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21E

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

The solution for the Laplace transformation is

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

See the step by step solution

## Step 1: Derive the given equation using Laplace transformation

Define $\mathcal{L}\left\{y\right\}\left(s\right)=Y\left(s\right)$

$\mathcal{L}\left\{y\text{'}\text{'}\right\}-2\mathcal{L}\left\{y\text{'}\right\}+\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{\mathrm{cos}\left(t\right)\right\}-\mathcal{L}\left\{\mathrm{sin}\left(t\right)\right\}$

Using the properties listed below; take the Laplace transform of the equation.

$\mathbf{L}\left\{\mathbf{y}\mathbf{\text{'}}\right\}\left(\mathbf{s}\right)\mathbf{=}\mathbf{sL}\left\{\mathbf{y}\right\}\left(\mathbf{s}\right)\mathbf{-}\mathbf{y}\left(\mathbf{0}\right)\phantom{\rule{0ex}{0ex}}\mathbf{L}\left\{\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\right\}\left(\mathbf{s}\right)\mathbf{=}{\mathbf{s}}^{\mathbf{2}}\mathbf{L}\left\{\mathbf{y}\right\}\left(\mathbf{s}\right)\mathbf{-}\mathbf{sy}\left(\mathbf{0}\right)\mathbf{-}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\phantom{\rule{0ex}{0ex}}\mathbf{L}\left\{\mathbf{cos}\left(\mathbf{bt}\right)\right\}\mathbf{=}\frac{\mathbf{s}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\mathbf{L}\left\{\mathbf{sin}\left(\mathbf{bt}\right)\right\}\mathbf{=}\frac{\mathbf{b}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}$

Substitute the properties into the equation.

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

## Step 2: Use initial condition and find the Y variable

Solve for the Laplace transform as:

${s}^{2}Y-s-3-2sY+2+Y=\frac{s-1}{{s}^{2}+1}$

Substitute the initial conditions:

$y\left(0\right)=1and{y}^{\text{'}}\left(0\right)=3$

Isolate the Y variable.

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