Eigenvectors/values and diff equations

[EvLer]

Homework Statement

the higher order equation y'' + y = 0 can be written as a first order system by introducing y' as another unknown besides y:

Homework Equations

d[y y']^t/dt = [y' y'']^t = [y' -y]^t

if this is du/dt = Au what is 2x2 matrix A?

The Attempt at a Solution

I think it's
[0 1]
[-1 0]
but then I get complex eigenvalues so I am a bit suspicious... could someone verify or help me out?
thanks.

 

[HallsofIvy]

You are exactly correct. If you let x= dy/dt, then d²y/dt²= dx/dt+ y= 0 or dx/dt= -y so your two equations are dy/dt= x and dx/dt= -1. Letting Y(t)= (y(t) x(t))^t, yes, your differential equation is
Y(t)= \left[\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right]Y(t)
and, yes, that has complex eigenvalues. Why does that bother you?
I notice that the problem only asks you to write the matrix equation, not to find the eigenvalues or solve the equation so I guess you will learn what complex eigenvalues mean in this situation later.

 

[tatianaiistb]

I'm trying to finish the very same problem. In addition to what is stated above, the problem says: "Find its eigenvectors, and compute the solution that starts from y(0), y'(0)=0."

I found the eigenvalues to be i and -i, and the eigenvectors to be
[1 -i]^T and [1 i]^T.

I wrote the solution to start as y(t)=(1/2)*e^it*[1 i]^T + (1/2)*e^-it*[1 -i]^T.

From there I don't know how to proceed. Any suggestions?

 

[Mark44]

e^it = cos(t) + isin(t)

Note that (1/2)(e^it + e^-it) = cos(t)
and that (1/(2i))(e^it - e^-it) = sin(t)