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### 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.${\mathbit{w}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbit{w}}{\mathbf{=}}{\mathbit{u}}\mathbf{\left(}\mathbf{t}\mathbf{-}\mathbf{-}\mathbf{2}\mathbf{\right)}{\mathbf{-}}{\mathbf{-}}{\mathbit{u}}\mathbf{\left(}\mathbf{t}\mathbf{-}\mathbf{-}\mathbf{4}\mathbf{\right)}{\mathbf{;}}{\mathbit{w}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{\mathbit{w}}{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{0}}$

On solving the given initial value problem using the method of Laplace transforms,the solution is$w\left(t\right)=\mathrm{cos}t+\left[1-\mathrm{cos}\left(t-2\right)\right]u\left(t-2\right)-\left[1-\mathrm{cos}\left(t-4\right)\right]u\left(t-4\right)$ 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 domain to a function of a complex variable s.

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

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

Where $w\left(0\right)=1andw\text{'}\left(0\right)=0$

$\mathcal{L}w\text{'}\text{'}\left(s\right)+\mathcal{L}w\left(s\right)=\mathcal{L}\left[u\left(t-2\right)-u\left(t-4\right)\right]\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}w\left(s\right)-s\mathcal{L}w\left(0\right)-w\text{'}\left(0\right)+\mathcal{L}w\left(s\right)=\frac{{e}^{-2s}}{s}-\frac{{e}^{-48}}{s}\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}w\left(s\right)-s-0+\mathcal{L}w\left(s\right)=\frac{{e}^{-2s}}{s}-\frac{{e}^{-48}}{s}\phantom{\rule{0ex}{0ex}}$

$\left({s}^{2}+1\right)\mathcal{L}y\left(s\right)-s=\frac{{e}^{-28}}{s}-\frac{{e}^{-18}}{s}\phantom{\rule{0ex}{0ex}}\mathcal{L}w\left(s\right)=\frac{s}{{s}^{2}+1}+\frac{{e}^{-2s}}{s\left({s}^{2}+1\right)}-\frac{{e}^{4s}}{s\left({s}^{2}+1\right)}$

## Step 3: By Partial fraction

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

Equation 1 becomes,

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

## Step 4: Taking inverse Laplace transform

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

Hence, $w\left(t\right)=\mathrm{cos}t+\left[1-\mathrm{cos}\left(t-2\right)\right]u\left(t-2\right)-\left[1-\mathrm{cos}\left(t-4\right)\right]u\left(t-4\right)$

, the graph is given below