text:
Facilitating Online/Meetings/Recording

This meeting was all chat based with no voices.

[6:32:20 PM] Heather Moran: FOC 09 course participant :)

[6:32:28 PM] Leigh Blackall: Leigh Blackall has shared contact details with Heather Moran.

[6:32:34 PM] Leigh Blackall: Hi Heather

[6:32:39 PM] Leigh Blackall: :)

[6:32:52 PM] Leigh Blackall: Elluminate appears to be not working tonight

[6:33:04 PM] Leigh Blackall: would you like to try talking here on Skype?

[6:33:52 PM] Heather Moran: This is the first time I've used skype so may not be able too. :)

[6:34:02 PM] Leigh Blackall: we'll see

[6:34:03 PM] *** Call from Heather Moran, duration 00:49. ***

[6:34:13 PM] Leigh Blackall: I can hear you

[6:34:20 PM] Leigh Blackall: but you don't hear me?

[6:34:24 PM] Leigh Blackall: yes, am talking

[6:34:30 PM] Leigh Blackall: I think my new computer has a problem

[6:34:34 PM] Leigh Blackall: its not you

[6:34:42 PM] Leigh Blackall: had the same issue with Herve

[6:34:49 PM] Leigh Blackall: have to chat sorry

[6:34:54 PM] *** Call ended ***

[6:35:01 PM] Leigh Blackall: r you a good typer

[6:35:03 PM] Leigh Blackall: ?

[6:35:14 PM] Heather Moran: no worries, I am good at chat

[6:35:22 PM] Leigh Blackall: great :)

[6:35:29 PM] Leigh Blackall: how r you going with the course?

[6:35:40 PM] Heather Moran: Who else is online? are there many?

[6:35:52 PM] Leigh Blackall: Herve and Kylee were, but have left

[6:36:04 PM] Leigh Blackall: actually, they are still on, but we've said goodbye

[6:36:07 PM] Leigh Blackall: Contact sent to Heather Moran

[6:36:11 PM] Leigh Blackall: Contact sent to Heather Moran

[6:36:43 PM] Heather Moran: Thanks, have added them

[6:36:59 PM] Heather Moran: So are you in Canberra at the moment Leigh?

[6:37:05 PM] Leigh Blackall: Yes I am :)

[6:37:11 PM] Leigh Blackall: its nice actually

[6:37:17 PM] Heather Moran: Lucky devil.....wish I was

[6:37:26 PM] Heather Moran: Well, Queensland anyway

[6:37:28 PM] Leigh Blackall: Palmy North is nice

[6:37:57 PM] Leigh Blackall: Did we meet when I popped in on that conference there last week?

[6:38:04 PM] Heather Moran: I'm in Bulls actually......which is sometimes nicer (weatherwise) than PN....but Queensland is better

[6:38:26 PM] Leigh Blackall: Kiwis love QL.. may as well give it to them lol

[6:39:20 PM] Heather Moran: haha..... my daughter is over there....that 's my fascination with the place

[6:39:42 PM] Leigh Blackall: Well, it is a nice place

[6:39:50 PM] Leigh Blackall: how are you going with FOC

[6:39:57 PM] Leigh Blackall: esspecially assignment 2?

[6:40:04 PM] Heather Moran: Blahhhhhhh

[6:40:10 PM] Leigh Blackall: :)

[6:40:26 PM] Leigh Blackall: you where gunna say..?

[6:40:31 PM] Leigh Blackall: were

[6:40:38 PM] Heather Moran: Assignment two is a little bit of a worry for me but will get there in the end

[6:41:10 PM] Leigh Blackall: is it stage fright? not sure you understand? unclear in any way.. or just reluctant?

[6:41:26 PM] Heather Moran: All of the above.......

[6:41:33 PM] Leigh Blackall: hmm..

[6:41:37 PM] Leigh Blackall: well.. http://en.wikiversity.org/wiki/Facilitating_Online#Assignment_2:_Facilitate_an_online_event

[6:41:47 PM] Leigh Blackall: lets look at it

[6:42:11 PM] Heather Moran: I dont think its stage fright, probably more to do with the content of the mini conference and who i can rope in to be the guest speaker

[6:42:20 PM] Leigh Blackall: I see..

[6:42:27 PM] Leigh Blackall: what interests you?

[6:42:32 PM] Leigh Blackall: topic wise

[6:42:41 PM] Leigh Blackall: r u a teacher?

[6:42:47 PM] Leigh Blackall: or a pro facilitator?

[6:43:21 PM] Heather Moran: yes I am a teacher, I teach business and computing subjects but was wanting to get away from that as much as I can

[6:43:55 PM] Leigh Blackall: hmm.. any contacts in local government? the people who try to facilitate community consultation?

[6:44:07 PM] Leigh Blackall: people like that always have interesting takes on "facilitation"

[6:44:58 PM] Leigh Blackall: or perhaps you already have another idea

[6:45:50 PM] *** Leigh Blackall added Hervé Carpentier ***

[6:46:13 PM] Leigh Blackall: I've just added Herve, incase he'd like to just watch the chat

[6:46:26 PM] Hervé Carpentier: ok

[6:46:49 PM] Leigh Blackall: Herve is a slow typer (so he says) so will probably remain silent

[6:47:12 PM] Leigh Blackall: you still there Heather?

[6:47:15 PM] Heather Moran: Yes,

[6:47:19 PM] Leigh Blackall: ok :)

[6:47:23 PM] Heather Moran: Was just thinking?

[6:47:35 PM] Leigh Blackall: hmm.. any contacts in local government? the people who try to facilitate community consultation?

people like that always have interesting takes on "facilitation"

[6:47:44 PM] Leigh Blackall: or perhaps you already have another idea

[6:48:12 PM] Heather Moran: I know those sorts of people have interesting takes on lots of '

[6:48:20 PM] Heather Moran: things'

[6:48:25 PM] Leigh Blackall: :)

[6:48:40 PM] Heather Moran: ;)

[6:49:06 PM] Leigh Blackall: so perhaps something else?.. anything crop up in the course as a topic you would like to hear someone talk about more?

[6:49:21 PM] Heather Moran: Sometimes, I feel that I may be too boring becasue I dont get into those types of talks,

[6:49:38 PM] Leigh Blackall: not boring.. just other interests

[6:50:03 PM] Leigh Blackall: so you're here to experience what it means to facilitate online

[6:50:07 PM] Leigh Blackall: we know that much

[6:50:17 PM] Heather Moran: So how long does the mini conference have to be? or should be?

[6:50:34 PM] Leigh Blackall: you mean your part in it?

[6:51:27 PM] Heather Moran: well, yes....do you expect that we will facilitate a meeting/confference for a minimum period or is it see how it goes?

[6:52:16 PM] Leigh Blackall: not really.. but usually 30 mins to an hour (if live).. or equal if a chat or forum based meeting

[6:52:48 PM] Leigh Blackall: hopefully we'll get around 5-10 events like that, and then we have a mini conference

[6:53:01 PM] Leigh Blackall: (your's being 1 event)

[6:53:53 PM] Heather Moran: So ideally, the topics should all be focussed around the same type of subject?

[6:54:09 PM] Leigh Blackall: yes, loosely based on facilitating online

[6:54:22 PM] Hervé Carpentier: what platform would we be using?

[6:54:30 PM] Leigh Blackall: the topics covered in the course are a good guide - but outside topics are welcome

[6:54:32 PM] Hervé Carpentier: would we all use the same?

[6:54:42 PM] Hervé Carpentier: ie Elluminate?

[6:54:43 PM] Leigh Blackall: no, preferably different topics

[6:54:47 PM] Leigh Blackall: oh yes

[6:54:52 PM] Leigh Blackall: Elluminate is handy

[6:55:01 PM] Leigh Blackall: (although not tonight it isn't)

[6:55:03 PM] Hervé Carpentier: (when working)

[6:55:06 PM] Leigh Blackall: lol

[6:55:09 PM] Hervé Carpentier: :)

[6:55:42 PM] Leigh Blackall: http://en.wikiversity.org/wiki/Facilitating_Online /course_mini_conference

[6:55:45 PM] Heather Moran: and its not handy at work either because IT cont seem to get it working for more than one person at a time!! :)

[6:56:01 PM] Leigh Blackall: take a look at that link

[6:56:08 PM] Hervé Carpentier: It would make sense to all use the same platform; it would help for consistency

[6:56:09 PM] Leigh Blackall: it is the organising page for the conference

[6:56:28 PM] Hervé Carpentier: I cannot see the link

[6:56:39 PM] Leigh Blackall: http://en.wikiversity.org/wiki/Facilitating_Online /course_mini_conference

[6:56:51 PM] Leigh Blackall: Heather?

[6:57:06 PM] Hervé Carpentier: :)

[6:57:07 PM] Leigh Blackall: Herve? still trouble?

[6:57:29 PM] Heather Moran: reading now

[6:57:41 PM] Hervé Carpentier: I have it now

[6:57:48 PM] Leigh Blackall: at the bottom is a link to the FOC08 conference

[6:58:05 PM] Leigh Blackall: "FO08 - This conference program is better viewed on Wikieducator, where the 2008 course was held"

[6:58:25 PM] Leigh Blackall: click the " better viewed on Wikieducator" link

[6:58:30 PM] Leigh Blackall: ok?

[6:58:56 PM] Leigh Blackall: this is the link: http://wikieducator.org /Facilitating_online_communities/Conference/FOC08

[7:00:00 PM] *** Leigh Blackall added Debra Maddocks ***

[7:00:09 PM] Leigh Blackall: Hi Deb, I'm adding you to a group chat

[7:00:19 PM] Heather Moran: I remember reading this a few weeks ago.... I see that some of the previous participants have doubled-up and dual facilitated, how does that work and is it an effective way of doing the assignment?

[7:00:19 PM] Leigh Blackall: we are looking at last years conference page: http://wikieducator.org /Facilitating_online_communities/Conference/FOC08

[7:00:50 PM] Leigh Blackall: is it an effective way of doing the assignment?personally, I don't think it is.. but some people seem to prefer it

[7:01:00 PM] Heather Moran: hi Debs ;)

[7:01:25 PM] Heather Moran: Ok.

[7:02:21 PM] Hervé Carpentier: what do you mean 'doubled-up and dual facilitated'?

[7:02:29 PM] Leigh Blackall: I reckon it is easier to settle on your own topic, find your own guest, and facilitate on your own

[7:03:03 PM] Leigh Blackall: Herve, Heather means some people teamed up and both did 1 event

[7:03:18 PM] Hervé Carpentier: stange idea...

[7:03:24 PM] Heather Moran: Herve, some people partnered up with others on last years course and facilitated an event together.

[7:03:26 PM] Leigh Blackall: eg: # 8 (Think, Learn & Create) Using Mind Maps

[7:03:33 PM] Leigh Blackall: that was 2 people

[7:04:08 PM] Hervé Carpentier: 2 facilitator in an event? (yawn)

[7:04:34 PM] Heather Moran: Interesting that you chose the PN one Leigh.....haha

[7:06:01 PM] Leigh Blackall: Deb seems to be having trouble navigating Skype

[7:06:24 PM] Leigh Blackall: trying to get her into this chat (she's chatting with me in another line)

[7:07:19 PM] Heather Moran: She's into private chats...lol

[7:07:46 PM] Leigh Blackall: oh no!

[7:07:53 PM] Leigh Blackall: inefficient

[7:08:18 PM] Leigh Blackall: here she is! :)

[7:08:31 PM] Debra Maddocks: ahh- 2 lines- or is that double lines.... will try and catch up (happy)

[7:08:44 PM] Heather Moran: There's some interesting topics in last years selection...I suppose the biggest challenge for me will be to find something as interesting and siomeone interesting to talk with/to.

[7:09:04 PM] Leigh Blackall: deb, we're just looking at last year's conference page: http://wikieducator.org/Facilitating_online_communities/Conference/FOC08

[7:09:10 PM] Heather Moran: 'someone'

[7:09:25 PM] Leigh Blackall: oh Heather! it'll be interesting!

[7:09:32 PM] Leigh Blackall: because you chose it!

[7:10:21 PM] Leigh Blackall: So the idea is to get this year's conference page, looking something like last years: http://en.wikiversity.org/wiki/Facilitating_Online/course_mini_conference

[7:10:44 PM] Debra Maddocks: ok, I had some very loosely defined thoughts about following on from the twitter-blogging debate and bringing in maybe someone like Joyce from EIT (being that I was at the T&L conf last week?

[7:10:46 PM] Heather Moran: Well, to be fair Leigh, having you as the course guru is a little intimidating....I saw and heard you at the teaching and learning conference debate last week!!!!

[7:10:56 PM] Hervé Carpentier: I think I have found a topic, if I can convince a friend to take part

[7:11:02 PM] Debra Maddocks: we almost missed leigh last week !!!!

[7:11:14 PM] Hervé Carpentier: and if the time difference allows it to happen

[7:12:23 PM] Leigh Blackall: You saw me there! How emberressing! you should have said earlier! I can't remember what I said there.. 3 minutes is impossible (as is joking about sucha serious issue ;)

[7:12:39 PM] Leigh Blackall: Herve, what's your topic?

[7:12:41 PM] Leigh Blackall: idea

[7:12:43 PM] Hervé Carpentier: What dates are we talking about for the miniconference?

[7:13:16 PM] Leigh Blackall: DATES: Week of the 9th of Nov

[7:13:20 PM] Heather Moran: Yeah, i was going to...... as you were running out of the theatre to catch the plane lol

[7:13:22 PM] Debra Maddocks: you quoted illich ? a lot, can't spell. the debate was a 'draw', not enoght time

[7:13:34 PM] Leigh Blackall: a draw!

[7:13:41 PM] Leigh Blackall: down with Palm North

[7:14:03 PM] Leigh Blackall: So, there's an event you might faciluitate? A debate

[7:14:03 PM] Hervé Carpentier: my idea is to invite someone who is experienced in delivering blended learning courses

[7:14:06 PM] Debra Maddocks: think that was right- heather? a draw (down-haha)

[7:14:18 PM] Hervé Carpentier: to talk about about his experiences.

[7:14:24 PM] Debra Maddocks: Yes, a debate on the use of twitter in an educational context!!!!

[7:14:26 PM] Hervé Carpentier: He is based inthe UK

[7:15:01 PM] Leigh Blackall: Herve, so you want to bring in a "teacher" to talk about "delivery".. hope you'll draw out aspects of facilitation (vs crowd control ;)

[7:15:07 PM] Hervé Carpentier: Is twitter your topic Debra?

[7:15:11 PM] Heather Moran: Didnt all participants get a prize though?

[7:15:17 PM] Hervé Carpentier: I cannot wiat to hear about it...

[7:15:46 PM] Debra Maddocks: Hi herve- not sure yet- i am mulling over, and realise that time is marching on, there aren't too many topics on the wiki yet....

[7:16:10 PM] Leigh Blackall: there aren't too many topics on the wiki yet....I'm concerned by that also :(

[7:16:10 PM] Hervé Carpentier: Leigh, this is the idea. I just need to dig it further

[7:16:18 PM] Debra Maddocks: herve- i thought you would love that topic :P

[7:16:44 PM] Leigh Blackall: Sure Herve.. be careful not to miss an opportunity to explore facilitation in its own right is all

[7:17:11 PM] Leigh Blackall: Twitter is a wild one.. how does a facilitator work in that space? a good question

[7:18:06 PM] Debra Maddocks: i haven't got my head around that yet- i am not sure if its possibel, that might have to be the debate- can it be done? facilitating with a twitter group..

[7:18:23 PM] Leigh Blackall: its a good question - for sure

[7:18:38 PM] Leigh Blackall: perhaps you might facilitate a discussion that is open to anyone to participate

[7:18:49 PM] Leigh Blackall: your challenge is to promote it well enoigh

[7:19:05 PM] Heather Moran: a facilitator can and should be able to work with any given tools. Its just that the tools can sometimes dictate the outcome.

[7:19:13 PM] Leigh Blackall: speaking of which - arriving at an idea and event is one thing - you MUST promote it and get people there too

[7:19:22 PM] Leigh Blackall: this side of the coin was very weak last year

[7:19:40 PM] Debra Maddocks: yea- that was a thought that popped up. theFOC-09 group shuld be a good learning ground...we have a wide range of thoughts on twitter...hehe

[7:20:09 PM] Debra Maddocks: can we get a copy of this conversation at all??

[7:20:12 PM] Heather Moran: "speaking of which - arriving at an idea and event is one thing - you MUST promote it and get people there too"  this year will be no different I imagine ;)

[7:20:55 PM] Leigh Blackall: it should be - you're forewarned ;)

[7:21:19 PM] Leigh Blackall: your plans need:

[7:21:20 PM] Leigh Blackall:    * Title for the session

* The aim of the session

* Promotions for the session (contact list, flier, bios, summary)

* Key links for the event, dates and times, background information, and discussion starters

* A description of technical support services available

* A contingency plan for technical problems, poor or over attendance, and other disruptions

* An indication of whether recording will be done and where it will be made available after the event

[7:21:22 PM] Heather Moran: lol, ok

[7:21:58 PM] Debra Maddocks: hmmm- will have to get a bit more tech savvy.....

[7:22:22 PM] Leigh Blackall: the tech support is more about what you will do if it fails.. a bit like tonight

[7:22:45 PM] Debra Maddocks: ha! yes, good experience to have anyway.

[7:23:16 PM] Hervé Carpentier: Technically, could you get several people at the same time to talk in skype?

[7:23:22 PM] Leigh Blackall: when we stop comparing online communication to face to face - its all good experience to have

[7:23:44 PM] Leigh Blackall: Technically, could you get several people at the same time to talk in skype?yes, about 5 is the limit

[7:23:49 PM] Leigh Blackall: and all 5 need broadband

[7:23:54 PM] Leigh Blackall: limitless chat tho

[7:24:01 PM] Hervé Carpentier: thanks

[7:24:23 PM] Leigh Blackall: WizIQ is pretty good - broadband need though

[7:24:47 PM] Hervé Carpentier: what is your definition of broadband?

[7:24:54 PM] Hervé Carpentier: 128K +?

[7:25:04 PM] Debra Maddocks: i think there are some options on the course wiki for this year???

[7:25:49 PM] Leigh Blackall: 128K +?at least

[7:25:54 PM] Heather Moran: `for what debra? topics or communication mediums/

[7:26:27 PM] Debra Maddocks: comm mediums- think stephen put some up earlier??? not sure- will have a catch up 2morrow

[7:27:18 PM] Leigh Blackall: So how r we then? winding down? Would it be ok if I published this text for others?

[7:27:35 PM] Hervé Carpentier: no pb (y)

[7:27:49 PM] Debra Maddocks: I might have to leave now, finish dinner etc yes, please do publish this- will you send a link?

[7:28:02 PM] Leigh Blackall: I will send link if Herve gives ok

[7:28:19 PM] Leigh Blackall: sorry

[7:28:23 PM] Leigh Blackall: Heather I meant

[7:28:30 PM] Debra Maddocks: :) thanks

[7:28:39 PM] Leigh Blackall: Heather?

[7:28:47 PM] Hervé Carpentier: ok

[7:29:09 PM] Heather Moran: Yeah, do it...i have no probs with that.

[7:29:24 PM] Leigh Blackall: :) great. Thakns. Very nice chatting to you 3

[7:29:24 PM] Debra Maddocks: thanks all, see you next time

[7:29:36 PM] Heather Moran: Bugger I should have negiotiated a pass for my consent!!!

[7:29:43 PM] Leigh Blackall: lol

[7:29:48 PM] Leigh Blackall: it all helps ;)

[7:29:53 PM] Debra Maddocks: good one heather!!!

[7:29:53 PM] Hervé Carpentier: :)

[7:30:53 PM] Leigh Blackall: good luck with your preps.. I hope you get a lot out of the exercise (ie rich learning) assessment is just a formality really.

[7:31:12 PM] Debra Maddocks: lol bye!!

[7:31:53 PM] Hervé Carpentier: bye

[7:32:13 PM] Heather Moran: I'd better not comment on that ;) Bye and thanks

[7:32:42 PM] Leigh Blackall: oh dear.. all 3 responses were ominous.. we'll see in the end

[7:32:52 PM] Leigh Blackall: cuz

title: Facilitating Online/Meetings/Recording


text:
User:Ignastan

hello

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Ignastan


text:
User talk:Silosteve5101

 Hello Silosteve5101, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! Jtneill - Talk - c 11:27, 7 October 2009 (UTC)

 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

title: User talk:Silosteve5101


text:
User:Jack Russell~enwikiversity

My name is Jack Russell. I used to be on Dogbook and Dogspace, but they've got too many ads now. Then a teacher at school told me that this wiki is not just for university students, it's for just about everyone. So I thought I might give it a try.

I enjoy going to the aquarium because I like dolphins and coral.

I was once a dolphin trainer. Then I became the captain of a small yacht that sailed rich people all around the world. I made quite a bit of money doing that. I eventually retired, and now I'm pretty much just spending my time eating dog food, chasing cars, and making trouble here. Oh, and I flew planes.

It seems that I've now been entered into the mascot contest.

Here are some of my favorite links so far:
 * Life sciences
 * Zoology
 * Other nice media about my cousins...

You may wonder how I type with paws. It is actually not that difficult. It just takes practice. Believe it or not, I can type at speeds of 80-95 words per minute.

title: User:Jack Russell~enwikiversity


title: File:HW3-10plot.png


text:
User:TerekV

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

Voodoo Dolls
 * Research Component

Voodoo Dolls
 * week 1

Voodoo
 * week 2

Voodoo
 * week 3

Voodoo Dolls; Begin making one
 * week 4

Voodoo Dolls
 * week 5

Voodoo Dolls; Continue making one
 * week 6

Continue making Voodoo Doll. Paper.
 * week 7

Finish Doll, Have nice draft done.
 * week 8

Doll and Paper and Presentation.

Midterm Assignment 1: Taste & Aesthetics
Taste and Aesthetics General Education class, AIS, Fall Quarter, 2009 Terek W. Varland 11\23\2009

Piece: Persephone Unbound (1999) Artist: Beverly Pepper Born: Dec. 20th, 1922 Nationality: American Field: Painting, Sculpture, Abstract

It was difficult finding a piece of art at the Olympic Sculpture Park I did not like. I seemed to enjoy and appreciate everything I laid eyes on (except for the art piece made of logs, metal bars, and chain—but that one was off-limits, as everyone seemed to dislike that one). Like my musical tastes—ranging from Jack Johnson to Marilyn Manson, from the Ink Spots to Ludacris, and from the Dead Weather to Beethoven—it is extremely rare when I do not like and cannot find some redeeming or alluring qualities in a person or a work of art. However, on my way out of the park, I did find another.

The art piece was called “Persephone Unbound.” It was created in 1999, by the American-born painting and sculpture artist, Beverly Pepper (born in 1922). It basically looked like a bunch of mirrored and elongated picture frames cascading out of each other.

To be honest, when I first initially looked it over, I did not mind it. However, the more I scrutinized it, the more I found myself in a disposition towards it. It just began to bother me, I cannot tell you why exactly, but I assure you, it was definitely looking at me funny.

Maybe it was the way it just seemed so simple… that this “Beverly Pepper” gets paid lavishly to do something that I could have done. Why was I not being made rich and renowned yet? I could just as easily put on display some of my past homework drawings, and still I would not draw multitudes of people from all over the world to want to look at them. What made her so special? Luck of the draw?

I suppose this loathing I suddenly had for lack of effort could have been expelled at almost any of the pieces present. The way the park was laid out bothered me too; that I had to walk everywhere to see piece after piece of “what-in-the-hell” art sculptures. I must have had one of my bi-polar moments or something.

What infuriates me most is the fact that I never really disliked the piece itself. Instead—I hated that I could like such a simplistic piece. There was obviously no real skill or idea involved. I could hear her thought process for the piece exaggerated in my head:

“Hmm… the Olympic Sculpture Park could use another piece of work from me. They liked ‘Perre’s Ventaglio III’ I made in 1967, but they could really use something more modern. More simple, but not overly cliché. Oh! I know! How about I take a big slab of something—why not a rock!—and put it on a base, give it a title, and throw in a bunch of ‘inspiration’ and ‘what-motivated-me-to-make-such-a-piece’ crap to make it seem like I am actually very deep and philosophical! Haha, stupid art critics will see it as an outstanding work of art, and will throw in their own ‘obviously-this-piece-has-many-profound-ideas-behind-it” two-cents to make they look like they know what they’re talking about too! I have a big, recognizable name now, and I could put up a bag of rotting turtle eggs if I wanted to, and it would still be marveled and revered. Life as a lucky artist rocks!”

Even after all that though, I cannot help but liking it against my conscious’s will. The heart wants what the heart wants I suppose. Like a song that repeats the same verse over and over again throughout the course of its play (I can think of many unimaginable songs like this), some of them, I still cannot help but hate to love.

http://media.photobucket.com/image/beverly%20pepper%20olympic%20sculpture%20park/amdraheim/P1030096.jpg (A picture of the piece) http://www.beverlypepper.net/biography.html http://images.google.com/imgres?imgurl=http://www.seattlepi.com/dayart/20070118/226sculpture_pepperpersephone.jpg&imgrefurl=http://www.seattlepi.com/visualart/299887_the_art18.html&usg=__v7RMaOA3hhij-lu2SDPgk2bpHeI=&h=226&w=151&sz=8&hl=en&start=59&itbs=1&tbnid=X0Se2ZOc2umJ6M:&tbnh=108&tbnw=72&prev=/images%3Fq%3Dbeverly%2Bpepper%2Bolympic%2Bsculpture%2Bpark%26gbv%3D2%26ndsp%3D21%26hl%3Den%26sa%3DN%26start%3D42

Midterm Assignment 2: Manipulation
http://www.youtube.com/watch?v=Xu8y3cAhB6U&feature=related Skittles "Tailor" commercial.

Truth Through Humor. The people in the mirrors are not the man being tailored--one of them is not even pretending to be.

http://www.youtube.com/watch?v=P_SFrBe9fSo Geico "Caveman Loyalty" commercial.

False Dilemma (Either/Or reasoning). The cavemen are having a dilemma about loyalty. Basically, if you're a caveman and with Geico, you are the bad guy.

Midterm Assignment 3: Always Wanted To Progress Report (100 words or so)
Midterm Report General Education class, AIS, Fall Quarter, 2009 Terek W. Varland 11\23\2009

I have managed to get a lot done over the course of the seven weeks that I have lived here. I have managed to settle in with my roommate, get comfortable with this strange new city environment, being miles upon miles away from any family, seeing my best friend Jessica every weekend, not get a vehicle, get drunk for the first time (you can blame my bestie for that one), get a crazy girlfriend, not get a job, and get (most) of my homework done in-time. But when it comes to final projects? Not so much….

I have successfully managed to fail at getting past my procrastination tendencies, and now it is crunch-time. Which is okay by me, I tend to work best under pressure. I bought more burlap this weekend for the voodoo doll. It took forever to find it, and with the help of Jessica’s willingness to drive me everywhere to look for it, I eventually found it at a feed store in Sammamish/Issaquah area called The Grange.

Now I intend to go home, eat (finally got groceries after living on two weeks of candy, ramen, and graham crackers and frosting!) and work on that voodoo doll. Wish me luck.

~Terek.

Final Presentation

 * Final Research Topic (10%)

<>

 * Research (20%)

<>

 * 1) Internet site: "How to Use a Voodoo Doll - wikiHow." WikiHow - The How-to Manual That You Can Edit. Web. 08 Dec. 2009.

 * 2) Book: Hurbon, Laënnec. Voodoo: Search for the Spirit. New York: H.N. Abrams, 1995. Print.

 * 3) Scholarly Journal:

 * 4) Other (interview, TV show,etc): Voodoo doll I got as a present from my mom when I was twelve, which she bought from New Orleans. In it's coffin-case, the hand-crafted voodoo doll also came with a little note on how to use it. It agrees with the pin-stabbing directions and color-coding of the pins found on WikiHow. Normally I wouldn't use a site like that, but honestly, it's all I could find. It (the note) also explains how to ensnare spiritual energy from a person and insert it into the voodoo doll.

 * Your accomplishment (20%)

<> I made two "self" voodoo dolls--one more than I originally planned. But they didn't come out the way I planned, so being the perfectionist that I am, I am still not exactly happy with them, but still proud I made them. I made a speech outline too, and did some research, but did not get to class in-time to present anything. So I felt I came pretty close to accomplishing my goal, and am pleased with the results.

 * Project Review (50%)
 * 1)What went well:

Burlap is cheap.

I already had antique buttons and sewing equipment, so I didn't have to buy any of that.

I liked my voodoo doll concept art for it.

I had a lot of buttons to choose from.

 * 2)What went poorly:

Took forever to find a place that had burlap.

Burlap edges are very fragile and shred way too easily.

Had to learn how to sew--which I did not do well.

Neither did appropriately cutting out the right sizes of burlap.

Research was almost impossible. Seattle Library had nothing. AIS Library had nothing. Internet has nothing. I basically know what I know about Voodoo dolls from the Voodoo religion and what I've learned over time.

Time management. Waited until the last minute basically. No sleep.

Voodoo dolls did not look as awesome as their drawn versions/blueprints.

 * 3)Future strategies:

I plan to reinforce the first Voodoo doll.

As for future strategies, I would definitely manage my time better, and not consider it "homework," but as fun. As a hobby or something. Because it is something I wanted to do, on my bucket list, but as soon as something I love becomes work or required, I procrastinate and run from it until the last minute, which I frustrate myself in because at the same time, if someone's not making me do something, even if it's something I love, I'm often too lazy to do it.

And sorry about the run-on by-the-way.

(Face of Voodoo Doll attempt #1.) (My mini-me.)  (Yay, my arm fell off!)  (I put my hair and a small weight in this doll's chest.)  (Profile shot.)  (Voodoo Doll attempt #2.)  (Hood, eyes and mouth.)  (Antique button eyes and stitched mouth for added awesomeness.)  (Denim broken heart.)  (Smaller, more compact, and less fragile than first attempt.)  (Extremely deformed.)  (I put my hair in the spinal column of this one.)  (Mug shot.)  (Hooded doll.)  (Another profile.)  (Together forever.)  (Cat eye.)  (Back.)  (Accented shoulder-blades and spine.)  (Showing off flexible neck and arms.)  (Presentation speech outline, page 1.)  (Presentation speech outline, page 2.)

Self-Assessment
Self Assessment General Education class, AIS, Fall Quarter, 2009 Terek W. Varland 12\15\2009

I cannot remember how long this section is supposed to be, but I figure I will just start writing, and see where and how far it takes me.

Altogether, I think I had a pretty nice project going.

I liked my choice of buttons and my original idea/concepts for my voodoo doll. The first voodoo doll, the more fragile and larger one, was an attempt to make it look like I planned. I originally intended to make and give this one to my best friend Jessica, if I ever ended up joining the Marines like I dreamed of.

Since it is my hair inside it (a part of my soul/essence of my being), and she is extremely worried about me joining up, it would be like she still had me in a way, and could watch out for me. She could maybe sleep at night, and if taught how to use the pins (she used to be Mormon, but now she is not and refuses to pray), maybe she would poke the mini-me with one of those good colored pins and wish good things unto me and feel like she is contributing to my safe return. She worries about me like no other (almost as much as I worry about her), so I thought the thought would count on this one, even though we both do not participate in pagan rituals. I know that sounds pretty crazy, but hey—we are. So we get special crazy privileges. Even if I were to die, she would have a part of me; some of my imagination, my love, my soul. We are pretty mushy best friends.

That last paragraph probably seems like all needless information, but that was the sole intent driving me behind this project at the time. Kind of like the idea-to-make-a-voodoo-doll’s creation story or something. Then this class offered told us to get up and do something, and I figured I would kill two stones with one bird.

I know it probably seems unfair that my family was not getting one, but I figured they would be the ones with all my other stuff if I died. But now that I have two “self” voodoo dolls, maybe I will give them one too.

The second one was just made out of scrap burlap and spur-of-the-moment imagination.

I liked my deformed-looking Voodoo dolls, and I think I might continue to make various looking ones of different shapes, sizes and detail—not for voodoo practice or witchcraft, just because they look cool and compliment my curiosity shop of a bedroom quite nicely. Or even stuffed animals or something, and give them to future girlfriends. Would totally melt their hearts!

I am not sure if this is at all what you were looking for in a self assessment, but this is what came out.

I really enjoyed this class, and think you were one of my most understanding and influential teachers in a long time. I have all these things I want to do, all these “needless” skills or things I want to get done in my life, and seeing that you seem to pick something and go with it, rather than procrastinating like I do, has really kind of woke me up.

Cheers, ~Terek

PS: I totally want a concertina now. …but I should probably get off my lazy ass and mess with the guitar and violin I told myself I would learn to play. Or attempt to ride my unicycle… haha.

title: User:TerekV


text: File:PEA1 HW1 Problem13.PNG

title: File:PEA1 HW1 Problem13.PNG


text:
User:AudreyW

=Humanities I/The Commons=

title: User:AudreyW


text:
User:JFulton

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

Plan a family vacation to a European country.
 * Research Component

I will research cost, passport requirements, and the country my family would most want to visit in the future.
 * Week 1

•Determine country •Look up passport requirements •Look up flight cost

 * Week 2

•Find book about history of country •Find book with historical sites in country

 * Week 3

•Find scholarly journal •Find means of transportation in country

 * Week 4

•Find reliable internet site with information on country •Find internet site that recommends things to do

 * Week 5

•Make savings plan •Plan budget for trip •Research advise for best places to visit(Videos)
 * Week 6

•Go through Information pull it all together •Double check sources

 * Week 7

•Create final presentation •Write about what went well, what went wrong, and future strategies. •Write about accomplishments

 * Week 8

•Catch up on any late work

Midterm Assignment 1: Taste & Aesthetics
While walking through the Olympic Sculpture Park I came across several statues that were interesting to look at. There were two statues in particular that I did not really care for. Eagle by Alexander Calder and Sky Landscape by Louise Nevelson were my least favorite sculptures in the park. In the end I decided to research the Eagle by Alexander Calder and attempt to see it in a different perspective. Abstract art has never been one of my favorite types of art. Eagle for me looked nothing like the bird, but instead curved metal painted red. My research of Calder was surprisingly interesting though. Calder came from a family of artist, and from a young age was encouraged to be artistic. Though Calder never intended on becoming a world renowned artist, he originally was and engineer and did various jobs until 1923 when he became dedicated to art. One thing in particular that I was surprised to find out about Calder was that he loved the circus, something I can relate to also. Calder became famous for his Cirque Calder, his large scale sculptures such as Flamingo, and his mobiles. Looking again at Eagle, I still am indifferent about the sculpture. Other pieces of Calder’s work such as Crinkly avec disc Rouge I find much more interesting to look at. The size of Eagle mixed in with its solid color to me seems boring to look at despite the curvy lines of the sculpture. Crinkly avec disc Rouge on the other hand has more than multiple colors, shapes, and patterns. Resources: http://calder.org/home.html

Midterm Assignment 2: Manipulation
Links: http://www.youtube.com/watch?v=RQ0A0xbORPU http://www.youtube.com/watch?v=MDUQW8LUMs8

Midterm Assignment 3: Always Wanted To Progress Report (100 words or so)
In the beginning of this project I stuck to my planned schedule like glue. Over the last couple of weeks I unfortunately have drifted away from my schedule. Luckily for me I planned for my procrastination in week eight, which is dedicated to catching up on any and all late work. Hopefully I will be able to catch up before week eight. Besides falling behind on my planned schedule my project is going well. I believe I made it clear that I will not actually be going to England anytime soon.

Final Presentation

 * Topic (10%)

Plan a future trip to London, England with family.

 * Research (20%)

<>

 * 1) Internet site:http: //www.state.gov/; Priceline.com; http://www.ricksteves.com/home.htm

 * 2) Book:A traveler’s history of England

Fifth edition Christopher Daniell

 * 3) Scholarly Journal:Wood, Michael. "Birth of England: The Wessex Kings." (2004): n. pag. Web. 8 Dec 2009. .

 * 4) Other (interview, TV show,etc):http://www.youtube.com/watch?v=OPNJV-SK--k

http://images.google.com/imghp?hl=en&tab=wi http://www.visitlondon.com/attractions/culture/top-ten-attractions

 * Your accomplishment (20%)

My plan for vacationing to London is still a few years away. That is why I created a savings plan that will help me reach my goal. Starting after Christmas I plan on stashing away roughly one hundred and fifty dollars a month in my savings account. If everything goes according to plan in twenty-four months I should, hopefully, have three thousand and six hundred dollars saved. Assuming my cousin and nephew pitch in the money I save should be enough. Not to mention my dad will also save money for the trip, which will be a big help. Also I created a general list of ideal places to visit while in London. Of course there are the must sees like the Tower of London and Big Ben, but there are also the smaller places like local shops and pubs like the Cheshire Cheese. A general ideal list was all I was able to come up with due to the fickle nature of my relatives. One moment they might want to see the Eye of London, the next they might just want to go to the theater and watch a movie instead. So a list of suggested ideal places to visit seemed like the wisest choice and the best way to avoid arguments.

 * Project Review (50%)
 * 1.Information about London was the best part of this project in my opinion. In the beginning I had a general idea about the size of London and the many interesting historical sites. But by the end of this project I was nearly overwhelmed by all of the information. There were just so many suggested places to see and events to do, picking through the information on what to present and what to leave out was almost difficult.

 * 2. Planning is what went poorly in this project. Instead of spreading my workload throughout the weeks, I should have attempted to get the majority of it out of the way at the beginning. Also with getting a new job my project schedule was interrupted, leaving me less free time to get my work done. Besides poor time management I feel this project went alright.

 * 3. Some future strategies I have in mind are planning my time better. Normally I start off strong on a project then drift off course over the weeks. In the future I can plan to do the majority of the project in the beginning versus spreading it out over time, saving me from last minute work.

Self-Assessment
Although I felt this class was mostly review on facts that I had previously learned, I did manage to learn a few new tidbits of information. The hip-hop video in particular was a good learning experience. While I had known some of the fact in the movie I never thought or knew about the extent of violence and discrimination in the music. I thought the whole video was interesting, but did wish that they had interviewed female artist and got their take on the situation. Another interesting topic that we went over in class was the author discussion. Even now I can only name a couple of female or minority authors, but white male authors I can list without a problem. It had never occurred to me before then that a majority of the books I read where by white males. I am unsure about what to do with that information though. In general I pick my reading literature by reading the back of the book and the first couple of pages to see if it is interesting or not. If not obviously I don’t read it. The author of the book never crosses my mind, even when I have finished reading. Even though a large majority of this class was review, I do feel like I have grown somewhat by attending it.

title: User:JFulton


text:
User:Egm6321.f09.Team2/HW2

Problem #1
Problem Statement: Find the Euler Integrating Factor h(y) for a non-linear 1st order ODE for the particular case where we assume hxN=0

Two conditions must be satisfied in order for a nonlinear 1st order ODE to be exact.

The first condition is that $$F(x,y,y')=0$$ must be in the form $$M(x,y)+N(x,y)y'=0$$

The second condition is that $$M_y=N_x$$. This example will show how to satisfy the second condition.

Start with equation $$M(x,y)+N(x,y)y'=0$$ which already satisfies the first condition.

If we assume the equation is not exact we can multiply it by an unkown factor $$h(x,y)$$ (Euler Integrating Factor) to attempt to find a factor that will make the equation exact.

The equation becomes:

$$h(x,y)[M(x,y)dx+N(x,y)dy]=0$$

$$(hM)dx+(hN)dy=0, where\ hM=\overline M, and\ hN=\overline N$$

$$\overline M_y=(hM)_y=h_yM+hM_y$$

$$\overline N_x=(hN)_y=h_xN+hN_x$$

Set $$\overline M_y=\overline N_x$$ and rearrange,

$$h_xN-h_yM+h(N_x-M_y)=0$$

To solve for $$h_y$$ we will let $$h_xN=0$$ which implies that $$h_x=0$$ since $$\scriptstyle M\ne 0$$

$$-h_yM+h(N_x-M_y)=0$$, rearrage terms

$${h_y\over h}dy={1\over M}(N_x-M_y)$$ take the integral of both sides,

$$\int {h_y\over h}dy=\int {1\over M}(N_x-M_y)dy$$

If the right hand side of the integral is only a function of y then it can be written as $$g(y)$$

$${1\over M}(N_x-M_y)=g(y)$$

$$ln\mid h\mid=\int g(y)dy$$

$$h(y)=exp\int_{}^{y}g(s)ds$$  You have lost a negative sign in your expression for $$g(y)$$--Egm6321.f09.TA 02:38, 28 September 2009 (UTC)

Problem 2
Problem Statement: Given the Non-homogeneous linear 1st order ODE with VC $$y'+{1\over x}y=x^2$$ show the steps to obtain $$y={x^3\over 4}+{C\over x}$$ where C=Const.

The first step is to check if the given equation is in the proper form:

$$P_{(x)}y'+Q_{(x)}y=R_{(x)}$$ ; Yes [pg.(8-1) Eg.(1)]

If $$P_{(x)}\ne 0$$ for all (x) then:

$$y'+{Q_{(x)}\over P_{(x)}}y={R_{(x)}\over P{(x)}}$$

In this form we can use the Integration Factor Method to solve for y

$$N_{(x,y)}{dy\over dx}+M_{(x,y)}=0$$

Multiply by $$h_{(x,y)}$$

$$h_{(x,y)}[N_{(x,y)}dy+M_{(x,y)}dx]$$

$$(hM)dx+(hN)dy=0$$ where $$(hM)=\overline M;\ (hN)=\overline N$$

Satisfy the exactness condition $$\overline M_y=\overline N_x$$ where:

$$\overline M_y=(hM)_y=h_yM+hM_y$$

$$\overline N_x=(hN)_x=h_xN+hN_x$$

$$h_xN-h_yM+h(Nx-My)=0$$

To solve for h(x,y) we must make an assumption that h is a function of (x) only.  You do not  have to make this assumption. You have chosen $$h(x,y)=h(x)$$ instead of $$h(x,y)=h(y)$$. --Egm6321.f09.TA 02:42, 28 September 2009 (UTC)

Assume hyM=0, so our equation becomes;

$$h_xN+h(N_x-M_y)=0$$

$${h_x\over h}=-{1\over N_{(x,y)}}(N_x-M_y)$$

Let $$N_{(x,y)}=1$$ and $$M_{(x,y)}={Q_{(x)}\over P_{(x)}}y={1\over x}y$$

$$N_x=0$$ and $$M_y={1\over x}$$

Substitute back into the right hand side of the equationto verify h is a function of (x) only,

$${1\over 1}(0-{1\over x})={1\over x}=f(x)$$

Using that result we can now solve for $$h$$

$$\int {h_x\over h}dx=\int {1\over x}dx$$

$$ln|h|=ln|x|$$

$$h_x=x$$

Now that we have $$h_x$$ we can multiply through our original equation to obtain,

$$(x)y'+(x){1\over x}y=(x)x^2\ =xy'+y=x^3$$

Integrating yields:

$$xy={x^4\over 4}+C$$ divide by x,  Show the steps - the terms must be grouped together as the derivative of a product before you can integrate this. It is not directly integrable in this form.--Egm6321.f09.TA 02:42, 28 September 2009 (UTC)  $$y={x^3\over 4}+{C\over x}$$

Problem #3
Problem Statement: Show that $$ \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$  is exact.

In this problem, we are asking if: (1) this equation is only a function of x, and (2) it is "exact", meaning exact or exactly integrable by the Integrating Factors Method.

To be exact, it must meet 2 conditions:

(1)it must meet the form $$M(x,y) + N(x,y)y' = 0$$

(2)$$M_y = N_x$$

If $$M(x,y)dy = a(x)$$, then $$M(x,y) = a(x)y + k$$ where k is a constant.

Similarly:

$$N(x,y)dx = b(x)$$, then $$N(x,y) = \int_{}^{x}b(s)\, ds = c(x) = \overline{b}(x)$$. Replacing M and N in our initial form

$$M(x,y)dx + N(x,y)dy = [a(x)y + k]dx + c(x)dy = 0$$

Dividing by $$dx$$ and re-arranging:

$$[a(x)y +k] + \overline{b}(x) y' = 0$$. Our initial problem meets condition 1, where

$$\overline{b}(x) = \frac{1}{2}x^2 = N(x,y)$$

$$a(x) + k = x^4y + 10 = M(x,y)$$ where $$k = 10$$

For condition 2, $$M_y = \frac{d}{dy}[a(x) + k] = \frac{d}{dy}(x^4y + 10) = x^4$$ and $$N_x = \frac{d}{dx}(\frac{1}{2}x^2) = x$$

But these two are not equal. Therefore, using the Integrating Factors Method, we must find $$h(x,y)$$ such that $$ (hM)_y = (hN)_x$$ and thereby satisfy condition 2.

Differentiating and rearranging terms gives us:

$$h_xN - h_yM + h(N_x - M_y) = 0$$. If we assume $$h_y=0$$ since $$M_y$$ and $$N_x$$ are both functions of x, and rearrange terms, we get:

$$\frac{h_x}{h} = \frac{-1}{N}(N_x - M_y)dx$$. Integrating both sides creates $$log (h)$$. Solving for h:

$$ h = e$$^$$(-2)(\int_{}^{x}N_x - M_y,\ dx$$. We know $$N_x$$ and $$M_y$$.  Therefore $$h = e$$^$$(-2)(\frac{1}{2}x^2 - \frac{1}{5}x^5)$$.

Multiplying $$M(x,y)$$ and $$N(x,y)$$ by $$h$$ causes our function to meet the 2nd condition and therefore be exact.

$$(e$$^$$(\frac{2}{5}x^5 - \frac{1}{2}x^2) (\frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$ is exact.  This equation is incomplete. There are parenthesis missing. Where are the differentials?--Egm6321.f09.TA 02:47, 28 September 2009 (UTC)  You need to show that you have arrived at the correct integrating factor. Specifically: $$\frac{\partial\overline{M}}{\partial y}=\frac{\partial\overline{N}}{\partial x}$$.--Egm6321.f09.TA 02:47, 28 September 2009 (UTC)

Problem #4
Problem Statement: Show that $$ (\frac{1}{3}x^{3})(y^{4})y'+(5x^{3}+2)(\frac{1}{5}y^{5})=0$$ is an exact nonlinear, first order ODE.

To prove this equation is an exact, non-linear, 1st order ODE, it must meet 2 conditions:

(1) it must fit the form $$M(x,y)y' + N(x,y) = 0$$ and

(2) $$M_y = N_x$$

It is obvious from simple observation that our equation meets condition 1; however it fails to meet condition 2.

$$M(x,y) = \frac{1}{3}x^3y^4$$  and   $$N(x,y) = (5x^3 - 2)(\frac{1}{5}y^5)$$  then

$$M_y = \frac{4}{3}x^3y^3$$ and $$N_x = 5x^3y^4$$. Clearly $$M_y \ne N_x$$.

To meet condition 2, we must find h(x,y) such that $$(hM)_y = (hN)_x$$. As $$M_y$$ and $$N_x$$ only differ in $$y$$, let us assume that $$h_x = 0$$. Differentiating and rearranging terms gives us:

$$\frac{h_y}{h} = \frac{1}{M}(N_x - M_y)dx$$. Integrating both sides creates $$log h$$. Solving for h:

$$ h = e$$^$$(\frac{1}{M}\int_{}^{y}N_x - M_y,\ dy)$$. We know $$N_x$$ and $$M_y$$. Substituting for $$N_x$$ and $$M_y$$ and simplifying:

$$h = e$$^$$(15y-4(log(y))$$. This, multiplied by our original equation will cause it to be an exact, nonlinear, first order ODE.

 The $$\frac{1}{M}$$ should be inside the integral.--Egm6321.f09.TA 02:51, 28 September 2009 (UTC) The simpliest method to prove this equation as an exact, nonlinear, first order ODE is to show that (1) it fits the form for a class of exact, non-linear 1st order ODE's. Specifically:

$$\overline{b}(x)c(y)y' + a(x)\overline{c}(y) = 0$$

where $$\overline{b}(x) = \int_{}^{x} b(x)\, dx$$

and $$\overline{c}(y) = \int_{}^{y} c(y)\, dy$$

If we assume:

$$a(x) = 5x^{3}+2$$

$$b(x) = x^{2}$$ and

$$c(y) = y^{4}$$ then

$$\overline{b}(x) = \int_{}^{x} x^{2}\, dx = \frac{1}{3}x^{3}$$ and $$\overline{c}(y) = \int_{}^{y} y^{4}\, dy = \frac{1}{5}y^{5}$$

Since our equation does fit the form for this class, it is an exact, nonlinear, first order ODE.

 Need to verify that $$\overline{N}_x=\overline{M}_y$$--Egm6321.f09.TA 02:51, 28 September 2009 (UTC)

Problem #5
Problem Statement: Show that the second exactness condition for $$xyy''+x(y')^{2}+yy'=0$$ is satisfied.

The 2nd exactness condition actually consists of 2 parts, and our equation must meet both:

(1) fxx + (2p) fxy + (p2) fyy = gxp + p(gyp) - gy

(2) fxp + p (fyp) + 2 (fy) = gpp

where p is y'

In our equation,

f(x,y,p) = $$xy$$ and  g(x,y,p) = $$x(y')^{2}+yy'$$

Looking at our equation, for the first half of the condition:

fxx = $$\frac{d}{dx}$$[$$\frac{d}{dx}$$($$xy$$)] = $$\frac{d}{dx}$$($$y$$) = 0

fxy = $$\frac{d}{dy}$$[$$\frac{d}{dx}$$($$xy$$)] = $$\frac{d}{dy}$$($$y$$) = 1

fyy = $$\frac{d}{dy}$$[$$\frac{d}{dy}$$($$xy$$)] = $$\frac{d}{dy}$$($$x$$) = 0

gxp = $$\frac{d}{dp}$$[$$\frac{d}{dx}$$($$x(p)^{2}+yp$$)] = $$\frac{d}{dp}$$($$p^{2}$$) = 2p

gyp = $$\frac{d}{dp}$$[$$\frac{d}{dy}$$($$x(p)^{2}+yp$$)] = $$\frac{d}{dp}$$($$p$$) = 1

gy = $$\frac{d}{dy}$$($$x(p)^{2}+yp$$) = p

Plugging back into the first half of our condition,

$$0 + 2p(1) + p2(0) = 2p + p(1) - p$$

$$2p = 2p$$  and therefore the first half of the condition is satisfied.

 What is sup? Check to verify that your code has rendered properly.--Egm6321.f09.TA 02:57, 28 September 2009 (UTC)

Looking at the 2nd half of the condition:

fxp = $$\frac{d}{dp}$$[$$\frac{d}{dy}$$($$xy$$)] = $$\frac{d}{dp}$$($$y$$) = 0

fyp = $$\frac{d}{dp}$$[$$\frac{d}{dx}$$($$xy$$)] = $$\frac{d}{dp}$$($$x$$) = 0

fy = $$\frac{d}{dy}$$($$xy$$)] = x

gpp = $$\frac{d}{dp}$$[$$\frac{d}{dp}$$($$x(p)^{2}+yp$$)] = $$\frac{d}{dp}$$($$2xp + y$$) = $$2x(p')$$

Plugging back into the second half of our condition (remembering that p = y'),

$$0 + p(0) + 2(x)p' = 2x(p')$$ thereby satisfying the second half of the condition.  What is $$p'$$ ? This would be $$ y''$$ and this should not appear in this problem.--Egm6321.f09.TA 02:57, 28 September 2009 (UTC)

Problem #6
Problem Statement: Derive $$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$ using the following relations:

(1)  $$f(x,y,p):=\phi_{p}(x,y,p)$$ (2)  $$g(x,y,p):=\phi_{x}+\phi_{y}p$$ (3)  $$\phi_{xp}=\phi_{px}$$ (4)  $$\phi_{yp}=\phi_{py}$$

Start by taking the partial derivative of g with respect to p twice: $$g_{pp}=\frac{\partial^2 g}{\partial p^2}$$

$$g_{p}=\phi_{xp}+\phi_{yp}p+\phi_{y}$$ $$g_{pp}=\phi_{xpp}+\phi_{ypp}p+\phi_{yp}+\phi_{yp}$$ $$=>g_{pp}=\phi_{xpp}+\phi_{ypp}p+2\phi_{yp}$$

Now use (1), (3), and (4) to substitute $$\phi_{xpp}$$, $$\phi_{ypp}$$, and $$\phi_{yp}$$

$$f_{x}=\phi_{px}=\phi_{xp}$$ by (1) and (3) $$f_{xp}=\phi_{xpp}$$ $$f_{y}=\phi_{py}=\phi_{yp}$$ by (1) and (4) $$f_{yp}=\phi_{ypp}$$

Therefore,  $$g_{pp}=f_{xp}+f_{yp}p+2f_{y}=f_{xp}+pf_{yp}+2f_{y}$$  Clear and concise.--Egm6321.f09.TA 02:58, 28 September 2009 (UTC)

Problem #7
Problem Statement: Derive $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_{y}$$ using the following relations: (1) $$f(x,y,p):=\phi_{p}(x,y,p)$$ (2) $$g(x,y,p):=\phi_{x}+\phi_{y}p$$ (3) $$\phi_{px}=\phi_{xp}$$ (4) $$\phi_{py}=\phi_{yp}$$ (5) $$\phi_{xy}=\phi_{yx}$$

Solving (1) for $$\phi_{p}$$ and taking the partial with respect to $$x$$ yields $$\phi_{px}=f_{x}$$ Then solving (2) for $$\phi_{x}$$ and taking the partial with respect to $$p$$ yields $$\phi_{xp}=g_{p}-\phi_{yp}p-\phi_{y}$$ Equating these two by (3) and solving for $$\phi_{y}$$ yields $$\phi_{y}=-f_{x}+g_{p}-\phi_{yp}p$$ Performing a similar task, take (1), solve for $$\phi_{p}$$ and take the partial with respect to $$y$$ $$\phi_{py}=f_{y}$$ Then solving (2) for $$\phi_{y}$$ and taking the partial with respect to $$p$$ yields $$\phi_{yp}=\frac{g_{p}}{p}-\frac{g}{p^2}-\frac{\phi_{xp}}{p}+\frac{\phi_{x}}{p^2}$$ Equating these two by (4) and solving for $$\phi_{x}$$ yields $$\phi_{x}=f_{y}p^2-g_{p}p+g+\phi_{xp}p$$

Finally, $$fx=\phi_{px}=\phi_{xp}$$ and $$fy=\phi_{py}=\phi_{yp}$$ as previously shown and so the two equations are (a) $$\phi_{y}=-f_{x}+g_{p}-f_{y}p$$ (b) $$\phi_{x}=f_{y}p^2-g_{p}p+g+f_{x}p$$ Taking the partial with respect to $$x$$ for (a) and the partial with respect to $$y$$ for (b) and equating based on (5) yields $$-f_{xx}+g_{px}-f_{yx}p=f_{yy}p^2-g_{py}p+g_{y}+f_{xy}p$$

Noting $$f_{yx}=f_{xy}$$, $$g_{px}=g_{xp}$$, $$g_{py}=g_{yp}$$, and moving the f terms to the left and the g terms to the right

 $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_{y}$$  You have performed much unnecessary rearranging of $$\Phi$$ terms to get here. You need to rearrange until the end. Equate $$\Phi_{xy}=\Phi_{yx}$$ using the appropriate relations and your solution will be must more straight forward.--Egm6321.f09.TA 03:01, 28 September 2009 (UTC)

Problem #8
Problem Statement: Equations 4&5 on p.([[media:Egm6321.f09.p10-2.png|10-2]]).

Given:

$$

(8x^{5}y^{'})y^{''} + 2x^{2}y^{'} + 20x^{4}(y^{'})^{2} + 4xy = 0

$$

Show that the Nonlinear 2nd-Order ODE is exact

Solution
Second Condition of Exactness: 10-2 Eq (4),(5)

$$

\ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \

$$

$$

\ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \

$$

$$

\ F(x,y,p)y^{''} + G(x,y,p) = 0

$$

$$

\ F(x,y,p) = 8x^{5}p \qquad \qquad G(x,y,p) = 2x^{2}p + 20x^{4}p^{2} + 4xy

$$

The Following partial derivatives are then identified:

$$

F_{x}=40x^{4}p  \qquad \qquad  G_{x}= 4xp + 80x^{3}p^{2} + 4y

$$

$$

F_{xx}=160x^{3}p \qquad \qquad  G_{xp}= 4x + 160x^{3}p

$$

$$

F_{xy}=0 \qquad \qquad  G_{y}= 4x

$$

$$

F_{xp}=40x^{4}  \qquad \qquad  G_{yp}= 0

$$

$$

F_{y}=0 \qquad \qquad  G_{p}= 2x^{2} + 40x^{4}p

$$

$$

F_{yy}=0 \qquad \qquad  G_{pp}= 40x^{4}

$$

$$

\ 160x^{3}p +2p(0) +p^{2}(0) = 4x + 160x^{3}p + (0) -4x

\ 160x^{3}p = 160x^{3}p $$

Page 10-2 Eq(4) is then satisfied.

Applying the results in Page 10-2 Eq(5):

$$

\ 40x^{4} + p(0) + 2(0) = 40x^{4}

$$

$$

\ 40x^{4} = 40x^{4}

$$

Page 10-2 Eq(5) is also satisfied. Because Eq(4) and Eq(5) are both satisfied, then the Secound Exactness Condition is satisfied.

Therefore, $$

\ F(x,y,p) = 8x^{5}p \qquad \qquad G(x,y,p) = 2x^{2}p + 20x^{4}p^{2} + 4xy

$$

is Exactness Second ODE is an exact, nonlinear second ODE
 Nice work. One caveat: For your final statement, express the equation in terms of $$x,y,y'$$. The equation involves an unknown function $$y(x)$$, where $$p$$ was a dummy variable which helped us perform differentiations. --Egm6321.f09.TA 03:05, 28 September 2009 (UTC)

Problem #9
Problem Statement: Verify the exactness of the ODE $$\sqrt{x}y''+2xy'+3y=0$$.

Given

$$

\sqrt[]{x}y^{''} + 2xy^{'} + 3y = 0

$$

Prove that the equation is not exact.

Solution
First Condition of Exactness Page10-1 Eq(1):

$$

\ F(x,y,p)y^{''} + G(x,y,p) = 0

$$

$$

\ F(x,y,p) = \sqrt[]{x} \qquad \qquad G(x,y,p) = 2xp + 3y

$$

This satisfy satisfies the First Condition of Exactness. The Second Exactness Condition for a second order ODE is as follows:

$$

\ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Page 10-2 Eq(4)

$$

$$

\ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Page 10-2 Eq(5)

$$

The following partial derivatives are found:

$$

F_{x}= -\frac{1}{2}x^{\frac{-1}{2}}  \qquad \qquad   G_{x}= 2p

$$

$$

F_{xx}= -\frac{1}{4}x^{\frac{-3}{2}}  \qquad \qquad  G_{xp}= 2

$$

$$

F_{xy}= 0 \qquad \qquad \qquad  G_{y}= 3

$$

$$

F_{xp}= 0  \qquad  \qquad \qquad  G_{yp}= 0

$$

$$

F_{y}= 0 \qquad \qquad \qquad  G_{p}= 2x

$$

$$

F_{yy}=0 \qquad \qquad \qquad  G_{pp}= 0

$$

$$

F_{yp}=0 \qquad \qquad

$$

Using these values in Eq(4):

$$

\frac{-1}{4}x^{\frac{-3}{2}} + 2p(0) + p^{2}(0) = 2 + p(0) -3

$$

$$

\frac{-1}{4}x^{\frac{-3}{2}} = -1

$$

Page 10-2 Eq(4) is not satisfied therefore the ODE is not exact.
 Add equation numbers to your solution here so that you can reference the equations in your solution, rather than a transparency that is located somewhere else on the web. This will make your solution self contained and complete in itself.--Egm6321.f09.TA 03:09, 28 September 2009 (UTC)

Page 10-2 Eq(5) can be populated to find:

$$

\ 0 = 0 + p(0) + 2(0)

$$

$$

\ 0 = 0

$$

Eq (5) is satisfied.

In order for the ODE to be exact it must satisfy both Eq(4)and Eq(5), since it fails to do this it is concluded that the ODE is not exact.

Contributing Team Members
Egm6321.f09.Team 2.walker 20:58, 20 September 2009 (UTC) (Walker, Matthew)

Joe Gaddone 14:04, 21 September 2009 (UTC)

Egm6321.f09.Team2.sungsik 19:20, 23 September 2009 (UTC)

Kumanchik 19:53, 23 September 2009 (UTC)

title: User:Egm6321.f09.Team2/HW2


text:
User:Egm6321.f09.Team2/HW3

Homework Assignment #3 - due Wednesday, 10/7, 21:00 UTC

Problem 1
Find $$(m,n)$$ such that eqn. 1 on ([[media:Egm6321.f09.p13-1.png|p.13-1]]) is exact. A first integral is $$ \Phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2 $$ where $$k_1,k_2$$ are constants.

Problem Statement : Given a L2_ODE_VC $$\sqrt{x}y''+2xy'+3y=0$$

Find (m,n) from the integrating factor (xm,yn) that makes the equation exact.

A first integral is $$\phi(x,y,p)=xp+\big(2x^{3/2}\big)y+k_1=k_2$$

 * $$\phi_p=x$$
 * $$\phi_x=p+3x^{1/2}$$
 * $$\phi_y=2x^{3/2}-1$$

$$\phi(x,y,p)=h(x,y)+\int_{}f(x,y,p)dp, f=\phi_p$$

$$\phi(x,y,p)=h(x,y)+\int_{}xdp={h(x,y)+xp}$$

$$g(x,y,p)=\phi_x+\phi_yp=h_x+\phi_x+(h_y+\phi_y)p$$

$$g(x,y,p)=p+3x^{1/2}+(2x^{3/2}-1)p=h_x+x+(h_y+0)p$$

 * $$h_y=2x^{3/2}$$
 * $$h_x=3x^{1/2}-x$$

$$y^n=h_y\Rightarrow lny^n=ln(2x^{3/2})$$  $$n=ln(2x^{3/2})$$ $$x^m=h_x\Rightarrow lnx^m=ln(3x^{1/2}-x)$$  $$m=ln(3x^{1/2}-x)$$  You were asked to find the $$m,n$$ that make the ODE exact. Your expressions for m,n are functions of $$x$$ which is not constant. Start with the ODE and multiply by $$h(x,y)=x^my^n$$ and then enforce the exactness conditions to solve for $$m,n$$. --Egm6321.f09.TA 05:11, 15 October 2009 (UTC)

Problem 2
Solve eqn. 2 on ([[media:Egm6321.f09.p13-1.png|p.13-1]]) for $$y(x)$$.

Problem Statement: Given a first integral$$\phi$$ of a L2_ODE_VC, solve for $$y(x)$$.

$$\phi(x,y,p)=xp+(2x^{3\over 2}-1)y+k_1=k_2$$ (1)

where k1 and k2 are const, and $$p=y'$$

Eq. (1) is in the form $$M(x,y)+N(x,y)y'$$ where

$$M(x,y)=(2x^{3\over 2}-1)y$$

$$N(x,y)=x$$

so it satisfies the 1st condition of exactness.

Check if $$M_y=N_x$$ for the 2ndcondition of exactness

$$M_y=2x^{3\over2}-1$$

$$N_x=1$$

$$M_y\ne N_x$$ so we do not satisfy the 2nd condition of exactness.

We must apply the integrating factor method for a L1_ODE_VC.

$$xp+(2x^{3\over 2}-1)y=k_2-k_1$$, divide by x to obtain the form:

$$y'+a_0(x)y=b(x)$$ where:

$$a_o(x)={1\over x}(2x^{3\over 2}-1)=2\sqrt{x}-{1\over x}$$

$$b(x)={k_2-k_1\over x}$$

From our solution of a general non-homogeneous L1_ODE_VC p.8-1

$$h(x)=exp\int_{}^{x}a_0(s)ds$$

$$h(x)=exp\int_{}^{x}2\sqrt{x}-{1\over x}=exp\bigg({4\over 3}x^{3\over 2}-ln\left|x\right|\bigg)\bigg({{k_2-k_1}\over x}\bigg)=exp{4\over 3}x^{3\over 2}-x$$

 Something isn't right here. I think this is where you got off track.--Egm6321.f09.TA 23:49, 27 October 2009 (UTC)

From p.8-2 Eq. (4)

$$y(x)={1\over h(x)}\int_{}^{x}h(s)b(s)ds$$

Use the product rule of integration $$\int_{} ab=a\int_{} b-\int_{}a'\int_{}b$$

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-\int_{}^{x}h'(x)\int_{}^{x}b(x)\bigg]$$

In our example $$\int_{}^{x}h'(x)=h(x)$$ so,

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-h(x)\int_{}^{x}b(x)\bigg]$$

$$y(x)=0$$

 $$y(x)\neq 0$$. Check your steps. Egm6321.f09.TA 23:49, 27 October 2009 (UTC)

Problem 3
From ([[media:Egm6321.f09.p13-1.png|p.13-1]]), find the mathematical structure of $$\Phi$$ that yields the above class of ODE.

$$F(x,y,y',y'')={d\phi\over dx}=\phi_x(x,y,p)+\phi_y(x,y,p)p+\phi_p(x,y,p)p'\ where\ p=y'$$

$$\phi=h(x,y)+\int_{}\phi_pdp=h(x,y)+\phi_pp$$

$$\phi_x=h_x+\phi_{px}p$$

$$\phi_y=h_y+\phi_{py}p$$

$$g=\phi_x+\phi_yp=h_x+\phi_{px}p+(h_y+\phi_{py}p)p$$

$$h_y=\phi_y-\phi_{px}$$

Take the integral of $$h_y$$

$$h(x,y)=(\phi_y-\phi_{px})y+k_1$$

Substitute back into the equation for $$\phi$$

$$\phi=(\phi_y-\phi_{px})y+k_1+\phi_pp$$

Rearrange the terms to obtain

 $$\phi(x,y,p)=\phi_pp+(\phi_y-\phi_{px})y+k$$ where,

 * $$P(x)=\phi_p$$
 * $$T(x)=(\phi_y-\phi_{px})y$$
 * $$k=k_1$$

$$\phi(x,y,p)=P(x)p+T(x)y+k$$

Your solution is not easy to follow. The reader has to guess why you are taking the steps you chose. Egm6321.f09.TA 01:53, 28 October 2009 (UTC)

Problem 4
From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for the case $$n=1$$ (N1_ODE) $$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$$. Show that $$f_0-\frac{df_1}{dx}=0 \Leftrightarrow\Phi_{xy}=\Phi_{yx}$$. Hint: Use $$f_1=\Phi_y$$. Specifically: 4.1) Find $$f_0$$ in terms of $$\Phi$$ 4.2) Find $$f_1$$ in terms of $$\Phi$$($$f_1=\Phi_y$$) 4.3) Show that $$ f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$$.

Problem Statement: Given a N1_ODE, for the case n=1 $$F(x,y,y')=0\Leftrightarrow {d\phi\over dx}(x,y)$$

Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$ Hint:$$f_1=\phi_y$$

$$F={d\phi\over dx}(x,y^{(0)},....y^{(n-1)}=\phi_x+\phi_y6{(0)}y^{(1)}$$  Why is there a 6 in here ? --Egm6321.f09.TA 14:03, 15 October 2009 (UTC)

$$F=\phi_x+\phi_yy'$$

$$f_i:={\partial F\over \partial y^i}$$

4.1
Find $$f_0$$ in terms of $$\phi$$.

$$f_0={\partial F\over \partial y}={\partial (\phi_x+\phi_yy')\over \partial y}=\phi_{xy}$$  $$f_0=\phi_{xy}$$

 $$f_0=\phi_{xy}+\Phi_{yy}y'$$ You are missing terms.

4.2
Find $$f_1$$ in terms of $$\phi_y$$

$$f_1={\partial F\over \partial y'}={\partial (\phi_x+\phi_yy')\over \partial y'}=\phi_y$$  $$f_1=\phi_y$$

4.3
Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$

$$\phi_{xy}-{d\phi_y\over df_x}=\phi_{xy}-\phi_{yx}=0$$  $$\phi_{xy}=\phi_{yx}$$

 You should get $$\phi_{xy}+\phi_{yy}y'=\Phi_{yx}+\phi_{yy}y'$$. --Egm6321.f09.TA 14:03, 15 October 2009 (UTC)

Problem 5
Problem statement : From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for a non-linear, second order, ordinary differential equation (N2_ODE), $$n=2$$:
 * 1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$
 * 2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$
 * 3) Show $$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$
 * 4) Relate eqn. 5 to eqs. 4&5 ([[media:Egm6321.f09.p10-2.png|p.10-2]]).

Before solving, there are a few definitions required. First,
 * $$F(x,y,y',y)=\frac{d}{dx}\phi(x,y,y')=\phi_x+\phi_{y}y'+\phi_{y'}y$$

Second,
 * $$f_i:=\frac{\partial F}{\partial y_i}$$

where
 * $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

Therefore, we can write out $$f_0$$, $$f_1$$, and $$f_2$$ as,
 * $$f_0=\frac{\partial F}{\partial y}=\phi_{xy}+\phi_{yy}y'+\phi_yy'_y+\phi_{y'y}y+\phi_{y'}y_y\!\,$$
 * $$f_1=\frac{\partial F}{\partial y'}=\phi_{xy'}+\phi_{yy'}y'+\phi_y+\phi_{y'y'}y+\phi_{y'}y_{y'}\!\,$$
 * $$f_2=\frac{\partial F}{\partial y}=\phi_{xy}+\phi_{yy}y'+\phi_yy'_{y}+\phi_{y'y}y+\phi_{y'}\!\,$$

Since $$\phi$$ is only a function of $$x$$, $$y$$, and $$y'$$, any partial with respect to $$y''$$ is zero. Additionally, partials of $$y$$ and its derivatives with respect to a different $$y$$ or its derivatives are zero. Therefore,
 * $$f_0=\frac{\partial F}{\partial y}=\phi_{xy}+\phi_{yy}y'+\phi_{y'y}y''\!\,$$
 * $$f_1=\frac{\partial F}{\partial y'}=\phi_{xy'}+\phi_{yy'}y'+\phi_{y'y'}y''+\phi_y\!\,$$
 * $$f_2=\frac{\partial F}{\partial y''}=\phi_{y'}\!\,$$

Taking the total derivative of $$f_2$$ with respect to $$x$$,
 * $$\frac{df_2}{dx}=\phi_{y'x}+\phi_{y'y}y'+\phi_{y'y'}y''$$

 Notice that $$f_1-\frac{df_2}{dx}=\phi_y$$, therefore (1) is satisfied. Taking the total derivative of $$\phi_y$$ with respect to x we get,
 * $$\frac{d\phi_y}{dx}=\phi_{yx}+\phi_{yy}y'+\phi_{yy'}y''$$

 Comparing to $$f_0$$ we see that, $$f_0=\frac{d\phi_y}{dx}$$, therefore (2) is satisfied. Combining the derivative of (1) with (2) we get,
 * $$\frac{df_1}{dx}-\frac{d^2f_2}{dx^2}=\frac{d\phi_y}{dx}=f_0$$

 Rearranging leads to $$\frac{df_1}{dx}-\frac{d^2f_2}{dx^2}=f_0$$ therefore (3) is satisfied. Finally, we can relate (3) to equations 4 & 5 ([[media:Egm6321.f09.p10-2.png|p.10-2]]). By redefining $$F\!\,$$ in terms of new functions $$f\!\,$$ and $$g\!\,$$ (used in eq. 4 & 5) yields,
 * $$F=fy''+g\!\,$$

Following all the same steps to get $$f_i\!\,$$, for $$i=0,1,2\!\,$$, in terms of these new variables yields,
 * $$f_0=\frac{\partial F}{\partial y}=f_yy''+g_y\!\,$$
 * $$f_1=\frac{\partial F}{\partial y'}=f_{y'}y''+g_{y'}\!\,$$
 * $$f_2=\frac{\partial F}{\partial y''}=f\!\,$$

Taking the total derivative of $$f_1\!\,$$ with respect to $$x\!\,$$ once, and taking it twice for $$f_2\!\,$$ yields,
 * $$\frac{df_1}{dx}=\frac{df_{y'}y}{dx}+g_{y'x}+g_{y'y}y'+g_{y'y'}y\!\,$$
 * $$\frac{d^2f_2}{dx^2}=\frac{df_{y'}y}{dx}+f_{xx}+f_{xy}y'+f_{xy'}y+(f_{yx}+f_{yy}y'+f_{yy'}y)y'+f_yy\!\,$$

Now plugging $$f_0\!\,$$, $$\frac{df_1}{dx}\!\,$$, and $$\frac{d^2f_2}{dx^2}\!\,$$ into (3) yields,
 * $$2f_yy+g_y-g_{y'x}-g_{y'y}y'-g_{y'y'}y+f_{xx}+2f_{xy}y'+f_{xy'}y+f_{yy}y'^2+f_{yy'}yy'=0\!\,$$
 * $$=>(2f_y-g_{y'y'}+f_{xy'}+f_{yy'}y')y''+g_y-g_{y'x}-g_{y'y}y'+f_{xx}+2f_{xy}y'+f_{yy}y'^2=0\!\,$$

Since $$f\!\,$$ and $$g\!\,$$ are not functions of $$y\!\,$$, $$y\!\,$$ can be arbitrary without changing their values. Therefore, to ensure the equation equals zero, the equation can be split into two parts, each equaling zero independently allowing $$y''\!\,$$ to be arbitrary.
 Therefore,
 * $$2f_y-g_{y'y'}+f_{xy'}+f_{yy'}y'=0\!\,$$
 * $$g_y-g_{y'x}-g_{y'y}y'+f_{xx}+2f_{xy}y'+f_{yy}y'^2=0\!\,$$

These equations are equal to 4 & 5 ([[media:Egm6321.f09.p10-2.png|p.10-2]]) if setting $$y'=p\!\,$$.

 Good work. Consider using the notation $$y'=p$$ and $$y''=q$$. Then you do not have to have primes in your subscripts and you can take partials with respect to $$p,q$$. --Egm6321.f09.TA 01:27, 16 October 2009 (UTC)

Problem 6
Problem statement : From ([[media:Egm6321.f09.p14-2.png|p.14-2]]), for the Legendre differential equation $$F=(1-x^2)y''-2xy'+n(n+1)y=0$$, 6.1 Verify exactness of this equation using two methods: 6.1a.) ([[media:Egm6321.f09.p10-3.png|p.10-3]]), Equations 4&5. 6.1b.) ([[media:Egm6321.f09.p14-1.png|p.14-1]]), Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

First, no matter which method is used, the form of the equation meets the first exactness criteria since, $$F$$ can be written as,
 * $$F(x,y,y',y)=f(x,y,y')y+g(x,y,y')=0\!\,$$

Next we test the second set of exactness relations. The first two for part 1 are,
 * $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y\!\,$$
 * $$f_{xp}+pf_{yp}+2f_y=g_{pp}\!\,$$

where,
 * $$f=1-x^2\!\,$$
 * $$g=-2xy'+n(n+1)y\!\,$$:$$\!\,$$

Therefore we can calculate all the partials required to test exactness,
 * $$f_{xx}=-2, f_{xy}=f_{yy}=f_{xp}=f_{yp}=f_y=0\!\,$$
 * $$g_{xp}=-2, g_y=n(n+1), g_{yp}=g_{pp}=0\!\,$$

 Plugging into the exactness relations we see,
 * $$n(n+1)=0\!\,$$ (satisfied for $$n=-1$$ and $$n=0$$)

and,
 * $$0=0\!\,$$ (satisfied identically)

Last, we test the second exactness relation for part 2. That is,
 * $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0\!\,$$

where $$f_i=\frac{\partial F}{\partial y^i}\!\,$$, $$i=0,1,2\!\,$$. Writing out each $$f_i\!\,$$,
 * $$f_0=n(n+1)\!\,$$
 * $$f_1=-2x=>\frac{df_1}{dx}=-2\!\,$$
 * $$f_2=1-x^2=>\frac{df_2}{dx}=-2x=>\frac{d^2f_2}{dx^2}=-2\!\,$$

 Plugging into the relationship yields,
 * $$n(n+1)+2-2=0\!\,$$ (satisfied for $$n=-1$$ and $$n=0$$)

 Nice work. What about the case when $$n\neq 0,1$$? Can you make the equation exact using an integration factor?--Egm6321.f09.TA 01:29, 16 October 2009 (UTC)

Problem 7
Problem statement: From ([[media:Egm6321.f09.p14-3.png|p.14-3]]). Show that equations 1 and 2, namely 7.1 $$\forall u(x)\!\,$$ and $$v(x)\!\,$$, $$L(u+v)=L(u)+L(v)\!\,$$. and 7.2 $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\!\,$$ with $$u\!\,$$ a function of $$x\!\,$$. are equivalent to equation 3 on [[media:Egm6321.f09.p3-3.png|p.3-3]].

An operator is linear if it follows the relationship $$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)\!\,$$. This is equivalent to saying that the operator must satisfy simultaneously.
 Given $$L(\alpha u+\beta v)\!\,$$, it can be split into two parts using (1),
 * 1) $$L(u+v)=L(u)+L(v)\!\,$$
 * 2) $$L(\alpha u)=\alpha L(u)\!\,$$
 * $$L(\alpha u+\beta v)=L(\alpha u)+L(\beta v)\!\,$$

Then each part can be used with (2)
 * $$L(\alpha u)=\alpha L(u)\!\,$$
 * $$L(\beta v)=\beta L(v)\!\,$$

Therefore, the two relations (1) and (2) are equivalent to the composite relation if taken together only. <
 clear explanation. --Egm6321.f09.TA 04:02, 16 October 2009 (UTC)

Problem 8
From ([[media:Egm6321.f09.p15-2.png|p.15-2]]), plot the shape function $$N_{j+1}^{2}(x)$$.

[[Media:Graph1.pdf]]

correct. Egm6321.f09.TA 03:08, 28 October 2009 (UTC)

Problem 9
Problem Statement: From (| p.16-2), show that $$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t\right)$$ $$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t\right)$$

$$y_{xxx}=(d/dx)\bigg[{d/dx\over (d/dx)y}\bigg]=(dt/dx)(d/dt)(dt/dx)(d/dt)(dt/dx)(d/dt)y$$

What does this notation mean? Egm6321.f09.TA 03:20, 28 October 2009 (UTC)

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$ 'Chain Rule'

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(t_{tt}))$$

$$y_{xxx}=(e^{-t})(d/dt)(-e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

$$y_{xxx}=(e^{-t})(2e^{-2t}(y_t)-e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))$$

$$y_{ttt}=2e-3t(y_t)-e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}9y_{ttt})$$

Factor out $$e^{-3t}$$ and re-arrange terms in ordre of derivative,

$$y_{xxx}=(e^{-3t})(y_{ttt}-3y_{tt}+2y_t)$$

$$y_{xxxx}=(d/dx)(d/dx)(d/dx)(d/dx)y$$

$$y_{xxxx}=(dt/dx)(d/dt)[(dt/dx)(d/dt)]((dt/dx)(d/dt))\langle(dt/dx)(d/dt)y\rangle$$

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(-2e^{-2t}(y_t)+e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(-2e^{-3t}(y_t)+e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}(y_{ttt}))$$

$$y_{xxxx}=(e^{-t})(6e^{-3t}(y_t)-2e^{-3t}(y_{tt})-3e^{-3t}(y_{tt})+e^{-3t}(y_{ttt})+6e^{-3t}(y_{tt})-2e^{-3t}(y_{ttt})-3e^{-3t}(y_{ttt})+e^{-3t}(y_{tttt}))$$

$$y_{xxxx}=6e^{-4t}(y_t)+2e^{-4t}(y_{tt})+3e^{-4t}(y_{tt})-e^{-4t}(y_{ttt})+6e^{-4t}(y_{tt})-2e^{-4t}(y_{ttt})-3e^{-4t}(y_{ttt})+e^{-4t}(y_{tttt}))$$

Factor out $$e^{-4t}$$ and re-arrange terms in order of derivative.

$$y_{xxxx}=(e^{-4t})(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

It is not clear at all here which terms your differential operators are acting on. At a minimum, you need some parenthesis or brackets to clarify the grouping of terms. Egm6321.f09.TA 03:20, 28 October 2009 (UTC)

Problem 10
Problem Statement: Solve equation 1 on ([[media:Egm6321.f09.p16-1.png|p.16-1]]),
 * $$ x^2y''-2xy'+2y=0 \!\,$$

using the method of trial solution $$y=e^{rx}\!\,$$ directly for the boundary conditions $$\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$$ Compare the solution with equation 10 on [[media:Egm6321.f09.p16-3.png|p.16-3]]. Use MATLAB to plot the solutions.

We solve by trial solution. First attempt $$y(x)=e^{rx}\!\,$$. The derivatives are,
 * $$y'=re^{rx}\!\,$$
 * $$y''=r^2e^{rx}\!\,$$

Plugging into the differential equation we obtain,
 * $$x^2r^2e^{rx}-2xre^{rx}+2e^{rx}=0\!\,$$

Dividing by $$e^{rx}\!\,$$ yields,
 * $$x^2r^2-2xr+2=0\!\,$$

which cannot be solved for uniquely. Therefore we change the trial solution to, $$y(x)=x^r\!\,$$. The derivatives are,
 * $$y'=rx^{r-1}\!\,$$
 * $$y''=r(r-1)x^{r-2}\!\,$$

And plugging into the differential equation yields,
 * $$r(r-1)x^r-2rx^r+2x^r=0\!\,$$

Dividing by $$x^r\!\,$$ since $$x=0\!\,$$ is a trivial solution yields,
 * $$r^2-3r+2=(r-2)(r-1)=0\!\,$$

The roots are then $$r=1\!\,$$ and $$r=2\!\,$$, exactly the same answer as equation 10 on [[media:Egm6321.f09.p16-3.png|p.16-3]]. The form of the answer is then,
 * $$y(x)=C_1x+C_2x^2\!\,$$

 The boundary conditions can be used to obtain $$C_1=4\!\,$$ and :$$C_2=-1\!\,$$. Therefore the solution is,
 * $$y(x)=4x-x^2\!\,$$

very good. Egm6321.f09.TA 03:45, 28 October 2009 (UTC)

Problem 11
Problem Statement: From (| p.17-4) obtain equation 2 from p.17-3 $$

Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$ using the integrator factor method.

Solution:

$$u_{1}(x) Z'+\left[a_{1}u_{1}(x)+2u_{1}' (x) \right]Z=0 $$

$$h(x)Z'+h(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]Z=0 $$

$$\frac{d}{dx}\left[h(x)Z(x) \right]=h(x)Z'(x)+h'(x)Z$$

Let $$h'(x)= h(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

$$\frac{h'(x)}{h(x)}= \left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

Integrating both sides

$$ \int{\frac{h'(x)}{h(x)}} = \int\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

$$ ln|h| = \left[\int_{}^{x}a_{1}(s)ds+ln\left|u_{1} \right|^{2} \right]$$

$$ h = \exp\left[\int^x a_1(s)ds\right]{u_{1}^2}$$

$$ [h(x)Z(x)]' = 0$$ ,where hZ(x) is

$$ hZ(x) = \exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)$$

$$ [hZ(x)]' = [\exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)]'$$

$$ \int[hZ(x)]' = \int[\exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)]'=\int0$$

$$ \exp\left[\int^x a_1(s)ds\right]{u_{1}^2}Z(x)= C $$

Where C is constant

Therefore,

$$ Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$

very good. Egm6321.f09.TA 04:03, 28 October 2009 (UTC)

Problem 12
Problem Statement: From (| p.18-1), develop reduction of order method using the following algebraic options

$$y(x)=U(x)\pm u_1 (x)$$

$$y(x)=\frac{U(x)}{u_1 (x)}$$

$$y(x)=\frac{u_1 (x)}{U(x)}$$

Solution:

a. $$ y(x) = U(x) \pm u_1(x) $$  <eq(1)>

$$y'(x) = U' \pm u_1'$$        <eq(2)>

$$y(x) = U \pm u_1''$$        <eq(3)>

Substituding Eq(1),(2) and(3) into following Eq

$$ y'' + a_1y' + a_0y = 0 $$,

$$ y + a_1y' + a_0y = a_0U \pm a_0u_1 + a_0U' \pm a_1u_a' + U \pm u_1'' $$

Therefore

$$ a_0U \pm a_0u_1 + a_0U' \pm a_1u_a' + U \pm u_1=0 $$

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.

b.

$$y(x)=\frac{U(x)}{u_1 (x)}$$   <eq(1)>

$$y'(x) = \frac{U'u_{1}-Uu_1'}{u_{1}^2}$$   <eq(2)>

$$y''(x) = \frac{(U'u_{1}-Uu_{1}')'u_{1}^2-(U'u_{1}-Uu_{1}')(u_{1}^2)'}{u_{1}^4}$$    <eq(3)>

Substituding eq(1),(2) and (3) into following Equation

$$ y'' + a_1y' + a_0y = 0 $$

$$ y''+a_1y'+a_0y = \frac{(U'u_{1}-Uu_{1}')'u_{1}^2-(U'u_{1}-Uu_{1}')(u_{1}^2)'}{u_{1}^4}

+a_1{\frac{U'u_{1}-Uu_1'}{u_{1}^2}}

+a_0{\frac{U(x)}{u_1 (x)}}$$

Multify equation by $$ u_{1}^4 $$

Rearrage

$$ y''+a_1y'+a_0y = {(U'u_{1}-Uu_{1}')'u_{1}^2-(U'u_{1}-Uu_{1}')(u_{1}^2)'}

+a_1{U'u_{1}-Uu_1'}{u_{1}^2}{u_{1}^2}

+a_0{U(x)}{u_1 (x)}{u_{1}^3}=0$$

The above y(x) here is not valid because the equation above is not missing U(x). The Reduction of Order Method is possible only when dependent variable is missing.

c.

$$y(x)=\frac{u_1 (x)}{U(x)}$$    <eq(1)>

$$y'(x) = \frac{u_{1}'U-u_1U'}{U^2}$$    <eq(2)>

$$y''(x) = \frac{(u_{1}'U-u_{1}U')'U^2-(u_{1}'U-u_{1}U')(U^2)'}{U^4}$$   <eq(3)>

Substituding eq(1),(2) and (3) into following Equation

y'' + a_1y' + a_0y = 0

$$ y''+a_1y'+a_0y = \frac{(u_{1}'U-u_{1}U')'U^2-(u_{1}'U-u_{1}U')(U^2)'}{U^4}

+a_1\frac{u_{1}'U-u_1U'}{U^2}

+a_0\frac{u_1 (x)}{U(x)}$$

Multify equation by $$ U^4 $$

Rearrage

$$ y''+a_1y'+a_0y = [(u_{1}'U-u_{1}U')'U^2-(u_{1}'U-u_{1}U')(U^2)']

+a_1[u_{1}'U-u_1U']U^2

+a_0[u_1]U^3 = 0 $$

In the above equation, the dependent variable U(x) is present and not missing. The Reduction of Order Method is possible only when dependent variable is missing.

very good. Egm6321.f09.TA 04:15, 28 October 2009 (UTC)

Problem 13
Problem Statement: From (| p.18-1), Find $$u_{1}(x)$$ and $$ u_{2}(x)$$ of equation 1 on p.18-1 using 2 trial solutions:

$$ y=ax^b$$

$$ y=e^{rx}$$

Compare the two solutions using boundary conditions $$y(0)=1$$ and $$ y(1)=2$$ and compare to the solution by reduction of order method 2. Plot the solutions in Matlab.

Solution:

1. When $$ y=ax^b $$

$$ (1-x^{2})y''-2xy'+2y=0 $$ <eq(1)>

$$ y=ax^{b} $$ <eq(2)>

$$ y'=bax^{b-1} $$  <eq(3)>

$$ y''=ab(b-1)x^{(b-2)} $$  <eq(4)>

Substituting eq(2),(3) and (4) into Eq(1)

$$ (1-x^2)ab(b-1)x^{(b-2)} - 2xabx^{(b-1)} + 2ax^b = 0 $$ $$ ab(b-1)x^{(b-2)}-ab(b-1)x^b - 2abx^b + 2ax^b = 0 $$

$$ ab(b-1)x^{(b-2)}-\left[ab(b-1) + 2ab - 2a \right]x^b = 0 $$

In the first Term of LHS $$ x^{(b-2)}$$,

$$ ab(b-1) = 0 $$

a and b both cannot be zero, Therefore b=1

In the second Term of LHS, $$ x^b$$

$$ ab(b-1) + 2ab - 2a = 0$$

$$ a(b^2+b-2) = 0$$

a can not be zero ,hence $$ b =1,-2$$

$$ u_1(x) = ax$$

$$ u_2(x) = \frac{a}{x^2} $$

The trail solution is

$$ y = c_1u_1(x) + c_2u_2(x) $$

$$ y = C_1ax + C_2\frac{a}{x^2} $$

Using the boundary conditions,

$$ y(0) = 1 $$

$$ y(1) = 2 $$

$$ 1 = C_1a(0)+C_2\frac{a}{0} $$

$$ 2 = C_1a(1)+C_2\frac{a}{1}$$

This Equation cannot be established.

Therefore

$$ y=ax^b$$ cannot be a solution.

2. When $$ y=e^{rx}$$ $$ (1-x^{2})y''-2xy'+2y=0 $$ <eq(1)>

$$ y=e^{rx} $$  <eq(2)>

$$ y'= re^{rx} $$  <eq(3)>

$$ y''= r^{2}e^{rx} $$   <eq(4)>

Substituting Eq(2),(3) and(4) into Eq(1)

$$ (1-x^{2}) r^{2}e^{rx} -2x re^{rx}  +2 e^{rx}  =0  $$

Missing a term from the line above. Egm6321.f09.TA 04:52, 28 October 2009 (UTC)

Rearrange

$$ -x^2r^2-2xr + r^2 +2 = 0$$ $$ r^2 +2 = 0$$

$$ r = \pm \sqrt2i$$

How can you simply this? Why did you set $$x=0$$? Egm6321.f09.TA 04:52, 28 October 2009 (UTC)

The trial solution becomes,

$$\displaystyle y = e^{\sqrt2ix}$$

$$\displaystyle y = C_1 cos(\sqrt2x) + C_2 sin(\sqrt2x) $$

Boundary Condition y(0)=1 and y(1)=2

$$\displaystyle 1 = C_1 cos(0) + C_2 sin(0)$$

$$ C_{1}= 1$$

$$\displaystyle 2 = cos(\sqrt2) + C_2 sin(\sqrt2)$$

$$ C_2= \frac{2-cos(\sqrt2)}{sin(\sqrt2)}$$

The solution is

$$ y= cos(\sqrt2x) + \frac{2-cos(\sqrt2)}{sin(\sqrt2)}sin(\sqrt2x)$$

[[Media:Graph13.pdf|Graph]]

Contributing Team Members
Joe Gaddone 16:46, 3 October 2009 (UTC) Kumanchik 20:37, 7 October 2009 (UTC) Matthew Walker Egm6321.f09.Team2.sungsik 21:16, 7 October 2009 (UTC)

title: User:Egm6321.f09.Team2/HW3


text: File:Egm6321 hw3 team4.PNG

title: File:Egm6321 hw3 team4.PNG


text:
User:Egm6321.f09.team3.Kuljevan.Francisco/HW3

= Problem 1 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$ x^{m+ \frac{1}{2}}y^ny'' + 2x^{1+m}y^ny' + 3x^my^{n+1} $$

Find
m & n such that the ODE is exact.

Solution
$$f(x,y,p) = x^{m+ \frac{1}{2}}y^n$$

$$g(x,y,p) = 2x^{1+m}y^ny' + 3x^my^{n+1} $$

which satisfies the first condition of exactness.

Condition 2 is met by satifaction of the following equations:

$$ f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y \qquad{...(i)}$$

$$ f_{xp} + pf_{yp} + 2f_y = g_{pp} \qquad{...(ii)}$$

The partial derivatives are given below:

$$ \begin{align} f_x &= \left(m+ \frac{1}{2}\right)x^{m- \frac{1}{2}}y^n \\ f_{xx} &= \left(m- \frac{1}{2}\right) (m+ \frac{1}{2})x^{m- \frac{3}{2}}y^n \\ f_{xy} &= n \left(m+ \frac{1}{2}\right)x^{m- \frac{1}{2}}y^{n-1} \\ f_y &= nx^{m+ \frac{1}{2}}y^{n-1} \\ f_{yy} &= n(n-1)x^{m+ \frac{1}{2}}y^{n-2} \\ f_p &= f_{xp} = f_{yp} = f_{pp} = 0 \\ g_x &= 2(m+1)x^my^np + 3mx^{m-1}y^{n+1} \\ g_{xx} &= 2m(m+1)x^{m-1}y^np + 3m(m-1)x^{m-2}y^{n+1} \\ g_{xy} &= 2(m+1)nx^my^{n-1}p + 3m(n+1)x^{m-1}y^n \\ g_y &= 2nx^{m+1}y^{n-1}p + 3(n+1)x^my^n \\ g_{yy} &= 2n(n-1)2nx^{m+1}y^{n-2}p + 3n(n+1)x^my^{n-1} \\ g_p &= 2x^{m+1}y^n \\ g_{px} &= 2(m+1)x^my^n \\ g_{py} &= 2nx^{m+1}y^{n-1} \\ g_{pp} &= 0 \\ \end{align} $$

Substitute into i. to get:

$$ \left(m- \frac{1}{2}\right) \left(m+ \frac{1}{2}\right)x^{m- \frac{3}{2}}y^n + 2p\cdot n\left(m+ \frac{1}{2}\right)x^{m- \frac{1}{2}}y^{n-1} + p^2 n(n-1)x^{m+ \frac{1}{2}}y^{n-2} = 2(m+1)x^my^n + p \cdot 2nx^{m+1}y^{n-1} - 2nx^{m+1}y^{n-1}p + 3(n+1)x^my^n $$

Substitute into ii. to get:

$$ 0 + p \cdot 0 + 2 \cdot nx^{m+ \frac{1}{2}}y^{n-1} = 0 $$

Equation ii. suggests that n must be zero in order for the equation to be satisfied.

$$ n = 0 $$

Simplify equation i to get:

$$ \left(m- \frac{1}{2}\right) \left(m+ \frac{1}{2}\right)x^{m- \frac{3}{2}} + 0 + 0 = 2(m+1)x^m + 0 - 0 + 3x^m $$

Note the powers of x need to be the same to simplify, which suggests $$ m = \pm \frac{1}{2} $$.

$$ m = \frac{1}{2} $$ satisfies equation i.

Therefore,

$$ n = 0 $$

and

$$ m = \frac{1}{2} $$

The final ODE is

$$ xy'' + 2x^{\frac{3}{2}}y' + 3x^{\frac{1}{2}}my $$

or,

$$ y'' + 2x^{\frac{1}{2}}y' + 3x^{\frac{-1}{2}}my $$

 * }

= Problem 2 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$ xp - (2x^{\frac {3}{2}} - 1)y + k_1 = k_2 $$

Find
y(x)

Solution
The ODE is

$$ y' - \frac{2x^{\frac{3}{2}}-1}{x}y = k_2 - k_1 $$

where

$$ a_0(x) = - \frac{2x^{\frac{3}{2}}-1}{x} \qquad{\text{and}} $$

$$ b(x) = k_2 - k_1 $$

This is a linear, first order ODE with varying coefficients.

In standard form, $$ N = 1 $$ and $$ M = a_0 \cdot y $$.

This makes

$$ \begin{align} h(x) &= \exp \left( \int a_0(s) ds\right) \\ &= \exp \left( \int \left(-2 \sqrt{s} + s^{-1}\right) ds\right) \\ &= x \cdot \exp \left(- \frac{4}{3} x^{\frac {3}{2}}\right) \end{align} $$

Multiply the ODE by the integrating factor to get:

$$ (hy)' = hb $$

Integrate to get:

$$ y(x) = \frac{1}{h(x)} \int h(s) b(s) ds $$

$$ y(x) = \frac{1}{x \cdot \exp \left(- \frac{4}{3} x^{\frac {3}{2}}\right)} \int s \cdot \exp \left(- \frac{4}{3} s^{\frac {3}{2}}\right) \cdot (k_2 - k_1) ds $$

 * }

= Problem 3 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
A class of exact linear second order ODEs with varying coefficients

Find
The mathematical structure of $$\phi$$

Solution
Since $$ F = \frac {d \phi}{dx} = \phi_x(x,y,p) + \phi_y(x,y,p) \cdot y' + \phi_p(x,y,p) \cdot y' $$

and $$ F = P(x) y'' + Q(x) y' + R(x) y $$

This suggests

i. $$ P(x) = \phi_p $$ ii. $$ Q(x) = \phi_y $$

iii. $$ R(x)y = \phi_x $$

From i.,

$$ \phi = \int P(x) dp = P(x) \cdot p + C(x,y) $$

From ii.,

$$ \phi = \int Q(x) dy = Q(x) \cdot y + C(x,p) $$

This suggests $$ \phi $$ has a structure of:

$$ \phi = P(x) \cdot p + Q(x) \cdot y + C(x,y) $$

Take the parital derivative with respect to x,

$$ \frac{\partial{\phi}}{dx} = \frac{\partial{P}}{dx} \cdot p + \frac{\partial{Q}}{dx} \cdot y + \frac{\partial{C}}{dx} $$

and compare to iii, which suggests C is not a function of x!

The final form is:

$$ \phi = P(x) \cdot p + T(x) \cdot y + k $$

where P(x) and T(x) are functions, and k is a constant.

 * }

= Problem 4 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The conditions for exactness:

i. $$ F = \frac{d\phi}{dx} = \phi_x + \phi_y \cdot y' + ... + \phi_{y^{n-1}} \cdot y^n $$

ii. $$ f_i = \partial{F}{y^{(i)}} $$ such that $$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} - ... + (-1)^n \frac{d^nf_n}{dx^n} = 0 $$

Find
the n= 1 case

Solution
$$ f_1 = \frac{\partial{F}}{dy'} = \frac{\partial{}}{dy'}(\phi_x + \phi_y y') = \phi_y $$

Therefore,

$$ f_0 - \frac{df_1}{dx} = \frac{\partial {F}}{dy} - \frac{d}{dy}(\phi_y) = \frac{\partial{(\phi_x + \phi_y y')}}{dy'} - \frac{d}{dx}(\phi_y) $$

$$ \frac{\partial{}}{dy'}(\phi_x + \phi_y y') - \frac{d}{dx}(\phi_y) = \phi_{xy} + \phi_{yy}y' - \phi_yx - \phi_{yy}y' $$

$$ \frac{\partial{}}{dy'}(\phi_x + \phi_y y') - \frac{d}{dx}(\phi_y) = \phi_{xy}  - \phi_yx $$

Therefore,

$$ f_0 - \frac{df_1}{dx} = 0 $$ is equivalent to $$ \phi_{xy}  = \phi_yx $$, which is the condition of exactness for the linear first order ODE!

 * }

= Problem 5 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The conditions for exactness:

i. $$ F = \frac{d\phi}{dx} = \phi_x + \phi_y \cdot y' + ... + \phi_{y^{n-1}} \cdot y^n $$

ii. $$ f_i = \partial{F}{y^{(i)}} $$ such that $$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} - ... + (-1)^n \frac{d^nf_n}{dx^n} = 0 $$

Find
the n=2 case

Solution
$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} $$

Integrate with respect to x to get:

$$ \int f_0 dx - f_1 + \frac{df_2}{dx} $$

where $$ f_0 = \frac{\partial{F}}{dy} = \phi_{xy} + \phi_{yy} y' + \phi_{y'y} y'' $$

Therefore, $$ \int f_0 dx = \phi_y $$

Now, take the derivative with respect to x to get the original equation,

$$ \frac{d}{dx}(\phi_y) = \phi_{xy} + \phi_{yy} y' + \phi_{yy'} y'' $$

which is, of course, $$ f_0 $$!

The final equations is validated:

$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} $$

 * }

= Problem 6 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The Legendre linear, second order ODE with varying coefficients: $$ F = (1-x^2)y'' - 2xy' + n(n+1)y = 0 $$

Find
Verify the exactness of the equation using:

a. equations 4 and 5 from 10-2

b. equation 5 from 14-1

Solution
a. Using the following 2 equations:

i. $$ f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y $$

ii. $$ f_{xp} + pf_{yp} + 2 f_y = g_{pp} $$

where F = f(x,y,p)y'' + g(x,y,p)

$$f(x,y,p) = 1-x^2$$

$$g(x,y,p) = -2xp + n(n-1) y$$

The partial derivatives are:

$$f_x = -2x$$

$$f_{xx} = -2$$

$$f_{xy} = f_y = f_{yy} = f_{xp} = f_{yp} = f_p = f_{pp} = 0$$

$$g_x = -2p$$

$$g_{xp} = -2$$

$$g_y = n(n+1)$$

$$g_p = -2x$$

$$ g_{xx} = g_{xy} = g_{yy} = g_{yp} = g_{pp} = 0 $$

Substitute to get:

i. $$ -2 + 2p \cdot 0 + p^2 \cdot 0 = -2 + p \cdot 0 - n(n+1) $$

ii. $$ 0 + p \cdot 0 + 2 \cdot 0 = 0 $$

The ODE is not exact because the equation i is not satisfied.

b. Using the following equation:

$$ f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} = 0 $$

where

$$ f_0 = \frac{\partial{F}}{dy} = n(n+1) $$

$$ f_1 = \frac{\partial{F}}{dy'} = -2x $$

$$ f_2 = \frac{\partial{F}}{dy''} = 1-x^2 $$

Combining,

$$n(n+1) - (-2) + (-2) = 0$$

Which is exact for the solution of $$ n = 0 $$ (trivial) or $$ n=-1 $$

However, the ODE can be made exact using the integrating factors method:

$$h(x,y) = x^my^n$$

Multiply by the ODEto get:

$$ x^my^n [ (1-x^2)y'' - 2xy' + n(n+1) y = 0 ] $$

which makes $$ f(x,y,p) = x^my^n - x^{m+2}y^n $$ and $$ g(x,y,p) = - 2x^{m+1}y^np + \alpha(\alpha+1)x^m y^{n+1} $$

The partial derivatives are:

$$f_x = mx^{m-1}y^n - (m+2)x^{m+1}y^n$$

$$f_{xx} = m(m-1)x^{m-2}y^n - (m+2)(m+1)x^{m}y^n$$

$$f_{xy} = mnx^{m-1}y^{n-1} - (m+2)nx^{m+1}y^{n-1}$$

$$f_y = nx^my^{n-1} - nx^{m+2}y^{n-1} $$

$$ f_{yy} = n(n-1)x^my^{n-2} - n(n-1)x^{m+2}y^{n-2} $$

$$ f_{xp} = f_{yp} = f_p = f_{pp} = 0 $$

$$ g_x = -2(m+1)x^my^np + m\alpha (\alpha + 1)x^{m-1}y^{n+1} $$

$$ g_{xx} = -2m(m+1)x^{m-1}y^np + m(m-1)\alpha (\alpha + 1)x^{m-2}y^{n+1} $$

$$ g_{xy} = -2(m+1)nx^my^{n-1}p + m\alpha (\alpha + 1)(n+1)x^{m-1}y^{n} $$

$$ g_y = - 2nx^{m+1}y^{n-1}p + (n+1)\alpha(\alpha+1)x^m y^{n} $$

$$ g_{yy} = - 2n(n-1)x^{m+1}y^{n-2}p + n(n+1)\alpha(\alpha+1)x^m y^{n-1} $$

$$ g_p = -2x^{m+1}y^n $$

$$ g_{yp} = -2nx^{m+1}y^{n-1} $$

$$ g_{xp} = -2(m+1)x^my^n $$

$$ g_{pp} = 0 $$

Substituting into equation i:

$$ m(m-1)x^{m-2}y^n - (m+2)(m+1)x^{m}y^n + 2p \cdot (mnx^{m-1}y^{n-1} - (m+2)nx^{m+1}y^{n-1}) + p^2 \cdot (n(n-1)x^my^{n-2} - n(n-1)x^{m+2}y^{n-2}) = -2(m+1)x^my^n + p \cdot (-2nx^{m+1}y^{n-1}) - (- 2nx^{m+1}y^{n-1}p) + (n+1)\alpha(\alpha+1)x^m y^{n}) $$

and into equation ii:

$$ 0 + p \cdot 0 + 2 \cdot (nx^my^{n-1} - nx^{m+2}y^{n-1}) = 0 $$

Equation ii can only be solved by $$ n = 0 $$

which leaves equation i as:

$$ m(m-1)x^{m-2} - (m+2)(m+1)x^{m} + 2p \cdot (0 + 0) + p^2 \cdot (0 - 0) = -2(m+1)x^m + p \cdot (0) - (-0) + \alpha(\alpha+1)x^m) $$

Which may be satisfied for $$ m = 1 $$

(Incidentally, the solution requires $$ \alpha = 1 $$ to be true.)

 * }

= Problem 7 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The definitions of linearity:

a. $$ L(\alpha u + \beta v) = \alpha L(u) + \beta L(v)$$

b. i. $$ L(u + v) = L(u) + L(v) $$ and ii. $$ L( \lambda u ) = \lambda L(u) $$

Find
Show the two definitions are true.

Solution
Apply b. ii to get:

$$ L( \alpha u ) = \alpha L(u)$$

and

$$ L( \beta v ) = \beta L(v) $$

Combine using b. i to get:

$$ L( \alpha u + \beta v) = \alpha L(u) + \beta L(v) $$

which matches the definition of linearity given in a.

 * }

= Problem 8 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
Knowing the plots of the other shape functions

Find
Plot the shape function $$N_{j+1}^{2}(x)$$.

Solution

[[media:HW3-8.jpg|Graph]]

 * }

= Problem 9 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
x^2 y'' - 2xy' + 2y = 0

Find
$$y_{xxx}$$ and $$y_{xxxx}$$

Solution
Recall that $$ \frac{d}{dt} y = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{dy}{dx} \cdot e^t $$

Which yields $$ y_t = y_x \cdot e^t $$

Rearrange to get: $$ y_x = y_t \cdot e^{-t} $$

Take the third derivative:

$$ \frac{d^3}{dx^3}y = \frac{d}{dx}(\frac{d}{dx}(\frac{d}{dx}(y))) = \frac{dt}{dx}\frac{d}{dt}(\frac{dt}{dx}\frac{d}{dt}(\frac{dt}{dx}\frac{d}{dt}(y))) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(e^{-t}\frac{d}{dt}(e^{-t}\frac{d}{dt}(y))) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(e^{-t}\frac{d}{dt}(e^{-t} y_t)) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(e^{-t}(-e^{-t} y_t + e^{-t}y_{tt})) $$

$$ \frac{d^3}{dx^3}y = e^{-t}\frac{d}{dt}(-e^{-2t} y_t + e^{-2t}y_{tt}) $$

$$ \frac{d^3}{dx^3}y = e^{-t}(2e^{-2t} y_t - e^{-2t}y_{tt} - 2e^{-2t}y_{tt} + e^{-2t}y_{ttt}) $$

$$ \frac{d^3}{dx^3}y = 2e^{-3t} y_t - 3e^{-3t}y_{tt} + e^{-3t}y_{ttt} $$

$$ y_{xxx} = e^{-3t} (2y_t - 3y_{tt} + y_{ttt}) $$

To find $$ y_{xxxx} $$ take an additional derivative:

$$ y_{xxxx} = \frac{dt}{dx}\frac{d}{dt}(2e^{-3t} y_t - 3e^{-3t}y_{tt} + e^{-3t}y_{ttt}) $$

$$ y_{xxxx} = e^{-t} (-3 e^{-3t}y_{ttt} + e^{-3t}y_{tttt} + 9 e^{-3t}y_{tt} - 3 e^{-3t}y_{ttt} - 6 e^{-3t}y_t + 2e^{-3t}y_{tt} $$

$$ y_{xxxx} = e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) $$

 * }

= Problem 10 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution Solve equation 1 on p.16-1, $$ x^2y''-2xy'+2y=0 $$ using the method of trial solution $$ y=e^{rx}$$ directly for the boundary conditions $$\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$$ Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

Given
x^2y'' - 2xy' + 2y = 0

Find
The solution using the method of trial solution $$ y = e^{rx} $$

Part 1
Assume a trial solution of the given ODE to be $$ y = e^{rx} $$, where r is a constant and x is the independent variable. By differentiating the trial solution we get the equations:

$$ y = e^{rx} $$

$$ y' = r*e^{rx} $$

$$ y'' = {r}^{2}*e^{rx} $$

Therefore when we plug the trial solution into the given equation we get the solution

$$ {x}^{2}*{r}^{2}*e^{rx} - 2*x*r*e^{rx} + 2*e^{rx} = 0 $$

Simplifying we can eliminate the $$ e^{rx} $$ because it appears in every term, leaving us with:

$$ {x}^{2}*{r}^{2} - 2*x*r + 2 = 0 $$

$$ {r}^{2}*{x}^{2} - 2*x*r + 2 = 0 $$ This solution of r must a function of x that cannot be solved directly, and thereby violating our initial assumptions. This departure from the assumption in the solution means that we must try another solution to find agreement. Therefore we should chose another guess and continue. The next most likely trial solution to try would be $$ y = {x}^{r} $$, which as shown in the note produces a viable solution.

Part 2
Assume the solution $$ y = {x}^{r} $$, where r is a constant, and the full solution is $$ y(x) = {C}_{1}*{x}^{{r}_{1}} + {C}_{2}*{x}^{{r}_{2}} $$

$$ y = {x}^{r} $$

$$ y' = r*{x}^{r-1} $$

$$ y''= r*(r-1)*{x}^{r-2} $$

Plugging in these values

$$ {x}^{2}*r*(r-1)*{x}^{r-2} - 2*x*r*{x}^{r-1} + 2*{x}^{r} = 0 $$

Simplifying with standard algebraic techniques we get the solution

$$ {r}^{2} - 3*r + 2 = 0 $$, which determines $$ {r}_{1} = 1 $$ and $$ {r}_{2} = 2 $$.

Therefore we have our solution

$$ y(x) = {C}_{1}*{x} + {C}_{2}*{x}^{2} $$

using the boundary conditions we obtain values for $$ {C}_{1} $$ and $$ {C}_{2} $$.

$$ 3 = {C}_{1} + {C}_{2} $$

$$ 4 = {C}_{1}*2 + {C}_{2}*4 $$

And solving for the constants:

$$ {C}_{1} = 4 $$

$$ {C}_{2} = -1 $$

The solutions is therefore: $$ y(x) = 4*x - {x}^{2} $$

The plot of this solution is shown below.

 * }

= Problem 11 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$ u_1(x) z' + [a_1(x)u_1(x) + 2u_1'(x)]z = 0 $$

Find
Solution using the integrating factor method.

Solution
$$ z' + [a_1(x) + 2 \frac {u_1'(x)}{u_1(x)}]z = 0 $$

$$ a_0 = a_1 + 2 \frac {u_1'}{u_1}] $$ and $$ b = 0 $$

The integrating factor h(x) is:

$$ h(x) = exp( \int a_0(s) ds = exp( \int a1(s) + 2\frac {u_1'(s)}{u_1(s)} ds) = exp( \int a1(s) ds + 2 ln(x)) $$

$$ (hz)' = 0 $$

Integrate to solve:

$$ hz = constant $$

Rearrange to solve for z:

$$ z = \frac {constant}{(u_1(x))^2} \cdot exp[- \int a_1(s) ds] $$
 * }

= Problem 12 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
The reduction of order method 2 using $$ y(x) = U(x)u_1(x) $$

Find
Alternate definitions of y(x) that create reduction of order method 2

a. $$ y(x) = U(x) +/- u_1(x) $$

b. $$ y(x) = U(x)/u_1(x) $$

c. $$ y(x) = u_1(x)/U(x) $$

Solution
a.

Take the first and second derivatives of y(x) to get:

$$y'(x) = U' +/- u_1'$$

$$y(x) = U +/- u_1''$$

The desired form is $$ y'' + a_1y' + a_0y = 0 $$, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0U +/- a_0u_1 + a_0U' +/- a_1u_a' + U +/- u_1'' $$

Which is $$ 0 = a_0U +/- a_0u_1 + a_0U' +/- a_1u_a' + U +/- u_1 $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.

b.

Take the first and second derivatives of y(x) to get:

$$y'(x) = U'/u_1 + U/u_1'$$

$$y(x) = U/u_1 + 2U'/u_1' + U/u_1''$$

The desired form is $$ y'' + a_1y' + a_0y = 0 $$, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0 U/u_1 + a_1 U'/u_1 + a_1 U/u_1' + U/u_1 + 2U'/u_1' + U/u_1'' $$

Which is $$ 0 = a_0 U/u_1 + a_1 U'/u_1 + a_1 U/u_1' + U/u_1 + 2U'/u_1' + U/u_1 $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.

c.

Take the first and second derivatives of y(x) to get:

$$y'(x) = u_1'/U + u_1/U'$$

$$y(x) = u_1/U + 2u_1'/U' + u_1/U''$$

The desired form is $$ y'' + a_1y' + a_0y = 0 $$, multiply the above equations appropriately to get:

$$ a_0y + a_1y' + y = a_0 u_1/U + a_1 u_1'/U + a_1 u_1/U'+ u_1/U + 2u_1'/U' + u_1/U''$$

Which is $$ 0 = a_0 u_1/U + a_1 u_1'/U + a_1 u_1/U'+ u_1/U + 2u_1'/U' + u_1/U $$

The assumed form of y(x) here is not valid because the equation above is not missing U. The method cannot be developed further.
 * }

= Problem 13 = {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Please click on show to see the problem solution

Given
$$ (1-x^2)y'' - 2xy' + 2y = 0 $$

Find
u_1 and u_2 using the following two methods:

$$ y = ax^b \qquad{...(i)} $$

$$ y = e^{rx} \qquad{...(ii)}$$

and compare the two solutions and the solution using the reduction of order method 2 using the boundary conditions $$y(0) = 1$$ and $$y(1) = 2$$

Solution
Part 1

Given

$$ (1-x^{2})y''-2xy'+2y=0 $$ (Equation 1)

Find solution using the trial solution

$$ y=ax^{b} $$

$$ y'=bax^{b-1} $$

$$ y''=b(b-1)ax^{b-2} $$

Substituting into Equation 1

$$ (1-x^{2}) ax^{b-2} -2x bax^{b-1} +2 ax^{b}  =0 $$ $$ (1 - {x}^{2})*(a{b}^{2} - ab)*{x}^{b-2} -2 abx^{b} +2 ax^{b}  =0 $$

Dividing through by $$ {x}^{b} $$ $$ (1 - {x}^{2})*(a{b}^{2} - ab)*{x}^{-2} -2 ab +2 a  =0 $$ This results in the following equation:

$$ ({x}^{-2} - 1)*a*({b}^{2} - b) - 2 ab + 2 a  =0 $$

We can divide through by a, leaving us with:

$$ ({x}^{-2} - 1)*({b}^{2} - b) - 2 b + 2 = 0 $$

Therefore by comparison of forms we know:

$$ {u}_{1}(x) = {x}^{-2} - 1 $$ ,and $$ {u}_{2}(x) = 1 $$

$$ {k}_{1} = {b}^{2} - b $$, and $$ {k}_{2} = -2*b + 2 $$

Therefore the solution is $$ y(x) = ({b}^{2} - b)*({x}^{-2} - 1) + (-2*b + 2)*1 $$

using the boundary conditions to solve for b we obtain:

$$ 1 = -1*({b}^{2} - b) + (-2*b + 2) $$

$$ 2 = -2*b + 2 $$

Given these parameters $$ b = 0 $$, making the trial solution $$ a*{x}^{0} $$ , this trivial solution illustrates this is not a solution to the ODE. We must therefore try another solution.

Part 2 Given

$$ (1-x^{2})y''-2xy'+2y=0 $$ (Equation 1)

Find solution using the trial solution

$$ y=e^{rx} $$

$$ y'= re^{rx} $$

$$ y''= r^{2}e^{rx} $$

Substituting into Equation 1

$$ (1-x^{2}) r^{2}e^{rx} -2x re^{rx}  +2 e^{rx}  =0  $$ Assuming $$ y=er^{x}\neq0$$, divide through by $$  er^{x}$$ $$ (1-x^{2}) r^{2} -2x r  + 2 = 0 $$          Following this results we can construct our solution:

$$ {u}_{1}(x) = 1 - {x}^{2} $$ ,and $$ {u}_{2}(x) = -x*r + 1 $$

$$ {k}_{1} = {r}^{2} $$, and $$ {k}_{2} = 2 $$ Therefore the solution is $$ y(x) = {r}^{2}*(1 - {x}^{2}) + 2*(-x*r + 1) $$

using the boundary conditions to solve for r we obtain:

$$ 1 = {r}^{2} + 2 $$

Therefore $$ r = \sqrt{-3} $$

$$ 2 = -2*r + 2 $$

Therefore $$ r = 0 $$

Since one of the roots of r is zero (similar to the previous case) making the solution of the form, $$ y(x) = e^{0} $$, it is shown that this is also not the solution.
 * }

= Contributing Team Members =

Angela Diggs Egm6321.f09.team3.diggs 01:36, 7 October 2009 (UTC) Barrett Mooney Egm6321.f09.team3.mooney 17:30, 6 October 2009 (UTC) Catherine Snow EGM6321.f09.Team3.cgs 16:27, 7 October 2009 (UTC)


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