Which of the following functions is a solution to the differential equation y" + 2y' + y = 0

possible answers:

y = et

y = -e t

y = e -t

y = e -2t

i tryed solving it put i only get :

y(x) = c1e^(-x)+c2e^(-x) x

Question

Which of the following functions is a solution to the differential equation y" + 2y' + y = 0

possible answers:
	
y = et
		
y = -e t
		
y = e -t
		
y = e -2t


i tryed solving it put i only get :

y(x) = c1e^(-x)+c2e^(-x) x

anonymous · Answer

$$y''+2y'+y=0$$
gives the characteristic equation
$$r^2+2r+1=(r+1)^2=0$$
which has root $r=-1$ of multiplicity 2, which means the solution would be of the form
$$y=C_1e^{-x}+C_2xe^{-x}$$
By the principle of superposition, either one of these terms is considered a solution, so $y=e^{-x}$ is a solution if $C=1$.