mersenneforum.org solving 2nd order differential equations
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 2009-10-29, 01:39 #1 Joshua2     Sep 2004 13×41 Posts solving 2nd order differential equations so I have two part question. I have a 2nd order linear homogeneous equation but the initial values aren't at y(0) and y'(0) like normal. They are @ -2 instead. So it seems there may be three ways to go about it. The first I thought of was to pretend it is at 0 and then try to shift the answer, but that is hard for complicated equations. I could also just plug everything in and solve two equations with two unknowns, but it is a lot of algebra. I think the teacher had a third way that might be the easiest similar to my 2nd way, but I couldn't understand it. 2nd question is solving inhomogeneous 2nd order linear so our teacher has us solving these with an alpha beta method I don't really understand. But its solving to easier equations. let u=y'+ay and u' + Bu = the left hand side of the equation y'' + cy' + d. I couldn't find anyone on the net who used this method and explained. Everyone seems to use complicated stuff like Lagrange or Fourier or Wronskian Last fiddled with by Joshua2 on 2009-10-29 at 01:58
 2009-10-29, 04:40 #2 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 39316 Posts First question: It does not matter as long as your initial conditions are the same i.e. y(-2) and y'(-2) are fine. Second question: I have never seen that method used. Your alpha and beta parts I believe are exponents in a binomial in either your particular or complimentary portion of the answer. Method of undetermined coefficients is one of my preferred methods of solution for a second order nonhomogeneous equation though variation of parameters can work as well.
 2009-10-29, 05:06 #3 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 39316 Posts Edit: as for my response to the first question...you might want to double check with someone else first. If I am wrong, someone will undoubtedly correct me on the forum. However, for some reason that is the way I remember it- that your initial conditions must be the same (can be any value). I do not remember the proof for this though, or the logic.
2009-10-29, 06:17   #4
Joshua2

Sep 2004

10000101012 Posts

Quote:
 Originally Posted by Primeinator First question: It does not matter as long as your initial conditions are the same i.e. y(-2) and y'(-2) are fine.
I think I figured out a really easy way to do it. I know they have to be the same, but its harder when they aren't at zero. Pretend initial conditions are at t = 0. After you find it replace all the t's with (t + negative initial condition offset) in this case (t+2). Its worked so far on the homework but we only had 2 of them. I think it will always work though?

Last fiddled with by Joshua2 on 2009-10-29 at 06:18

2009-10-29, 10:08   #5
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by Joshua2 so I have two part question. I have a 2nd order linear homogeneous equation but the initial values aren't at y(0) and y'(0) like normal. They are @ -2 instead. So it seems there may be three ways to go about it. The first I thought of was to pretend it is at 0 and then try to shift the answer,
You need to go back and learn some linear algebra. In particular,
look up "affine linear transform".

Quote:
 but that is hard for complicated equations. I could also just plug everything in and solve two equations with two unknowns, but it is a lot of algebra. I think the teacher had a third way that might be the easiest similar to my 2nd way, but I couldn't understand it. 2nd question is solving inhomogeneous 2nd order linear so our teacher has us solving these with an alpha beta method I don't really understand. But its solving to easier equations. let u=y'+ay and u' + Bu = the left hand side of the equation y'' + cy' + d. I couldn't find anyone on the net who used this method and explained. Everyone seems to use complicated stuff like Lagrange or Fourier or Wronskian

One difficulty you are having is simply sloppy thinking in general.
You seem unable to express what you really mean. The

The fact that you characterize various methods as "complicated stuff" really
shows that you lack fundamental understanding of what you are doing. I
don't know you personally, but from the comments you have made, it seems
that you shouldn't be taking this class. You have insufficient background.
Further, you seem to be learning the material as "pseudo-mathematics".
(according to Andre Toom). You are trying to learn the material as
"plug and chug" mechanical manipulations without any UNDERSTANDING
of what is going on.

If you came to me as a student, asking for office hours help, I would give

Quit this class. Go back and take a class in linear algebra. You don't
seem to understand e.g. what the Wronskian is, where it arises, what it
is used FOR, and why it can be used in solving DiffEq's.

I am not sure that I understand your use of Fourier's name.
Certainly you can't be using Fourier Transforms?? Their use is way

There are also several different things that can be ascribed to Lagrange
in this context.

You need to be able to express yourself PRECISELY if you are to
have any hope of understanding this material. Stop the handwaving.

Last fiddled with by R.D. Silverman on 2009-10-29 at 10:10 Reason: fix quotes and typos

 2009-10-29, 20:41 #6 Joshua2     Sep 2004 10000101012 Posts I did take Linear Algebra last year. I am not going to be a math major and the Dif Eq is the last math class I need. Our teacher is going to cover the Wronskian next week after midterms. We haven't covered solving 2nd order equations with Lagrange or Fourier transforms. I have been watching youtube videos from MIT and KhanAcademy and they use these methods. The method our teacher has been using solving two first order equations is much simpler. Why isn't it a more common method?
2009-10-30, 02:03   #7
flouran

Dec 2008

72·17 Posts

Quote:
 Originally Posted by Joshua2 I did take Linear Algebra last year. I am not going to be a math major and the Dif Eq is the last math class I need. Our teacher is going to cover the Wronskian next week after midterms. We haven't covered solving 2nd order equations with Lagrange or Fourier transforms. I have been watching youtube videos from MIT and KhanAcademy and they use these methods. The method our teacher has been using solving two first order equations is much simpler. Why isn't it a more common method?
If you took Linear Algebra last year, then Wronskian should be rather elementary. I took Linear Algebra back in high school (but it wasn't taught at my high school, obviously) when I was 15, and I got an A. It really was not that hard.

Math is fun partly because it can challenge the person who is doing it in developing new ways of thinking. If you keep depending on people to supply answers that you can easily figure out (assuming you have taken the proper pre-requisites), then you are not fully engaging in mathematics for what it is (at least at your stage). Enjoy math. Don't just toss it aside like another class. Think about the problem for a few days, and if you cannot figure it out, then I would be glad to help you in private (feel free to PM me ).

Last fiddled with by flouran on 2009-10-30 at 02:06

 2009-10-30, 02:50 #8 Joshua2     Sep 2004 13·41 Posts I can do the Wronskian no problem. He just hasn't told us why we want to yet and for the midterm wants us to be using the ways he has already taught us. I think my shifting will work the way I'm doing it. I'm pretty sure I understand it. Thanks!
 2009-10-30, 05:13 #9 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 3×5×61 Posts Are these second order equations constant coefficient?
 2009-10-30, 07:37 #10 Joshua2     Sep 2004 53310 Posts yes, they are for now and linear 2.

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