text:
User talk:Rlyon

 Hello Rlyon, 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
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 * 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 22:23, 15 October 2009 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

User:Rlyon/NYU Stern Office Hours
Hi Robert, FYI I moved your office hours signup page into a user subpage, since it seemed to belong better there than in the mainspace. Sincerely, James. -- Jtneill - Talk - c 22:23, 15 October 2009 (UTC)
 * Page protection is a bit tricky. Custodians can protect pages for such needs but this is used rarely. The options really only allow prevention of editing by new and unregistered users, or to Administrators only. In your case, I imagine we do want to allow new editors at least (e.g., your students) to edit and in some cases they may be unregistered e.g., . In general, you won't get unwanted edits and usually if you do they are reverted pretty fast. This isn't foolproof, but I'm guessing that it should be quite satisfactory. Otherwise you'd probably need to use an institution-specific system which could track/limit editing to an enrolled cohort. I hope this helps. Let me know if I can do anything else. Sincerely, James. -- Jtneill - Talk - c 22:43, 15 October 2009 (UTC)
 * Just correcting/clarifying above - Semi-protection: protection of a page against editing by an unregistered user is an option, but that would force your students to register and login in order to signup. So, edits such as wouldn't be possible. If you want semi-protection then let me know. -- Jtneill - Talk - c 22:56, 15 October 2009 (UTC)
 * You're doing great, don't worry, it takes a while to get oriented. There is no single way that everyone follows for such discussion threads. But a common way is for a topic, to have a heading, and then to indent replies using colons e.g., :, ::, etc. Some people like to keep the thread on a single page, others post across each others' talk pages (like we're doing). I prefer the latter (because it alerts each other), but don't really mind. I'll put semi-protection on your user hours page - see how it goes - it can be undone any time. -- Jtneill - Talk - c 23:08, 15 October 2009 (UTC)
 * I've undone that protection - perhaps the semi-protection on WV is protecting against both unregistered and new accounts after all - and yours is still considered a new account. So, yes, that is too much, I've undone it.
 * Thanks, James. But doesn't that require both conversants to be "watching" the userspace talk page? Or is that the default? -- Robert --Rlyon 23:13, 15 October 2009 (UTC)
 * By default, if someone edits your user talk page, you get the orange message box appearing at the top of your screen (which I like). Otherwise, yes, for any other pages, you probably need to have them on your watchlist if you want to track edits there. -- Jtneill - Talk - c 23:22, 15 October 2009 (UTC)

title: User talk:Rlyon


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User:Rlyon/NYU Stern Office Hours

 Professor Robert Lyon Fall 2010 Office Hours Sign-up: Available Hours: Mondays: 10:00 a.m.-4:30 p.m. Tuesdays: 3:30 -5:30 p.m. Wednesdays: 10:00 a.m.-4:30 p.m. Thursdays: 3:30 -5:30 p.m.

Sign Up Instructions: User:Rlyon/Sign_Up_Instructions

Wednesday, Oct. 6: (Time) / (Your name):

10:00 / _______________ 10:20 / 10:40 / _______________ 11:00 / _______________ 11:20 / _____________ 11:40 / Rishi Rajani 12:00 / LUNCH 1:00 / Elizabeth Taus 1:20 / Lin Ye 1:40 / _______ 2:00 / Priya Kumar 2:20 / _______________ 2:40 / _______________ 3:00 / _______________ 3:20 / Amy Luo 3:40 / _______________ 4:00 / Jason Jang 4:20 / Ash Bose 4:40 / _Tae Jun (T.J.) Kim 5:00 / Ryan Kim

Friday, Oct. 8: (Time) / (Your name):

10:00 / Jennie Zhou 10:20 /   Ryan Kim 10:40 / Melanie Ferreira 11:00 / Benjamin Strulowitz 11:00 11:20 / John Gonsalves 11:40 / Lulu Peng 12:00 / Karan 12:20 / Daniel 12:40 / _______________ 1:00 / LUNCH 2:00 / _______________ 2:20 / Preeyam 2:40 / Casey Woo 3:00 / Saanjali Kotahwala 3:20 / Steven Wu 3:40 / Josh Martone 4:00 / Ashwini Menon 4:20 / Geo Abraham 4:40 / Amanda Siciliano 5:00 / Jason Jang

Wednesday, Oct. 13: (Time) / (Your name):

11:00 / _______________ 11:30 / _______________ 12:00 / LUNCH 1:00 / _______________ 1:30 / _______________ 2:00 / _______________ 2:30 / Shiyuan Zhuang 3:00 / Teri Amy 3:30 / _______________ 4:00 / _______________ 4:30 / _______________

Thursday, Oct. 14: (Time) / (Your name):

3:30 / _________________ 4:00 / _________________ 4:30 / _________________ 5:00 / ___Eun Ju Oh ____ 5:30 / _________________ Monday, Oct. 18: (Time) / (Your name):

11:00 / _______________ 11:30 / _______________ 12:00 / LUNCH 1:00 / _______________ 1:30 / _______________ 2:00 / _______________ 2:30 / _______________ 3:00 / _______________ 3:30 / _______________ 4:00 / _______________ 4:30 / _______________

Tuesday, Oct. 19: (Time) / (Your name):

3:30 / _________________ 4:00 / _________________ 4:30 / _________________ 5:00 / _________________ 5:30 / _________________ Wednesday, Oct. 20: (Time) / (Your name):

11:00 / _______________ 11:30 / _______________ 12:00 / LUNCH 1:00 / _______________ 1:30 / _______________ 2:00 / _______________ 2:30 / _______________ 3:00 / _______________ 3:30 / _______________ 4:00 / _______________ 4:30 / _______________

Thursday, Oct. 21: (Time) / (Your name):

3:30 / _________________ 4:00 / _________________ 4:30 / _________________ 5:00 / _________________ 5:30 / _________________

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

title: User:Rlyon/NYU Stern Office Hours


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User talk:TarekaBrown

 * Tareka

Fill out the schedule as much as you can, and let's see how it looks. If the process isn't an 8-week one,remember that you can always increase the research aspect of the assignment. Best, Stevenarntson

Progress Report
Please post your progress report today.
 * Best, Stevenarntson

title: User talk:TarekaBrown


text:
User talk:VictorPorton

No Solicitation
Wikiversity's mission is to create and host free learning materials and activities. Solicitation for external organizations and websites is not accepted. I've moved the external link on Automatic transformation of XML namespaces to the talk page. Users who are interested can still find the information, but because of the Donate link on the destination page, this is solicitation and not appropriate for the article. -- Dave Braunschweig (discuss • contribs) 13:21, 28 August 2014 (UTC)

Can Human Rights Be Accomplished Experiment
This resource is not accepted at Wikiversity. A blog would be a better choice for this type of resource. -- Dave Braunschweig (discuss • contribs) 12:43, 21 June 2019 (UTC)

Social Crypto Libertarianism
This project is a violation of Deletions policy regarding solicitation. It has been deleted. This is your second or third attempt to use Wikiversity to "generate direct financial benefit to the contributor." Your account is now blocked. You are welcome to appeal this block, but you will need to provide specific goals for your contributions that do not mention you personally, and do not link to any external resources which would directly or indirectly provide benefit to you, the contributor. -- Dave Braunschweig (discuss • contribs) 23:51, 16 September 2020 (UTC)

 * I disagree with you that this project is a violation of Deletions policy regarding solicitation. This policy is "Solicitation for products, services, companies, events, people or other things with no educational merit or which generate direct financial benefit to the contributor." It is not a violation because it a. educates about a software and b. helps to conduct a scientific experiment that is provides an educational merit. 2. My service may generate indirect financial benefit to the contributor, not the direct one. Namely my service is intended to change the global world in such a way that everyone (including me) would benefit from his or her contributions. So your statement saying that I generate direct benefit to myself is quite like saying: We don't allow posting from Wright brothers, because it generates a direct benefit for them, as they after the experiment gain a new financial ability, to buy air tickets. Please explain what you mean by "providing specific goals for your contributions that do not mention me personally". I didn't understand this your phrase. I do not link to any external resources which would directly or indirectly provide benefit to me, the contributor. (I link to https://reward.portonvictor.org but it does not benefit me personally as all the money accepted for the project is to be used for the research, not my personal benefit.) --VictorPorton (discuss • contribs) 00:05, 17 September 2020 (UTC)

 * Dave, also please take into account that I just forgot that my site takes donations before posting a link to it. Is it enough and/or necessary to remove the donation link from my site to unblock my account and restore the deleted pages? --VictorPorton (discuss • contribs) 00:17, 17 September 2020 (UTC)

 * Your contributions have indicated a clear intent to use Wikiversity for direct financial gain. For example, the first paragraph of the deleted Social Crypto Libertarianism/Proposed Committee Rules stated, "The committee members can be either volunteer or paid (contact [VictorPorton's personal email address] if you want to participate). In any case the committee member can vote to allocate funds to themselves to make a business of this work." This is a violation of Wikiversity policy.
 * The requirements for being unblocked are stated above. "... you will need to provide specific goals for your contributions that do not mention you personally, and do not link to any external resources which would directly or indirectly provide benefit to you, the contributor."
 * Dave Braunschweig (discuss • contribs) 16:19, 17 September 2020 (UTC)

 * Dave, it seems that you just misunderstood. I am going to pay to committee members, not they to me. (I am financing a research.) It is a direct financial lost, not a direct financial gain. Committee members can allocate money to themselves, I didn't say that they allocate money to me. The goal of the project on the Wikiversity is to check whether my model of committees is sustainable (whether the committees will reproduce). They provide benefits equally to all the humanity as every scientific research. So I complied to your requirement "to provide specific goals for your contributions that do not mention you personally, and do not link to any external resources which would directly or indirectly provide benefit to you, the contributor". If I didn't comply to something in it, please explain what exactly. If I need to remove the donation link at my site before being unblocked, please tell me so. So, I expect that you unblock me and restore the deleted articles. If not, explain why exactly.

 * You cannot use Wikiversity for direct financial benefit. Not for you, not for them. This research project is not compatible with Wikiversity policy.
 * Your account will remain blocked until sometime in the future, when and if you return with a different project and a different approach. As noted, it must be a project that does not mention you personally, and does not link to any external resources which would directly or indirectly provide benefit to you. -- Dave Braunschweig (discuss • contribs) 18:24, 17 September 2020 (UTC)

 * I am surprised that your rules disallow me to offer my money to a third party (never heard about any set of rules that would disallow this, I also don't understand the rationale behind this rule). Well, I will remove the proposal to pay them money. I will also remove the donation link from my site. Now OK? Will you unblock me now and allow me to restore the pages (I need your copy.) and remove the proposal to pay money to the participants? Or do I need to do anything else to comply? --VictorPorton (discuss • contribs) 18:33, 17 September 2020 (UTC)

The proposed research is not appropriate for Wikiversity. It is based on financial reward and involves experimenting on human psychology. We cannot approve or support that. You will need to find another venue. I will email a copy of the deleted content so you can create it elsewhere. -- Dave Braunschweig (discuss • contribs) 23:00, 17 September 2020 (UTC)

 * Dave, could you explain which rules of Wikiversity disallow financial rewards during an experiment and which disallow experimenting on human psychology? (I realize that an experiment should not harm, but there is no foreseen harm of this experiment.) OK, as you insist that this experiment is not appropriate for Wikiverity, I stop using Wikiversity for this. But please unblock my account for my Automatic_transformation_of_XML_namespaces project. --VictorPorton (discuss • contribs) 23:09, 17 September 2020 (UTC)

title: User talk:VictorPorton


text:
User talk:Patrickmd

 Hello Patrickmd, 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! --AFriedman 22:25, 15 October 2009 (UTC)

 Hello , we are the Baah and Bahh twins and we'd both like to welcome you to Wikiversity. While you sit back and explore Wikiversity, enjoy some of our fresh goat milk. You can help us learn by creating new learning resources or help us by improving the guided tours and other exploration tools. We look forward to seeing you around. Baaaahbye for now.

title: User talk:Patrickmd


text:
User:Egm6321.f09.team5/HW4

 See further comments in google doc for Team 5. Egm6321.f09 21:03, 23 November 2009 (UTC)

Problem Statement
Pg. 19-1. Lengendre Differential Equation (1) Pg. 14-2 w/ n=0, so that the homogenous solution $$ U_{1}(x) = 1 $$

Use Reduction of order method 2 (undetermined factor) to find $$U_{2}(x)$$, 2nd homogeneous solution.

Solution
When n=0, the Legendre equation becomes:

$$ F = (1-x^{2})y^{''} - 2xy^{'} + 0 = 0 \qquad \qquad Equation \quad 1$$

Using the Undetermined Factor as follows:

Assume the full homogeneous solution is of the form:

$$ y(x) = U(x)u_{1}(x) $$

rearranging the differential equation 2 to fit the standard form of equation 2 as follows:

$$ y^{''} + a_{1}(x)y^{'} + a_{0}(x)y = 0 \qquad \qquad Equation \quad 2$$

Equation 1 becomes:

$$ y^{''} - \frac{2x}{(1-x^{2})}y^{'} = 0 \qquad \qquad Equation \quad 3$$

The undetermined factor method allows for the following form:

$$ 0 = U^{'}\left [ a_{1}u_{1} + 2u_{1}^{'} \right ] + U^{''}u_{1} $$

Equation 3 becomes:

$$ 0 = U^{'}\left [ \frac{-2x}{(1-x^{2})} \right ] + U^{''} \qquad \qquad Equation \quad 4 $$

Letting $$ z = U^{'} $$ Equation 4 becomes a L1.ODE.VC which can be solved as follows:

$$ z^{'} + z \left [ \frac{-2x}{(1-x^{2})} \right] = 0 $$

Integrating by parts yields to solve for z yields:

$$ z(x) = c(1-x^{2}) = U^{'}(x) \qquad \qquad where \quad c=\exp(k) $$

Integrating again the result is:

$$ U(x) = c \left [ x - \frac{x^{3}}{3} \right ] + c2 \qquad \qquad. $$                where c and c2 are constants

The expression for the second homogeneous solution is given as:

$$ u_{2} = u_{1} \int_{}^{x} \exp (\ln(1-t^{2})) dt \qquad \qquad Equation \quad 5 $$

Using Equation 5 the second homogeneous solution yields:

$$ u_{2} = \left [ x - \frac{x^{3}}{3} \right ] $$

The Full homogeneous solution can be give as follows:

$$ y(x) = K_{1} + K_{2} \left [ x - \frac{x^{3}}{3} \right ] $$

Problem Statement
King et al. Pg 28. Problem 1.1b

$$ xy^{''} + 2y^{'} + xy = 0 $$          Equation 0

1) Check for exactness, If not exact can it be made exact using integration factor.

2) Use the following trial solutions to find the homogeneous solution

3) Use the undetermined factor method to find the second homogeneous solution, knowing a homogeneous solution

1)
The First step is to check for exactness using the following conditions:

Condition 1: The ODE must be of the form:

$$F=f(x,y,p)y^{''}+g(x,y,p)$$            Equation 1

Where $$ P=y^{'} $$

Condition 2: The following Conditions must be met:

$$    \ f_{xx} + 2pf_{xy} + p^{2}f_{yy} = g_{xp} + pg_{yp} - g_{y}  \qquad $$      Equation 2

$$    \ f_{xp} + pf_{yp} + 2f_{y} = g_{pp}  \qquad $$         Equation 3

$$ f(x)=x $$

$$ g(x)= 2p + xy $$

$$ f_{x}= 1 \qquad \qquad  g_{y}= x $$

$$ f_{xx}= 0  \qquad \qquad g_{yp}= 0 $$

$$ f_{xy}= 0  \qquad \qquad  g_{x}= y  $$

$$ f_{y}= 0        \qquad \qquad \qquad  g_{xp}= 0 $$

$$ f_{yy}= 0    \qquad \qquad  g_{p}= 2 $$

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

Equation 2 yields:

$$ 0 + 2p(0) + p^{2}(0) = 0 + p(0) - x $$

$$ 0 \neq -x $$

This shows that the ODE is not exact.

The next step is to check whether it can be made exact by using the integration factor method:

The integration factor to be used will be:

$$ h(x,y) = x^{m}y^{n} $$

multiplying the integration factor by the ODE yields the following:

$$ \left [ x^{m}y^{n} \right ] \left[ xy^{''} + 2y^{'} + xy \right ] = 0 $$

$$ x^{m+1}y^{n}y^{''} + 2x^{m}y^{n}y{'} + x^{m+1}y^{n+1} = 0 $$

The next step is to check once again for exactness:

$$ f = x^{m+1}y^{n} $$ $$ g= 2x^{m}y^{n}p + x^{m+1}y^{n+1} $$

$$ f_{x}= y^{n}(m+1)x^{m} \qquad \qquad  g_{y}= 2x^{m}ny^{n-1}p + x^{m+1}(m+1)y^{n} $$

$$ f_{xx}= y^{n}(m+1)mx^{m-1}  \qquad \qquad g_{yp}= 2x^{m}ny^{n-1} $$

$$ f_{xy}= ny^{n-1}(m+1)x^{m}  \qquad \qquad  g_{x}= 2y^{n}pmx^{m-1} + y^{n+1}(m+1)x^{m} $$

$$ f_{y}= x^{m+1}ny^{n-1}        \qquad \qquad \qquad  g_{xp}= 2y^{n}mx^{m-1} $$

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

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

Using Equation 3 in order to check for exactness yields the following:

$$ 0 + p(0) + 2(x^{m+1}y^{n-1}n \neq 0 $$

Therefore the new ODE is also concluded not to be exact.

2)
The next step is to find a homogeneous solution by using different trial solutions as follows:

Trial Solution 1

$$ y(x)=\exp^{rx} $$

$$ y^{'}(x) = r\exp^{rx} $$

$$ y^{''}(x) = r^{2}\exp^{rx} $$

using these results back in Equation 0 the following is obtained:

$$ x(r^{2}\exp^{rx}) + 2r\exp^{rx} + x\exp^{rx} = 0 $$

$$ \exp^{rx} (r^{2}(x) + 2r + x) = 0 $$

$$ r^{2}x + 2r + x = 0 $$

This trial solution is abandoned as it does not yield a root for the homogeneous solution

Trial Solution 2

$$ y(x)=x\exp^{rx} $$

$$ y^{'}(x) = (xr\exp^{rx} + x^{'}\exp^{rx}) $$

$$ y^{}(x) = r^{2}x\exp^{rx} + rx^{'}\exp^{rx} + x^{'}r\exp^{rx} + x^{}\exp^{rx} $$

Using these results in Equation 0 the following is obtained:

$$ x^{2}r^{2} + 2rxx^{'} + xx^{''} + 2xr + 2x^{'} = 0 $$

This trial solution is abandoned as it will not yield a root for the homogeneous solution

Trial Solution 3

$$ y(x)=x^{-1}\exp^{rx} $$

$$ y^{'}(x) = (rx^{-1}\exp^{rx} - x^{-2}\exp^{rx}) $$

$$ y^{''}(x) = \exp^{rx}(-2rx^{-2} + r^{2}x^{-1} + 2x^{-3}) $$

Using these results in Equation 0 the following is obtained:

$$\exp^{rx} \left [ r^{2} + 1 \right ] = 0 $$

From the previous equation the following result is found:

$$ r = \sqrt[]{-1} = i $$

$$ y(x) = x^{-1}\exp^{ix} $$

Using the de Moivre relationships the following form is obtained for the homogeneous solution:

$$ y(x)= x^{-1}(cosx + isinx) $$

3)
Given a homogeneous solution the full solution can be found using the undetermined factor method. The full solution is assumed to be of the form:

$$ y(x) = U(x)u_{1}(x) $$

Where $$ u_{1}(x) $$ is the known homogeneous solution.

$$ u_{1}(x)= \frac{sin(x)}{x} $$

Equation 0 is organized as follows:

$$ y^{''} + \frac{2}{x}y^{'} + y = 0 $$

where:

$$ a_{0}=1 \qquad \qquad a_{1} = 2x^{-1} $$

Because of the undetermined factor method the following can be written:

$$ 0 = U^{'} \left [ x^{-2}\sin(x) + 2x^{-1}\cos(x) \right ] + U^{''}\frac{\sin(x)}{x} $$  Equation 3

A variable is defined as follows:

$$ z=U^{'} $$

Equation 3 can be rewritten as follows:

$$ z\left [ x^{-2}\sin(x) + 2\cos(x)x^{-1} \right] + z^{'} \frac{\sin(x)}{x} $$

Solving the previous L1.ODE.VC using integration by parts and then solving for $$ U $$ the following is identified:

$$ u_{2}(x) = u_{1} \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt $$

$$ \int{}^{t} a_{1}(s)ds = 2ln(x) $$

$$ \int{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp^{-\int{}^{t} a_{1}(s)ds} dt = \int{}^{x}\frac{t^{2}}{\sin(t)^{2}}t^{-2} = -cot(x)  $$

$$ u_{2}(x) = \frac{\sin(x)}{x}\frac{-\cos(x)}{\sin(x)} = -\frac{\cos(x)}{x}  $$

This result confirms the results found in part 2 for the full homogeneous solution.

Problem Statement
Pg. 21-3

$$ U_{1}(x)= \exp^{x} $$

$$ \qquad  (x-1)y^{''}-xy^{'} + y =0   \qquad \qquad Equation \quad 1$$

Explain how do you know the homogeneous solution given.

Use the following trial solution:

$$ y(x)= \exp^{rx} \qquad \qquad r \quad is \quad constant$$

Use the Reduction of order method 2 (undetermined factor) to find 2nd equation.

NOTE:

-Find r1,r2. How many valid homogeneous solutions?

If only one valid homogeneous solution, $$\quad U_{1}(x)=\exp^{r_{1}x} \quad$$, find $$\quad U_{2}(x) \quad$$ using undetermined factor.

Solution
The first step is to try to find a homogeneous solution by the trial solution technique:

The trial solution used will be $$ y(x) = \exp^{rx} $$

Using the trial solution the following is obtained:

$$ y(x) = \exp^{rx} $$

$$ y^{'}(x) = r\exp^{rx} $$

$$ y^{''}(x) = r^{2}\exp^{rx} $$

Using the previous values and substituting them to Equation 1, the following is found:

$$ (x-1)r^{2}\exp^{rx} - x(r\exp^{rx}) + \exp^{rx} = 0 $$

$$ \exp^{rx} (r^{2}x - r^{2} - xr + 1) = 0 $$

$$ r^{2}x - r^{2} - xr + 1 $$

It is observed that when r = 1 a solution is found. Therefore the following is obtained:

$$ u_{1}(x) = \exp^{x} $$

With the homogeneous solution found the next step is to find second homogeneous solution. The first step is to rearrange Equation 1 as follows:

$$ y^{''} - (\frac{x}{x-1}y^{'} + (\frac{1}{x-1})y = 0 $$

Using the undetermined factor method the full solution is said to be of the following form:

$$ y(x) = U(x)u_{1}(x) $$

with this method the following can be written:

$$ 0= U^{'} \left[ a_{1}u_{1} + 2u_{1}^{'} \right ] + U^{''}u_{1} $$

A variable Z is now defined as follows:

$$ z=U^{'} $$

In adddition the following is also identified:

$$ a_{1} = (\frac{-x}{(x-1)}) $$

$$ a_{0} = (\frac{1}{(x-1)} $$

With this form an expression can be written as follows:

$$ u_{2} = u_{1} \int_{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp (- a*_{1}(t)dt) \qquad \qquad Equation \quad 2$$

The value for a* is found as follows:

$$ a* = \int{}^{t} a_{1}(s)ds = - \left [x + ln(x-1) \right ] $$

Using this value into Equation 2 yields and analyzing the integral as a first step yields:

$$ \int{}^{x} \frac{1}{\exp^{2t}} \left[ \exp^{t} + (t-1) \right ] dt $$

$$ = - \exp^{-x} + \frac{\exp^{-2x}}{-4} \left [ 2x-1 \right ] $$

$$ u_{2} = \exp^{x} \left [ \exp^{-x} + \frac{\exp^{-2x}}{-4} \left [ 2x-1 \right ] \right ] $$

$$ u_{2} = 1 + \frac{\exp^{-x}\left[2x-1 \right ]}{-4} \qquad \qquad Equation \quad 3$$

It is then concluded that the full homogeneous solution to Equation 1 is as follows:

$$ y(x) = K_{1}\exp^{x} + K_{2} (1 + \frac{\exp^{-x}\left[2x-1 \right ]}{-4}) $$

$$ K_{1} \quad and \quad K_{2} $$ are constants to be found using boundary conditions or initial conditions.

Problem Statement
Pg. 22-1 King et al. Pg. 28, Pb 1.3a

Find the General Solution to the following Non-homogeneous L2.ODE.VC:

$$ y^{''} - 2y^{'} + y = x^{1.5}\exp^{x} $$ Equation 0

Solution
Because of linearity and superposition the following can be stated about the full solution:

$$ y(x) = y_{h} + y_{p} $$

where:

$$y_{h} = homogeneous \quad solution $$

$$y_{p} = particular\quad solution $$

The first task is to find the Homogeneous solution. This will be done by using the trial solution method as follows:

$$ y(x) = \exp{rx} $$

$$ y^{'}(x) = r\exp{rx} $$

$$ y^{''}(x) = r^{2}\exp{rx} $$

The Homogeneous ODE would be of the following form:

$$ y^{''} - 2y^{'} + y = 0 $$

Plugging the trial solution values to this equation the following is obtained:

$$ r^{2}\exp{rx} - 2r\exp{rx} + \exp{rx} = 0 $$

This simplifies to:

$$ (r-1)^{2} = 0 $$

$$ r = 1 $$

From this the following is identified:

$$ u_{1}(x) = \exp{x} $$

The next step is then to use the method of undetermined factor to identify the full solution:

$$ f(x) = U^{'}(a_{1}u_{1} + 2u_{1}^{'}) + U^{''}u_{1} $$

where:

$$ a_{1} = -2 \qquad \qquad a_{0} = 1 \qquad \qquad f(x)= x^{1.5}\exp^{x} $$

note: $$ a*_1 = \int{}^{t} a_{1}(s)ds = -2t $$

$$u_{2} = u_{1} \int_{}^{x} \frac{1}{(u_{1}(t))^{2}} \exp (- a*_{1}(t)dt)$$

$$u_{2} = exp(x) \int_{}^{x} \frac{1}{(exp(2t))} \exp (2t)dt$$

$$u_{2} = \int_{}^{x}\ 1dt \exp(x)$$

$$u_{2} = x\exp(x)$$

$$y_{p} = \int_{}^{x} \ s^{1.5} \ exp(s) \frac{(exp(s) \ x \ exp(x) - exp(x) \ s \ exp(s)}{exp(2s)} ds$$

$$y_{p} = 1.5 \ x^{0.5} \ exp(x)-2.5 \ x^{1.5} \ exp(x)$$

$$y = y_h+y_p$$

$$y(x) = A\ exp(x) + B\ x\ exp(x) -1.5\ x^{0.5}\ exp(x) - 2.5\ x^{1.5}\ exp(x)$$

Problem Statement
Pg. 22-3

Describe in words, a step by step method of attacking similar problems when given only homogeneous or non-homogeneous L2.ODE.VC

Solution
In order to solve the homogeneous or non-homogeneous L2.ODE.VC the following method may be employed.

For homogeneous ODE's simply apply the following method

1) Verify exactness of the ODE.

Test for exactness. Use the Reduction of Order Method 0 if missing dependent variable. If that works, then there is an exact homogeneous solution. If the ODE is not exact apply the Integrating Factor Method to obtain an exact equation.

2) Apply trial solutions.

This step involves guessing a value for y(x) and applying values of y(x) and it's derivatives to the ODE. Once the correct form of y(x) is assumed, the roots of the characteristic are determined. If one of these roots is a constant, we have solved for our first homogeneous solution. The first homogeneous solution takes the form of the y(x) which yielded the characteristic equation. The first homogeneous solution contains the constant root of the characteristic equation as its coefficient.

3) Apply Undetermined Factor Method to find the second homogeneous solution.

4) Use Variation of Parameter Method to find the complete solution.

For a non-homogeneous L2.ODE.VC

Need to find homogeneous solution and particular solution. Find homogeneous solution using the previous steps. Find particular solution by using Trial Solution, Varying Parameters, or Alternative Method. Alternative Method can be used in Reduction of Order Method 2 if we know only one homogeneous solution obtained from trial solution.

Problem Statement
Pg. 23-1

Non-homogeneous L2.ODE.VC

a.) $$ (x-1)y^{''} - xy^{'} + y = x $$

b.) $$ xy^{''} + 2y^{'} + xy = x $$

Two Methods:

-Variation of Parameters (after knowing $$ U_{1}(x) and U_{2}(x) $$

-Alternative Method (bypass $$ U_{2} $$

Solution a.):
For a.) $$ (x-1)y^{''} - xy^{'} + y = x $$

Variation of Parameters:
For $$ (x-1)y^{''} - xy^{'} + y = x $$

From problem 3 we know u1 and u2:

$$ u_{1}(x) = \exp(x) $$

$$ u_{2} = 1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4} $$

So the homogenous part would be:

$$ y(x) = K_{1}\exp(x) + K_{2} (1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}) $$

Next we need to find the particular solution:

$$y_p=P_1(x)u_1(x)+P_2(x)u_2(x)$$

We need to use the Wronskian matrix:

$$\underline{W}=\begin{bmatrix} u_1 & u_2\\ u'_1 & u'_2 \end{bmatrix}$$

$$W=det(\underline{W})$$

$$W=u_1u'_2-u'_1u_2$$

$$W=exp(x)*\frac{\partial }{\partial x}[ 1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4} ]-exp(x)*[ 1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}] $$

$$W=exp(x)*(\frac{xexp(-x)}{2}-\frac{exp(-x)}{2}-\frac{exp(-x)}{4})-exp(x)-\frac{x}{2}+\frac{1}{4}$$

$$W=\frac{x}{2}-\frac{1}{2}-\frac{1}{4}-exp(x)-\frac{x}{2}+\frac{1}{4}$$

$$W=\frac{1}{2}-exp(x)$$

from 19-3,

$$P_1(x)=-\int \frac{u_2(s)f(s)}{W}ds+A$$

$$P_2(x)=\int \frac{u_1(s)f(s)}{W}ds+B$$

The constants are arbitrary, and can be set to 0

$$P_1(x)=)=-\int \frac{x[1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}]}{\frac{1}{2}-exp(x)}dx$$

$$P_2(x)=\int \frac{xexp(x)}{\frac{1}{2}-exp(x)}dx$$

The full solution is:

$$ y(x) = y_h+y_p = K_1u_1+K_2u_2 + P_1u_1 + P_2u_2 $$

$$ y(x)= K_{1}\exp(x) + K_{2} (1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4})-exp(x)\int \frac{x[1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4}]}{\frac{1}{2}-exp(x)}dx+(1 + \frac{\exp(-x)\left[2x-1 \right ]}{-4})\int \frac{xexp(x)}{\frac{1}{2}-exp(x)}dx$$

Alternative Method
This is the form of the equation we are using:

$$y''+a_1(x)y'+a_0(x)y=f(x)$$

$$y''+\frac{x}{(x-1)}y'+\frac{1}{(x-1)}y=\frac{x}{(x-1)}$$

We know the value of $$u_1$$

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

We don't know the value of U(x)

$$U'(a_1u_1+2u_1')+U''u_1=f(x)$$

Missing dependent variable U, so use Reduction of Order Method 0

let Z(x) = U'(x)

$$Z(a_1u_1+2u_1')+Z'u_1=f(x)$$

Solve for Z(x):

$$Z(x)=\frac{}{}\int^xh(s)f(s)ds$$

$$h(x)=exp(\int^x a_0(s)ds)$$

$$h(s)=h(x)=exp(\int^x(a_1u_1+2u_1')dx)=exp(\int^x(\frac{x}{(x-1)}exp(x)+2exp(x))dx$$

$$h(x)=exp(x)-exp(x)(x-1)+2exp(x)$$

$$h(s)=exp(s)-exp(s)(s-1)+2exp(s)$$

$$Z(x)=\frac{}{}\int^x(exp(s)-exp(s)(s-1)+2exp(s))(\frac{s}{(s-1)})ds=\int^x(exp(s)-sexp(s)+exp(s)+2exp(s)-sexp(s)+s^2exp(s)-sexp(s)-2sexp(s))ds$$

$$Z(x)=\int^x(4exp(s)-5sexp(s)+s^2exp(s))ds$$

$$Z(x)=4exp(x)-5exp(x)(x-1)+exp(x)(x^2-2x+2)$$

$$U(x)=\int^x(Z(s))ds$$

$$U(x)=\int^x(4exp(s)-5exp(s)(s-1)+exp(s)(s^2-2s+2))ds$$

$$U(x)=4exp(x)-5exp(x)(x-2)+exp(x)(x^2-4x+6)$$

The solution is:

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

$$y(x)=(4exp(x)-5exp(x)(x-2)+exp(x)(x^2-4x+6))exp(x)$$

Solution b.):
For b.) $$ xy^{''} + 2y^{'} + xy = x $$

Variation of Parameters:
For $$ xy^{''} + 2y^{'} + xy = x$$

From problem 2 we know u1 and u2:

$$ u_{1}(x)= \frac{sin(x)}{x} $$

$$ u_{2}(x)= -\frac{cos(x)}{x} $$

So the homogenous part would be:

$$ y(x) = K_{1}(\frac{sin(x)}{x}) - K_{2}(\frac{cos(x)}{x}) $$

Next we need to find the particular solution:

$$y_p=P_1(x)u_1(x)+P_2(x)u_2(x)$$

We need to use the Wronskian matrix:

$$\underline{W}=\begin{bmatrix} u_1 & u_2\\ u'_1 & u'_2 \end{bmatrix}$$

$$W=det(\underline{W})$$

$$W=u_1u'_2-u'_1u_2$$

$$W=(\frac{sin(x)}{x})\frac{\partial }{\partial x}(-\frac{cos(x)}{x})-\frac{\partial }{\partial x}(\frac{sin(x)}{x})(-\frac{cos(x)}{x}) $$

$$W=(\frac{sin(x)}{x})(\frac{cos(x)}{x^2}+\frac{sin(x)}{x})-(\frac{cos(x)}{x}-\frac{sin(x)}{x^2})(-\frac{cos(x)}{x})$$

$$W=(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})$$

from 19-3,

$$P_1(x)=-\int \frac{u_2(s)f(s)}{W}ds+A$$

$$P_2(x)=\int \frac{u_1(s)f(s)}{W}ds+B$$

The constants are arbitrary, and can be set to 0

$$P_1(x)=-\int \frac{-cos(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx$$

$$P_2(x)=\int \frac{sin(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx$$

The full solution is:

$$ y_p=(-\int \frac{-cos(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx)(\frac{sin(x)}{x})+(\int \frac{sin(x)}{(\frac{sin(x)cos(x)}{x^3}+\frac{sin^2(x)}{x^2})+(\frac{cos^2(x)}{x^2}-\frac{sin(x)cos(x)}{x^3})}dx)(\frac{-cos(x)}{x})$$

Alternative Method:
This is the form of the equation we are using:

$$y''+a_1(x)y'+a_0(x)y=f(x)$$

$$y''+\frac{2}{x}y'+y=1$$

We know the value of $$u_1$$

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

We don't know the value of U(x)

$$U'(a_1u_1+2u_1')+U''u_1=f(x)$$

Missing dependent variable U, so use Reduction of Order Method 0

let Z(x) = U'(x)

$$Z(a_1u_1+2u_1')+Z'u_1=f(x)$$

Solve for Z(x):

$$Z(x)=\frac{}{}\int^xh(s)f(s)ds$$

$$h(x)=exp(\int^x a_0(s)ds)$$

$$h(s)=h(x)=exp(\int^x(a_1u_1+2u_1')dx)=exp(\int^x(\frac{2sin(x)}{x^2}+\frac{2cos(x)}{x}-\frac{2sin(x)}{x^2})dx$$

$$h(x)=\int^x(\frac{2cos(x)}{x})dx$$

$$h(s)=\int^x(\frac{2cos(s)}{s})ds$$

$$Z(x)=\int^x(\int^s(\frac{2cos(s)}{s})ds)dx$$

$$U(x)=\int^x(Z(s))ds$$

$$U(x)=\int^x(\int^s(\int^s(\frac{2cos(s)}{s})ds)ds)dx$$

The solution is:

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

$$y(x)=(\int^x(\int^s(\int^s(\frac{2cos(s)}{s})ds)ds)dx)(\frac{sin(x)}{x})$$

Contributing Authors
--Egm6321.f09.team5.GV 20:24, 21 October 2009 (UTC)

--Egm6321.f09.team5.risher 20:55, 21 October 2009 (UTC)

--Egm6321.f09.team5.bear 20:55, 21 October 2009 (UTC)

title: User:Egm6321.f09.team5/HW4


title: File:Dc987-exp00.png


text:
Actions influence attitudes

One of social psychology’s most significant findings is that action shapes attitude, a principle that can be illustrated further in class.

For example, the low-ball technique, a device presumably used by some new-car dealers, demonstrates the powerful effects of action on attitude. After a customer has signed on to buy a new car because of its very low price, the salesperson reports that the sales manager won’t agree because “we’d be losing money". Reportedly, more customers stick with their purchase, even at the higher price, than would have agreed if the full price had been disclosed from the start.

In another investigation of the technique, introductory psychology students were asked to participate in a laboratory experiment at 7am. Only 24% of those asked came. When students agreed to participate without knowing the time and then were asked if they would come at 7am, 53% of those who had been asked showed up!

Brainwashing, a technique first used on American prisoners during the Korean War, also demonstrates how attitudes may follow actions. The captors would have the prisoners write a series of essays, each one representing a more serious attack on the U.S. government. Slowly the writer’s attitude would become consistent with his words (Myers, 2008).

In his delightful book, Influence, Robert Cialdini gives numerous examples of how other compliance professionals use the written statement to shape attitudes. The Amway Corporation, for example, promotes sales by asking its personnel to set individual sales goals and then record them on paper: “Whatever the goal, the important thing is that you set it ... and write it down. There is something magical about writing things down ... when you reach that goal, set another, and write that down. You’ll be off and running.”

Similarly, some door-to-door sales companies have used a “write-it-down” technique to counteract legislation that allows customers a few days to cancel their order. By having the customer rather than the salesperson fill out the original sales agreement, they push them to commit to the purchase. As one encyclopedia company official notes, this change alone has proved to be “a very important psychological aid in preventing customers from backing out of their agreements.”

The actions-shaping-attitude principle may also operate in the “50 words or less” testimonial contests. Why do manufacturers of toothpaste, breakfast cereal, and chewing gum pay thousands of dollars to a contestant who composes a short personal statement that begins, “Why I like…”? To get as many people as possible to go on record as favoring the product. Saying is believing!

title: Actions influence attitudes


text:
Social norms

Discuss about social norms which we might “violate”.

Solomon Asch demonstrated that most people will yield to a majority opinion even when it conflicts with their own. We believe we are different, but that’s usually no the case. We underestimate the power of social forces.

Stanley Milgram described a class project. He asked his students to violate a simple social constraint – to get on a subway or city bus and ask another passenger for his or her seat. Students who tried found the task to be nearly impossible. One graduate student reported “I just couldn’t go on. It was one of the most difficult things I ever did in my life”.

Unconvinced, Milgram tried himself. He approached a seated passenger but the words, he said, “seemed lodged in my trachea and would simply not emerge. I stood there frozen, then retreated, the mission unfulfilled.” After several unsuccessful attempts he final choked out the request “Excuse me sir, may I have your seat?” “Amazingly” Milgram reported “ the man immediately stood up and gave me his seat!” But the experience wasn’t over. In taking the man’s seat, Milgram observed, “I was overwhelmed by the need to behave in a way that would justify my request. My head sank between my knees… I actually felt as if I was going to perish.” Not until he left the train did his tension dissipate.

Milgram concluded that this simple experiment demonstrated several social principles. First, enormous inhibitory anxiety ordinarily prevents us from breaking social norms. Second, we have a powerful need to justify our actions after violating a norm. Third, the power of immediate circumstances on our feelings and behaviour is immense.

Discuss how difficult it is to break social norms. Do any students think they could do this without feeling the need to justify their actions? Does this account for the popularity of Chaser (Chaser’s war on everything – TV show) – because they break social norms in a way we’d like to be able to do, and they get away with it?

title: Social norms


text:
User:Egm6321.f09.team7/HW4

 Do remember to log in with your wiki username before editing so that your IP address is not recorded; see the history page. It would be helpful to give a title to each problem; see Team 3. See also the links to the lecture transparencies where HW problems were assigned in Team 1. Do create an internal link to each HW report subpage in your team wiki page to allow easy access to your HW reports.

See further comments below.

Egm6321.f09 13:18, 24 October 2009 (UTC)

See further comments in google doc for Team 7. Egm6321.f09 21:03, 23 November 2009 (UTC)

HW4: Problem 1
Problem Statment and Given:  p.19-1 For the Legendre equation 1 on p.14-2 with $$n=0$$, the homogeneous solution $$u_1(x)=1$$. Use reduction of order method 2 (undetermined factor) to find $$u_2(x)$$, the second homogeneous solution.

Solution:  In general for any given L2-ODE-VC Non Homogenous: $$y'' + a(x)y' + b(x)y = f(x)$$ Apply the following transformation where u1(x) is a homogenous solution: $$y(x) = U(x)u_1(x)$$ $$y(x) = U(x)u_1(x)$$ $$y' = U'u_1 + Uu_1'$$ $$y = Uu_1 + 2U'u_1' + Uu_1''$$ $$Uu_1 + U'(2u_1'+ u_1 a(x)) + U(u_1+ u_1' a(x) + u_1 b(x)) = f(x)$$
 * {| style="width:100%" border="0" align="left"

U''u_1 + U'(2u_1'+ u_1 a(x)) = f(x)$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

For this problem: $$u_1(x) = 1$$ $$u_1'(x) = 0$$ $$f(x) = 0$$ $$a(x) = -2x/(1-x^2)$$

We obtain the following ODE: $$U''-2x/(1-x^2) U' = 0$$

Multiply by $$(1-x^2)$$ and let $$U' = Z$$: $$(1-x^2) Z'-2xZ = 0$$ $$\frac{d((1-x^2) Z)}{dx} = 0$$ $$(1-x^2) Z= C_1$$ $$U'= \frac{C_1}{1-x^2}$$ $$U= \int \frac{C_1}{(1+x)(1-x)} dx$$ $$U= \frac{C_1}{2}[ \int ( \frac{1}{1+x} + \frac{1}{1-x} ) dx]$$

$$U= \frac{C_1}{2}[ ln( (1+x)(1-x) ) + k]$$ We obtain the following Solution:

 * {| style="width:100%" border="0" align="left"

y(x) = U(x)= ln( (1-x^2)^\frac{C_1}{2} ) + C_2$$ Apply  $$y(x) = U(x)u_1(x)$$ You get $$y(x) = U(x)$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

 But you have not identified the second homogeneous solution $$\displaystyle u_2(x)$$ yet. See Team 1. Egm6321.f09 16:59, 10 November 2009 (UTC)

HW4: Problem 2
Problem Statement and Given:

Solve problem 1.1b page 28 from King book, see slide 19-1 in the notes.

Show that the function $$ u_{1}$$ is a solution of the differential equation given below. Use the reduction of order method to find the second independent solution, $$ u_{2}$$.

$$ xy''+2y'+xy=0$$

$$ u_{1}=x^{-1}sin(x)$$

Solution:

Find the first and second derivative of $$ u_{1}$$ and plug them into the equation to check if the equation is equal to zero.

$$ u_{1}'=x^{-1}cos(x)-x^{-2}sin(x)$$

$$ u_{1}''=-x^{-1}sin(x)-x^{-2}cos(x)-x^{-2}cos(x)+2x^{-3}sin(x)$$

$$ xu_{1}''+2u_{1}'+xu_{1}=0$$

$$ 2x^{-2}sin(x)-sin(x)-2x^{-1}cos(x)+2x^{-1}-2x^{-2}sin(x)+sin(x)=0$$

$$0=0$$

The function $$ u_{1}$$ is a solution of the differential equation given.

Now, we use the reduction of order method to find the second independent solution, $$ u_{2}$$.

Put the equation in the form:

$$ y''+a_{1}(x)y'+a_{0}y=0$$

$$ y''+\frac{2}{x}y'+y=0$$

From the notes in class, slides 17-1 to 17-4, we get the formula:

$$ u_{2}=u_{1}\int_{}^{x} \frac{1}{(u_{1}(t))^{2}}exp(-a_{1}(t))dt$$ Where: $$ -a_{1}(t)=-\int_{}^{t} a(s)ds $$

$$ -a_{1}(t)=-\int_{}^{t} \frac{2}{s}ds = -2 ln (t)=ln(t)^{-2}$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{(t^{-1}sin(t))^{2}}exp(ln t^{-2})dt$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{t^{-2}sin^{2}(t)}t^{-2}dt$$

$$ u_{2}=x^{-1}sin(x)\int_{}^{x} \frac{1}{ sin^{2}(t)} dt$$

$$ u_{2}=x^{-1}sin(x)(-cot(x)) $$

$$ u_{2}=-x^{-1}cos(x) $$

HW4: Problem 3
Problem Statement and Given:  $$ xy''+2y'+xy=x$$ from problem 2: $$ y''+\frac{2}{x}y'+1y=x$$ $$ a_1=\frac{2}{x}$$ $$ a_0=1$$ $$ f(x)=x$$ $$u_1(x)=\frac{sin(x)}{x}$$ $$u_2(x)=-\frac{cos(x)}{x}$$

Solution: The solution is in the form $$y(x)=c_1(x)u_1(x)+c_2(x)u_2(x)$$ The terms $$y(x)=c_1(x)$$ and $$y(x)=c_2(x)$$ can be found by using the Wronskian matrix.

$$ u'_2(x)=\frac{cos(x)}{x^2}+\frac{sin(x)}{x}$$ $$ u'_1(x)=\frac{cos(x)}{x}-\frac{sin(x)}{x^2}$$

$$ W={\left[\begin{matrix}u_1(x) & u_2(x) \\ u'_1(x) & u'_2(x)\end{matrix}\right]}$$ $$detW=u_1(x)u'_2(x)-u'_1(x)u_2(x)=(\frac{sin(x)}{x})(\frac{cos(x)}{x^2}+\frac{sin(x)}{x})-(\frac{cos(x)}{x}-\frac{sin(x)}{x^2})(-\frac{cos(x)}{x})$$

$$detW=\frac{cos(x)sin(x)}{x^3}+\frac{sin^2(x)}{x^2}+\frac{cos^2(x)}{x^2}-\frac{cos(x)sin(x)}{x^3}=\frac{1}{x^2}(sin^2(x)+cos^2(x))=\frac{1}{x^2}$$

$$ W^{-1}=\frac{1}{detW}{\left[\begin{matrix}u_1(x) & -u_2(x) \\ -u'_1(x) & u'_2(x)\end{matrix}\right]}$$

$$ W^{-1}=\frac{1}{\frac{1}{x^2}}{\left[\begin{matrix}\frac{sin(x)}{x} & \frac{cos(x)}{x} \\ -(\frac{cos(x)}{x}-\frac{sin(x)}{x^2}) & (\frac{cos(x)}{x^2}+\frac{sin(x)}{x})\end{matrix}\right]}$$

$$ {\left[\begin{matrix}c'_1 \\ c'_2 \end{matrix}\right]}= W^{-1}{\left[\begin{matrix}0 \\ f(x)\end{matrix}\right]}$$

$$ {\left[\begin{matrix}c'_1 \\ c'_2 \end{matrix}\right]}= x^2{\left[\begin{matrix}\frac{sin(x)}{x} & \frac{cos(x)}{x} \\ -(\frac{cos(x)}{x}-\frac{sin(x)}{x^2}) & (\frac{cos(x)}{x^2}+\frac{sin(x)}{x})\end{matrix}\right]} {\left[\begin{matrix}0 \\ f(x)\end{matrix}\right]}$$

$$ {\left[\begin{matrix}c'_1 \\ c'_2 \end{matrix}\right]}= {\left[\begin{matrix}x^2cos(x) \\ x^2sin(x) \end{matrix}\right]}$$

Integrate to find $$ c'_1$$ and $$c'_2 $$

$$c_1=\int_{}^{x} s^2cos(s)ds=(x^2-2)sin(x)+2xcos(x)+C_1$$ $$c_2=\int_{}^{x} s^2sin(s)ds=2xsin(x)-(x^2-2)cos(x)+C_2$$

Plug this into the general solution: $$y(x)=[(x^2-2)sin(x)+2xcos(x)+C_1][\frac{sin(x)}{x}]-[2xsin(x)-(x^2-2)cos(x)+C_2][\frac{cos(x)}{x}]$$

$$y(x)=\frac{(x^2-2)sin^2(x)}{x}+\frac{2xcos(x)sin(x)}{x}+\frac{C_1sin(x)}{x}-\frac{2xcos(x)sin(x)}{x}+\frac{(x^2-2)cos^2(x)}{x}-\frac{(C_2cos(x)}{x}$$

$$y(x)=\frac{x^2-2}{x}(cos^2(x)+sin^2(x))+\frac{C_1sin(x)}{x}-\frac{C_2cos(x)}{x}$$

$$y(x)=\frac{x^2-2}{x}+\frac{C_1sin(x)}{x}-\frac{C_2cos(x)}{x}$$

Method 2:  Assume the general solution is $$y(x)=U(x)u_1(x)$$ From the lecture notes let $$Z(x)=U'(x)$$ so $$u_1(x)Z'+[a_1(x)u_1(x)+2u'_1(x)]Z=f(x)$$ $$(\frac{sin(x)}{x})Z'+[(\frac{2sin(x)}{x^2})+2(\frac{cos(x)}{x}-\frac{sin(x)}{x^2})]Z=x$$ $$(\frac{sin(x)}{x})Z'+(\frac{2 cos(x)}{x})Z=x$$ $$Z'+(\frac{2 cos(x)}{sin(x)})Z=\frac{x^2}{sin(x)}$$

$$Z(x) = csc(x) (c_1 csc(x)-(x^2-2) cot(x)+2 x)$$

$$U(x)=\int_{}^{x}Z(s)ds$$ $$U(x)=csc(x) (-c_1 cos(x)+x^2-2)+c_2$$ Plug $$U(x)$$ into general solution to get: $$y(x)=(csc(x) (-c_1 cos(x)+x^2-2)+c_2)(\frac{sin(x)}{x})$$ $$y(x)=\frac{x^2-2}{x}+\frac{C_1sin(x)}{x}-\frac{C_2cos(x)}{x}$$

HW4: Problem 4
Problem Statement:

Describe in words, step by step, the method of attack given only the homogeneous or non-homogeneous L2. ODE. VC.

Solution:

Answer: Try reduction of method order 0, then order 1, and later integral factor method. If they do not work, use trial solution method. If both of the roots cannot be determine we can use the undetermined factor method to find u2 knowing u1. For the non-homogenous case solve the homogeneous solution and then solve the particular solution using variations of parameters.

 See my comments on the same problem in Team 3. See also the nice flow chart in Team 1. Egm6321.f09 13:24, 24 October 2009 (UTC)

HW4: Problem 5
Problem Statement:  1.3a in King Book. Solution: '''See attached file contains the solution! '''

Problem Statement
Pg. 21-3 confirm King pg. 28 pr 1.1 a

given the H-L-2ODE-VC below verify the first homogeneous solution and find the second homogeneous solution

$$(x-1)y^{''}-xy^{'} + y =0 $$

Problem Solution
For this problem the first homogeneous solution is given as: $$u_1=e^{x}$$

but how was this solution obtained? This solution can be obtained using the method of trial solution and starting with the familiar

form of $$ y(x)= e^{rx} \qquad \qquad r \quad is \quad constant$$

Now using this trial solution we will find the different differentiations of y:

$$ y(x) = e^{rx} $$

$$ y^{'}(x) = re^{rx} $$

$$ y^{''}(x) = r^{2}e^{rx} $$

Substituting these back into the original problem statement gives:

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

Taking out the common exponential term reduces the equation to:

$$ (r^{2}x - r^{2} - xr + 1) = 0 $$

This equation can be examined to find two solutions for r, the first and most obvious (and also given away by the original homogeneous solution) is that r=1. This leads us back to the problem statement and answer to our question:

$$ u_{1}(x) = e^{x} $$

unfortunately the second solution for r2 would be a function of x which violates the original trial solution condition that r be a constant. Therefore we will have to rely on other methods to find the second homogeneous solution.

To find a second solution knowing the first the method of undetermined factors can be utilized

The first step is to rearrange the original equation to give the form:

$$ y^{''} - (\frac{x}{x-1})y^{'} + (\frac{1}{x-1})y = 0 $$

making the identification:

$$ a_{1} = (\frac{-x}{(x-1)}) $$

$$ a_{0} = (\frac{1}{(x-1)} $$

Now assume and answer with the form:

$$ y(x) = U(x)u_{1}(x) $$

after taking the first and second derivatives of y and plugging them back in the following is found:

$$ 0= U^{'} \left[ a_{1}u_{1} + 2u_{1}^{'} \right ] + U^{''}u_{1} $$

After some rearranging and using the integrating factor method as described in the homework from page 17-4 it is found that the second homogeneous solution takes the following for:

$$ u_{2} = u_{1} \int_{}^{x} \frac{1}{(u_{1}(t))^{2}} e (-\bar{a}_{1}(t)dt) $$

with the following resulting from the derivation:

$$ \bar{a}_{1} = \int{}^{t} a_{1}(s)ds = - \left [x + ln(x-1) \right ] $$

Using this value into Equation 2 yields and analyzing the integral as a first step yields:

$$ \int{}^{x} \frac{1}{e^{2t}} [ e^{t} + (t-1) ] dt $$

$$ u_{2} = e^{x} [ \frac13e^{-x} + \frac{x}2e^{-2x} ] $$

$$ u_{2} =-\frac13e^{-x} + \frac{x}2$$

knowing u1 and u2 and adding in two constants of integration the full homogeneous solution is:

$$ y(x) = K_{1}e^{x} + K_{2} (-\frac13 e^{-x} + \frac{x}2) $$

Problem Statement
Pg. 23-1 []

Part A

Given the H-L-2ODE-VC below find the full solution using both the Variation of Parameters, and the Alternative Method:

$$(x-1)y^{''}-xy^{'} + y = x $$

Problem Solution
The first method to be used is the Variation of Parameters, to use this method it is necessary to have both homogeneous solutions. These solutions were found above in the problem from 21-3 and are as follows:

$$u_1=\exp^{x}$$

$$ u_{2} =-\frac13e^{-x} + \frac{x}2$$

The first step in our solution is to assume that the solution takes the following form:

$$ y(x) = c_1(x)u_1(x) + c_2(x) u_2(x)$$

from here the first and second derivatives are found and added back to the equation taking out the common terms gives us

$$ c_1({u_1}^{}+{a_1}{u_1}^{'}+{a_0}u_1)+c_2({u_2}^{}+{a_1}{u_2}^{'}+{a_0}u_2)+{c_1}^'{u_1}^'+{c_2}^'{u_2}^'=f $$

The first two terms in the above equation are homogeneous solutions are therefore = 0 so this leaves us with

$${c_1}^'{u_1}^'+{c_2}^'{u_2}^'=f $$

combining this with the "breakthrough assumption" of:

$${c_1}^'{u_1}+{c_2}^'{u_2}=0 $$

allows to construct a matrix known as the wronskion matrix which has the following form:

$$\left|  \begin{matrix} u_1 & u_2 \\ {u_1}^' & {u_2}^' \end{matrix}  \right|$$

Let the wronskion matrix be denoted as W and its determinent be known as the wronskion and be denoted by W

Then using the definition of the Wronskion and solving a linear first order ODE with constant coefficients as shown on page 19-3 we can find two new variables:

$$ d_1(x) =  -\int{}^{x} \frac{u_2(s)f(s)}{W(s)}  ds $$

$$ d_2(x) =  \int{}^{x} \frac{u_1(s)f(s)}{W(s)}  ds $$

for our application all of the above are known quantities except for W:

$$ W =  \frac13 + \frac12 e^x  -  \frac{x}2  e^x + \frac13 $$

$$ W  =  \frac23  +  \frac12 e^{x} - \frac{x}2  e^x  $$

so we can determine the values of d_1 and d_2

$$ d_1 = -\frac14 [5xe^{-x} - 8 e^{-x} + x^3]$$

$$ d_1 = 6x e^x - 6e^x  - 3e^{-x}  +  3xe^{-x}  $$

The final form of the solution is in the form:

$$ y(x) = Au_1 +  Bu_2  +d_1u_1  + d_2u_2$$

with A and B being constants of integration and coming from IC or BC's solving for our problem gives:

$$ y(x) = Ax + B(-\frac13e^{-x} + \frac{x}2) + \frac14[5x-8+x^3e^x] + [6x e^x - 6e^x  - 3e^{-x}  +  3xe^{-x}]*[-\frac13e^{-x} + \frac{x}2]$$

The next step is to work the the same problem but using the alternate method. This process is shown in the following papers:

Contributing Authors
Egm6321.f09.team7.hua 01:27, 21 October 2009 (UTC) Egm6321.f09.team7.mm 02:19, 21 October 2009 (UTC) --Egm6321.f09.team7.benedict 23:05, 21 October 2009 (UTC)

 Only 3 contributing team members ? Egm6321.f09 13:24, 24 October 2009 (UTC)

title: User:Egm6321.f09.team7/HW4


text:
User:Egm6321.f09.team7.mm

Homework Four: Problem 1
Problem Statment and Given:  p.19-1 For the Legendre equation 1 on p.14-2 with $$n=0$$, the homogeneous solution $$u_1(x)=1$$. Use reduction of order method 2 (undetermined factor) to find $$u_2(x)$$, the second homogeneous solution.

Solution:  In general for any given L2-ODE-VC Non Homogenous: $$y'' + a(x)y' + b(x)y = f(x)$$ Apply the following transformation where u1(x) is a homogenous solution: $$y(x) = U(x)u_1(x)$$ $$y(x) = U(x)u_1(x)$$ $$y' = U'u_1 + Uu_1'$$ $$y = Uu_1 + 2U'u_1' + Uu_1''$$ $$Uu_1 + U'(2u_1'+ u_1 a(x)) + U(u_1+ u_1' a(x) + u_1 b(x)) = f(x)$$
 * {| style="width:100%" border="0" align="left"

U''u_1 + U'(2u_1'+ u_1 a(x)) = f(x)$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

For this problem: $$u_1(x) = 1$$ $$u_1'(x) = 0$$ $$f(x) = 0$$ $$a(x) = -2x/(1-x^2)$$

We obtain the following ODE: $$U''-2x/(1-x^2) U' = 0$$

Multiply by $$(1-x^2)$$ and let $$U' = Z$$: $$(1-x^2) Z'-2xZ = 0$$ $$\frac{d((1-x^2) Z)}{dx} = 0$$ $$(1-x^2) Z= C_1$$ $$U'= \frac{C_1}{1-x^2}$$ $$U= \int \frac{C_1}{(1+x)(1-x)} dx$$ $$U= \frac{C_1}{2}[ \int ( \frac{1}{1+x} + \frac{1}{1-x} ) dx]$$

$$U= \frac{C_1}{2}[ ln( (1+x)(1-x) ) + k]$$ We obtain the following Solution:

 * {| style="width:100%" border="0" align="left"

y(x) = U(x)= ln( (1-x^2)^\frac{C_1}{2} ) + C_2$$ Apply  $$y(x) = U(x)u_1(x)$$ You get $$y(x) = U(x)$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

title: User:Egm6321.f09.team7.mm


text:
Talk:Selecting a university

Page Expansion
I chanced across this page and I thought more information could be filled into it. I have introduced an arrangement by country, and if you have yours to offer please do not hesistate to fill it in, perhaps create a page under your university to give a little more information to people local and abroad who may consider studying at your university. Lysander89 03:32, 16 October 2009 (UTC)

title: Talk:Selecting a university


text:
User:Munxayavong

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

I had to edit my project a little, Baby bella decided she wanted to come earlier than planned!

witness and closely participate in the birth of a human being
 * Research Component

watch and participate in a birthing of a child (My Best friend/God Daughter)
 * week 1

Was not in Class yet, Did not have assignment.
 * week 2

Stages of pregnancy
 * week 3

Mother had gestational Diabetes, so I will research a little about Gestational Diabetes
 * week 4

(Bella arrived at the beginning of this week!) Research the Hospital Bella was born in and write up about my experience!
 * week 5

Research post pregnancy care procedures that should be taken to care for baby.
 * week 6

I stayed with my Friend for the last couple months of her pregnancy since her husband is gone for border patrol, attended all of her appointments with her and all of the above! I will document "the life of a pregnant women" for the last two weeks or so up till giving birth!
 * week 7

Collect photos and extras needed to put together for presentation. Also research "parenthood" a little bit.
 * week 8

Put together presentation and make sure I have everything I need!

Final Presentation

 * Topic (10%)

< >

 * Research (20%)

<>

 * 1) Internet site:
 * 2) Book:
 * 3) Scholarly Journal:
 * 4) Other (interview, TV show,etc):

 * Your accomplishment (20%)

<>

 * Project Review (50%)
 * 1)What went well
 * 2)What went poorly
 * 3)Future strategies

title: User:Munxayavong


text:
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User talk:Torigrace89

Hello Former Student!

I'm writing to you because you took a class from me in the past few years at the Art Institute. As of Spring, 2010, I'll not be teaching further classes at AIS for the foreseeable future. (We're parting ways on good terms!) Until now, you've been able to reach my Wikiversity educational resources from my website, but I've decided to delete that link. You can still reach all of those materials here.

If you're interested in what I'm up to in the future (which will include an album in 2010 and a novel in 2011) you can friend my facebook page or drop me an email at info@stevenarntson.info and you'll receive a periodic update about performances, classes, and publications.

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I hope all is well for you. Best! Stevenarntson

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Personal Essay
Good draft. The piece has a nice flow to it now, and a clear form. In fact, it seems almost a little too focused right now--I'd like to see it sprawl out a bit more. Whenever I hear about homeless people stealing, I always think of Les Miserables, in which Jean Valjean is arrested for stealing--but he has to steal to survive. In the case of some people who are criminals, should they be stopped, or allowed to do it so they can survive? Especially with regard to coffee, given that coffee moguls are getting rich off of a substance that people are picking for slave wages in foreign countries, there's a global aspect to the situation that is also interesting.
 * Best, Stevenarntson

title: User talk:Torigrace89


text:
User talk:SusyG

Hello Former Student!

I'm writing to you because you took a class from me in the past few years at the Art Institute. As of Spring, 2010, I'll not be teaching further classes at AIS for the foreseeable future. (We're parting ways on good terms!) Until now, you've been able to reach my Wikiversity educational resources from my website, but I've decided to delete that link. You can still reach all of those materials here.

If you're interested in what I'm up to in the future (which will include an album in 2010 and a novel in 2011) you can friend my facebook page or drop me an email at info@stevenarntson.info and you'll receive a periodic update about performances, classes, and publications.

If you know any current or former students of mine who might be interested in any of this, please forward this information if you have a moment.

I hope all is well for you. Best! Stevenarntson

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Personal Essay
I like the story of this essay, but right now I feel like it doesn't have a conclusion. The story gets started, and it's a mystery story: where did the recycling bin go? Maybe time to start asking around. Then, that will become the story.
 * Good start! Stevenarntson

title: User talk:SusyG


text:
User talk:AshleyLincoln

Hello Former Student!

I'm writing to you because you took a class from me in the past few years at the Art Institute. As of Spring, 2010, I'll not be teaching further classes at AIS for the foreseeable future. (We're parting ways on good terms!) Until now, you've been able to reach my Wikiversity educational resources from my website, but I've decided to delete that link. You can still reach all of those materials here.

If you're interested in what I'm up to in the future (which will include an album in 2010 and a novel in 2011) you can friend my facebook page or drop me an email at info@stevenarntson.info and you'll receive a periodic update about performances, classes, and publications.

If you know any current or former students of mine who might be interested in any of this, please forward this information if you have a moment.

I hope all is well for you.
 * Best! Stevenarntson

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Personal Essay
There are some good ideas in here, but I'd like to see this more developed--at times it seems a little more like a rant than a considered essay. There are aspects to the Arti Institute that should certainly be critiqued as emblematic of higher education in general, and specifically of for-profit higher education. I see some examples of these issues here, but the examples are not enlarged upon as issues. Look at your story as something that means something larger about educational trends, and talk somewhat about those trends, using your experiences as examples of them.
 * Best, Stevenarntson

title: User talk:AshleyLincoln


text:
User talk:PavelF

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Progress Report
Please post your progress report today.
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text:
User talk:GarrettAbbey

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Progress Report
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User talk:Meganc

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User talk:JLSinegal

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User talk:JamiB

Hello Former Student!

I'm writing to you because you took a class from me in the past few years at the Art Institute. As of Spring, 2010, I'll not be teaching further classes at AIS for the foreseeable future. (We're parting ways on good terms!) Until now, you've been able to reach my Wikiversity educational resources from my website, but I've decided to delete that link. You can still reach all of those materials here.

If you're interested in what I'm up to in the future (which will include an album in 2010 and a novel in 2011) you can friend my facebook page or drop me an email at info@stevenarntson.info and you'll receive a periodic update about performances, classes, and publications.

If you know any current or former students of mine who might be interested in any of this, please forward this information if you have a moment.

I hope all is well for you. Best! Stevenarntson

 Hello JamiB, 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! --Mu301Bot 07:19, 16 October 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:19, 16 October 2009 (UTC)

Personal Essay
Please post your personal essay, and let me know with an email--I'll take a look at it and offer any suggestions I might have.
 * Best, Stevenarntson

title: User talk:JamiB


text:
User talk:Poonag~enwikiversity

 Hello Poonag, 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! --Mu301Bot 07:19, 16 October 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:19, 16 October 2009 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Poonag. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Poonag~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:41, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:18, 19 April 2015 (UTC)

title: User talk:Poonag~enwikiversity


text:
User talk:Paulmiller~enwikiversity

 Hello Paulmiller, 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! --Mu301Bot 07:19, 16 October 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:19, 16 October 2009 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Paulmiller. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Paulmiller~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:47, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:17, 19 April 2015 (UTC)

title: User talk:Paulmiller~enwikiversity


text:
User talk:AmandaMcCracken

 Hello AmandaMcCracken, 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! --Mu301Bot 07:20, 16 October 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:20, 16 October 2009 (UTC)

title: User talk:AmandaMcCracken


text:
Application Layer

The application layer is the highest layer in the protocol stack, and contains many of the technologies that power the web, yet we take it for granted nearly every day. In this topic, we examine in great detail the architectures contained within this layer, and the how we use them.

Architectures
The Application Layer is used in three ways:
 * Client-Server
 * Peer to Peer
 * Hybrid

Client-server
The Client Server Model operates mainly on the concept of an always-on server, to whom a client may connect to. The server normally has a static IP address, and an associated DNS record to allow hosts to look it up. Clients may come and go, and are never directly connected to each other. Applications that communicate between hosts using the client server model must send information through the server, which may be problem if the server happens to go offline.

An example of the Client Server Architecture is the web server and browser. Wikiversity is hosted on a number of mediawiki servers, which are always on, allowing you and other people from across the globe to view this web page.

Peer to Peer
Also known as P2P, the peer to peer architecture does not require the use of centralised servers. Instead, hosts communicate directly with each other. Hosts may have dynamic IP addresses and may join or leave such networks at any time.

A greatly known example of P2P is the bittorrent file sharing protocol.

Hybrid
Hybrid architectures use a combination of both P2P and client server architecture.

An example of this is the Skype protocol. A central server stores each user's address book and IP address. When one person wants to make a call to another, Skype provides the IP address to the destination host and a peer to peer connection is established between the two users.

Technologies used in the Application Layer

 * Domain Name System
 * Voice over Internet Protocol

title: Application Layer


text:
Voice over Internet Protocol

Introduction
Voice over Internet Protocol (VoIP) is used as another medium for communication. Phone calls are through the data network instead of the Public Switched Telephone Network or analog lines.

Voice Over IP is the technology that uses communications protocol to carry voice packets over an IP link. The end-point are typically IP phones or can be computers that uses headset and speaker. Voice before being sent to destination via IP link is converted from analog to IP by means of DSP resources that reside in the IP phones. VoIP technology has brought a revolution in the networking industry by enabling voice packets to traverse through the same link as the data link, thus helping organization to reduce call cost. Companies are able to remove the need to use a separate voice path, remove recurring cost of PBX installation, since VoIP and more preferably it's subset IP Telephony provides centralised call controlling system. The voice packets traverse over the internet/IP link using UDP protocol riding on RTP packets. The stages that are involved from converting the analog to IP voice and then transmitting it through the IP link are mentioned below:

1. Analog signal (Voice that user generates to speak) is sampled as per Nyquist theorem

2. The sample voice is Quantised (Called Quantization)

3. Quantised sample is encoded using 8 bit code

4. The encoded voice is compressed (optional)

5. The compressed voice is sent across the IP link to the remote site

Protocol Breakdown
VoIP phone calls are broken down into packets and sent across the data network. Usually voice packets are tagged with a Virtual LAN (VLAN) tag so that the routers can tell the difference between a voice packet and a data packet. Then the router will redirect the voice traffic to the proper place. For more information visit the Wikipedia page].

Sampling Theorem
Sampling is a process of converting Analog signals into a set of numeric value that can be further transformed into bits for the communication systems to understand. It is basically the conversion or representation of voltages into a digitised format.

Sampling theory is based on Nyquist-Shannon's theorem which states that the sampling rate should be twice the rate of highest frequency of Audible voice so that a perfect reconstruction is possible. The maximum frequency of Human voice is approximately 9000 Hz, although the majority of normal human speech falls in the 3500 hz range. Industry standard value for sampling is considered as 4000 Hz (4KHz). So the sampling rate would be 2 times of 4000 Hz = 8000 samples per second.

Public VoIP Services
Public VoIP service consist of vendors such as [Skype] or [Vonage]. AT&T and TimeWarner are also providing VoIP services through their broadband services. As with public services the customer does not have the phone switch at their place of residence.

Private VoIP Services
Private VoIP Services such as Cisco's CallManager] are private to a company. The company owns all the equipment from phone to phone switch. This kind of implementation is very expensive but also some what more secure because the company has control over every end point before the phone call exits to the PSTN.

title: Voice over Internet Protocol


text:
Talk:Network+/Standards/TCP/IP Model/Introduction

Purpose of the page
I see this page as a gateway to introducing the overall structure of the TCP/IP model, without going too far in depth, since topics for examining each specific layer exist. Basic explanations and applications should be described here, more detailed theory should be left to the other pages.

Nomination for renaming
Why people seem to insist on calling this "TCP/IP" I have no clue. IP is on its own, TCP happens to be a protocol one layer deeper. Can it even be honestly said it's the predominant use of IP anymore, what with the increased use of things like RTP (which uses UDP)? Networking by its nature is layered, and calling it "TCP/IP" suggests TCP and IP are on equal levels when they are very clearly not.

How about something more like "IP Fundamentals" or "IP Networking Fundamentals?" Rchandra (discuss • contribs) 17:06, 18 June 2013 (UTC)

 * Wikipedia calls it Internet protocol suite.  Something like that makes sense to me.  Internet Protocol Fundamentals or Internet Protocol Suite Fundamentals or Internet Protocol Networking Fundamentals.  -- Dave Braunschweig (discuss • contribs) 18:24, 19 June 2013 (UTC)


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Mobile Networks

This topic page is for organizing the development of content related to wireless systems. If you are knowledgeable in any area Wireless systems, feel free to improve upon what you see, we would greatly appreciate your contributions.

Introduction
Wireless networks have significantly impacted the world as far back as World War II. With the use of wireless networks, information could be sent overseas or behind enemy lines easily and quickly and was more reliable. Since then wireless networks have continued to develop and its uses have significantly grown. Cellular phones are part of huge wireless network systems. People use these phones daily to communicate with one another.

Sending information over seas is only possible through wireless network systems using satellites and other signals to communicate across the world otherwise getting information Emergency services such as the police department utilize wireless networks to communicate important information quickly. People and businesses use wireless networks to send and share data quickly whether it be in a small office building or across the world. Another important use for wireless networks is as an inexpensive and rapid way to be connected to the Internet in countries and regions where the telecom infrastructure is poor or there is a lack of resources, like most Developing Countries.

Wireless networks allow you to eliminate messy cables. Wireless connections offer more mobility, the downside is there can sometimes be interference that might block the radio signals from passing through. One way to avoid this is by putting the source of your wireless connection in a place where the signal will have as little interference as possible. Sometimes nearby networks are using the same frequencies, this can also cause interference within the network and can reduce its performance.

Compatibility issues also arise when dealing with wireless networks. Different components not made by the same company may not work together, or might require extra work to fix compatibility issues. To avoid this, purchase products made by the same company so that there are fewer compatibility issues.

Wireless networks, in terms of internet connections, are typically slower than those that are directly connected through an Ethernet cable. Though the speed is slower, most things will still move at the same speed except for things like video downloads. Though wireless technology continues to develop, it is now easier to get networks up and running cheaper and faster than ever before.

A wireless network is more vulnerable because anyone can try to break into a network broadcasting a signal. Many networks offer WEP - Wired Equivalent Privacy - security systems which have been found to be vulnerable to intrusion. Though WEP does block some intruders, the security problems have caused some businesses to stick with wired networks until security can be improved. Another type of security for wireless networks is WPA - Wi-Fi Protected Access. WPA provides more security to wireless networks than a WEP security set up. The use of firewalls will help with security breaches which can help to fix security problems in some wireless networks that are more vulnerable.

Wireless networking can entail a number of technologies. Some are listed below.

Bluetooth
Typically used to create Personal Area Networks (PANs), Bluetooth has a distance limitation of up to 100 meters. The Bluetooth standard defines 3 classes of transmitters and receivers.
 * Class 1 defines a maximum permitted power of 100mW (20 dBm) and can transmit up to approximately 100 meters.
 * Class 2 defines a maximum permitted power of 2.5mW (4 dBm) and can transmit up to approximately 10 meters.
 * Class 3 defines a maximum permitted power of 1mW (0 dBm) and can transmit up to approximately 1 meter.

IEEE 802.11
As described in Network communication medium, IEEE 802.11 refers to the standards of wireless technology most commonly used today in network connections.

The current standard for this the IEEE 802.11x standards also called Wi-Fi for Wireless Fidelity. Wi-Fi is being used by both business and consumer as a way to share a internet connection as well as P2P connections. Currently there are three Wi-Fi Standards a, b and g, the n and i standards are currently being developed.

802.11b - 802.11b was the first wi-fi standard to be introduced and has a maximum speed of 11Mbps.

802.11a - 802.11a was the second wi-fi standard and had a maximum speed of 54Mbps with the same range transmission and is not compatible with 802.11b because it used the 5GHz band and which has less interference.

802.11g - 802.11g was the third standard and had a maximum speed of 54Mbps and much farther range than both a and b but is backward compatible with 802.11b.

Security

 * Wired Equivalent Privacy (WEP)
 * Wi-Fi Protected Access (WPA)
 * Wi-Fi Protected Access Version 2 (WPA2)
 * 802.1x

Learning resources
A good beginning is ftp://ftp.3gpp.org Specs Year 2010 Dec for instance Chose Release 8 or above Pick up a maiden document such as 36.300 Subsequently as you delve deeper on a specific subject such as PHY layer, you might want to download specs 36.211,36.212.36.213 or 36.214

Wikipedia

 * Ralink

Related news

 * April 2007 Spectrum formally devoted to television signals to be auctioned by Feds...
 * March 2007 Mesh networking patents to be sold at auction.

title: Mobile Networks


text:
Talk:Network Classifications

Deletion
This page only contains links to wikipedia pages and does not have any course content worth merging with the developing networking syllabus. Unless there is a way to fit this page into the syllabus I do not see a use for it. Please share your thoughts. Lysander89 08:19, 16 October 2009 (UTC)

title: Talk:Network Classifications