Use Laplace transform to solve the folowing initial value problem y"+2y'+y=4e^{-t} y(0)=2 y'(0)=-1

Use Laplace transform to solve the folowing initial value problem $y"+2{y}^{\prime }+y=4{e}^{-t}y\left(0\right)=2{y}^{\prime }\left(0\right)=-1$
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Step 1
Given : $y"+2{y}^{\prime }+y=4{e}^{-t}y\left(0\right)=2{y}^{\prime }\left(0\right)=-1$
Applying Laplace Transform
$L\left(y"\right)+2L\left({y}^{\prime }\right)+L\left(y\right)=4L\left({e}^{\left(}t\right)\right)$
$⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+2\left[sY\left(s\right)-y\left(0\right)\right]+Y\left(s\right)=\frac{4}{\left(s+1\right)}$
$⇒\left({s}^{2}+2s+1\right)Y\left(s\right)-2s+1-4=\frac{4}{\left(s+1\right)}\left[y\left(0\right)=2,{y}^{\prime }\left(0\right)=-1\right]$
$⇒\left(s+1{\right)}^{2}Y\left(s\right)=\frac{4}{\left(s+1\right)}+2s+3$
$⇒Y\left(s\right)=\frac{4}{\left(s+1{\right)}^{3}}+\frac{\left(2s+3\right)}{\left(s+1{\right)}^{2}}$
$⇒Y\left(s\right)=\frac{4}{\left(s+1{\right)}^{3}}+\frac{2\left(s+\frac{3}{2}+1-1\right)}{\left(s+1{\right)}^{2}}$
$⇒Y\left(s\right)=\frac{4}{\left(s+1{\right)}^{3}}+\frac{2\left(s+1\right)}{\left(s+1{\right)}^{2}}+\frac{1}{\left(s+1{\right)}^{2}}$
$Y\left(s\right)=\frac{4}{\left(s+1{\right)}^{3}}+\frac{2\left(s+1\right)}{\left(s+1{\right)}^{2}}+\frac{1}{\left(s+1{\right)}^{2}}$
Step 2
Now taking inverse Laplace Transform,
${L}^{-1}\left\{Y\left(s\right)\right\}=4{L}^{-1}\left\{\frac{4}{\left(s+1{\right)}^{3}}\right\}+2{L}^{-1}\left\{\frac{2\left(s+1\right)}{\left(s+1{\right)}^{2}}\right\}+{L}^{-1}\left\{\frac{1}{\left(s+1{\right)}^{2}}\right\}$
Using the identity ${L}^{-1}\left\{\frac{1}{\left(s+a{\right)}^{n}}\right\}=\frac{{t}^{n-1}{e}^{-at}}{\left(n-1\right)!}$ , we get
$y\left(t\right)=2{t}^{2}{e}^{-t}+2{e}^{-t}+t{e}^{-t}$