# Find the inverse Laplace transform f(t)=L^)-1 {F(s)} of each of the following functions.

Find the inverse Laplace transform $f\left(t\right)={L}^{-1}\left\{F\left(s\right)\right\}$ of each of the following functions.
$\left(i\right)F\left(s\right)=\frac{2s+1}{{s}^{2}-2s+1}$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$\left(ii\right)F\left(s\right)=\frac{3s+2}{{s}^{2}-3s+2}$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$\left(iii\right)F\left(s\right)=\frac{3{s}^{2}+4}{\left({s}^{2}+1\right)\left(s-1\right)}$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.

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$\left(i\right)F\left(s\right)=\frac{2s+1}{{s}^{2}-2s+1}$
$\frac{2s+1}{{\left(s-1\right)}^{2}}=\frac{A}{s-1}+\frac{B}{{\left(s-1\right)}^{2}}$
$2s+1=\frac{A}{s-1}+\frac{B}{{\left(s-1\right)}^{2}}$
$2s+1=A\left(s-1\right)+B$
$A=2,-A+B=1$
$B=3$
$F\left(s\right)=\frac{2}{s-1}+\frac{3}{{\left(s-1\right)}^{2}}$
$f\left(t\right)=2{e}^{t}+3t{e}^{t}$
$\left(ii\right)F\left(s\right)=\frac{3s+2}{{s}^{2}-3s+2}$
$\frac{3s+2}{\left(s-2\right)\left(s-1\right)}=\frac{A}{s-2}+\frac{B}{s-1}$
$3s+2=A\left(s-1\right)+B\left(s-2\right)$
$A+B=3$
$A=8,B=-5$
$-A-2B=2$
$F\left(s\right)=\frac{8}{s-2}-\frac{5}{s-1}$
$f\left(t\right)=8{e}^{2t}-5{e}^{t}$
$\left(iii\right)F\left(s\right)=\frac{3{s}^{2}+4}{\left({s}^{2}+1\right)\left(s-1\right)}$
$\frac{3{s}^{2}+4}{\left(s-1\right)\left({s}^{2}+1\right)}=\frac{As+B}{{s}^{2}+1}+\frac{C}{s-1}$
$⇒3{s}^{2}+4=A\left({s}^{2}-s\right)+B\left(s-1\right)+C\left({s}^{2}+1\right)$
$A+B=3$
$-A+B=0$
$-B+C=4$
$A=-\frac{1}{2},B=-\frac{1}{2},C=\frac{7}{2}$
$F\left(s\right)=\frac{-1s}{2\left({s}^{2}+1\right)}-\frac{1}{2\left({s}^{2}+1\right)}+\frac{7}{2\left(s-1\right)}$

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