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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. Sketch the graph of the solution.${\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbit{y}}{\mathbf{=}}{\mathbf{3}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{2}}{\mathbit{t}}{\mathbf{-}}{\mathbf{-}}{\mathbf{3}}\left(sin2t\right){\mathbit{u}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{-}}{\mathbf{2}}{\mathbit{\pi }}{\mathbf{\right)}}{\mathbf{;}}{\mathbit{y}}\left(0\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{\mathbit{y}}{\mathbf{\text{'}}}\left(0\right){\mathbf{=}}{\mathbf{-}}{\mathbf{2}}$

On solving the given initial value problem using the method of Laplace transforms, the solution is $y\left(t\right)=\mathrm{cos}t-\mathrm{sin}2t-\left[2\mathrm{sin}t-\mathrm{sin}2t\right]u\left(t-2\pi \right)$ and the corresponding graph is, 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 domainto a function of a complex variables.

## Step 2: Taking Laplace Transform of initial value Problem

Given initial value problem

$y\text{'}\text{'}+y=3\mathrm{sin}2t-3\left(\mathrm{sin}2t\right)u\left(t-2\pi \right)$

Where $y\left(0\right)=1$ and $y\text{'}\left(0\right)=-2$

Laplace Transform for the initial value problem

$\mathcal{L}y\text{'}\text{'}\left(s\right)+\mathcal{L}y\left(s\right)=\mathcal{L}\left[3\mathrm{sin}2t-3\mathrm{sin}2tu\left(t-2\pi \right)\right]\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+\mathcal{L}y\left(s\right)=\frac{3·2}{{s}^{2}+4}-\frac{3·2{e}^{-2\pi s}}{{s}^{2}+4}\phantom{\rule{0ex}{0ex}}$

${s}^{2}\mathcal{L}y\left(s\right)-s+2+\mathcal{L}y\left(s\right)=\frac{6}{{s}^{2}+4}-\frac{6{e}^{-2\pi s}}{{s}^{2}+4}\phantom{\rule{0ex}{0ex}}\left({s}^{2}+1\right)\mathcal{L}y\left(s\right)-\left(s-2\right)=\frac{6}{{s}^{2}+4}-\frac{6{e}^{-2\pi s}}{{s}^{2}+4}$

$\mathcal{L}y\left(s\right)=\frac{s-2}{{s}^{2}+1}+\frac{6}{\left({s}^{2}+4\right)\left({s}^{2}+1\right)}-\frac{6{e}^{-2\pi s}}{\left({s}^{2}+4\right)\left({s}^{2}+1\right)}$

## Step 3: By Partial function method

$\frac{1}{\left({s}^{2}+4\right)\left({s}^{2}+1\right)}=\frac{\frac{1}{3}}{{s}^{2}+1}-\frac{\frac{1}{3}}{{s}^{2}+4}$

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

## Step 4: Laplace transform function

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

Hence,$y\left(t\right)=\mathrm{cos}t-\mathrm{sin}2t-\left[2\mathrm{sin}t-\mathrm{sin}2t\right]u\left(t-2\pi \right)$

and the graph is  ### Want to see more solutions like these? 