# Find the inverse of Laplace transform frac{2s+3}{(s-7)^4}

Find the inverse of Laplace transform
$\frac{2s+3}{\left(s-7{\right)}^{4}}$
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Step 1
Find
${L}^{-1}\left\{\frac{2s+3}{\left(s-7{\right)}^{4}}\right\}$
For this take partial fraction
$\frac{2s+3}{\left(s-7{\right)}^{4}}=\frac{2}{\left(s-7{\right)}^{3}}+\frac{17}{\left(s-7{\right)}^{4}}$
Step 2
Use linearity property of Inverse Laplace transformation
That is , for function f(s),g(s) and a,b be any constants.
${L}^{-1}\left\{af\left(s\right)+bf\left(s\right)\right\}=a{L}^{-1}\left\{f\left(s\right)\right\}+b{L}^{-1}\left\{g\left(s\right)\right\}$
Step 3
Now equation (i) becomes
${L}^{-1}\left\{\frac{2s+3}{\left(s-7{\right)}^{4}}\right\}={L}^{-1}\left\{\frac{2}{\left(s-7{\right)}^{3}}\right\}+{L}^{-1}\left\{\frac{17}{\left(s-7{\right)}^{4}}\right\}={e}^{7t}{t}^{2}+\frac{17{e}^{7t}{t}^{3}}{6}$
Formula used
${L}^{-1}\left(F\left(s\right)\right)=f\left(t\right)$
${L}^{-1}\left(F\left(S-a\right)\right)={e}^{at}f\left(t\right)$
${L}^{-1}\left(\frac{n!}{{S}^{n+1}}\right)={t}^{n}$