 # DIFFERENTIAL EQUATIONGiven f(t)=-frac{1}{2t}+8 , 0leq t<4 , f(t+4)=f(t)Find F(s)=Lleft{f(t)right} of the Periodic Function SchachtN 2020-10-26 Answered

DIFFERENTIAL EQUATION
Given $f\left(t\right)=-\frac{1}{2t}+8,0\le t<4,f\left(t+4\right)=f\left(t\right)$
Find $F\left(s\right)=L\left\{f\left(t\right)\right\}$ of the Periodic Function

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Step 1
According to the given information it is required to find
$F\left(s\right)=L\left\{f\left(t\right)\right\}$ Where, L represents Laplace Transformation and given:
$f\left(t\right)=-\frac{1}{2t}+8,0\le t<4,f\left(t+4\right)=f\left(t\right)$
Step 2
Laplace transformation of a periodic function with period $p>0$ is given by
$L\left\{f\left(t\right)\right\}=L\left\{{f}_{1}\left(t\right)\right\}\cdot \frac{1}{1-{e}^{-sp}}$
$L\left\{f\left(t\right)\right\}=\frac{{\int }_{0}^{p}f\left(t\right){e}^{-st}dt}{1-{e}^{-sp}}$
here, ${f}_{1}\left(t\right)$ is one period of the function.
Step 3
Calculate the required integral to find the Laplace transformation of given periodic function, as
${\int }_{0}^{p}f\left(t\right){e}^{-st}dt={\int }_{0}^{4}\left(-\frac{1}{2}t+8\right){e}^{-st}dt$
$=-\frac{1}{2}{\int }_{0}^{4}\left(t-16\right){e}^{-st}dt$
$=-\frac{1}{2}{\left[\left(t-16\right)\int {e}^{-st}dt-\int \left(\frac{d}{dt}\left(t-16\right)\int {e}^{-st}dt\right)dt\right]}_{0}^{4}$
$=-\frac{1}{2}{\left[\frac{\left(t-16\right){e}^{-st}}{-s}-\int \left(\frac{{e}^{-st}}{-s}\right)dt\right]}_{0}^{4}=-\frac{1}{2}{\left[-\frac{t-16}{s{e}^{-st}}-\frac{1}{{e}^{st}{s}^{2}}\right]}_{0}^{4}$
$=-\frac{1}{2}\left[-\frac{4-16}{s{e}^{4s}}+\frac{0-16}{s{e}^{0-s}}-\frac{1}{{e}^{4s}{s}^{2}}+\frac{1}{{e}^{0-s}{s}^{2}}\right]=-\frac{1}{2}\left[-\frac{12}{s{e}^{4s}}-\frac{16}{s}-\frac{1}{{e}^{4s}{s}^{2}}+\frac{1}{{s}^{2}}\right]$
$=\frac{16s-1}{2{s}^{2}}-\frac{\left(12-1\right){e}^{-4s}}{{s}^{2}}$
Step 4
Thus,
$F\left(s\right)=L\left\{f\left(t\right)\right\}=L\left\{{f}_{1}\left(t\right)\right\}\cdot \frac{1}{1-{e}^{-sp}}=\frac{{\int }_{0}^{p}f\left(t\right){e}^{-st}dt}{1-{e}^{-sp}}$
$=\frac{16s-1}{2{s}^{2}\left(1-{e}^{-4s}\right)}-\frac{\left(12-1\right){e}^{-4s}}{{s}^{2}\left(1-{e}^{-4s}\right)}$
$⇒F\left(s\right)=\frac{16s-1}{2{s}^{2}\left(1-{e}^{-4s}\right)}-\frac{\left(12-1\right){e}^{-4s}}{{s}^{2}\left(1-{e}^{-4s}\right)}$