# L^{-1}bigg(frac{5}{s^{2}(s^{2}-4s+5)}bigg)=5int_{0}^{infty}(t-T)e^{2T}sin 2T dT Select one: True or False

Select one: True or False
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Step 1
Given:
Let us consider,
${L}^{-1}\left(\frac{5}{{s}^{2}\left({s}^{2}-4s+5\right)}\right)$
Step 2
${L}^{-1}\left(\frac{5}{{s}^{2}\left({s}^{2}-4s+5\right)}\right)$
$={L}^{-1}\left(\frac{5}{{s}^{2}}\cdot \frac{1}{{s}^{2}-4s+5}\right)$
$\text{Let us consider,}$
$H\left(s\right)=\frac{5}{{s}^{2}}$
$\text{By Applying the inverse laplace transform as below,}$
${L}^{-1}\left(H\left(s\right)\right)=5{L}^{-1}\left(\frac{1}{{s}^{2}}\right)$
$h\left(t\right)=5t$
$\text{Step 3}$
$\text{Let us consider,}$
$G\left(s\right)=\frac{1}{{s}^{2}-4s+5}$
$G\left(s\right)=\frac{1}{{s}^{2}-4s+4+1}$
$G\left(s\right)=\frac{1}{\left(s-2{\right)}^{2}+1}$

${L}^{-1}\left(G\left(s\right)\right)={L}^{-1}\left(\frac{1}{\left(s-2{\right)}^{2}+1}\right)$
$={e}^{2t}\mathrm{sin}t$
$\text{Step 4}$
${L}^{-1}\left(H\left(s\right)G\left(s\right)\right)={\int }_{0}^{\mathrm{\infty }}h\left(t-T\right)\cdot g\left(T\right)$

${L}^{-1}\left(\frac{5}{{s}^{2}\left({s}^{2}-4s+5\right)}\right)=5{\int }_{0}^{\mathrm{\infty }}\left(t-T\right){e}^{2T}\mathrm{sin}T$
Hence ,
The given Equation is False.