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Expert-verified 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 ${Y}{\left(}{s}{\right)}$, the Laplace transform of the solution ${y}{\left(}{t}{\right)}$ to the given initial value problem.${17}{.}{{y}}^{\text{'}\text{'}}{+}{{y}}^{{\text{'}}}{-}{y}{=}{{t}}^{{3}}{;}{}{}{}{y}{\left(}{0}{\right)}{=}{1}{,}{}{}{}{{y}}^{{\text{'}}}{\left(}{0}{\right)}{=}{0}$

The Initial value for ${y}{\text{'}}{\text{'}}{+}{y}{\text{'}}{-}{y}{=}{{t}}^{{3}}$is $Y=\frac{{s}^{5}+{s}^{4}+6}{{s}^{4}\left({s}^{2}+s-1\right)}$

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

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

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

$\mathcal{L}\left\{y\text{'}\right\}\left(s\right)=s\mathcal{L}\left\{y\right\}\left(s\right)-y\left(0\right)\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}\left(s\right)={s}^{2}\mathcal{L}\left\{y\right\}\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\phantom{\rule{0ex}{0ex}}C\left\{{t}^{n}\right\}\left(s\right)=\frac{n!}{{s}^{n+1}}\phantom{\rule{0ex}{0ex}}\mathcal{L}\left\{y\text{'}\text{'}\right\}+\mathcal{L}\left\{y\text{'}\right\}-\mathcal{L}\left\{y\right\}=\mathcal{L}\left\{{t}^{3}\right\}$

Substitute the properties into the equation.

$\left[{s}^{2}Y-sy\left(0\right)-y\text{'}\left(0\right)\right]+\left[sY-y\left(0\right)\right]-Y=\frac{3!}{{s}^{4}}$

Substitute the initial conditions

$y\left(0\right)=1andy\text{'}\left(0\right)=0\phantom{\rule{0ex}{0ex}}\left({s}^{2}Y-s\right)+\left(sY-1\right)-Y=\frac{6}{{s}^{4}}$

Isolate the Y variable and solve:

${s}^{2}Y+sY-Y=\frac{6}{{s}^{4}}+s+1\phantom{\rule{0ex}{0ex}}Y\left({s}^{2}+s-1\right)=\frac{{s}^{5}+{s}^{4}+6}{{s}^{4}}\phantom{\rule{0ex}{0ex}}Y=\frac{{s}^{5}+{s}^{4}+6}{{s}^{4}\left({s}^{2}+s-1\right)}$

Therefore, the initial value for $y\text{'}\text{'}+y\text{'}-y={t}^{3}$is $Y=\frac{{s}^{5}+{s}^{4}+6}{{s}^{4}\left({s}^{2}+s-1\right)}$ ### Want to see more solutions like these? 