Solving Characteristic Equation: y'''-y''+y'-y=0

[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?

 

[SteamKing]

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.

 

[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

 

[pmsrw3]

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

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" :-)

 

[hunt_mat]

I never really understood long division.

 

[newtomath]

Got it now, thanks