# Use Laplace transform to solve the folowing initial value problem y"+2y'+2y=0 y(0)=2 y'(0)=-1

Use Laplace transform to solve the folowing initial value problem $y"+2{y}^{\prime }+2y=0$
$y\left(0\right)=2$
${y}^{\prime }\left(0\right)=-1$
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berggansS
Step 1Given $y"+2{y}^{\prime }+2y=0$
$y\left(0\right)=2$
${y}^{\prime }\left(0\right)=-1$Taking Laplace transform on both sides$L\left\{y"\right\}+2L\left\{{y}^{\prime }\right\}+2L\left\{y\right\}=0$
$⇒{s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+2\left[sy\left(s\right)-y\left(0\right)\right]+2y\left(s\right)=0$
given$y\left(0\right)=2$
${y}^{\prime }\left(0\right)=-1$
$⇒{s}^{2}y\left(s\right)-2s+1+2\left[sy\left(s\right)-2\right]+2y\left(s\right)=0$
$⇒\left[{s}^{2}+2s+2\right]y\left(s\right)-2s-3=0$
$⇒y\left(s\right)=\frac{2s+3}{{s}^{2}+2s+2}$Step 2${L}^{-1}y\left(s\right)={L}^{-1}\left\{\frac{2s+3}{{s}^{2}+2s+2}\right\}$
$={L}^{-1}\left\{2\frac{s+1}{\left(s+1{\right)}^{2}+1}+1\frac{1}{\left(s+1{\right)}^{2}+1}\right\}$
$=2{L}^{-1}\left\{\frac{s+1}{\left(s+1{\right)}^{2}+1}\right\}+{L}^{-1}\left\{\frac{1}{\left(s+1{\right)}^{2}+1}\right\}$
$⇒y\left(t\right)=2{e}^{-t}\mathrm{cos}t+{e}^{-t}\mathrm{sin}t$
$⇒y\left(t\right)=2{e}^{-t}\mathrm{cos}t+{e}^{-t}\mathrm{sin}t$