Question

how to solve this system:
x'=x-y+z
y'=x+y-z
z'=2z-y

Answer 1

Gaussian elimination

Answer 2

This system has repeated eigenvalues 2 and 1(multiplicity 2)

Answer 3

so the result should be something like: e^t(c1+tc2)eigenvector+e^2tc3eigenvector

Answer 4

In general, yes.
The surest way is to substitute the solution and differentiate. It takes only seconds to do.

Answer 5

substitute with what?

Answer 6

Put the solution back into the original equations and differentiate to see if they come out equal.
For the repeated eigenvalue, there is one more step to do to find an additional eigenvector from the one with multiplicity two.

Answer 7

Try Paul's online notes:
http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx
for tackling the repeated eigenvalues.

Answer 8

if i solve this for 1: i get eigenvector [1,1,1]
and x1= c1e^t[1,1,1]+t*c2e^t[1,1,1]?

Answer 9

and if i solve this for 2: [1,0,1,]
x2 = c3*e^(2t)[1,0,1]

Answer 10

so the result should be for x(t) = e^t(c1+tc2)+e^(2t)c3

Answer 11

cuz i dont know if i made a mistake....we said in school it should be x(t)=e^t(c1+c2+tc2)+e^(2t)c3

Answer 12

"2" is a distinct eigenvalue, so it requires only a single term.
"1" is a repeated eigenvalue, but you need two eigenvectors.
There are times you get two eigenvectors out of one single eigenvalue, which would make life easier. You'll find out when you put it to the echelon form.

Unfortunately the echelon form of this system only gives you one eigenvector, so you'd have to calcuate another one. See Paul's online notes for the procedure.

Answer 13

ok, thaks

Answer 14

For the eigenvalue with multiplicity 2, it would be in the form:
X=c1v1e^-t + c2[ v1 te^-t + v2 e^-t]
where v1,v2 are the eigenvectors.
It is similar but not identical to what you wrote.