Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1-10, determine the inverse Laplace transform of the given function.

$\frac{\mathbf{s}\mathbf{-}\mathbf{1}}{\mathbf{2}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{s}\mathbf{+}\mathbf{6}}$

Step-by-Step Solution

Answer

The inverse Laplace transform for the given function is 

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

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 $[0,\mathrm{\infty })$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
• $\mathrm{f}={\mathcal{L}}^{-1}\left\{\mathrm{F}\right\}$
  
• ${\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{(\mathrm{s}-\mathrm{a})}^{\mathrm{n}+1}}\right\}={\mathrm{e}}^{\mathrm{at}}{\mathrm{t}}^{\mathrm{n}},\mathrm{n}=1,2,\dots$
  

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:

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

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

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

Therefore, the inverse Laplace transform for the given function is 

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