permaneceerc

2021-09-19

Derive Laplace Transform of the following:
a)$f\left(t\right)={e}^{-at}$
b) $f\left(t\right)=t$c) $f\left(t\right)={t}^{2}$
d) $\mathrm{cosh}t$

Sally Cresswell

Step 1
we have given f(t).
We have to derive the following laplace transform
Step 2
1) Derive Laplace transform of the following
$f\left(t\right)={e}^{-at}$
We know that
$L\left\{f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
Now, $L\left\{{e}^{-dt}\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\cdot {e}^{-2t}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+a\right)t}dt={\left[{\frac{{e}^{-\left(s+a\right)t}}{-\left(s+2\right)}}_{0}^{\mathrm{\infty }}\right]}_{0}^{\mathrm{\infty }}$
Hence, $L\left\{{e}^{-dt}\right\}=\left[0-\frac{{e}^{0}}{-\left(s+2\right)}\right]=\frac{1}{\left(s+2\right)}$
b) f(t)=t
Now, $L\left\{f\left(t\right)\right\}=L\left\{t\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}tdt$
$={\left[t\frac{{e}^{-st}}{-s}\right]}_{0}^{\mathrm{\infty }}-{\int }_{0}^{\mathrm{\infty }}\frac{dt}{dt}\frac{{e}^{-st}}{-s}dt$
$=\left(0+0\right)+\frac{1}{s}{\left[\frac{{e}^{-st}}{\left(-s\right)}\right]}_{0}^{\mathrm{\infty }}=\frac{1}{{s}^{2}}\left(0+1\right)=\frac{1}{{s}^{2}}$
Hence, $L\left\{f\left(t\right)\right\}=L\left\{t\right\}=\frac{1}{{s}^{2}}$
c)$f\left(t\right)={t}^{2}$
Now,$L\left\{f\left(t\right)\right\}=L\left\{{t}^{2}\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{t}^{2}dt$
$={\left[{t}^{2}\frac{{e}^{-st}}{\left(-s\right)}\right]}_{0}^{\mathrm{\infty }}-{\int }_{0}^{\mathrm{\infty }}2t\frac{{e}^{-st}}{\left(-s\right)}dt$
$=\left(0-0\right)+\frac{2}{s}\left[{\left[t\frac{{e}^{-st}}{\left(-s\right)}\right]}_{0}^{\mathrm{\infty }}-{\int }_{0}^{\mathrm{\infty }}\frac{dt}{dt}\frac{{e}^{-st}}{\left(-s\right)}dt\right]$

Do you have a similar question?