# Invert the following system to the time domain using partial fraction. f(s)=frac(6)(s^2(s+2)(s+4))

Invert the following system to the time domain using partial fraction
$f\left(s\right)=\frac{6}{{s}^{2}\left(s+2\right)\left(s+4\right)}$

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Step 1
By using partial fraction method and the table of Laplace transform we solve the given problem as follows:
Step 2
$f\left(s\right)=\frac{6}{{s}^{2}\left(s+2\right)\left(s+4\right)}$ By using partial fraction
$\frac{6}{{s}^{2}\left(s+2\right)\left(s+4\right)}=\frac{A}{s}+\frac{B}{{s}^{2}}+\frac{C}{s+2}+\frac{D}{s+4}$
$⇒6=A\left({s}^{3}+6{s}^{2}+8s\right)+B\left({s}^{2}+6s+8\right)+C\left({s}^{3}+4{s}^{2}\right)+D\left({s}^{2}+2{s}^{2}\right)$
$⇒6=\left(A+C+D\right){s}^{3}+\left(6A+B+4C+2D\right){s}^{2}+\left(8A+6B\right)s+8B$
$⇒A+C+D=0,6A+B+4C+2D=0$
$8A+6B=0,8B=6⇒B=\frac{3}{4}$
on solving thus we get
$A=\frac{-9}{16},B=\frac{3}{4},C=\frac{3}{4},D=\frac{-3}{16}$
$⇒\frac{6}{{s}^{2}\left(s+2\right)\left(s+4\right)}=\frac{\frac{-9}{16}}{s}+\frac{\frac{3}{4}}{{s}^{2}}+\frac{\frac{3}{4}}{s+2}+\frac{\frac{-3}{16}}{s+4}$
Now apply inverse laplace we get
${L}^{-1}\left\{\frac{6}{{s}^{2}\left(s+2\right)\left(s+4\right)}\right\}=\frac{-9}{16}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{3}{4}{L}^{-1}\left\{\frac{1}{{s}^{2}}\right\}+\frac{3}{4}{L}^{-1}\left\{\frac{1}{s+2}\right\}-\frac{3}{16}{L}^{-1}\left\{\frac{1}{s+4}\right\}$
$⇒f\left(t\right)=\frac{-9}{16}+\frac{3}{4}t+\frac{3}{4}{e}^{-2t}-\frac{3}{16}{e}^{-4t}$