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Expert-verified Found in: Page 375 ### 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 1-10, determine the inverse Laplace transform of the given function.$\frac{\mathbf{s}\mathbf{-}\mathbf{1}}{\mathbf{2}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{s}\mathbf{+}\mathbf{6}}$

The inverse Laplace transform for the given function is

${\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}-1}{2{\mathrm{s}}^{2}+\mathrm{s}+6}\right\}=\frac{1}{2}{\mathrm{e}}^{-\frac{1}{4}\mathrm{t}}\mathrm{cos}\frac{\sqrt{47}}{4}\mathrm{t}-\frac{5\sqrt{47}}{94}{\mathrm{e}}^{-\frac{1}{4}\mathrm{t}}\mathrm{sin}\frac{\sqrt{47}}{4}\mathrm{t}$

See the step by step solution

## Step 1: Determining the inverse laplace transform

• For a given transfer function H, the Inverse Laplace Transform takes the output Y(s) and determines what X(s) it is in terms of (s).
• Consider a function $\mathrm{F}\left(\mathrm{s}\right)$, if there is a function $\mathrm{f}\left(\mathrm{t}\right)$ that is continuous on $\left[0,\mathrm{\infty }\right)$and satisfies $\mathcal{L}\left\{\mathrm{f}\right\}=\mathrm{F}$then we say that $\mathrm{f}\left(\mathrm{t}\right)$ is the inverse Laplace transform of $\mathrm{F}\left(\mathrm{s}\right)$ and employ the notation
• ${\mathrm{f}}{=}{{\mathcal{L}}}^{-1}{\left\{}{\mathrm{F}}{\right\}}$
• ${{\mathcal{L}}}^{-1}\left\{\frac{\mathrm{n}!}{{\left(\mathrm{s}-\mathrm{a}\right)}^{\mathrm{n}+1}}\right\}{=}{{\mathrm{e}}}^{{\mathrm{at}}}{{\mathrm{t}}}^{{\mathrm{n}}}{,}{\mathrm{n}}{=}{1}{,}{2}{,}{\dots }$

## Step 2: Find inverse laplace transform for the given function

The given function is $\frac{\mathrm{s}-1}{2{\mathrm{s}}^{2}+\mathrm{s}+6}$

Simplify $\frac{\mathrm{s}-1}{2{\mathrm{s}}^{2}+\mathrm{s}+6}$ as:

$\begin{array}{c}\frac{\mathrm{s}-1}{2{\mathrm{s}}^{2}+\mathrm{s}+6}=\frac{\mathrm{s}-1}{2\left({\mathrm{s}}^{2}+\frac{1}{2}\mathrm{s}+3\right)}\\ =\frac{1}{2}\left(\frac{\mathrm{s}-1}{{\mathrm{s}}^{2}+\frac{1}{2}\mathrm{s}+3}\right)\\ =\frac{1}{2}\left(\frac{\mathrm{s}-1}{\left({\mathrm{s}}^{2}+\frac{1}{2}\mathrm{s}+\frac{1}{16}\right)+3-\frac{1}{16}}\right)\\ =\frac{1}{2}\left(\frac{\mathrm{s}-1}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\sqrt{\frac{47}{16}}\right)}^{2}}\right)\end{array}$

Further simplify the equation as follows:

$\begin{array}{c}\frac{s-1}{2{s}^{2}+s+6}=\frac{1}{2}\left(\frac{s+\frac{1}{4}-\frac{5}{4}}{{\left(s+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)\\ =\frac{1}{2}\left(\frac{\left(s+\frac{1}{4}\right)-\frac{5}{4}}{{\left(s+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)\\ =\frac{1}{2}\left(\frac{s+\frac{1}{4}}{{\left(s+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)-\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(s+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)\end{array}$

Find the inverse Laplace transform of

$\frac{\mathrm{s}-1}{2{\mathrm{s}}^{2}+\mathrm{s}+6}=\frac{1}{2}\left(\frac{\mathrm{s}+\frac{1}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)-\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)$using ${\mathcal{L}}^{-1}\left\{\frac{\mathrm{b}}{{\left(\mathrm{s}-\mathrm{a}\right)}^{2}+{\left(\mathrm{b}\right)}^{2}}\right\}={\mathrm{e}}^{\mathrm{at}}\mathrm{sinbt}$ and ${\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}-\mathrm{a}}{{\left(\mathrm{s}-\mathrm{a}\right)}^{2}+{\left(\mathrm{b}\right)}^{2}}\right\}={\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}$ as:

$\begin{array}{c}{\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}-1}{2{\mathrm{s}}^{2}+\mathrm{s}+6}\right\}={\mathcal{L}}^{-1}\left\{\frac{1}{2}\left(\frac{\left(\mathrm{s}+\frac{1}{4}\right)}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)-\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)\right\}\\ ={\mathcal{L}}^{-1}\left\{\frac{1}{2}\left(\frac{\left(\mathrm{s}+\frac{1}{4}\right)}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)\right\}-{\mathcal{L}}^{-1}\left\{\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right)\right\}\\ =\frac{1}{2}{\mathcal{L}}^{-1}\left\{\frac{\left(\mathrm{s}+\frac{1}{4}\right)}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right\}-\frac{5\sqrt{47}}{94}{\mathcal{L}}^{-1}\left\{\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^{2}+{\left(\frac{\sqrt{47}}{4}\right)}^{2}}\right\}\\ =\frac{1}{2}{\mathrm{e}}^{-\frac{1}{4}\mathrm{t}}\mathrm{cos}\frac{\sqrt{47}}{4}\mathrm{t}-\frac{5\sqrt{47}}{94}{\mathrm{e}}^{-\frac{1}{4}\mathrm{t}}\mathrm{sin}\frac{\sqrt{47}}{4}\mathrm{t}\end{array}$

Therefore, the inverse Laplace transform for the given function is

${\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}-1}{2{\mathrm{s}}^{2}+\mathrm{s}+6}\right\}=\frac{1}{2}{\mathrm{e}}^{-\frac{1}{4}\mathrm{t}}\mathrm{cos}\frac{\sqrt{47}}{4}\mathrm{t}-\frac{5\sqrt{47}}{94}{\mathrm{e}}^{-\frac{1}{4}\mathrm{t}}\mathrm{sin}\frac{\sqrt{47}}{4}\mathrm{t}$ ### Want to see more solutions like these? 