Q10RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

Step-by-Step Solution

Answer

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

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

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

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

$\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)=\frac{{\displaystyle {\int }_0^T{e}^{-st}f\left(t\right)dt}}{1-e^{-sT}}$

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

$\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)=\frac{{\displaystyle {\int }_0^{\pi }{e}^{-st}f\left(t\right)dt}}{1-e^{-s\pi }}$

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}{\underbrace{{\int }_0^{\frac{\pi }{2}}{e}^{-st}f\left(t\right)dt}}-\underset{I_2}{\underbrace{{\int }_{\frac{\pi }{2}}^{\pi }{e}^{-st}f\left(t\right)dt}}\end{array}$$\dots .\left(1\right)$

Step 3:Find I 1 and I 2

The expression for $I_1$ is obtained as follows:

$\begin{array}{c}{I}_1={\int }_0^{\frac{\pi }{2}}{e}^{-st}f\left(t\right)dt\\ ={\int }_0^{\frac{\pi }{2}}{e}^{-st}\cos tdt\\ =\left|\begin{array}{cc}u=\cos t& dv=e^{-st}dt\\ du=-\sin tdt& v=-\frac{e^{-st}}{s}\end{array}\right|\\ ={[-\frac{e^{-st}\cos t}{s}]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\frac{e^{-st}\sin t}{s}dt\end{array}$ 

Simplify further as:

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

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

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

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:

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