# Direction of oscillations

1. Apr 8, 2005

### stunner5000pt

how would one go about finding the direction of oscillation for a differntial equation?

for example $$\frac{d^2 y}{dt^2} = -2y$$
has eigenvalues $\pm \sqrt{2}i$
and the corresponding matrix is $$\left(\begin{array}{cc}0&2\\-2&0\end{array}\right)$$ so the solution curves will form closed loops around the origin. But what about the direction -i.e. counter clockwise or clockwise, how would one go about figuring this out??

please help!!! I have an exam tomorrow on this stuff and this is where i am snagged!

2. Apr 9, 2005

### stunner5000pt

the two associated eigenvectors would be computed as follows

$AX=\lambda X$

$$\left(\begin{array}{cc}0&2\\-2&0\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = i \sqrt{2} \left(\begin{array}{cc}x\\y\end{array}\right)$$

which gives teh equation
$$2y = i \sqrt{2} x$$ and
$$-2x = i \sqrt{2} y$$
which gives me eigenvector which is $2y = i \sqrt{2} x$ which gives me $$V_{1} = \left(\begin{array}{cc}2\\i \sqrt{2}\end{array}\right)$$ and $$V_{2} = \left(\begin{array}{cc}2\\-i \sqrt{2}\end{array}\right)$$
is this correct?? Am i supposed to simplify the eigenvectors??
Consider the general example: $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$.
The characteristic equation is $$\left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|= \lambda^2- (a+d))\lamba+ ad-bc= 0$$.
The solutions to that depend on two numbers, not 3. Specifically, a+ d, the "trace" of the matrix and ad-bc, the "determinant" of the matrix. Any two matrices having the same trace and matrix will have the same eigenvalues. In particular, $$\left(\begin{array}{cc}0&-2\\2&0\end{array}\right)$$ has exactly the same eigenvalues but opposite rotation. In this case, since the eigenvectors are complex and you are working in the real plane, they don't help.