# Complex Eigenvectors

## 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.

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???

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.

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!

I like Serena
Homework Helper
Hi tatianaiistb!

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