# 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???

Related Calculus and Beyond Homework Help News on Phys.org
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.