# ODES with complex roots

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.

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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##.