Understanding Complex Roots in Differential Equations

[dan280291]

Hi, I am 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.

 

[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 $\lambda > 9$.

 

[dan280291]

Thanks a lot. I appreciate it.