- #1
adl2114
- 7
- 0
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?
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?