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15E

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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

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

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

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 as follows:

$\mathcal{L}\left\{y\text{'}\text{'}-3y\text{'}+2y\right\}=\mathcal{L}\left\{\mathrm{cos}t\right\}\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}-3\mathcal{L}\left\{y\text{'}\right\}+2\mathcal{L}\left\{y\right\}=\frac{s}{{s}^{2}+1}$

Solve for the transfer function as:

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

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

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

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