# Did I Inverse this Laplace correctly? L^(-1) (4s)/((s-6)^3)

${L}^{-1}\frac{4s}{\left(s-6{\right)}^{3}}$
$4{L}^{-1}\frac{s}{{s}^{3}}|s=s-6$
$4{L}^{-1}\frac{1}{{s}^{2}}|s=s-6$
$4{L}^{-1}\frac{1!}{{s}^{1+1}}|s=s-6$
$4t{e}^{6t}$
Is this correct?
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Phillip Fletcher
If you don't know residues, you can rewrite your function as
$\frac{4s}{\left(s-6{\right)}^{3}}=\frac{4\left(s-6\right)}{\left(s-6{\right)}^{3}}+\frac{24}{\left(s-6{\right)}^{3}}=\frac{4}{\left(s-6{\right)}^{2}}+\frac{24}{\left(s-6{\right)}^{3}}$
and use the "translation rule" to invert the two terms separately.
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Christopher Saunders
The ILT may be computed via the residue theorem as follows:
$\frac{4}{2!}\frac{{d}^{2}}{d{s}^{2}}{\left[s{e}^{st}\right]}_{s=6}=2\frac{d}{ds}{\left[\left(1+st\right){e}^{st}\right]}_{s=6}=2{\left[\left(t\left(1+st\right)+t\right){e}^{st}\right]}_{s=6}=\left(4t+12{t}^{2}\right){e}^{6t}$
You seem to be missing the ${t}^{2}$ term.