# Use the Laplace transform to solve the given initial value problem.y"+2y'+5y=0;y

Use the Laplace transform to solve the given initial value problem.y"+2y'+5y=0;y(0)=2,y'(0)=−1
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yagombyeR
${s}^{2}Y-sy\left(0\right)-{y}^{\prime }\left(0\right)+2sY-2y\left(0\right)+5Y=0$ Take the Laplace transform of both sides.
$Y\left({s}^{2}+2s+5\right)-2s+1-4=0$ Substitute in the intial conditions and solve for Y.
$Y\left({\left(s+1\right)}^{2}+4\right)=2s+3$
$Y=\frac{2s+3}{{\left(s+1\right)}^{2}+4}$
$Y=\frac{2\left(s+1-1\right)+3}{{\left(s+1\right)}^{2}+4}$ Add 1 to the s on top so that it matches the one on the bottom of the fraction, then split the fraction into two.
$Y=\frac{2\left(s+1\right)+1}{{\left(s+1\right)}^{2}+4}$
$Y=\frac{2\left(s+1\right)}{{\left(s+1\right)}^{2}+4}+\frac{1}{{\left(s+1\right)}^{2}+4}$
$y=2{e}^{-t}\mathrm{cos}2t+\frac{1}{2}{e}^{-t}\mathrm{sin}2t$ Take the inverse Laplace transform
Result:
$y=2{e}^{-t}\mathrm{cos}2t+\frac{1}{2}{e}^{-t}\mathrm{sin}2t$