# Solve the following IVP using Laplace Transformy′′+3y′+2y=e^(-t), y(0)=0 y′(0)=0

Solve the following IVP using Laplace Transform
$y\prime \prime +3y\prime +2y={e}^{-t},y\left(0\right)=0y\prime \left(0\right)=0$

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Step 1
We have to solve given IVP using Laplace transform.
Step 2
We have given
$y\prime \prime +3y\prime +2y={e}^{-t},y\left(0\right)=0y\prime \left(0\right)=0$
Taking Laplace transform on given differential equation
$L\left(y+3y\prime +2y\right)=L\left({e}^{-t}\right)$
$L\left(y\right)+3L\left(y\prime \right)+2L\left(y\right)=\frac{1}{s+1}\dots \left(i\right)$
We are using here

where $L\left(y\right)=Y\left(s\right)$
Then from equation (i) we get,
${s}^{2}Y\left(s\right)-sy\left(0\right)-y\prime \left(0\right)+3\left(sY\left(s\right)-y\left(0\right)\right)+2Y\left(s\right)=\frac{1}{s+1}$
${s}^{2}Y\left(s\right)+3sY\left(s\right)+2Y\left(s\right)=\frac{1}{s+1}$
$\left({s}^{2}+3s+2\right)Y\left(s\right)=\frac{1}{s+1}$
$\left(s\left(s+1\right)+2\left(s+1\right)\right)Y\left(s\right)=\frac{1}{s+1}$
$\left(s+1\right)\left(s+2\right)=\frac{1}{s+1}$
$Y\left(s\right)=\frac{1}{{\left(s+1\right)}^{2}\left(s+2\right)}\dots \left(ii\right)$
Now
$\frac{1}{{\left(s+1\right)}^{2}\left(s+2\right)}=\frac{A}{s+1}+\frac{B}{{\left(s+1\right)}^{2}}+\frac{C}{s+2}$
$\frac{1}{{\left(s+1\right)}^{2}\left(s+2\right)}=\frac{A\left(s+1\right)\left(s+2\right)+B\left(s+2\right)+C{\left(s+1\right)}^{2}}{{\left(s+1\right)}^{2}\left(s+2\right)}$