# Find inverse Laplace transformation L^(-1)=(s)/((s−3)(s−4)(s−12))

I have the following inverse laplace transformation:
${L}^{-1}=\frac{s}{\left(s-3\right)\left(s-4\right)\left(s-12\right)}$
After looking at the laplace transformations the closest I've found is:
$\frac{a{e}^{at}-b{e}^{bt}}{a-b}=\frac{s}{\left(s-a\right)\left(s-b\right)}$
I've been working out a solution for having three variables, but I cant seem to get the correct solution. Is there an identity for this type of transformation?
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Dobricap
Hint: The Partial Fraction Expansion is:
$\frac{s}{\left(s-3\right)\left(s-4\right)\left(s-12\right)}=-\frac{1}{2\left(s-4\right)}+\frac{1}{3\left(s-3\right)}+\frac{1}{6\left(s-12\right)}$
${L}^{-1}\left(\frac{s}{\left(s-3\right)\left(s-4\right)\left(s-12\right)}\right)=\left(-1/2\right){e}^{4t}+\left(1/3\right){e}^{3t}+\left(1/6\right){e}^{12t}$
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Emmy Swanson
Using partial-fraction expansion, we have
$\frac{s}{\left(s-3\right)\left(s-4\right)\left(s-12\right)}=\frac{A}{\left(s-3\right)}+\frac{B}{\left(s-4\right)}+\frac{C}{\left(s-12\right)}$
where A,B and C are obtained as follows:
$\begin{array}{rl}A& =\frac{s}{\left(s-4\right)\left(s-12\right)}{|}_{s=3}=\frac{1}{3},\\ B& =\frac{s}{\left(s-3\right)\left(s-12\right)}{|}_{s=4}=-\frac{1}{2},\\ C& =\frac{s}{\left(s-3\right)\left(s-4\right)}{|}_{s=12}=\frac{1}{6}.\end{array}$
As a result, we get
$\frac{s}{\left(s-3\right)\left(s-4\right)\left(s-12\right)}=\frac{1}{3\left(s-3\right)}-\frac{1}{2\left(s-4\right)}+\frac{1}{6\left(s-12\right)}$
Now, it is straightforward to use the Laplace inverse table, specifically, ${\mathcal{L}}^{-1}\left[\frac{1}{s+a}\right]={e}^{-at}u\left(t\right)$, so we get
$\begin{array}{rl}{\mathcal{L}}^{-1}\left[\frac{s}{\left(s-3\right)\left(s-4\right)\left(s-12\right)}\right]& ={\mathcal{L}}^{-1}\left[\frac{1}{3\left(s-3\right)}\right]-{\mathcal{L}}^{-1}\left[\frac{1}{2\left(s-4\right)}\right]+{\mathcal{L}}^{1}\left[\frac{1}{6\left(s-12\right)}\right]\\ & =\frac{1}{3}{\mathcal{L}}^{1}\left[\frac{1}{\left(s-3\right)}\right]-\frac{1}{2}{\mathcal{L}}^{1}\left[\frac{1}{\left(s-4\right)}\right]+\frac{1}{6}{\mathcal{L}}^{1}\left[\frac{1}{\left(s-12\right)}\right]\\ & =\frac{1}{3}{e}^{3t}-\frac{1}{2}{e}^{4t}+\frac{1}{6}{e}^{12t}\end{array}$