ok i'm only in cal 2 but we had to do second order differential equations and well my teacher just told us to take the general form of the solutions on faith

i was wanting someone to show me the theory behind the why those things work and maybe some proofs if they aren't too terribly difficult (i assume they would be)

4/28/2006 8:31:04 PM

or maybe you should wait until 341

4/29/2006 1:29:11 AM

^yeah, definitely

dont worry about that shit. when they say assume a solution, just fucking do it man

4/29/2006 2:08:14 AM

I added some photos of my notes from Hong's class in my gallery, but I sure as hell don't feel like explaining them.
To help you out, the first 3/4 of the first page is one problem and there's another squished down at the bottom (in two columns.) The second page is its own problem from start to finish.
The notes on the right hand side of the pages explain more of the theory behind some of the steps.
Mull over them to your heart's content, but there is a reason it's not covered in calc II.

You'd probably be better off trying to find someplace online that goes through the proof.

[Edited on April 29, 2006 at 2:33 AM. Reason : .]

[Edited on April 29, 2006 at 2:34 AM. Reason : .]

4/29/2006 2:32:41 AM

if you are the type of person that wants to understand theory and not just memorize patterns, don't take diffy q with hong

4/29/2006 1:10:56 PM

do a laplace transform
problem solved

4/29/2006 1:42:58 PM

it has to do with the roots of the characteristic solution, and if they're complex then euler's identiy comes in and that is where you get sines and cosines.

you can actually make it more complicated and start to think of the eigenvectors of the problem and things like that. in 341 you start to talk about that, but it really isn't until higher level math that you *really* see where these derivations come from. for now, take it on faith.

4/29/2006 2:31:06 PM

http://www4.ncsu.edu/%7Ejscook3/ma241_homogODEqs_183_187.pdf

It's not terribly profound or anything, basically I just assume that exp(rx) is a solution and see what that says about r. In all cases (but one) we get distinct roots (r1, r2) which give rise to the solution

y = C1*exp(r1*x)+C2*exp(r2*x)

Then when r1=r2 we are forced to give the solution

y = C1*exp(r*x)+C2*x*exp(r*x)

The "x" in "C2*x*exp(r*x)" is most curious. I prove in the notes that it works, but why put the x there to begin with? In other words how to derive the x? I'd like to know. Pragmatically speaking, from the general theory of linear ODEs we know we need 2 linearly independent fundamental solutions, and multiplying by x does give us the extra piece we need for the double root. But, besides the fact that it works, why do it to begin with.

Perhaps, like someone mentioned Laplace transforms give a derivation of the x.

More mundanely, we can expect it from the case y'' = 0. For that differential equation just integrate twice to get y = C1+C2*x = C1*exp(0*x)+C2*x*exp(0*x). So there it is, the mysterious x. But that's not a proof for the general repeated root. (hmmm it occurs to me we could complete the square and make it into a derivation for the general case... must calculate it... )

Let's calculate, double root case, consider (r-a)(r-a) = 0 aka r^2-2ra+a^2 = 0 which is the charactersitic eq-n for y''-(2a)y'+(a^2)y= 0. Wanna make it look like w'' = 0. How can we choose w to make this happen? This is in some sense analogus to what the integrating factor does for first order eq-ns of the form y'+p(x)y = q(x), the integrating factor kinda fixes the p(x) term so we can seperate variables. That involves calculus, so it would seem that whatever trickery I need to change the problem to w''=0 will involve calculus...

Maybe someone can answer this, where does the x in the double root case for linear, constant coefficient, second order, ordinary, homogeneous, differential equations come from?

4/29/2006 8:02:15 PM

I think he is asking how do we know the general form is THE solution to this type of diffeq.
So going on that...


When solving a diffeq. there are certain conditions that when met we can prove that the solution exist and is unique.

Thus, if you succeed in finding a solution that fits the boundary conditions (doesn't matter how) that is the one and only one soln.

In the case of the homogeneous second order, linear, constant coefficient diffeq, Leonhard Euler found a general form that can be used to solve this type.

The condition that must be met for Local existence and uniqueness of soln to
d/dt(x)=f(x,t) is a uniform Lipschitz condition in x in a ball around some IC for some finite time. Then the solution exist and is unique for some arbitrarily small delta in time.

For a homogeneous linear constant coeff diffeq this condition is met globally, and its pretty trivial to show that.

[Edited on April 29, 2006 at 8:46 PM. Reason : .]

4/29/2006 8:43:23 PM

^ sure. But, how do you find the solution even if you know it exists ( supposing that we didn't know the standard solutions already )?

4/29/2006 9:44:25 PM

Successive approximations

[Edited on April 29, 2006 at 10:49 PM. Reason : I'll explain more later]

4/29/2006 10:48:45 PM

Ok i'm back, and I wrote up a quick 2 min sketch of the idea.

Damn tww compressed it--looks like shit. I'll try something else.

OK try this
http://img91.imageshack.us/my.php?image=diffeq6bd.jpg




[Edited on April 29, 2006 at 11:42 PM. Reason : finally]

4/29/2006 11:34:06 PM

So we have to resort to generalized eigenvectors to explain it? I mean, it seems like in the easy case we ought to be able to derive it somehow w/o resorting to the matrix exponential. Nothing against the matrix exponential, it's one of my favorites, but...

Anyway, thanks for your reply, it ought to help me put the puzzle together more.

4/30/2006 10:14:32 AM

wow thanks guys this helps a lot even though i don't quite understand it all

4/30/2006 3:35:33 PM

this thread suddenly gave impressive relevance to the grad linear algebra class i just took. i saw the jordan form and i was like omgomg.

5/1/2006 1:51:13 AM

i second using a laplace transform

5/1/2006 11:24:20 AM

hahaha, both of you guys are EE/Comp E.




[Edited on May 1, 2006 at 5:14 PM. Reason : .]

5/1/2006 5:14:02 PM

yeah, but i am also *double* majoring in applied math. which is kinda like pure math, only cooler.

5/1/2006 5:27:01 PM

^I was referring to the two people who suggested the Laplace transform. I knew you were math, cuz you always ask those great abstract alg. questions.

And the only difference btw. applied and pure math at the undergrad level is like one math modeling class.

I just thought it was funny. Someone asked, "explain the theory and why the solution looks like exponentials."

Two EE engineers reply, "Laplace TF--Done!", and don't bother to show the theory of Laplace and why you get exponentials when taking inverse TF, don't bother about convergence, or a more general theory that can be extended to nonlinear and non-autonomous dynamical systems--not just LTI system.

While we mathematicians are trying to show him the theory.

[Edited on May 1, 2006 at 5:48 PM. Reason : .]

5/1/2006 5:47:09 PM

mathematicians win again, woo hoo.

I like applied abstract math, what's that make me ?

5/2/2006 2:22:04 PM