# Use the Laplace transform to solve the given initial-value problem {y}{''}+{2}{y}'+{y}={0},{y}{left({0}right)}={1},{y}'{left({0}right)}={1}

Use the Laplace transform to solve the given initial-value problem
$y{}^{″}+2{y}^{\prime }+y=0,y\left(0\right)=1,{y}^{\prime }\left(0\right)=1$
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Step 1
It is given that the initial value problem is $y{}^{″}+2{y}^{\prime }+y=0,y\left(0\right)=1,{y}^{\prime }\left(0\right)=1$
Step 2
Take Laplace transformation as,
$L\left\{y\right\}+2L\left\{{y}^{\prime }\right\}+L\left\{y\right\}=L\left\{0\right\}$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+2\left[sY\left(s\right)-y\left(0\right)\right]+Y\left(s\right)=0$
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+2sY\left(s\right)-2y\left(0\right)+Y\left(s\right)=0$
${s}^{2}Y\left(s\right)-s-1+2sY\left(s\right)-2+Y\left(s\right)=0$
$Y\left(s\right)\left({s}^{2}+2s+1\right)-s-3=0$
$Y\left(s\right)\left({s}^{2}+2s+1\right)=s+3$
Step 3
On further simplification,
$Y\left(s\right)=\frac{s+3}{{\left(s+1\right)}^{2}}$
$y\left(t\right)={L}^{-1}\left\{\frac{s+3}{{\left(s+1\right)}^{2}}\right\}$

$y\left(t\right)={L}^{-1}\left\{\frac{2}{{\left(s+1\right)}^{2}}\right\}+{L}^{-1}\left\{\frac{1}{s+1}\right\}$
$y\left(t\right)=\frac{2t}{{e}^{t}}+{e}^{-t}$
$y\left(t\right)={e}^{-t}\left(2t+1\right)$
Step 4
The solution of the initial value problem is $y\left(t\right)={e}^{-t}\left(2t+1\right)$