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

Step 1
We have to solve given IVP using Laplace transform.
Step 2
We have given
$y\prime \prime +3y\prime +2y=e^{-t},y(0)=0y\prime (0)=0$
Taking Laplace transform on given differential equation
${\displaystyle L(y+3y\prime +2y)=L\left(e^{-t}\right)}$
${\displaystyle L\left(y\right)+3L\left(y\prime \right)+2L\left(y\right)=\frac{1}{s+1}\dots \left(i\right)}$
We are using here
${\displaystyle L\left(y\right)=s^{2}Y\left(s\right)-sy\left(0\right)-y\prime \left(0\right),L\left(y\prime \right)=sY\left(s\right)-y\left(0\right)\text{ and }L\left(e^{-t}\right)=\frac{1}{s+1}}$
where $L(y)=Y(s)$
Then from equation (i) we get,
${\displaystyle s^{2}Y\left(s\right)-sy\left(0\right)-y\prime \left(0\right)+3(sY\left(s\right)-y\left(0\right))+2Y\left(s\right)=\frac{1}{s+1}}$
${\displaystyle s^{2}Y\left(s\right)+3sY\left(s\right)+2Y\left(s\right)=\frac{1}{s+1}}$
${\displaystyle (s^2+3s+2)Y\left(s\right)=\frac{1}{s+1}}$
${\displaystyle (s(s+1)+2(s+1))Y\left(s\right)=\frac{1}{s+1}}$
${\displaystyle (s+1)(s+2)=\frac{1}{s+1}}$
${\displaystyle Y\left(s\right)=\frac{1}{{(s+1)}^2(s+2)}\dots \left(ii\right)}$
Now
${\displaystyle \frac{1}{{(s+1)}^2(s+2)}=\frac{A}{s+1}+\frac{B}{{(s+1)}^2}+\frac{C}{s+2}}$
${\displaystyle \frac{1}{{(s+1)}^2(s+2)}=\frac{A(s+1)(s+2)+B(s+2)+C{(s+1)}^2}{{(s+1)}^2(s+2)}}$