• Support PF! Buy your school textbooks, materials and every day products Here!

Complex Eigenvectors

  • #1

Homework Statement



The higher order equation y"+y=0 can be written as a unknown d/dt[y y']=[y' y"]=[y' -y]

If this is du/dt=Au, what is the 2x2 matrix A? Find its eigenvectors and eigenvalues, and compute the solution THAT STARTS FROM y(0)=2, y'(0)=0.

Homework Equations



y'=Ay
y(0)=y0

The Attempt at a Solution



I found matrix A
[0 1
-1 0].
The eigenvalues are i and -i, and the eigenvectors
[1 -i]^T
[1 i]^T

I found the geneal solution to be:
y(t) = c1eit[1 i]^T+c2e-it[1 -i]^T

Which is equivalent,
y(t)=c1[cos(t) -sin(t)]^T + c2[sin(t) cos(t)]^T

I just don't know how to incorporate the initial conditions that y(0)=2 and y'(0)=0???

Any ideas???
 

Answers and Replies

  • #2
254
0
You know the function y(t), and you know what y(0) equals (and you know y'(t) from differentiating your original y(t) equation), so then try plugging in for t and see what you get.
 
  • #3
That's what I tried doing but I'm getting a funky solution:

y(0)=2=c1*[1 0]^T + c2*[0 1]^T, so this is saying that [c1 c2]^T=2 ?????

When I differentiate y(t) as in previous post,
y'(t)=[c2*cos(t)-c1*sin(t) -c1*cos(t)-c2*sin(t)]^T
y'(0)=0=[c2 -c1]^T=0

Totally lost!
 
  • #4
I like Serena
Homework Helper
6,577
176
Hi tatianaiistb! :smile:

Your solution is not y(t), but [y(t) y'(t)] which is a vector as it should be.
 

Related Threads on Complex Eigenvectors

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
7K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
4
Views
893
  • Last Post
Replies
3
Views
970
  • Last Post
Replies
2
Views
2K
Replies
5
Views
1K
Replies
7
Views
548
  • Last Post
Replies
11
Views
1K
Top