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Found in: Page 415

### 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 3-10, determine the Laplace transform of the given function.$\mathbit{f}\mathbf{\left(}\mathbit{t}\mathbf{\right)}\mathbf{=}\mathbit{c}\mathbit{o}\mathbit{s}\mathbit{t}\mathbf{,}\mathbf{-}\mathbit{\pi }\mathbf{/}\mathbf{2}\mathbf{\le }\mathbit{t}\mathbf{\le }\mathbit{\pi }\mathbf{/}\mathbf{2}$and ${\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}$has period $\mathbit{\pi }$

Therefore, the solution is.$\text{L}\left\{f\left(t\right)\right\}=\frac{s}{{s}^{2}+\text{1}}+\frac{\text{2}{e}^{-\frac{\pi }{2}s}}{\left({s}^{2}+\text{1}\right)\left(\text{1}-{e}^{-\pi s}\right)}$

See the step by step solution

## Step 1: Given Information

The given function value is.$f\left(t\right)=cost,-\pi /2\le t\le \pi /2$

## Step 2: Find the Laplace transform

On the interval we can write the given function as

$f\left(t\right)=|\mathrm{cos}t|=\left\{\begin{array}{ll}\mathrm{cos}t,& 0\le t\le \frac{\pi }{2}\\ -\mathrm{cos}t,& \frac{\pi }{2}\le t\le \pi \end{array}$

We will find the Laplace transform of given function using the following equation

Since .$T=\pi$ So, plug $T=\pi$ into above equation as:

Let's find $I={\int }_{0}^{\pi }{e}^{-st}f\left(t\right)dt.$

$\begin{array}{c}I={\int }_{0}^{\pi }{e}^{-st}f\left(t\right)dt\\ =\underset{{I}_{1}}{\underset{⏟}{{\int }_{0}^{\frac{\pi }{2}}{e}^{-st}f\left(t\right)dt}}-\underset{{I}_{2}}{\underset{⏟}{{\int }_{\frac{\pi }{2}}^{\pi }{e}^{-st}f\left(t\right)dt}}\end{array}$$\dots .\left(1\right)$

## Step 3:Find I1and I2

The expression for ${I}_{1}$ is obtained as follows:

$\begin{array}{c}{I}_{1}={\int }_{0}^{\frac{\pi }{2}}{e}^{-st}f\left(t\right)dt\\ ={\int }_{0}^{\frac{\pi }{2}}{e}^{-st}\mathrm{cos}tdt\\ =|\begin{array}{cc}u=\mathrm{cos}t& dv={e}^{-st}dt\\ du=-\mathrm{sin}tdt& v=-\frac{{e}^{-st}}{s}\end{array}|\\ ={\left[-\frac{{e}^{-st}\mathrm{cos}t}{s}\right]}_{0}^{\frac{\pi }{2}}-{\int }_{0}^{\frac{\pi }{2}}\frac{{e}^{-st}\mathrm{sin}t}{s}dt\end{array}$

Simplify further as:

$\begin{array}{c}{\text{I}}_{\text{1}}\text{=}\frac{\text{1}}{\text{s}}\text{-}\frac{\text{1}}{\text{s}}{\int }_{\text{0}}^{\frac{\text{π}}{\text{2}}}{\text{e}}^{\text{-st}}\text{sintdt}\\ \text{=}|\begin{array}{cc}\text{u=sint}& {\text{dv=e}}^{\text{-st}}\text{dt}\\ \text{du=costdt}& \text{v=-}\frac{{\text{e}}^{\text{-st}}}{\text{s}}\end{array}|\\ \text{=}\frac{\text{1}}{\text{s}}\text{-}\frac{\text{1}}{\text{s}}\left({\left[\text{-}\frac{{\text{e}}^{\text{-st}}\text{sint}}{\text{s}}\right]}_{\text{0}}^{\frac{\text{π}}{\text{2}}}\text{+}{\int }_{\text{0}}^{\frac{\text{π}}{\text{2}}}\frac{{\text{e}}^{\text{-st}}\text{cost}}{\text{s}}\text{dt}\right)\\ \text{=}\frac{\text{1}}{\text{s}}\text{-}\frac{\text{1}}{\text{s}}\left(\text{-}\frac{{\text{e}}^{\text{-}\frac{\text{π}}{\text{2}}\text{t}}}{\text{s}}\text{+}\frac{\text{1}}{\text{s}}{\text{I}}_{\text{1}}\right)\\ \text{=}\frac{\text{1}}{\text{s}}\text{+}\frac{{\text{e}}^{\text{-}\frac{\text{π}}{\text{2}}\text{t}}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{{\text{s}}^{\text{2}}}{\text{I}}_{\text{1}}\end{array}$

## Step 4: Simplify the resulting equation

From the resulting equation that

${I}_{1}=\frac{s+{e}^{-\frac{\pi }{2}s}}{{s}^{2}+\text{1}}$…… (2)

Similarly, We find

${I}_{2}=\frac{{e}^{-\frac{\pi }{2}s}-{e}^{-\pi s}s}{{s}^{2}+1}$…… (3)

On substituting equation (2) and (3) in equation (1) gives

$I=\frac{s\left(1-{e}^{-\pi s}\right)+\text{2}{e}^{-\frac{\pi }{2}s}}{{s}^{2}+\text{1}}$

Thus the Laplace transform of given function is obtained as

$\text{L}\left\{\text{f(t)}\right\}\text{=}\frac{\text{s}\left({\text{1-e}}^{\text{-πs}}\right){\text{+2e}}^{\text{-}\frac{\text{π}}{\text{2}}\text{s}}}{\left({\text{s}}^{\text{2}}\text{+1}\right)\left({\text{1-e}}^{\text{-πs}}\right)}$

Rewrite above equation as:

$\text{L}\left\{f\left(t\right)\right\}\left(s\right)=\frac{s}{{s}^{2}+\text{1}}+\frac{\text{2}{e}^{-\frac{\pi }{2}s}}{\left({s}^{2}+\text{1}\right)\left(\text{1}-{e}^{-\pi s}\right)}$