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

Step 1
According to the given information it is required to find
$F(s)=L\{f(t)\}$ Where, L represents Laplace Transformation and given:
$f(t)=-\frac{1}{2t}+8,0\le t<4,f(t+4)=f(t)$
Step 2
Laplace transformation of a periodic function with period $p>0$ is given by
$L\{f(t)\}=L\{f_1(t)\}\cdot \frac{1}{1-e^{-sp}}$
$L\{f(t)\}=\frac{{\int }_0^{p}f(t)e^{-st}dt}{1-e^{-sp}}$
here, $f_1(t)$ is one period of the function.
Step 3
Calculate the required integral to find the Laplace transformation of given periodic function, as
${\int }_0^{p}f(t)e^{-st}dt={\int }_0^4(-\frac{1}{2}t+8)e^{-st}dt$
$=-\frac{1}{2}{\int }_0^4(t-16)e^{-st}dt$
$=-\frac{1}{2}{[(t-16)\int e^{-st}dt-\int (\frac{d}{dt}(t-16)\int e^{-st}dt)dt]}_0^4$
$=-\frac{1}{2}{[\frac{(t-16)e^{-st}}{-s}-\int \left(\frac{e^{-st}}{-s}\right)dt]}_0^4=-\frac{1}{2}{[-\frac{t-16}{s{e}^{-st}}-\frac{1}{e^{st}{s}^2}]}_0^4$
$=-\frac{1}{2}[-\frac{4-16}{s{e}^{4s}}+\frac{0-16}{s{e}^{0-s}}-\frac{1}{e^{4s}{s}^2}+\frac{1}{e^{0-s}{s}^2}]=-\frac{1}{2}[-\frac{12}{s{e}^{4s}}-\frac{16}{s}-\frac{1}{e^{4s}{s}^2}+\frac{1}{s^2}]$
$=\frac{16s-1}{2s^2}-\frac{(12-1)e^{-4s}}{s^2}$
Step 4
Thus,
$F(s)=L\{f(t)\}=L\{f_1(t)\}\cdot \frac{1}{1-e^{-sp}}=\frac{{\int }_0^{p}f(t)e^{-st}dt}{1-e^{-sp}}$
$=\frac{16s-1}{2s^2(1-e^{-4s})}-\frac{(12-1)e^{-4s}}{s^2(1-e^{-4s})}$
$\Rightarrow F(s)=\frac{16s-1}{2s^2(1-e^{-4s})}-\frac{(12-1)e^{-4s}}{s^2(1-e^{-4s})}$