# y''+y=e^{-2t}sin t , y(0)=y′(0)=0 solution of given initial value problem with Laplace transform

$y+y={e}^{-2t}\mathrm{sin}t,y\left(0\right)=y\prime \left(0\right)=0$
solution of given initial value problem with Laplace transform
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Nathanael Webber
Given:
$y+y={e}^{-2t}\mathrm{sin}t,y\left(0\right)=y\prime \left(0\right)=0$
To find: solution of given initial value problem with Laplace transform
Step 2 Solution
$y+y={e}^{-2t}\mathrm{sin}t,y\left(0\right)=y\prime \left(0\right)=0$
Taking Laplace transform of the differential equation , we get:
$\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-y\prime \left(0\right)\right]+Y\left(s\right)=L\left[{e}^{-2t}\mathrm{sin}t\right]$ where $Y\left(s\right)=L\left(y\left(t\right)\right)$
using initial values we get
$\left[{s}^{2}Y\left(s\right)-0\right]+Y\left(s\right)=L\left[{e}^{-2t}\mathrm{sin}t\right]$

$L\left(\mathrm{sin}t\right)=\frac{1}{{s}^{2}+1}$
$L\left({e}^{-2t}\mathrm{sin}t\right)=\frac{1}{\left(s+2{\right)}^{2}+1}$
Step 3
$⇒Y\left(s\right)\left({s}^{2}+1\right)=\frac{1}{\left(s+2{\right)}^{2}+1}$
$⇒Y\left(s\right)=\frac{1}{\left({s}^{2}+1\right)\left(\left(s+2{\right)}^{2}+1\right)}$
after partial fraction we get
$Y\left(s\right)=\frac{1}{8}\left[-\frac{s}{{s}^{2}+1}+\frac{1}{{s}^{2}+1}+\frac{s}{\left(s+2{\right)}^{2}+1}+\frac{3}{\left(s+2{\right)}^{2}+1}\right]$
using inverse Laplace transform to get y(t) :
$y\left(t\right)={L}^{-1}\left[Y\left(s\right)\right]$
$=\frac{1}{8}\left[-\mathrm{cos}t+\mathrm{sin}t+{e}^{-2t}\mathrm{cos}t+3{e}^{-2t}\mathrm{sin}t\right]$
required solution of differential equation .