Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz4 => Topic started by: Zhangxinbei on October 18, 2019, 02:27:42 PM

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''=4Acos2t4Bsin2t
Plug in:
A= 3/10
B= 3/20
Yp(t)=3/10cos2t3/20sin2t
Y(t) = c1+c2e^(4t)3/10cos2t3/20sin2t
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)3/10cos2t3/20sin2t