geduiwelh

2021-09-10

Find the inverse Laplace transform of
a) $\mathrm{sin}\left(at\right)+{e}^{-bt}$
b) $\frac{t{e}^{-bt}}{{\left(a-b\right)}^{2}}+\mathrm{sin}\left(at\right)$
c) $\frac{\mathrm{sin}\left(at\right)}{{\left(b-a\right)}^{2}}+\frac{{e}^{-bt}}{{\left(a-b\right)}^{2}}$
d) $\frac{{e}^{-bt}}{{\left(a-b\right)}^{2}}+\frac{t{e}^{-at}}{b-a}-\frac{{e}^{-at}}{{\left(b-a\right)}^{2}}$
e)$\mathrm{sin}\left(at\right)+\frac{t{e}^{-bt}}{{\left(b-a\right)}^{2}}-\frac{t{e}^{-at}}{{\left(a-b\right)}^{2}}$

Pohanginah

Step 1
Here,
$F\left(s\right)=\frac{1}{{\left(s+a\right)}^{2}\left(s+b\right)}$
step 2
Rewrite the function.
$F\left(s\right)=\frac{1}{a-b}×\frac{\left(s+a\right)-\left(s+b\right)}{{\left(s+a\right)}^{2}\left(s+b\right)}$
$=\frac{1}{a-b}×\left[\frac{\left(s+a\right)}{{\left(s+a\right)}^{2}\left(s+b\right)}-\frac{\left(s+b\right)}{{\left(s+a\right)}^{2}\left(s+b\right)}\right]$
$=\frac{1}{a-b}×\left[\frac{1}{\left(s+a\right)\left(s+b\right)}-\frac{1}{{\left(s+a\right)}^{2}}\right]$
$=\frac{1}{a-b}×\left[\frac{1}{a-b}×\frac{\left(s+a\right)-\left(s+b\right)}{\left(s+a\right)\left(s+b\right)}-\frac{1!}{{\left(s+a\right)}^{1+1}}\right]$
$=\frac{1}{{\left(a-b\right)}^{2}}\left[\frac{1}{s+b}-\frac{1}{s+a}\right]-\frac{1}{a-b}×\frac{1!}{{\left(s+a\right)}^{1+1}}$
step 3
Apply inverse laplace transform formula.
$f\left(t\right)=\frac{1}{{\left(a-b\right)}^{2}}\left[{e}^{-bt}-{e}^{-at}\right]-\frac{1}{a-b}×{t}^{1}{e}^{-at}$
$=\frac{{e}^{-bt}}{{\left(a-b\right)}^{2}}+\frac{t{e}^{-at}}{b-a}-\frac{{e}^{-at}}{{\left(b-a\right)}^{2}}$
Step 4