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

On solving the given initial value problem using the method of Laplace transforms the solution is $y\left(t\right)=t-\left(t-4+2\mathrm{cos}\left(t-2\right)-\mathrm{sin}\left(t-2\right)\right)u\left(t-2\right)$and the 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: Applying Laplace transform and using its linearity

$\mathcal{L}\left\{y\text{'}\text{'}+y\right\}=\mathcal{L}\left\{t-\left(t-4\right)u\left(t-2\right)\right\}\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{t\right\}-\mathcal{L}\left\{\left(t-4\right)u\left(t-2\right)\right\}$

$\left({s}^{2}Y\left(s\right)-sy\left(0\right)-y\left(0\right)\right)+Y\left(s\right)=\frac{1}{{s}^{2}}-\mathcal{L}\left\{\left(t-4\right)u\left(t-2\right)\right\}\phantom{\rule{0ex}{0ex}}{s}^{2}Y\left(s\right)-1+Y\left(s\right)=\frac{1}{{s}^{2}}-\mathcal{L}\left\{\left(t-4\right)u\left(t-2\right)\right\}$

Since,

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

We have that,

$\left({s}^{2}+1\right)Y\left(s\right)=\frac{1}{{s}^{2}}-{e}^{-2s}\left(\frac{1}{{s}^{2}}-\frac{2}{s}\right)+1\phantom{\rule{0ex}{0ex}}\left({s}^{2}+1\right)Y\left(s\right)=\frac{1+{s}^{2}}{{s}^{2}}-{e}^{-2s}\frac{1-2s}{{s}^{2}}\phantom{\rule{0ex}{0ex}}Y\left(s\right)=\frac{1}{{s}^{2}}-{e}^{-2s}\frac{1-2s}{{s}^{2}\left({s}^{2}+1\right)}$

Using partial fractions we get,

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

And inverse Laplace transform gives,

${\mathcal{L}}^{-1}\left\{\frac{{e}^{-2s}}{{s}^{2}}\right\}=\left(t-2\right)u\left(t-2\right)\phantom{\rule{0ex}{0ex}}{\mathcal{L}}^{-1}\left\{\frac{2{e}^{-2s}}{s}\right\}=2u\left(t-2\right)$

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

## Step 3: The solution of given IVP

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

Hence, $y\left(t\right)=t-\left(t-4+2\mathrm{cos}\left(t-2\right)-\mathrm{sin}\left(t-2\right)\right)u\left(t-2\right)$

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