# find the inverse of Laplace transform frac{3}{(s+2)^2}-frac{2s+6}{(s^2+4)}

find the inverse of Laplace transform
$\frac{3}{\left(s+2{\right)}^{2}}-\frac{2s+6}{\left({s}^{2}+4\right)}$
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Step 1
Given Laplace transform is,
$\frac{3}{\left(s+2{\right)}^{2}}-\frac{2s+6}{\left({s}^{2}+4\right)}$
Step 2
Taking inverse of Laplace transformation as,
${L}^{-1}\left\{\frac{3}{\left(s+2{\right)}^{2}}-\frac{2s+6}{\left({s}^{2}+4\right)}\right\}={L}^{-1}\left\{\frac{3}{\left(s+2{\right)}^{2}}\right\}-{L}^{-1}\left\{\frac{2s+6}{\left({s}^{2}+4\right)}\right\}$
$=3{L}^{-1}\left\{\frac{1}{\left(s+2{\right)}^{2}}\right\}-2{L}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}-6{L}^{-1}\left\{\frac{1}{{s}^{2}+4}\right\}$
Step 3
Some formulas of inverse of Laplace transformation are
1)
$⇒{L}^{-1}\left\{\frac{1}{\left(s+2{\right)}^{2}}\right\}={e}^{-2t}{L}^{-1}\left\{\frac{1}{{s}^{2}}\right\}={e}^{-2t}t$
2)${L}^{-1}\left\{\frac{a}{{s}^{2}+{a}^{2}}\right\}=\mathrm{sin}at$
$⇒{L}^{-1}\left\{\frac{1}{{s}^{2}+4}\right\}=\frac{1}{2}{L}^{-1}\left\{\frac{2}{{s}^{2}+4}\right\}=\frac{\left(\mathrm{sin}2t\right)}{2}$
3)${L}^{-1}\left\{sF\left(s\right)\right\}={f}^{\prime }\left(t\right)+f\left(0\right)$
Then ${L}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}=\frac{1}{2}{L}^{-1}\left\{s\frac{2}{{s}^{2}+4}\right\}=\frac{1}{2}\left(\left(\mathrm{sin}2t{\right)}^{\prime }+\mathrm{sin}\left(0\right)\right)=\mathrm{cos}2t$
Step 4
Then, consider equation (1),
$3{L}^{-1}\left\{\frac{1}{\left(s+2{\right)}^{2}}\right\}-2{L}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}-6{L}^{-1}\left\{\frac{1}{{s}^{2}+4}\right\}=3{L}^{-1}\left\{\frac{1}{\left(s+2{\right)}^{2}}\right\}-2{L}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}-\frac{6}{2}{L}^{-1}\left\{\frac{2}{{s}^{2}+4}\right\}=3{e}^{-2t}t-2\mathrm{cos}2t-3\mathrm{sin}2t$
Step 5
Therefore, the inverse of the given Laplace transformation is,
${L}^{-1}\left\{\frac{3}{\left(s+2{\right)}^{2}}-\frac{2s+6}{{s}^{2}+4}\right\}=3{e}^{-2t}t-2\mathrm{cos}2t-3\mathrm{sin}2t$