Find the Laplace transform Y(s), of the solution of the IVP y"+3y'+2y=cos(2t) y(0)=0 y'(0)=1 Do not solve the IVP

Step 1
The given initial value problem is $y^{″}+3y^{\prime }+2y=\cos (2t),y(0)=0,y^{\prime }(0)=1$
Find the Laplace transform Y(s) of the solution of the IVP as follows.
Step 2Consider the differential equation $y^{″}+3y^{\prime }+2y=\cos (2t)$
Take Laplace transform on both sides and obtain,
$L\{y"+3y^{\prime }+2y\}=L\{\cos (2t)\}$
$L\left\{y"\right\}+3L\left\{y^{\prime }\right\}+2L\left\{y\right\}=L\{\cos (2t)\}$
$s^{2}Y(s)-sy(0)-y^{\prime }(0)+3(sY(s)-y(0))+2(Y(s))=\frac{s}{(s^2+4)}$
$s^{2}Y(s)-sy(0)-1+3(sY(s)-y(0))+2(Y(s))=\frac{s}{(s^2+4)}$
$s^{2}Y(s)-1+3sY(s)+2Y(s)=\frac{s}{(s^2+4)}$
$Y(s)(s^2+3s+2)-1=\frac{s}{(s^2+4)}$
$Y(s)(s^2+3s+2)=\frac{s}{(s^2+4)+1}$
$Y(s)=\frac{s+(s^2+4)}{(s^2+4)(s^2+3s+2)}$
$Y(s)=\frac{(s^2+s+4)}{(s^2+4)(s^2+3s+2)}$
Step 3
Therefore, the Laplace transform Y(s) of the solution of the IVP is,$Y(s)=\frac{(s^2+s+4)}{(s^2+4)(s^2+3s+2)}$