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Q7.3 - 21E

Expert-verified
Found in: Page 365

### 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

# Given that ${\mathsc{L}}{\mathbf{\left\{}}{\mathbf{cosbt}}{\mathbf{\right\}}}{\mathbf{\left(}}{\mathbf{s}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{s}}{\mathbf{/}}\left({s}^{2}+{b}^{2}\right)$, use the translation property to compute ${\mathsc{L}}\left\{{e}^{\mathrm{at}}\mathrm{cosbt}\right\}$.

The value of $\mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}\right\}$ is $\frac{\mathrm{s}-\mathrm{a}}{{\left(\mathrm{s}-\mathrm{a}\right)}^{2}+{\mathrm{b}}^{2}}$.

See the step by step solution

## Define Laplace transform

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. Laplace transform $\mathcal{L}$of a function f(t) is defined as:

role="math" localid="1655792122645" ${\mathcal{L}}{\left\{}{f}{\left(}{t}{\right)}{\right\}}{=}{{\int }}_{{0}}^{\frac{<}{>}}{{e}}^{-st}{f}{\left(}{t}{\right)}{d}{t}$

In words, we can describe this expression as the Laplace transform of f(t) equals function F of s, that is, $\mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)$.

## Find the value of Leatcosbt

Given that $\mathcal{L}\left\{\mathrm{cos}bt\right\}\left(s\right)=s/\left({s}^{2}+{b}^{2}\right)$,

Find $\mathcal{L}\left\{{e}^{at}\mathrm{cos}bt\right\}\left(s\right)$ using to translation property $\mathcal{L}\left\{{e}^{at}f\left(t\right)\right\}\left(s\right)=F\left(s-a\right)$ as:

$\begin{array}{rcl}\mathcal{L}\left\{{e}^{at}\mathrm{cos}bt\right\}\left(s\right)& =& F\left(s-a\right)\\ & =& \mathcal{L}\left\{\mathrm{cos}bt\right\}\left(s-a\right)\\ & =& \frac{s-a}{{\left(s-a\right)}^{2}+{b}^{2}}\\ & & \end{array}$

Hence, the value of $\mathcal{L}\left\{{e}^{at}\mathrm{cos}bt\right\}$ is $\frac{s-a}{{\left(s-a\right)}^{2}+{b}^{2}}$.