The first one I'm having trouble with is:

y' = [2 1; -1 1]y

I found the eigenvalues to be 1.5 +/- 0.866i

I don't deal with complex numbers very often, so I'm a little rusty. I know the two eigenvectors are complex conjugates, but I'm having a difficult time finding them. A little help?

The second one is x" + 3x' +2x = 0

The professor mentioned looking in the book to see how a higher order ODE can be converted to a series of first order ODEs, but the book only has like half a page on it, and I don't quite understand it.

Thanks.

semiosFebruary 6 2009, 01:44:30 UTC 8 years ago

y1 = x

y2 = x'

Our system of equations is going to look like this:

yn' = ... with a first order derivate on the LHS and no derivates on RHS side of the equation

y1' = x' = y2

y2' = x'' = - 3x' - 2x = -3 y2 - y1

So our set of equivalent first order ODEs is,

y1' = y2

y2' = -3 y2 - y1

## Probably a bit late...

maynard_loverFebruary 9 2009, 07:50:14 UTC 8 years ago

So, we have the d.e. x" + 3x' + 2x = 0. Since it is a homogeneous equation (meaning equal to zero in this case, as in diff. eq. there is another type of d.e. that involves a homogeneous

function, which would require a different solution technique), we can make an auxiliary equation. Using m, we get:(m^2 + 3m + 2) = 0 (So, if it's the second derivative, i.e. x" in this case, it becomes a squared term, the first derivative becomes the m to the first power term, and the 2x just becomes a 2 in our auxiliary equation since it doesn't contain a derivative.)

Then, solve our new quadratic. Factoring, the easiest method, works here.

So, we get:

(m + 2) (m + 1) = 0

Set each part equal to zero, and we get m

_{1}= -2 and m_{2}= -1.Then, the solution to our auxiliary equation becomes x = c

_{1}e^(m_{1}x) + c_{2}e^(m_{2}x), where the c's represent constants. And it makes sense that we have two terms since our original differential equation is second order. If we had had a third order ODE, then we would add on a third term obviously.So, now we substitute our m values in and get x = c

_{1}e^(-2x) + c_{2}e^(-x). This is our solution, but we can now check it to make sure we are right.x" + 3x' + 2x = 0

So, x = c

_{1}e^(-2x) + c_{2}e^(-x). Take the derivative, twice.x' = -2*c

_{1}e^(-2x) + -c_{2}e^(-x),and x" = 4c

_{1}e^(-2x) + c_{2}e^(-x).(Just as a reminder, the derivative of e^(f(x)) is f'(x)*e^(f(x)).)

Now, plug in and multiply out.

[4c

_{1}e^(-2x) + c_{2}e^(-x)] + 3*[-2*c_{1}e^(-2x) + -c_{2}e^(-x)] + 2*[c_{1}e^(-2x) + c_{2}e^(-x)] = 0,which becomes:

4c

_{1}e^(-2x) + c_{2}e^(-x) + -6c_{1}e^(-2x) + -3c_{2}e^(-x) + 2c_{1}e^(-2x) + 2c_{2}e^(-x) = 0, and all terms cancel out, leaving us with zero. Thus, our solution ofx = cis indeed correct._{1}e^(-2x) + c_{2}e^(-x)Hope this helps! Any questions, feel free to ask.