# 3rd order ODE

1. Jul 17, 2011

### newtomath

I am stuck on solving for the roots of a charactristic equation:

y'''- y''+y'-y=0

where I set r^3-r^2+r-1=0 and factored out r to get r*[ r^2-r +1] -1 =0 to get the real root of 1. How can I solve for the compex roots?

2. Jul 17, 2011

### SteamKing

Staff Emeritus
By inspection, r = 1 is a root of your characteristic equation.
In order to find the other roots, you should factor (r - 1) from the char. eq.

3. Jul 17, 2011

### hunt_mat

So what I would do is write:
$$(r-1)(ar^{2}+br+c)=r^{3}-r^{2}+r-1$$
Expand and equate coefficients, then solve the quadratic

4. Jul 17, 2011

### pmsrw3

Typically people use long division. But in this case it's obvious that the other factor is r^2+1.

Or you could just go to Wolfram Alpha and say "factor r^3-r^2+r-1" :-)

Last edited: Jul 17, 2011
5. Jul 17, 2011

### hunt_mat

I never really understood long division.

6. Jul 18, 2011

### newtomath

Got it now, thanks