y'' +4y' = 3sin2t, y(0) = 0, y'(0)= -1
Sol:
r^2+4r=0
r(r+4)=0
r1=0,r2=-4
y=c1+c2e^(-4t)
y'' +4y' = 3sin2t
set Yp=Acos2t+Bsin2t
Yp'=-2Asin2t+2Bcos2t
Yp''=-4Acos2t-4Bsin2t
Plug in:
A= -3/10
B= -3/20
Yp(t)=-3/10cos2t-3/20sin2t
Y(t) = c1+c2e^(-4t)-3/10cos2t-3/20sin2t
plug in the initial values:
C1=1/8
C2=7/40
Therefore, the solution of this initial value problem is:
Y(t) = 1/8+7/40e^(-4t)-3/10cos2t-3/20sin2t
