Characteristic polynomial help

roryhand
Messages
2
Reaction score
0
y^(7)-y^(6)-2y^(4)+2y^(3)+dy-y=0

Note: There is exactly one real zero of the characteristic polynomial and it
has multiplicity 3 (it is a positive integer!). The other zeros are complex
and they have multiplicity 2.

Sadly I missed this lecture day, and am unsure of where to start. Any differential equation demi-gods out there?
 
Physics news on Phys.org
For an equation of order n, if a root (say r1) has a multiplicity s (s =< n), where x is the independent variable

e^{r_{1}x}, xe^{r_{1}x}, x^{2} e^{r_{1}x}, ..., x^{s-1} e^{r_{1}x}

For complex roots, let's say a+bi is repeated s times, then the complex conjugate a-bi is also repeated s times, therefore the solutions for real valued functions, where x is the independent variable:

e^{ax} \cos{bx}, e^{ax} \sin{bx}, xe^{ax} \cos{bx}, xe^{ax} \sin{bx},..., x^{s-1} e^{ax} \cos{bx}, x^{s-1} e^{ax} \sin{bx}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top