrm{sin}\left(t-4\pi \right)u\left(t-4\pi \right)$

$y\left(t\right)={e}^{-t}\mathrm{cos}t+2{e}^{-t}\mathrm{sin}t+\frac{1}{2}u\left(t-2\pi \right)-\frac{1}{2}{e}^{-t+2\pi }\mathrm{cos}tu\left(t-2\pi \right)-\frac{1}{2}{e}^{-t+2\pi }\mathrm{sin}tu\left(t-2\pi \right)\phantom{\rule{0ex}{0ex}}-\frac{1}{2}u\left(t-4\pi \right)+\frac{1}{2}{e}^{-t+4\pi }\mathrm{cos}tu\left(t-4\pi \right)+\frac{1}{2}{e}^{-t+4\pi }\mathrm{sin}tu\left(t-4\pi \right)\phantom{\rule{0ex}{0ex}}={e}^{-t}\mathrm{cos}t+2{e}^{-t}\mathrm{sin}t+\frac{1}{2}\left[1-{e}^{-t+2\pi }\left[\mathrm{cos}t+\mathrm{sin}t\right]\right]u\left(t-2\pi \right)-\frac{1}{2}\left[1-{e}^{-t+4\pi }\left[\mathrm{cos}t+\mathrm{sin}t\right]\right]u\left(t-4\pi \right)$

hence

$y\left(t\right)={e}^{-t}\mathrm{cos}t+2{e}^{-t}\mathrm{sin}t+\frac{1}{2}\left[1-{e}^{-t+2\pi }\left[\mathrm{cos}t+\mathrm{sin}t\right]\right]u\left(t-2\pi \right)-\frac{1}{2}\left[1-{e}^{-t+4\pi }\left[\mathrm{cos}t+\mathrm{sin}t\right]\right]u\left(t-4\pi \right)$ ### Want to see more solutions like these? 