solve the given initial value problem:
y''+4y'+4y=1, y(-1)=2, y'(-1)=1
solution:
r^2+4r+4=0
(r+2)^2=0
r1=r2=-2
Repeated roots
Homo solution is
y(t)=(c1)e^(-2t)+(c2)e^(-2t) (t)
y'(t)=-2(c1) e^(-2t)+(c2) e^(-2t)-2t(c2) e^(-2t)
Plug in initial value y(-1)=2 and y’(-1)=1
c2=5/(e^2)
c1=7/(e^2)
y(t)=(7/e^2)e^(-2t)+(5/e^2)e^(-2t) (t)
