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18E

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Found in: Page 350

### 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

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# In Problems $\mathbf{15}-\mathbf{24}$ , solve for ${\mathbf{Y}}\left(\mathbf{s}\right)$ , the Laplace transform of the solution ${y}{\left(}{t}{\right)}$ to the given initial value problem.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{e}}}^{\mathbf{2}\mathbf{t}}{\mathbf{-}}{{\mathbf{e}}}^{{\mathbf{t}}}{\mathbf{;}}{}{}{}{\mathbf{y}}\left(0\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{}{}{}{\mathbf{y}}{\mathbf{\text{'}}}\left(0\right){\mathbf{=}}{\mathbf{3}}$

The Initial value for $y\text{'}\text{'}-2y\text{'}-y={e}^{2t}-{e}^{t}$ is $Y\left(s\right)=\frac{{s}^{3}-2{s}^{2}-s+3}{\left({s}^{2}-2s-1\right)\left(s-2\right)\left(s-1\right)}$

See the step by step solution

## Step 1: Determine the Laplace Transform

• The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
• In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
• $\mathbf{F}\left(\mathbf{s}\right)\mathbf{=}{\int }_{\mathbf{0}}^{\infty }\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{t}\mathbf{\text{'}}$

## Step 2: Determine the Laplace transform

Applying the Laplace transform and using its linearity we get

$\mathcal{L}\left\{y\text{'}\text{'}-2y\text{'}-y\right\}=\mathcal{L}\left\{{e}^{2t}-{e}^{t}\right\}\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}-2\mathcal{L}\left\{y\text{'}\right\}-\mathcal{L}\left\{y\right\}=\frac{1}{s-2}-\frac{1}{s-1}$

Solve for the Laplace transform as:

$\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-2\left[sY\left(s\right)-y\left(0\right)\right]-Y\left(s\right)=\frac{1}{\left(s-2\right)\left(s-1\right)}\phantom{\rule{0ex}{0ex}}{s}^{2}Y\left(s\right)-s-3-2sY\left(s\right)+2-Y\left(s\right)=\frac{1}{\left(s-2\right)\left(s-1\right)}$

${s}^{2}Y\left(s\right)-2sY\left(s\right)-Y\left(s\right)=\frac{1}{\left(s-2\right)\left(s-1\right)}+s+1\phantom{\rule{0ex}{0ex}}\left({s}^{2}-2s-1\right)Y\left(s\right)=\frac{{s}^{3}-2{s}^{2}-s+3}{\left(s-2\right)\left(s-1\right)}$

Solve further as:

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

Therefore, the Initial value for $y\text{'}\text{'}-2y\text{'}-y={e}^{2t}-{e}^{t}$ is $Y\left(s\right)=\frac{{s}^{3}-2{s}^{2}-s+3}{\left({s}^{2}-2s-1\right)\left(s-2\right)\left(s-1\right)}$

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