find the general solution of the given differential equation
$y'' +2y'=3+4sin(2t)$
1. $y'' +2y'= 0$
$r^2+2r=0$
$r_1=-2$ and $r_2=0$
$y_1 = e^{-2t}$ and $y_2= = 1$
So the general solution of the homogenous SODE is $y = c_1e^{-2t}+c_2$
2. $y''+2y' = 3$
let
$Y(t) = At$
$Y'= A$
$Y''=0$
So, $Y''+2Y'=2A=3$
$A={3\over2}$
So $y={3\over2}t $
3.$y'' +2y'=4sin(2t)$
Consider $Y=Asin(2t)+Bcos(2t)$
$Y'=2Acos(2t)-2Bsin(2t)$
$Y''= -4sin(2t)-4Bcos(2t)$
So $Y''+2Y'=-4Asin(2t)-4Bcos(2t)+4Acos(2t) - 4Bsin(2t)
=(-4A-4B)sin(2t)+(-4B+4A)cos(2t)=4sin(2t)$
$( t = 0) \Rightarrow A=B$
$(t={\pi\over 4}) \Rightarrow -A-B = 1$
$A={1\over 2}$ and $B = {1\over2}$
So $y = {1\over 2} sin(2t) + {1\over 2}cos(2t)$
The general solution is $y = c_1e^{-2t}+c_2+{3\over2}t+{1\over 2} sin(2t) + {1\over 2}cos(2t) $
