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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. ${\mathbf{}}{\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbit{y}}{\mathbf{=}}{\mathbit{u}}\left(t--3\right){\mathbf{;}}{\mathbit{y}}\left(0\right){\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbit{y}}{\mathbf{\text{'}}}\left(0\right){\mathbf{=}}{\mathbf{1}}$

On solving the given initial value problem using the method of Laplace transforms, the solution is $y\left(t\right)=\mathrm{sin}t+\left[1-\mathrm{cos}\left(t-3\right)\right]u\left(t-3\right)$ 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

Given initial value problem

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

Where $y\left(0\right)=0andy\text{'}\left(0\right)=1\phantom{\rule{0ex}{0ex}}$

Laplace Transform for the initial value problem

$\mathcal{L}y\text{'}\text{'}\left(s\right)+\mathcal{L}y\left(s\right)=\mathcal{L}\left[u\left(t-3\right)\right]\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}y\left(s\right)-s\mathcal{L}y\left(0\right)-y\text{'}\left(0\right)+\mathcal{L}y\left(s\right)=\frac{{e}^{-3s}}{s}\phantom{\rule{0ex}{0ex}}{s}^{2}\mathcal{L}y\left(s\right)-0-1+\mathcal{L}y\left(s\right)=\frac{{e}^{-3s}}{s}$

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

## Step 3: By Partial fraction method

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

Equation first becomes,

$\mathcal{L}y\left(s\right)=\frac{1}{{s}^{2}+1}+\frac{{e}^{-3s}}{s}-\frac{s{e}^{-3s}}{{s}^{2}+1}$

## Step 4: Taking inverse Laplace transform

$y\left(t\right)={\mathcal{L}}^{-1}\frac{1}{{s}^{2}+1}+{\mathcal{L}}^{-1}\frac{{e}^{-3s}}{s}+{\mathcal{L}}^{-1}\frac{s{e}^{-3s}}{{s}^{2}+1}\phantom{\rule{0ex}{0ex}}=\mathrm{sin}t+u\left(t-3\right)-\mathrm{cos}\left(t-3\right)u\left(t-3\right)\phantom{\rule{0ex}{0ex}}=\mathrm{sin}t+\left[1-\mathrm{cos}\left(t-3\right)\right]u\left(t-3\right)$

Hence,

$y\left(t\right)=\mathrm{sin}t+\left[1-\mathrm{cos}\left(t-3\right)\right]u\left(t-3\right)$

The graph is given below