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Q7.4 - 7E

Expert-verified
Found in: Page 374

### 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

# Determine the inverse Laplace transform of the given function.$\frac{\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{16}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{13}}$.

The inverse laplace transform of the given function is $2{\mathrm{e}}^{-2\mathrm{t}}\mathrm{cos}3\mathrm{t}+4{\mathrm{e}}^{-2\mathrm{t}}\mathrm{sin}3\mathrm{t}$.

See the step by step solution

## 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 F(s), if there is a function f(t) that is continuous on $\left[0,\infty \right)$ and satisfies $\mathcal{L}\left\{f\right\}=F$ then we say that f(t) is the inverse Laplace transform of F(s) and employ the notation
• ${f}{=}{{\mathcal{L}}}^{-1}{\left\{}{F}{\right\}}$
• ${{\mathcal{L}}}^{-1}\left\{\frac{n!}{{\left(s-a\right)}^{n+1}}\right\}{=}{{e}}^{at}{{t}}^{{n}}{,}{n}{=}{1}{,}{2}{,}{\dots }$

## Find inverse laplace transform for the given function

The given function is $\frac{2s+16}{{s}^{2}+4s+13}$.

Simplify $\frac{2s+16}{{s}^{2}+4s+13}$ as follows:

$\begin{array}{rcl}\frac{2s+16}{{s}^{2}+4s+4+9}& =& \frac{2\left(s+8\right)}{\left({s}^{2}+4s+4\right)+9}\\ & =& \frac{2\left(s+2\right)+12}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}\\ & =& \frac{2\left(s+2\right)}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}+\frac{12}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}\\ & & \end{array}$

Find the inverse laplace transform of $\frac{2s+16}{{s}^{2}+4s+4+9}=\frac{2\left(s+2\right)}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}+\frac{12}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}$ using ${\mathcal{L}}^{-1}\left\{\frac{b}{{\left(s-a\right)}^{2}+{\left(b\right)}^{2}}\right\}={e}^{at}\mathrm{sin}bt$ and ${\mathcal{L}}^{-1}\left\{\frac{b}{{\left(s-a\right)}^{2}+{\left(b\right)}^{2}}\right\}={e}^{at}\mathrm{cos}bt$ as:

$\begin{array}{rcl}{\mathcal{L}}^{-1}\left\{\frac{2s+16}{{s}^{2}+4s+4+9}\right\}& =& {\mathcal{L}}^{-1}\left\{\frac{2\left(s+2\right)}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}+\frac{12}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}\right\}\\ & =& 2{\mathcal{L}}^{-1}\left\{\frac{s+2}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}\right\}+4{\mathcal{L}}^{-1}\left\{\frac{3}{{\left(s+2\right)}^{2}+{\left(3\right)}^{2}}\right\}\\ & =& 2{e}^{-2t}\mathrm{cos}3t+4{e}^{-2t}\mathrm{sin}3t\\ & & \end{array}$

Therefore, the inverse laplace transform of the given function is $2{e}^{-2t}\mathrm{cos}3t+4{e}^{-2t}\mathrm{sin}3t$.