Finding the period of a differential equation?

[adl2114]

find the period of the solution to this differential equation:

2y'' + 4y' + y = 0

I used the complementary equation 2r^2 + 4r + 1 = 0 and then used the quadratic equation to factor that into 2 roots of -2 + 2^(1/2) and -2 - 2^(1/2) and I know that 2 real, distinct roots means there is no period but If the problem had been different how would I go about finding the period after determining the roots?

 

[Char. Limit]

Well, first, your roots are off by a scaling factor of 1/2. And assuming that both roots are imaginary, well, e^(ix) is periodic, but I'm not sure of the period...