# ODES with complex roots

1. May 16, 2014

### dan280291

Hi, Im just having a little trouble with differential equations. I have y'' - 6y' + λy = 0
I know I need complex roots and setting e^$\alpha$x gives $\alpha$= 3+/-sqrt(9 - λ). Then I don't understand why set -ω^2= 9-λ.

How do you know if it is -ω^2 or w^2. Thanks for the help.

2. May 16, 2014

### LCKurtz

Complete the square on the characteristic equation:$$r^2 + 6r +\lambda = (r^2 + 6r + 9) +(\lambda - 9) = (r+3)^2 +(\lambda - 9) = 0$$So you have $(r+3)^2 = (9-\lambda)$. For complex roots you need the right side to be negative so set$$9-\lambda=-\omega^2.$$This corresponds to $\lambda > 9$.

3. May 16, 2014

### dan280291

Thanks a lot. I appreciate it.