find the general sollution for a 3rd order differential equation: y'''+y=0?

Question

Charateristic equation r^3+1=0 => r= -1 => y1= e^(-x). How to find y2 and y3 so that we can have a general solution y(x)= c1*y1+ c2*y2 + c3*y3?

Answer 1

r^3 + 1 = 0 => (r+1) (r^2 - r + 1) = 0
=> r = -1 or 1/2 ± √3/2 i
=> y1 = e^(-x)
y2 = e^(x/2) sin (√3 x / 2)
y3 = e^(x/2) cos (√3 x / 2)

Answer 2

You haven't found all the roots to your characteristic equation. r^3+1 = (r+1)(r^2-r+1). Find the imaginary roots to the resolvent quadratic and you've got it.

Answer 3

The general solution is Ae^(-x)