Comments | calculus: Systems of ODEs

otakuchibijosh in calculus

Systems of ODEs

Feb 03, 2009 18:04

I'm a graduate in chemical engineering taking a graduate ODE course. And, I'm having a little trouble with a couple homework problems. We're doing systems of first order ODEs which I've done previously in an undergrad ODE course a number of years ago, and slightly in a linear algebra course ( Read more... )

Comments 2

semios February 6 2009, 01:44:30 UTC

For the second problem, x'' + 3x' + 2x = 0, we can change it into a system of ODEs using the following substitutions.

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_lover February 9 2009, 07:50:14 UTC

For the second one, it's actually pretty easy, but I understand being a bit rusty!

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 m1 = -2 and m2 = -1.

Then, the solution to our auxiliary equation becomes x = c1e^(m1x) + c2e^(m2x), 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 ( ... )