# Find the inverse laplace transforms for the following functions F(s)=frac{e^{-3s}}{s^2} F(s)=frac{2e^{-5s}}{s^2+4}

Find the inverse laplace transforms for the following functions
$F\left(s\right)=\frac{{e}^{-3s}}{{s}^{2}}$
$F\left(s\right)=\frac{2{e}^{-5s}}{{s}^{2}+4}$
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Step 1
Given
the given functions are
$F\left(s\right)=\frac{{e}^{-3s}}{{s}^{2}}$
$F\left(s\right)=\frac{2{e}^{-5s}}{{s}^{2}+4}$
Apply formula ${L}^{-1}\left\{G\left(s\right){e}^{-as}\right\}=g\left(t-a\right)u\left(t-a\right)$
Step 2
$F\left(s\right)=\frac{{e}^{-3s}}{{s}^{2}}$
${L}^{-1}\left(\frac{1}{{s}^{2}}\right)=t$
Apply formula.
${L}^{-1}\left\{\left(\frac{1}{{s}^{2}}\right){e}^{-3s}\right\}=\left(t-3\right)u\left(t-3\right)$
Hence inverse Laplace transform of the function is $F\left(t\right)=\left(t-3\right)u\left(t-3\right)$
$F\left(s\right)=\frac{2{e}^{-5s}}{\left({s}^{2}+4\right)}$
${L}^{-1}\left(\frac{2}{\left({s}^{2}+4\right)}\right)=2{L}^{-1}\left(\frac{1}{\left({s}^{2}+4\right)}\right)=2\cdot \frac{\mathrm{sin}\left(2t\right)}{2}=\mathrm{sin}\left(2t\right)$
Apply formula.
${L}^{-1}\left\{\left(\frac{2}{{s}^{2}+4}\right){e}^{-5s}\right\}=\mathrm{sin}\left[2\left(t-5\right)\right]u\left(t-5\right)=\mathrm{sin}\left[2t-10\right]u\left(t-5\right)$
Hence inverse Laplace transform of the function is $F\left(t\right)=\mathrm{sin}\left[2t-10\right]u\left(t-5\right)$