Question

solve the DE y'''+2y'' +y=0

Answer 1

let y=e^(mx) be the trial solution
now here y' = my
and y'' =m^2 y
thus the auxillary eq is
m^2+m+1=0
or (m+1)^2 =0
because roots of above eq is -1,-1
thus the required solution
is y=(a+bx)*e^(-x)

Answer 2

wouldn't the auxillary eq be m^3 +m^2 +1=0?

Answer 3

yup
u r correct
i mistook the question.