# Inverse Laplace transform ,L^{-1}\left\{\frac{2s}{(s-3)^2}\right\}

Inverse Laplace transform
${L}^{-1}\left\{\frac{2s}{\left(s-3{\right)}^{2}}\right\}$

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Step 1
Here,
$f\left(s\right)=\frac{2s}{{\left(s-3\right)}^{2}}$
Step 2
Take the partial fraction of f(s)
${L}^{-1}\left\{\frac{2s}{\left(s-3{\right)}^{2}}\right\}$
$\frac{2s}{{\left(s-3\right)}^{2}}=\frac{a}{s-3}+\frac{b}{{\left(s-3\right)}^{2}}$
$=\frac{a\left(s-3\right)+b}{{\left(s-3\right)}^{2}}$
$=\frac{as-3a+b}{{\left(s-3\right)}^{2}}$
$2s=as-3a+b$
Step 3
Comparing coefficients of s
$a=2$
Comparing constant terms.
$-3a+b=0$
$-3\left(2\right)+b=0$
$b=6$
So, function will be,
$f\left(s\right)=\frac{2}{s-3}+\frac{6}{{\left(s-3\right)}^{2}}$
${L}^{-1}f\left(s\right)={L}^{-1}\left(\frac{2}{s-3}\right)+{L}^{-1}\left(\frac{6}{{\left(s-3\right)}^{2}}\right)$
Step 4
Solving first part.
${L}^{-1}\left(\frac{2}{s-3}\right)=2{L}^{-1}\left(\frac{1}{s-3}\right)$
$=2{e}^{3t}$
Step 5
Solving second part.
${L}^{-1}\left(\frac{6}{\left(s-3{\right)}^{2}}\right)={e}^{3t}{L}^{-1}\left\{\frac{6}{{s}^{2}}\right\}$
$=6{e}^{3t}{t}^{2-1}$
$=6t{e}^{3t}$
Step 6
Determine inverse laplace of the function.
${L}^{-1}\left\{f\left(s\right)\right\}=2{e}^{3t}+6t{e}^{3t}$