Author Topic: TUT0202Quiz1  (Read 655 times)

chaoy

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TUT0202Quiz1
« on: September 27, 2019, 06:31:12 PM »
(y)'-y=1+3sin(t),y(0)=y0
 
p(t)=-1,q(t)=1+3sin(t)
\mu (t)=e^{\int p(t)dt}=e^{\int (-1)dt}=e^{-t}
\mu (y)'-e^{-t}y=e^{-t}(1+3sin(t))
d(e^{-t}y)/dt=e^{-t}+3e^{-t}sin(t)
\int(d(e^{-t}y)/dt)=\int(e^{-t}+3e^{-t}sin(t)dt)
e^{-t}y=-e^{-t}+(-3/2)sin(t)e^{-t}+(-3/2)cos(t)e^{-t}+c
y(t)=-1+(-3/2)sin(t)+(-3/2)cos(t)+ce^{t}
y(0)=y0=-1+(-3/2)sin(0)+(-3/2)cos(0)+ce^{0}=-1+(-3/2)+c
c=y0+(5/2)
y(t)=-1+(-3/2)sin(t)+(-3/2)cos(t)+(y0+(5/2))e^{t}
if y0<-5/2,y0+(5/2)<0
   \lim_{t\to \infty}y(t)=\lim_{t\to \infty}(y0+(5/2))e^{-t}=-\infty
if y0>-5/2,y0+(5/2)>0
   \lim_{t\to \infty}y(t)=\lim_{t\to \infty}(y0+(5/2))e^{-t}=+\infty
if y0=-5/2,y0+(5/2)=0
   \lim_{t\to \infty}y(t)=\lim_{t\to \infty}(-1+(-3/2)sin(t)+(-3/2)cos(t))is finite
so y0=-5/2