text:
Talk:Motivation and emotion/Book/Extreme achievers

Hey EFA, I was just wondering if you happen to find anything to do with achievement and envy or something along those lines we could probably link both our pages together. Good luck with your chapter! --EssDubz 11:19, 31 October 2011 (UTC)

Hey, Liking the look of the page, maybe shrink the Rafael Nadal picture, as long term goals seems kinda weird looking there. I'm not sure if you can change it (as yours is the first quiz I have done), but it came up wrong if I didn't put answer A, and also gave me a mark out of 12. If you can't change that maybe just put a note somewhere saying ignore it. Otherwise good work Mlac 10:13, 5 November 2011 (UTC)

Hey there - this is a really interesting topic! I think your quiz is a great idea but you need to include clarification regarding the 'scoring system'. At the end of the quiz you say if you answered mostly 'c' etc...but you don't say which answers correspond to what. I hope that makes sense and is helpful! ShaunaB 00:36, 6 November 2011 (UTC)

Hi, just a couple of comments. As a non-sporting person I would probably be more inclined to 'read on' if there were a couple of non-sport related examples of extreme achievers in the Introduction. I also noticed there is a 'Summary' heading at the bottom of the chapter. Just a reminder to put in some text. Good luck. Crazydaisy 05:47, 6 November 2011 (UTC)

Thanks for the feedback everyone. Much appreciated. EFA 15:55, 6 November 2011 (UTC)

Liked the GCOS and 'Flow' diagram - excellent research and lay out - well done - Magnolia28

-- Jtneill - Talk - c 10:25, 28 November 2011 (UTC)

-- Jtneill - Talk - c 09:50, 10 December 2011 (UTC)


text:
Talk:Motivation and emotion/Book/2011/Managing life change

Comments
You have some interesting information in your chapter, great effort to have got so far. I have some thoughts for your chapter, I was initially thinking of doing this topic I hope you find this helpful. Self-determination theory for intrinsic goals and automated motivation maybe highly relevant, you could go with incremental theory (Dweck)also motivation, for managing challenges self-efficacy or some CBT (core beliefs) may provide insight and maybe a stress scale for life transitions. Maybe mentioning too managing expected changes, eg ageing and unexpected changes, Hope this helps (Susann 05:02, 1 November 2011 (UTC))

Hey there. I really like how you've covered managing change - although I understood implicitly that managing change is difficult, I've not thought about it in this much depth before. I like the part where you talk about depression as being beneficial during adolesence, almost as an adaptive function to help one realise which goals are worth directing energy towards. It reminds me of some of the evolutionary psychology stuff I've read; which basically covers off on why from an evolutionary perspective, some psychopathologies were adaptive back in the "cave man" days. Really interesting topic - I shall look forward to seeing the finished product :-). TabithaJ 08:45, 1 November 2011 (UTC)

Your chapter is looking very thorough. You have covered a lot of interesting content :) related chapters you may want to provide links for may be feedback, handling stress, and growth through adversity. Nice job :) some pictures/diagrams and other learning features may help break up the text a bit :) AngeM 23:05, 2 November 2011 (UTC)

Hey, firstly I love the colours also I like how you draw the reader in and make them think it's really creative. I really wanted to read it and learn more about the topic and you did this really well. The colour and getting the reader to think about themself is really good. Well done Rednoodles 00:40, 6 November 2011 (UTC)

Draft comments
I've had a quick look - here's some suggestions: Sincerely, James -- Jtneill - Talk - c 08:11, 2 November 2011 (UTC)
 * Overall - general structure looks good - look to enhance/develop and draw in the conclusion a few key themes/recommendations - look make these resonate e.g., through accompanying images/headings etc.
 * Crystallisation of desire - how does this differ from crystallisation of discontent?
 * Include review/content/findings from key article: Heatherton and Nichols (1994): http://wilderdom.com/psychology/articles/PersonalAccountsOfSuccessfulVsFailedLifeChange.html
 * Baumeister, R. F. (1994) is cited heavily, but appears to be a secondary reference/citation? Did you consult the primary source?
 * Paragraph length - some paragraphs are overly long - look to communicate one idea per paragraph using three to five sentences.

thanks James, I'm in the process of modifying it. Baumeister (1994) is a chapter in an editied book, perhaps I've referenced it incorrectly. However, given what I've read he was the 'founder' of crystallisaiton of discontent so it seemed adequate for me to reference him for that. however, I will search for more work regarding crystallisation of discontnet and desire to make it appear a little more balanced. Kindest regards, Sarah U3033746 01:30, 3 November 2011 (UTC)

G'day. This is an excellent looking chapter on what is a truly interesting topic. Great idea... kind of wish I'd thought of it myself. Regards U895075 02:06, 3 November 2011 (UTC)

Comment by Magnolia - you should get this published in a magazine - send it in - great article - excellent presentation and research

-- Jtneill - Talk - c 11:50, 24 November 2011 (UTC)

-- Jtneill - Talk - c 04:31, 9 December 2011 (UTC)

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


title: File:CLTI.DiffEQ.1.A.20111031.pdf


text:
United Kingdom Law/Reform/Great Reform Bill

This page is an attempt to collate into a workable document a number of proposals for reform of items of legislation in the UK.

Contributing
Please be specific when adding suggestions for reform, The table headings for each section is indicative of what suggestions should include:

in question.
 * Clause - A numbering which should be incremented if adding a new proposal.
 * Legislation - Ideally the short form title and year for specific Primary legislation (typically the Act of Parliament), or title, year and sequence number for secondary legislation (typically Statutory Instruments or Regulations) which contains the laws, regulations or provisions to be repealed. If you are able to assist by clarifying the title in existing proposals, please do so.
 * Portion - The portion of the legislation to be reformed. If you are able to assist other contributors by making explicit the specific section a proposal refers to, please do so.
 * Wording - The wording of the reform, the text to remove, amend or insert.
 * Reasoning - A brief explanation or justification as to why this bill should be concerned with reform of the legislation

 Do not delete without consensus Please do NOT simply delete proposals from this page. Debate them instead.

Preamble
A bill to reform and revise a number of statutory measures in the United Kingdom.

Extent, commencement, transitional provision and short title
(1) This Act extends to England and Wales, Scotland and Northern Ireland. (2) An amendment or repeal contained in this bill shall have the same extent as the Act or instrument or relevant part of the Act or instrument to which the amendment or repeal relates (ignoring extent by virtue of an Order in Council). (3) This Act comes into force at the end of the period of six months beginning from the date of Royal Assent. (This provision will give time for public and private sector organisations to adjust to the many reforms in this Act) (4) A Minister of the Crown, or two or more Ministers of the Crown acting jointly, may by order make transitional, transitory or saving provision in connection with the coming into force of any provision of this Act. (5) An order under subsection (4) is to be made by statutory instrument. (6) This Act may be cited as the  Reform Act 2012.

Applicability to Devolved assemblies
This bill shall neither add to nor subtract from the devolved powers of the National Assembly for Wales, the Scottish Parliament or the Northern Ireland Assembly.


text: United Kingdom Law/Reform/Great Reform Bill/set


text: United Kingdom Law/Reform/Great Reform Bill/entry


text:
United Kingdom Law/Reform/Great Reform Bill/Debate

Please engage in a discussion about the Great Reform Bill. What are the pros and cons of reforming the measures listed? Please sign your edits by including --Sfan00 IMG 13:39, 31 October 2011 (UTC) after your comment.


text:
User:Kaihsu/IMlawA2

European internal market.

Section A: The scope of the four freedoms.

Chapter 2: Material scope of the four freedoms – introduction.

Notes by Kaihsu Tai, 2011-11-16.

Overview, principles, structure
The substantive Treaty provisions are mostly in Part Three, Titles I to IV TFEU. Article 26 TFEU describes the internal market as “an area without internal frontiers”. The structure of each freedom starts with a general prohibition, followed by exceptions.

However, some principles are in Part Two: Discrimiation on grounds of nationality is prohibited: Article 18 TFEU. This applies to victim compensation and to university entrance: Case 186/97 Cowan, Case C-147/03 Austrian university entrance rules. Article 19 TFEU deals with discrimination on some other grounds.

Free movement of goods
Goods are “products which can be valued in money and which are capable, as such, of forming the subject of commercial transactions”: Case 7/68 Italian art treasures.

Article 30 (Part Three, Title II) TFEU prohibits customs duties between Member States. Articles 34 to 36 TFEU prohibits quantitative restrictions between Member States, with exceptions in Article 36. Article 110 TFEU prohibits discriminatory and protectionist taxes.

Free movement of persons, plus citizenship; workers
Part Three, Title IV TFEU deals with the movement of persons. In its first three chapters in turn: workers (Article 45 seq.), right of establishment (for self-employed persons), and freedom for service users and providers. These have spilled over to form citizenship rights: Articles 20 to 25 TFEU especially 21(1); Directive 2004/38; Case C-184/99 Grzelczyk. There has also been moves to remove obstacles against movement, even for third-country nationals: e.g. Schengen Agreement.

Definition of workers: “The essential feature of an employment relationship, however, is that for a certain period of time a person performs services for and under the direction of another person in return for which he receives remuneration”: Case 66/85 Lawrie-Blum. This has been extended to jobseekers: Article 7, Directive 2004/38.

Freedom of establishment, freedom to provide and receive services
Chapters 2 and 3 of Title IV are distinguished so: establishment covers permanent business in another Member State (Article 49 seq.); services involve temporary transactions (Article 56 seq.). Directive 2004/38 is relevant here. The “official authority” exception does not apply where the final discretion for decision making is elsewhere: Case 2/74 Reyners.

Freedom of establishment applies to a self-employed person, i.e. one who is “[1] outside any relationship of subordination concerning the choice of that activity, working conditions and conditions of remuneration; [2] under that person's own responsibility; and [3] in return for remuneration paid to that person directly and in full”: Case C-268/99 Jany. It applies for a company where there is “the actual pursuit of an economic activity through a fixed establishment in another Member State for an indefinite period”: Case C-221/89 Factortame II; “[1] actual establishment of the company concerned in the host Member State and [2] the pursuit of genuine economic activity there”: Case C-196/04 Cadbury Schweppes.

Freedom to provide or receive services includes the freedom to travel for such: Case 286/82 Luisi and Carbone. “Services” is defined residually: Article 57 TFEU. The freedom applies even where neither party travels: Case 352/85 Bond van Adverteerders. It can lead to third-country nationals being granted residence: Case C-60/00 Carpenter. It extends horizontally “to rules of any other nature aimed at regulating in a collective manner gainful employment and the provision of services”: Case 36/74 Walrave and Koch.

Free movement of capital and payments
Restriction of movement of capital and on payments are prohibited, with some exceptions: Chapter 4 (Article 63 to 66) plus Article 75 (freezing assets for security). This is the only freedom which deals with movement to and from third countries expressly. This has direct effect: Case C-163/94 Sanz de Lera. Double taxation is not contrary to free movement of capital: Case C-513/04 Kerckhaert and Morres v Belgium.

title: User:Kaihsu/IMlawA2


title: User:Callum6052


text:
Motivation and emotion/Book/2011/Disgust

Overview
Disgust is an emotion and a type of aversion to stimuli that is found to be extremely unpleasant or offensive whether physical or moral. There are evolutionary theories that seek to explain the phenomenon of disgust. These include survival by detecting tastes that could be potentially dangerous, avoiding a potentially disease-ridden rat or rejecting sexual partners based on poor genes. These genes can present as obesity, diseases and physical deformities often caused by genetic mutation. A disgust reaction can often cause the face to scrunch up and distort the mouth. Disgust has been discussed in the context of emotions particularly by renowned psycholgists Paul Ekman and Robert Plutchik. Ekman's focus was on emotions and emotive expression (His involvement in The Wizard Project), whereas Plutchik developed a Wheel of Emotions. Parrott managed to organise his list of emotions in the form of a table.

In the brain, the insular cortex is the part seen to be particularly active during disgust and observing expressions of disgust. Other physiological symptoms of a disgust reaction can include lowered heart rate and blood pressure, nausea and changes in respiratory behaviour. There has been evidence to suggest that women experience disgust more easily than men, especially when it comes to selecting and more specifically, rejecting a mate.

What is disgust?

Disgust is an emotion as well as a description of the physiological responses to a disturbing stimulus, whether that be a taste, a smell, a sight (visual), something touched (haptic) or even a sound (audio) such as dissonant chords. Other signs and symptoms apart from facial expression can include nausea, lowered blood pressure and lowered heart rate. Disgust can also be elicited from a moral stimulus such as the actions of others. This is known as moral disgust.

Moral disgust
Moral disgust refers to the way a person may feel towards another's actions. For example, "I am disgusted by your behaviour." Examples of such behaviour can be cruelty, selfishness, lack of consideration or almost any other negative behaviour. Moral disgust is also distinct from physical disgust in the way it is culturally defined, whereas physical disgust is prevalent across cultures and is therefore biological. It has been theorised that physical disgust and eventually moral disgust have evolved from a need for humans to avoid certain dangers.

Evolutionary theories
Potential hazards for humans can be found everywhere in nature. It is for this reason that humans have evolved disgust.

For example, rotting food, bodily fluids, animals (including parasites and insects) and signs of infection can all be hazardous to a human and also are likely to elicit a disgust reaction. A mutation may occur in a gene that causes a human to dislike the taste of a certain poisonous food. That human then passes on this mutation to his or her offspring who, therefore, also feel disgust with this food, they survive, and they pass on the helpful gene. And so on and so on. If they did not receive this mutation they might not have survived to pass along the gene and therefore the gene would have died out.

This phenomenon would not happen in the same way if the food were not hazardous to a human's health because the gene would not affect survival and therefore its own preservation by being passed down and causing non-recipients of the gene to die. Disgust has taken this hereditary path of passing genes down to offspring to become part of the human condition.

Paul Ekman
Paul Ekman was a psychologist who was named one of the top 100 influential people of 2009 by Time Magazine. He is widely known for his work in the field of emotions, and how facial expressions can express them to others.

His most famous piece of work was one named The Wizards Project. In this project he and Dr Maureen O'Sullivan tested participants' ability to detect lies in other people. From this study of 20,000 participants, they found that about 50 people of this sample (approx. 0.25% of the sample) showed great skill in detecting lies and would be able to correctly recognise a liar to at least 80% accuracy. This is a highly significant rate of accuracy compared with the rest of the sample who could only identify liars to about 50% accuracy, which is the expected accuracy if they had simply guessed. They coined the term 'Truth Wizards' for these seemingly gifted individuals. A television show was inspired by this research finding named Lie to Me.

In 1972 devised a list of six basic emotions. These included:

 * Anger
 * Disgust
 * Fear
 * Happiness
 * Sadness
 * Surprise

18 years later in 1990 he revised the list to include 11 more emotions. These were:
 * Amusement
 * Contempt
 * Contentment
 * Embarrassment
 * Excitement
 * Guilt
 * Pride in achievement
 * Relief
 * Satisfaction
 * Sensory pleasure
 * Shame
 * Shame

Ekman gained criticism in his work due to his theories not being subjected to scientific experimentation. Another limitation to Ekman's list is that he fails to take into account that many emotions are raw and many are derived from these raw emotions. Robert Plutchik attempted to make his own list that takes this problem into account.

Plutchik's wheel of emotions

Robert Plutchik devised a list of his own in 1980. This list was organised into a wheel and was comprised of 8 basic emotions and 8 advanced emotions that are made up of two basic ones. The basic ones include:

 * Joy
 * Trust
 * Fear
 * Surprise
 * Sadness
 * Disgust
 * Anger
 * Anticipation

The second list of 8 advanced emotions are made up of combinations of the previous 8 basic emotions. These advanced emotions are:

 * Optimism (Anticipation + Joy)
 * Love (Joy + Trust)
 * Submission (Trust + Fear)
 * Awe (Fear + Surprise)
 * Disappointment (Surprise + Sadness)
 * Remorse (Sadness + Disgust)
 * Contempt (Disgust + Anger)
 * Aggression (Anger + Anticipation)

Trying to organise emotions into a list is very difficult because some emotions are raw and some are derivatives of other emotions. This method of organisation has tried to address this problem whereas Paul Ekman's list does not.

Parrot's Emotions by Groups
W. Gerrod Parrott developed his own inventory of emotions presented in the form of a list sorted into a tree-like structure. Below is a table taken from Wikipedia:

As it can be seen in the table, Parrott names what seems like every single emotion thinkable which can be a strength and a weakness. It's a strength because it leaves no stone unturned and seeks to explain every aspect of emotion in as much detail as possible. However, it's also a weakness because there is a tendency for some of these many emotions to double up. For example some people might argue that amazement and astonishment are the same thing.

Cognitive theory of emotion
There have been numerous theories put forward to explain emotion with a cognitive approach. Many of these also share a physiological perspective. One such example is the Two-factor theory of emotion. This theory was tested in a study performed in 1962. Their paper concluded with 3 points:
 * 1) Given a state of physiological arousal for which an individual has no immediate explanation, he will label this state and describe his feelings in terms of the cognitions available to him. To the extent that cognitive factors are potent determiners of emotional states, it should be anticipated that precisely the same state of physiological arousal could be labeled [sic] "joy" or "fury" or "jealousy" or any of great diversity of emotional labels depending on the cognitive aspects of the situation. (p. 398)  This point explains that in their study they found that cognitive patterns determine how we label our physiological states with words such as 'joy, 'fury' or even 'disgust'.
 * 2) Given a state of physiological arousal for which an individual has a completely appropriate explanation, no evaluative needs will arise and the individual is unlikely to label his feelings in terms of the alternative cognitions available. (p. 398)
 * 3) Given the same cognitive circumstances, the individual will react emotionally or describe his feelings as emotions only to the extent that he experiences a state of physiological arousal. An experiment is described which, together with the results of other studies, supports these propositions. (p. 398) This third point illustrates how emotions are based on physiological factors, specifically the level to which the individual is physiologically aroused.

Gender differences in disgust responses

Women have been found to report stronger feelings of disgust than men according to a study in 2008. The responses in the study measured heart rate, electro-dermal activity, salivary cortisol and secretory immunoglobulin A. The study concluded that:

 * Replicating previous studies, women reported stronger feelings of disgust than men across all disgust inductions. Additionally, in Study 1, women showed a higher increase in skin conductance level than men. In conclusion, gender moderates subjective responses to disgust, whereas physiological disgust responses are only marginally moderated by gender.

Physiology of disgust

As mentioned in the introduction, the insular cortex is the most active part of the brain when an individual is experiencing disgust or is observing others expressing disgust. The Wikipedia page for Cognitive Neuroscience and Disgust lists physical, autonomic responses that can arise from feeling disgusted. They are listed as "reduced blood pressure, lowered heart rate and decreased skin conductance along with changes in respiratory behaviour."

A study in 2007 investigated activity in the brains of participants when presented with disgust- and fear-inducing stimuli. The study was comprised of 66 participants who were each presented with 50 images that elicited different emotional responses (10 neutral, 20 fear-inducing and 20 disgust-inducing). Each picture was presented for 4 seconds and each participant rated the picture on how fear-inducing or disgusting it was. Each participants brain activity was monitored using fMRI.

The study found that:

 * Both emotional stimulus categories resulted in activations in the extended occipital cortex, in the prefrontal cortex, and in the amygdala. However, insula [sic] activations were only significantly correlated with subjective ratings of disgust, pointing to a specific role of this brain structure in the processing of disgust. (p. 663)

This study aimed to settle discrepancies in earlier studies that presented unclear evidence in the distinction of structures engaged during fear and disgust. The study appears to have been successful in its goal.

Summary
Disgust has been given to the human race by the blind watchmaker known as evolution. It has helped us survive against poisonous substances that without it, would have killed us. Moral disgust has helped guide our morals and ethics to lead us to better interpersonal relationships. It is important to think of disgust as not just a word that describes a reaction, but as an emotion in itself as outlined by Paul Ekman in his list, or Robert Plutchik in his wheel or W. Gerrod Parrott's table. It is not just a series of cognitions that trigger an emotional response, but an intricate network of cognitive, emotive and physiological systems working together to produce what is known as disgust.


title: File:DigitalDesign.20111031.pdf


text:
Temperature

Temperature describes how much a substance, object or a body is hot or cold. Scientifically it is defined as the average kinetic energy of the molecules, ions or atoms of a substance or system.Temperature of a body can be measured with a thermometer.

Common temperature reference points
Some useful common references:

Conversion chart
Exact conversion chart:

title: Temperature


text:
Physics/Pressure

Pressure is defined as force per unit area. Pascal, PSI, bar, atmosphere, kg/square centimeter, etc are the units used for measuring pressure. Pressure is also expressed in mm/cm of Hg or water. 760mm of Hg equals 1 atmosphere, 1 bar, 100000 Pascals, 1 kg/sq. cm or 14.7 PSI(pounds per square inch.

title: Physics/Pressure


text:
User:Egm6321.f11.team2.rho/HW5

=R*5.4- Find $$\displaystyle y_{xxxxx} $$=

Given
A new variable (i.e. t) is introduced from current variable (i.e. x) in Euler Ln-ODE-CC. The transformation of variables is

Find
Find $$\displaystyle y_{xxxxx} $$ in terms of the derivatives of y with respect to t.

Solution
With the transformation of variables in Euler Ln-ODE-CC

We can obtain the derivatives of y with respect to x in terms of the derivatives of y with respect to t.

Finally,

title: User:Egm6321.f11.team2.rho/HW5


text: File:Fanwing Motor Placement.png

title: File:Fanwing Motor Placement.png


text:
User talk:JaeHaynes/enes100/WaterTreatment

10/10/20 .. please write next in week 3 .. good start getting these wiki pages together 1sfoerster 20:01, 31 October 2011 (UTC)


title: File:Pea1.f11.mtg33.djvu


text:
Wind shear

Wind shear is a difference in either wind speed and/or direction over a fairly short distance in the atmosphere. Wind shear can be divided into two different types: horiztontal and vertical wind shear.

The situations when the Wind shear is observed happen:
 * At weather fronts when the temperature difference across the front is 5 °C or more, and the front moves at 15 kt or faster.
 * At low level jets when a significant low level vertical wind shear can develop near the lower portion of the low level jet.
 * At mountains when winds blow over and create vertical shear on the lee side.
 * At inversions when on a clear and calm night a radiation inversion is formed near the ground.
 * At downbursts when an outflow boundary moves away from a thunderstorm.
 * At sailing when wind shear affects sailboats by presenting a different wind speed and direction at different heights along the sailing mast.

Other websites

 * National Science Digital Library - Wind shear

title: Wind shear


text:
Thunderstorm

Thunderstorms are small, intense weather systems that make strong winds, heavy rain, lightning, and thunder. Thunderstorms can happen anywhere with two conditions: the air near the Earth's surface must be warm and moist, and the atmosphere must be unstable. 100 lightning bolts hit the earth every second, and at any one moment, about 1,800 thunderstorms happen around the earth.

This energy moves to the air and makes it to spread quickly and send out sound waves. Thunder is the sound that comes from the rapid spread of air along the lightning strike. Thunder is slower than lightning, because light is faster than sound.

About 10% of thunderstorms are thought severe. Severe thunderstorms make things like high winds, hail, flash floods, and tornadoes. Hailstorms damage crops, damage the metal on cars, and break windows. Sudden flash floods that happen because of heavy rains is the biggest reason for weather-related deaths.

Lightning, which happens with all thunderstorms, makes thousands of forest fires each year in the United States. Lightning also kills or injures hundreds of people a year in the United States.

Thunderstorms do not do only damage, however: they can be a great help to man and all living creatures. We get lots of water for many continents during the summer. Plants receive lots of life-giving rain when they need it. Without the thunderstorms, many continents would become dry. Fish would die, crops would fail, and animals would perish.

Thunderstorms are also our natural air conditioners. Hot air at the surface rises up into the high atmosphere where it is put out into space. Clouds give us shades, and rain can cool down a hot day. Without thunderstorms, the earth would be as much as 20 F (11 C) warmer. In the summer, dust, haze, and other pollutants come together in the lower atmosphere. When the air rises, either in cumulus clouds or in thunderstorms, spreads the pollution higher up into the atmosphere. Rain from thunderstorms washes away many of these pollutants out of the air.

Other websites

 * Anatomy of a thunderstorm
 * Electronic Journal of Severe Storms Meteorology
 * Social & Economic Costs of Thunderstorms & High Winds NOAA Economics

title: Thunderstorm


text:
TAO/Handbook/Table of contents

To the German version of the table of content.

To do: Identify various content types. [Stijn] I pasted an English version of the Table of Contents. Introduction Description of the idea Examples References / Testimonials Practical and relevant considerations / theories Online communities (What's this?) Volunteers: Roles, transparency, recognition, motivation, care / management Ease of use (N) Onliner / > [Stijn] I will write about this for the survey anyway, so we'll include this here too. Target groups: what are seniors interested in? Pedagogy: Teaching methods approach Activities (case studies) with documentation and recommendations Development and how-to-use Activation and motivation (for example FFI) Courses / workshops for qualification (presence / online)
 * Online collaboration
 * Service (support, helpdesk)
 * Community activities (mentors, regular meetings)
 * Silver Knowledge
 * Free Cruise on the Internet
 * Silver Surfers' Social Networking Services (Stijn)
 * Volunteers
 * Creating a Glossary of Common Terms in a Community

Online tools Videoconference: Skype, FlashMeeting Virtual classrooms: Adobe Connect, Vitero, etc. e-Learning environments: Moodle, etc. Conditions Cooperation Public relations Sponsorship / Fundraising Methods Tools for practice Exploration and evaluation methods Creative and problem-solving methods Planning Methods Implementation and working methods Problem-oriented access Keywords and examples with practical solutions to problems Annotated literature review Annotated list of links

title: TAO/Handbook/Table of contents


text:
TAO/Handbook/Target audience

Target audience(s) of the TAO Handbook and the TAO Community of Practice
See: Personas

See also target audiences of particular modules: For example, the target audience of the silver knowledge module: These target groups need not only informations about Silver Knowledge, important is, that they get an imagination about, how they can involve and integrate elderly people into the wikimedia as wikipedia author. (See also the discussion on TAO/Silver_Knowledge respectively Talk:TAO/Silver_Knowledge.)
 * representatives and multiplicators of senior or education institutions, which will take part on the project silver knowledge or
 * representatives and volunteers of other wikimedia chapters, which will transfer silver knowledge

title: TAO/Handbook/Target audience


text:
TAO/Activities

Overview of the TAO Community Activities
This overview table gives you a general indication of the approaches being tested and implemented as part of the community activities in the project TAO. We want to describe these activities and support the transfer with our learning and/or research modules in this Wikiversity project.

title: TAO/Activities


text:
User:Egm6321.f11.team1/report5.bensplaypen.R*5.5

=Problem 2 - Solving Euler's 2nd-order differential equation using the method of trial solution=

Given
A second order linear ordinary differential equation with varying coefficients (L2-ODE-VC)

With Boundary Coditions,

Find
The solution using the method of trial solution

==Solution  == Let trial solution be

where r is constant. Then

(2.4) substitute into (2.1)

and yields,

A homogeneous solution take this form of

Therefore two roots,

Hence the general solution is

With the given boundary conditions, we get

Solving this gives c1=-6.5 and c2=4.5, therefore the solution of this BVP is

Plot homogeneous solution

Matlab code: hw5_problem2.m

Given
The Euler L2_ODE_VC:

The Euler L2_ODE_CC:

And the characteristic equation

Find
1.1) $$ \begin{align}{{a}_{2}},{{a}_{1}},{{a}_{0}} \end{align}$$ such that equation 3.3 is the characteristic equation of 3.1.

1.2) The 1st homogeneous solution

1.3) The second homogeneous solution

1.4) The general homogeneous solution

2.1) $$ \begin{align}{{b}_{2}},{{b}_{1}},{{b}_{0}} \end{align}$$ such that equation 3.3 is the characteristic equation of 3.2.

2.2) The 1st homogeneous solution

2.3) The 2nd homogeneous solution

2.4) The general homogeneous solution

1.1
The coefficients of 3.1 can be found by using the trial solution method. The trial solution for the Euler equation with variable coefficient is:

The derivatives of the trial solution is defined as:

Substituting the equations 3.1.1, 3.1.2 and 3.1.3 into 3.1 we get:

Therefore we need to solve the following equation to find the coefficients:

Comparing the equation 3.1.7 to the equation 3.3 we have:

Therefore, comparing the coefficients of both side of the equation 3.1.8 we get:

1.2
From the item 1.1, the characteristic equation of 3.1 is 3.3. Therefore, for 3.3 there are two identical roots equals to $$ \begin{align}r= \lambda \end{align}$$. The 1st homogeneous solution can be calculated by:

1.3
Since $$ \begin{align}\lambda \end{align}$$ has two identical roots the second homogeneous solution can be calculate using the form:

Calculating the derivatives of 3.1.17:

Substituting 3.1.17, 3.1.18 and 3.1.19 into 3.1 we have:

Rearranging the terms of 3.1.20:

The equation 3.1.21 is simplified as:

Substituting the coefficients $$ \begin{align}{{a}_{2}} \end{align}$$ (equation 3.1.10), $$ \begin{align}{{a}_{1}} \end{align}$$ (equation 3.1.13), $$ \begin{align}{{a}_{0}} \end{align}$$ (equation 3.1.14) and the 1st homogeneous solution with the respective derivatives presented in $$ \begin{align}{{y}_{1}} \end{align}$$ (equation 3.1.1), $$ \begin{align}{{y}_{1}}' \end{align}$$ (equation 3.1.2) and $$ \begin{align}{{y}_{1}}'' \end{align}$$ (equation 3.1.13), where r is replaced by $$ \begin{align}\lambda  \end{align}$$ into 3.1.22:

Therefore we will solve for:

Using the reduction of order:

Substituting 3.1.28 into 3.1.29:

Integrating both sides of 3.1.32:

Using the exponential function in 3.1.34

Where $$ \begin{align}{{k}_{2}} \end{align}$$ can be written as:

Therefore, z(x) can be written as:

Since the order of U(x) was reduced, we can get U(x) by integrating 3.1.38

Finally, substituting 3.1.42 and 3.1.16 into 3.1.17 the 2nd homogeneous solution is defined as:

1.4
The generalized homogeneous solution can be calculated by using the following equation:

Substituting equations 3.1.16 and 3.1.43 into 3.1.44 we have:

Rearranging the constants from equation 3.1.47, the generalized homogeneous solution can be expressed as:

2.1
The coefficients of 3.2 can be found by using the trial solution method. The trial solution for the Euler equation with constant coefficient is:

The derivatives of the trial solution is defined as:

Substituting the equations 3.2.1, 3.2.2 and 3.2.3 into 3.2 we get:

Therefore we need to solve the following equation to find the coefficients:

Comparing the equation 3.2.6 to the equation 3.3 we have:

Therefore, comparing the coefficients of both side of the equation 3.2.7 we get:

2.2
From the item 2.1, the characteristic equation of 3.2 is 3.3. Therefore, for 3.3 there are two identical roots equals to $$ \begin{align}r=\lambda  \end{align}$$. The 1st homogeneous solution can be calculated by:

2.3
Since $$ \begin{align}\lambda \end{align}$$ has two identical roots the 2nd homogeneous solution can be calculate using the form:

Calculating the derivatives of 3.2.15:

Substituting 3.1.15, 3.1.16 and 3.1.17 into 3.2 we have:

Rearranging the terms of 3.2.18:

The equation 3.2.19 is simplified as:

Substituting the coefficients $$ \begin{align}{{b}_{2}} \end{align}$$ (equation 3.2.9), $$ \begin{align}{{b}_{1}} \end{align}$$ (equation 3.2.11), $$ \begin{align}{{b}_{0}} \end{align}$$ (equation 3.2.12) and the 1st homogeneous solution with the respective derivatives presented in $$ \begin{align}{{y}_{1}} \end{align}$$ (equation 3.2.1), $$ \begin{align}{{y}_{1}}' \end{align}$$ (equation 3.2.2) and $$ \begin{align}{{y}_{1}}'' \end{align}$$ (equation 3.2.3), where r is replaced by $$ \begin{align}\lambda  \end{align}$$, into 3.2.20:

Therefore we need to solve for:

Integrating both sides of 3.2.24 we have:

Integrating 3.2.25 we have U(x) as:

Finally, substituting 3.2.26 and 3.2.14 into 3.2.15 the 2nd homogeneous solution is defined as:

2.4
The generalized homogeneous solution can be calculated by using the following equation:

Substituting equations 3.2.14 and 3.2.27 into 3.2.28 we have:

Rearranging the constants from equation 3.2.31, the generalized homogeneous solution can be expressed as:


text:
User talk:Amcilrick/Rodeo in Australia - Welfare of Animals Used in Rodeo Events

What a great choice of topic you have chosen. I will be back to read in more detail but wondered if you had seen this article


text:
User:Egm6321.f11.team3/Hwk5

Find
Find the following equation

Solution
We solved this problem on our own.

We know that Thus, put (5.1.2), (5.1.7) and (5.1.9) into (5.1.10) We can get

Find
(1)Using two methods to verify the exactness of the following L2-ODE-VC. Legendre: Bessel: Hermite: (2)If Hermite differential equation is not exact, check whether it is in power form, and see whether if it can be made exact using IFM with $$\displaystyle h(x,y)=x^{m}y^{n}$$. (3)The first few Hermite polynomials are Verify the (5.2.11), (5.2.12) and (5.2.13) are homogeneous solutions of the Hermite differential equation.

Solution
We solved this problem on our own.

Part 1.1.a
According to Legendre differential equation (5.2.8) Thus, put equation (5.2.15) and (5.2.16) into equation (5.2.6) and (5.2.7) we can get Thus, when only n = 0 or n = -1, Legendre differential equation will be exact.

Part 1.1.b
According R*5.1 work, we know that Put equation (5.2.15), (5.2.16), (5.2.21) and (5.2.22) into equation (5.2.2), (5.2.19) and (5.2.20)

Thus equation (5.2.5) becomes Thus, when only n = 0 or n = -1, Legendre differential equation will be exact.

Part 1.2.a
According to Hermite differential equation (5.2.10) Thus, put equation (5.2.25) and (5.2.26) into equation (5.2.6) and (5.2.7) we can get Thus, Hermite differential equation is exact only when n = -1.

Part 1.2.b
Put equation (5.2.25), (5.2.26), (5.2.29) and (5.2.30) into equation (5.2.2), (5.2.19) and (5.2.20) we can get Thus equation (5.2.5) becomes Thus, Hermite differential equation is exact only when n = -1.

Part 2
The power form for L2-ODE-VC is : We can choose Thus, Hermite differential equation is in power form In order to make Hermite differential equation be exact, multiply equation (5.2.10) by Put equation (5.2.37) and (5.2.38) into (5.2.7) Put equation (5.2.37) ,(5.2.38) and (5.2.39) into (5.2.6) Because b = 0, a cannot be zero so  Thus, Hermite differential equation becomes Thus after multiplying x, Hermite differential equation is exact.

Part3
The polynomial solution for Hermite differential equation is  Put equation (5.2.44), (5.2.45) and (5.2.46) into Hermite differential equation. Choose the initial condition as  Thus Choose the initial condition as  Thus, Choose the initial condition as  Thus,

Find
Find the expressions for X(x) in terms of cosKx,sinKx,coshKx,sinhKx.

Solution
We solved this problem on our own.

Recall Euler's formula:

Since X(x)is a real-valued function so the left hand side must also be real. Let:

The 4 real constants are c1,c2,b3,b4.

2.Another way:

We can use the same method above,so we can get:

R*5.4 - Finding the 5th derivatives of y with respect to t
From the lecture slide Mtg 31-1

Given
Transformation of variables is the first stage of the method 1 to solve Euler Ln-ODE-VC.

Euler Ln-ODE-VC:

$$\begin{align} x=\text{e}^{t}\end{align}$$

We need to transform x to new variable t.

Chain Rule:

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$$\begin{align} \underbrace{\frac{d y}{dt}}_{\displaystyle y_t}= {} & \underbrace{\frac{dy}{dx}}_{\displaystyle y_{x} }\underbrace{\frac{dx}{dt}}_{\displaystyle \text{e}^{t}} \\ \end{align}$$ (5.4.1)
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$$\begin{align} y_t = y_x e^{t} \end{align}$$

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and we obtain
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$$\begin{align} y_x = y_t e^{-t}\end{align}$$

(5.4.2)
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Find
Find the fifth derivative $$\begin{align}y(x(t))_{xxxxx}\end{align}$$ in terms of derivative of y with respect to t

Solution
Solved on our own

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$$\displaystyle y_{x}= \frac{dy}{dx} = \left(\frac{dt}{dx} \frac{d}{dt}\right)y=y_t e^{-t}$$ (5.4.3)
 *
 * }

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$$\begin{align} y_{xx}=\frac{d}{dx}\frac{d}{dx}y = \frac{dt}{dx}\frac{d}{dt}\left(y_t e^{-t}\right) \end{align}$$ (5.4.4)
 *
 * }

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$$\displaystyle y_{xx}= e^{-t}\left(-e^{-t}y_t+e^{-t}y_{tt}\right)=e^{-2t}\left(y_{tt}-y_t \right) $$ (5.4.5)
 *
 * }

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$$\begin{align} y_{xxx}= \frac{dt}{dx}\frac{d}{dt}\left( e^{-2t}\left(y_{tt}-y_t \right)\right) \end{align}$$ (5.4.6)
 *
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$$\begin{align} y_{xxx}= e^{-t}\left(-2e^{-2t}\left(y_{tt}-y_t\right)+e^{-2t}\left(y_{ttt}-y_{tt}  \right) \right)=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t \right) \end{align}$$ (5.4.7)
 *
 * }

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$$\begin{align} y_{xxxx}= \frac{dt}{dx}\frac{d}{dt}\left( e^{-3t}(y_{ttt}-3y_{tt}+2y_t) \right)\end{align}$$ (5.4.8)
 *
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$$\begin{align} y_{xxxx}= e^{-t}\left(-3e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+e^{-3t}(y_{tttt}-3y_{ttt}+2y_{tt})\right)=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t} \right) \end{align}$$ (5.4.9)
 *
 * }

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$$\begin{align} y_{xxxxx}= \frac{dt}{dx}\frac{d}{dt}\left( e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}) \right)\end{align}$$ (5.4.10)
 *
 * }

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$$\begin{align} y_{xxxxx}= e^{-t}\left(-4e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}+e^{-4t}(y_{ttttt}-6y_{tttt}+11y_{ttt}-6y_{tt})\right)

\end{align}$$ (5.4.11)
 *
 * }

Find
Solve 5.5.1 using Method 2 with a trail solution of 5.5.2 and boundary conditions: $$\displaystyle y(1)=-4, y(2)=7 $$ Plot the solution.

Solution
We solved this on our own. First we find the first and second derivative of the trial solution 5.5.2, And now we substitute 5.5.2, 5.5.3 and 5.5.4 into 5.5.1, $$\displaystyle x^2(r(r-1)x^{r-2})-2x(rx^{r-1})+2(x^r)=0 $$ And simplify, $$\displaystyle (r^2-r)x^r-2rx^r+2x^r=0 $$ $$\displaystyle r^2-3r+2=0 $$ $$\displaystyle r_1=1, r_2=2 $$ Therefore, the following form applies, $$\displaystyle y(x)=c_1x+c_2x^2 $$ Now we apply the given boundary conditions to solve the constants, $$\displaystyle y(1)=c_1(1)+c_2(1^2)=-4 $$ $$\displaystyle y(1)=c_1(2)+c_2(2^2)=7 $$ The equations are now solved simulataneously to detemine the constant values, $$\displaystyle c_1=-11.5 $$ $$\displaystyle c_2=7.5 $$ And the equation to plot becomes, $$\displaystyle y(x)=7.5x^2-11.5x $$ And the equation is plotted in Matlab over an arbitrary interval,

Find
Show the trial solution in Method 1 and Method 2 are equal

Solution
We solved this problem on our own.

Method 2: Characteristic equation:

So

Method 1

Let: substituting in 5.6.1

Let:

So So the two methods are the same.

Find
1.1 Find $$a_2, a_1, a_0$$ such that 5.7.1 is the characteristic equation of 5.7.2 1.2 1st homogeneous solution: $$y_1(x)=x^\lambda$$ 1.3 Complete solution: Find $$c(x)$$ such that $$y(x)=c(x)y_1(x)$$ 1.4 Find the 2nd homogeneous solution $$y_2(x)$$ 2 Repeat steps 1.1 through 1.4 for 5.7.3

Solution
We solved this on our own.

Part1.1
First we find the first and second derivative of the trial solution, Substituting these into 5.7.2 yields, $$\displaystyle a_2x^2(r(r-1)x^{r-2})+a_1x(rx^{r-1})+a_0x^r=0$$ $$\displaystyle a_2(r^2-r)x^r+a_1rx^r+a_0x^r=0 $$ $$\displaystyle a_2r^2x^r+(a_1-a_2)rx^r+a_0x^r=0 $$ $$\displaystyle a_2r^2+(a_1-a_2)r+a_0=0 $$ And using 5.7.1, $$\displaystyle a_2r^2+(a_1-a_2)r+a_0=(r-\lambda)^2=r^2-2\lambda r+\lambda^2 $$ The coefficients are matched, resulting in, $$\displaystyle a_2=1$$ $$\displaystyle (a_1-a_2)=2\lambda$$ $$\displaystyle a_0=\lambda^2 $$ Using $$\displaystyle\lambda=7$$, $$\displaystyle a_2=1$$ $$\displaystyle a_1=1-2\lambda=15$$ $$\displaystyle a_0=49 $$

Part1.2
The 1st homogeneous solution is found by evaluating 5.7.1 and seeing that $$\displaystyle r=\lambda$$. By using the trial solution of 5.7.4, we see that $$\displaystyle y_1(x)=x^\lambda$$.

Part1.3
$$\displaystyle y(x)=c(x)y_1(x)$$ Now we find the first and second derivative of the complete solution using the chain rule, $$\displaystyle y'(x)=c(x)y_1'(x)+c'(x)y_1(x)=c'x^\lambda + c\lambda x^{\lambda-1} $$ $$\displaystyle y(x)=c(x)y_1(x) +c'(x)y_1'(x)+c'(x)y_1'(x)+c''(x)y_1(x) $$ $$\displaystyle y(x)=c(x)y_1(x) +2c'(x)y_1'(x)+c(x)y_1(x)=cx^\lambda + 2c'\lambda x^{\lambda-1} + c(\lambda^2-\lambda)x^{\lambda-2} $$ And now subsituting these into 5.7.2 yields, $$\displaystyle a_2x^2(cy_1+2c'y_1'+cy_1)+a_1x(cy_1'+c'y_1)+a_0y_1=0$$ Distributed to, $$\displaystyle a_2x^2y_1c+a_2x^2cy_1+2a_2x^2c'y_1'+a_1xcy_1'+a_1xc'y_1+a_0cy_1=0$$ And grouped, $$\displaystyle (a_2x^2y_1)c+(2a_2x^2y_1'+a_1xy_1)c'+(a_2x^2y_1+a_1xy_1'+a_0y_1)c=0$$ The last term is equivalent to 5.7.2 and equals zero. Now we apply the first homoegeneous solution and the coefficients, $$\displaystyle (x^2x^\lambda)c''+(2x^2 \lambda x^{\lambda-1}+x(1-2\lambda)x^\lambda)c'=0$$ $$\displaystyle (x^{\lambda+2})c''+x^{\lambda+1}(2\lambda+(1-2\lambda))c'=0$$ $$\displaystyle xc''+c'=0$$ Integrating the equation yields, $$\displaystyle c(x)=k_1 ln(x)+k_2$$ And subsituted back into y(x) gives us,

Part1.4
The 2nd homogenous equation is found by inspected the complete solution. Since we know the the first homogeneous equation is, $$\displaystyle y_1=x^\lambda $$ We can see the 2nd equation must be the other term in the complete solution, $$\displaystyle y_2=x^\lambda ln(x) $$

Part2.1
First we find the first and second derivative of the trial solution with a constant coefficient, $$\displaystyle y=e^{rx}$$ $$\displaystyle y'=re^{rx}$$ $$\displaystyle y''=r^2e^{rx}$$ Substituting these into 5.7.3 yields, $$\displaystyle b_2(r^2e^{rx})+b_1(re^{rx})+b_0e^{rx}=0$$ $$\displaystyle b_2r^2+b_1r+b_0=0$$

And using 5.7.1, $$\displaystyle b_2r^2+b_1r+b_0=(r-\lambda)^2=r^2-2\lambda r+\lambda^2 $$ The coefficients are matched, resulting in, $$\displaystyle b_2=1$$ $$\displaystyle b_1=-2\lambda$$ $$\displaystyle b_0=\lambda^2 $$ Using $$\displaystyle\lambda=7$$, $$\displaystyle b_2=1$$ $$\displaystyle b_1=-14$$ $$\displaystyle b_0=49 $$

Part2.2
The 1st homogeneous solution is found by evaluating 5.7.1 and seeing that $$\displaystyle r=\lambda$$. By using the trial solution, we see that $$\displaystyle y_1(x)=e^{\lambda x}$$.

Part2.3
$$\displaystyle y(x)=c(x)y_1(x)$$ Now we find the first and second derivative of the complete solution using the chain rule, $$\displaystyle y'(x)=c(x)y_1'(x)+c'(x)y_1(x) $$ $$\displaystyle y(x)=c(x)y_1(x) +2c'(x)y_1'(x)+c''(x)y_1(x)$$ And now subsituting these into 5.7.3 yields, $$\displaystyle b_2(cy_1+2c'y_1'+cy_1)+b_1(cy_1'+c'y_1)+b_0y_1=0$$ Distributed to, $$\displaystyle b_2y_1c+b_2cy_1+2b_2c'y_1'+b_1cy_1'+b_1c'y_1+b_0cy_1=0$$ And grouped, $$\displaystyle (b_2y_1)c+(2b_2y_1'+b_1y_1)c'+(b_2y_1+b_1y_1'+b_0y_1)c=0$$ The last term is equivalent to 5.7.3 and equals zero. Now we apply the first homoegeneous solution and the coefficients, $$\displaystyle e^{\lambda x}c''+(2\lambda e^{\lambda x}-2\lambda e^{\lambda x})c'=0$$ $$\displaystyle e^{\lambda x}c''=0$$ Integrating the equation yields, $$\displaystyle c(x)=k_1x+k_2$$ And subsituted back into y(x) gives us,

Part2.4
The 2nd homogenous equation is found by inspected the complete solution. Since we know the the first homogeneous equation is, $$\displaystyle y_1=e^{\lambda x} $$ We can see the 2nd equation must be the other term in the complete solution, $$\displaystyle y_2=xe^{\lambda x} $$

Given
$$\begin{align} & y'+P(x)y=Q(x) \\ & {{y}_{h}}=exp [-\int{P(x)dx}] \\ \end{align}$$  (5.8.1)

Find
The particular solution using Variation of Parameters method.

Solution
I referred to http://en.wikiversity.org/wiki/User:EGM6321.f10.team6.cook/hw5 for the solution of this problem.

Then

Substitution gives

Therefore, the particular solution is

= References =

= Contributing Members =

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title: User:Egm6321.f11.team3/Hwk5


text:
User:Egm6321.f11.team2.Xia/RP5.3

=R*5.3 -Expressions for $$ \displaystyle X(x)$$=

Given
Given L4-ODE-CC,

$$\displaystyle X^{(4)} - K^4 X = 0 $$  (5.3.1)

Assuming

$$\displaystyle X(x) = e^{(rx)} $$  (5.3.2)

Where

$$\displaystyle r_{1,2} = \pm K $$

$$\displaystyle r_{3,4} = \pm i \, K $$  (5.3.3)

Find
In terms of $$ \displaystyle \cos Kx, \sin Kx, \cosh Kx, \sinh kh$$

Find expression for $$ \displaystyle X(x)$$

Solution
According to the (5.3.3),

$$\displaystyle X(x) = C_1 e^{Kx} + C_2 e^{-Kx} + C_3 e^{iKx} + C_4 e^{-iKx} $$  (5.3.4)

Since

$$ \displaystyle e^{iKx} = \cos Kx +i\sin Kx$$

$$ \displaystyle e^{Kx} = \cosh Kx + \sinh Kx$$

Substituting the above relations to (5.3.4)

$$\displaystyle X(x) = C_1 (\cosh Kx \ + \sinh Kx) + C_2 (\cosh Kx \ - \sinh Kx) + C_3 (\cos Kx + i \sin Kx) + C_4 (\cos Kx - i\sin Kx) $$  (5.3.5) $$\displaystyle X(x) = c_1 \cosh Kx \ + c_2 \sinh Kx + c_3 \cos Kx  + c_4 \sin Kx $$  (5.3.6) Where

$$\displaystyle c_1 = C_1 + C_2 \in \mathbb R $$

$$\displaystyle c_2 = C_1 - C_2 \in \mathbb R $$

$$\displaystyle c_3 = C_3 + C_4 \in \mathbb R $$

$$\displaystyle c_4 = i(C_3 - C_4) \in \mathbb R$$

title: User:Egm6321.f11.team2.Xia/RP5.3


text:
User:Chris Todd/Is V-League better than AVL

What is V-League
V-League - Chris Todd

For those who are not familiar with V-League or volleyball in generall there is a short video with some footage of the competition.

An athletes Opinion of V-League
An Interview with Simon Hone

An interview with one of Australia's up and coming volleyball players explores the opinion of athletes surrounding v-league. The focus is on the improvement of volleyball in Australia as V-League is now taking over from AVL. It is run by the clubs and the clubs determine the outcome of the competition. Instead of focussing on the AVF requirements V-League is hoping to improve volleyball as a whole.

Goals of V-League
The V-League has been founded by volleyballers, for volleyballers, in order to:

1/ provide an opportunity to play volleyball at an elite level,

2/ enhance the ability for clubs and teams to develop a strong supporter and funding base

3/ facilitate the development of volleyball players, coaches, officials and volunteers,

4/ help promote the sport of Volleyball in Australia.

(www.v-league.net)

Simons answers
Below are the answers to questions Simon was provided before the interview.

V-League is new in 2011, do you think that V-League is providing an elite level of volleyball in Australia stronger weaker or the same as AVL in the past? V-League is definitely providing the best level of volleyball in Australia. In AVL there was only 1 team per state and often some of the best players didn’t play because of the cost. This year there are 3 states with 2 teams each. We have the talent in Australia we just needed a competition to expose it and V-League gives us that opportunity. This year we have prize money for V-League, which has been attracting the big names as well. Former Australia player and long running captain, Ben Hardy, was lured back to Australia to participate in this lucrative competition. He played over 10 seasons in the best league in Europe. Its good to see Aussies returning to play in the national competition

Do you think that there are going to be more spectators watching V-League games than there were at AVL games? For sure, we have lots of home games because of the V-League format, plus in AVL we often had games where there was nobody from that state playing in the game, so nobody would come to watch, like when we flew to Perth and then would have to play the AIS, another Canberra team. This was frustrating as it meant we had to spend a lot of money on plane flights that we didn’t need. Most of the games I have played in this year have had bustling grandstands and support, which is great even if half of them are supporting other teams.

'''One of the goals of V-League is to develop players, coaches, officials and volunteers. Do you think this is happening, and how do you think this is this different to what was occuring in AVL?''' we often had young kids doing the lines at AVL last year and in crucial games or points, when there was a close call it was frustrating when the obvious decision was not made. V-League has a lot more support from officials to referee, plus with twice as many teams playing there are coaching opportunities as well. I’ve noticed most teams have two or three coaches at their games, compared to AVL last year where teams would only have their head coach travel away to matches.

What do you think will happen next year with V-League? I think it will continue to flourish moving into other states such as Queensland and South Australia. Both these states had good teams in the past but due to costs and inefficient running of AVL they have been absent for the last couple of years, V-League will be their opportunity to rejoin. Sponsors such as Testra have also shown interest in supporting V-League in the future.

Now most importantly, is V-League in 2011 better than AVL was in 2010? Yes for sure, more teams, better competition, and better coaching staff on all levels. Its provides a better base for volleyball in Australia for players, officials and fans. With sponsorships coming in and more teams wanting to join, this league could quickly become a talking point at dinner tables around Australia.

Acknowledgements
www.v-league.net

Karl Pearce Photography for images of Simon Hone

Les Young - Canberra Heat Volleyball for video footage used in the v-league video

Ella-Rose Ward & Simon Hone for their contributions to the interview

VVI - Volleyball Victoria Melbourne-Monash Uni Blues Australian Institute of Sport Canberra Heat Academy UTS-Sydney Uni Volleyball State Volleyball NSW Volleyball Western Australia


text:
User:Egm6321.f11.team2.Xia/RP5.9

=R*5.9 -=

Given
Nonhomogeneous L2-ODE-CC

$$\displaystyle a_2 y'' + a_1 y' +a_0 y = f(t)$$ (5.9.1)

Find
Using special IFM to solve (5.9.1) for general $$\displaystyle f(t)$$

Solution
Suppose the integrating factor has the form $$\begin{align} h(t,y) \end{align}$$

Multiply the integrating factor to the 5.9.1.

$$\displaystyle h(t,y) \cdot \left[ {{a}_{2}}y''+{{a}_{1}}y'+{{a}_{0}}y \right] = h(t,y) \cdot \left[{{f}_{\left( t \right)}}\right] $$  (5.9.2)

rewrite the 5.9.2 to check the exactness condition.

$$\displaystyle F(x,y,y',y) = \underbrace{h(t,y){a}_{2}}_{f(x,y,p)}y+ \underbrace{h(t,y){{a}_{1}}p+ h(t,y) {{a}_{0}}y - h(t,y)\cdot f(t)}_{g(x,y,p)} $$  (5.9.3)

where $$\begin{align} p:= y' \end{align} $$

therefore,

$$\displaystyle f(x,y,p) = h(t,y){a}_{2} $$

$$\displaystyle g(x,y,p) = h(t,y){{a}_{1}}p+ h(t,y) {{a}_{0}}y - h(t,y)\cdot f(t)$$

There are two relations for the 2nd condition of exactness.

$$\displaystyle {f}_{tt}+2p{f}_{ty}+{p}^{2}{f}_{yy}={g}_{tp}+p{g}_{yp}-{g}_{y} $$  (5.9.4)

$$\displaystyle {f}_{tp}+p{f}_{yp}+2{f}_{y}={g}_{pp} $$  (5.9.5)

From (5.9.5), we have

$$\displaystyle {h}_{y}=0$$

 (5.9.6) and from 5.9.4

$$\displaystyle {h}_{tt}{a}_{2}-{h}_{t}{a}_{1}+h{a}_{0}=0 $$  (5.9.7) So

$$\displaystyle {a}_{2}h''-{a}_{1}h'+{a}_{0}h=0 $$  (5.9.8)

thus, assuming $$ \displaystyle h(t)={e}^{\alpha t} $$

From 5.9.8, we have

$$\displaystyle {a}_{2}{\alpha}^{2}-{a}_{1}\alpha +{a}_{0}=0 $$  (5.9.10) then $$\displaystyle \frac{{e}^{-\alpha t}}{2\alpha a_{2}-a_{1}}\left({e}^{(\frac{a_{1}}{a_{2}}-2\alpha)t}\,\int{e}^{-\alpha t}\, f\left(t\right)dt-\int{e}^{\alpha t}\, f\left(t\right)dt\right)$$  ( 5.9.11) If $$\displaystyle a_{1}^{2}-4a_{0}a_{2}\neq0$$, substitute the solution of (5.9.10) to (5.9.11), we have

$$\displaystyle y(t)={C}_{1}\,{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}+{C}_{2}\,{e}^{\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}-{a}_{1}\right)\, t}{2\,{a}_{2}}+}$$

$$\displaystyle \frac{{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}}{\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}}\left({e}^{\frac{\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}\, t}{{a}_{2}}}\,\int{e}^{-\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}-{a}_{1}\right)\, t}{2\,{a}_{2}}}\, f\left(t\right)dt-\int{e}^{\frac{\left(\sqrt{{a}_{1}^{2}-4\,{a}_{0}\,{a}_{2}}+{a}_{1}\right)\, t}{2\,{a}_{2}}}\, f\left(t\right)dt\right)$$

title: User:Egm6321.f11.team2.Xia/RP5.9


text:
File:The Politics of the Olympic Games.ogv

Summary
A presentation on how the sports system in Australia has been impacted by politics in sport at all levels.


text:
User:Alex.Timms/Action Sports and their Rise to Popularity

I have changed my topic

=Introduction= Action or Extreme Sports have in recent years come to the forefront of events across the planet with the help of new tricks, new athletes and new venues. The popularity of these action sports will be discussed in depth in the article, but a great deal has to do with the competition, and having an elite sporting event every year as a pose to every four years keeps participants and spectators on the edge of their seats as developments and changes can happen so quickly. From rock climbing and marathon running the original action sports to today's challenge to the sporting community to devise new competitions and events to challenge the next generation who have access to better technology and more events to prove themselves worthy. With the help of major sponsors and partners to help lift a new event into the spotlight of the media jungle and compete against other more traditional sports that have been around for decades or centuries. And lastly to instill interest and fire in future generations to pick up where the sport left off and push the boundaries and take events to new levels.

=Action Sport History and Development=

Actions Sports are the public’s way of getting away from the safe and controlled environment that they reside in, the need to perform an Action Sport requires a higher level of mental and physical strains than would normally arise from traditional sports, aided by the development of better technologies surrounding the sport. By definition, an Action Sport is a sport that gives the partaker the feeling of an adrenalin rush, the activity being dangerous and hazardous to the health of the participant should things go wrong. For example, skydiving, jumping out of a perfectly good aeroplane, followed by lord know how long of free fall and then a chilled float to earth; or skateboarding, big hills, big ramps, sliding on hand rails; or big wave surfing, watching all the “lunatics” flit around the globe in search of the largest waves to conquer, and bailing will pretty much seal the riders doom. In the late 1970, upon the higher demand of safety surrounding daily life, Action Sports first came to light with the first two sports, rock climbing and marathon running shinning through. Rock climbing since its flare into the limelight of action sports has grown into many different forms of the sport like traditional climbing, bouldering, free climbing just to name a few. The need to jump into your car and cruise to the nearest mountain in order to get your daily dose of adrenalin has been greatly enhanced by the development of cars, and the construction of rock walls in just about every high school on the planet as well as some gyms. Now living in the city a rock climber can hit the gym for a fix! From the 70’s, the equipment surround climbing has also advanced, better ropes, better harnesses and better shoes. All of these aides have helped push rock climbing to where it is today, from a sport where only those with the equipment can pursue to getting the kids together and having a rock climbing birthday party! Marathon running was an easier action sport to get into that rock climbing simply because there was less specialist equipment that was required. Everyone out there can get their hands on a pair of joggers and then just his the pavement for some training before the big race. What was originally a 24.85-mile race in the first Greek Olympic games was later changed to 26.2 for the London Olympic games, all so the event would end right out the front of the Royal Families viewing box in White City Stadium. Rock climbing and marathon running are very similar in their challenges upon the athlete’s body and the pay off is similar, the adrenalin rush of not failing to finish the climb or run.

The action sport calendar includes the ESPN X-Games, of which there is a winter and summer component. The Winter X-Games have become so popular that there is now a North American competition Winter X and a European competition Winter X Europe, both of which are held early on in the year, around January, the 2012 Winter X-Games are to be held in Aspen, Colorado. The summer X-Games are usually held in the USA, in the middle of the year. And for true rock stars like Shaun White who stomps in the summer and winter X-Games, receiving medals for skating and snowboarding !

With the development of better equipments and newer techniques to approach these action sports, there came the development of sporting events, agendas and sponsorships deals to prove competitions of which to stage these new and exciting sports to the public, companies like RedBull and ESPN pursuing these challenges.

=Popularity, Events and Sponsorships=

Red Bull, an energy drink that “Gives you Wings” is a company at the forefront of the action sports world, pursuing new sports and events as well as sponsoring already developed sports. Red Bull own the rights to events like the Red Bull Air Race, an event to stretches the globe to the most glamorous cities in the world, to showcase true flying stars that battle each other and the clock for the championship. Red Bull Elevation BMX, a down hill BMX dirt racing competition of best tricks and most fluid run, with unclean landings and bails having an adverse effect. Street Style is a blend of soccer juggling and break dancing to produce an electrifying one on one knock out competition with one ball and the best tricks and skills excelling the winner to the next round.

Since it was first brainstormed in 1993 by ESPN managements and its creation in 1995, the Summer Extreme Games, now known as X-Games, as well as the Winter X-Games and the Winter X Europe. From the first event in Rhode Island on the east coast of the USA, approximately 198,000 spectators arrive to the venue to watch the event. Over the years, with the aide of the internet the fan base for the X Games has grown to over 240,000 at the venue and many more online or watching it live on ESPN. Through the years, the sports involved with the X-Games have come and gone, as seen entertaining to the public. Inline skating was a sport held at the first Extreme Games, which included events like speed, vert and street inline skating, has in recent years been eliminated due to its decline in popularity along with rock climbing and street luge. But new sports and events have been added to the agenda to fill the void left by the loss of these sport, such as Motor X, BMX and rally, and although some of these sports have been in existence from the first X-Games, new individual events under these sports have risen to the top of the entertainment ladder. For example, under the sport of skateboarding there are six disciplines, vert, park, street, big air best trick and S-K-A-T-E, of which there is the men’s and women’s events of street and vert competitions.

The X Games proved so popular that in 2002 the first X Games Global Championship was created, where five continents competed in winter and summer events in Whistler, Canada and Texas, USA respectively for the ultimate global champions. There were to nations from North America, Canada and USA competing at this event.

To help boost the youth of today into action sports, councils around the world are investing more and more money into skate parks where action sport nuts can descend to a venue and challenge one another into doing their best. Video games have also helped to boost the awesomeness of these action sports, with franchises like the Tony Hawk skater series, the Amped , , SSX and Shaun White Action Sport Franchise series for snowboarding.

=The Future=

Action Sport will always be on the forefront of the sporting world due to the appeal of younger audience and the younger participants in sporting events like Nyjah Huston, who in 2006 became the youngest athlete to compete at the X Games at just 11 years of age. Local community action sport events also help to fuel the fire that drives the passion for the ever-keen young enthusiasts in action sports. From local events where the competitors drop their name down to be able to throw down against one-another to amateur events with athletes, they cannot be pro's, but are allowed to have limited sponsorships, before being noticed and signed with pro teams and invites to pro events.

Skateboarding and BMX riding is local parks has been a common site over recent years and inline skating has severely declined in popularity, but a new face in the action sport world that is beginning to attract attentions and many young participants is riding a scooter. In local parks all across the country young kids are starting to pick up the art of riding a scooter in a bowl or half pipe, and with the development of new equipment like wider handlebars, better wheels and frames the pursuit of new tricks and ways of riding these scooters is pushed to the boundary. Accompanying the scooter is the ripstik, of which I’m sure almost everyone has seen one, a unique skate-like tool with a single wheel under each platform, and a strong spring that separates each platform driving the user to “wiggle” the board in order to get movement and momentum. These two new sports have risen over the years and with continued participation by the public and openings in local sporting events, these sports will most likely find their way into competitions around the globe like the X-Games. For Example, the newly refurbished Belconnen Skate Park down on the waterfront on Ginninderra Lake, which attracts the youth of today to challenge each other with tricks and skills and practice, all pointed towards the chance to demonstrate their own unique style upon competition day!

Kite Surfing has been a sport since French brothers Dominique and Bruno Legaignoux who were sailing nuts sat down and pondered the sport from 1984 until its development in 1993-94. Its true potential was unleashed off the coast of Hawaii by all round action sport junkie Laird Hamilton and drove this sport to gain keen new enthusiasts. With the development of the kite, new ideas surrounding the use of the kite have emerged, like snow kiting which in it has kite skiing and kite snowboarding. With the Winter X-Games constantly gaining interest, these winter kiting sports will attract pros that are out for a new way to get their adrenalin fix.

Ever heard of the Ultimate Boarder competition? A unique competition that incorporates snowboarding, skateboarding and surf boarding into three events by a single individual, “the only board sport triathlon in the world!”. This event was organised to first be held in 2008, over a week from the 13th to the 19th April in California in Squaw Valley for the snow event and Ventura for the skate and surf section. This event was created due to the continued interest that these three sports have attained over their years at the forefront of the action-sporting world. And in 2012 over the week of the 17th to the 24th of March the competition will see the fifth year of the event, with the setting to be held at Cardiff, Del Mar and Bear Mountain for the surfing, skating and snowboarding events respectively in California.

These newly developed sports in the action sport world continue to provide a platform for new ideas and more new events and competitions to bring together the best of the best. The X-Games will see new line-ups in competitors as well as events and locations as deals run out with Aspen and California and more and more international superstars rise to the challenge. But nothing is as important as the build up for local talent and interest to further these challenging and technical sports.


text:
User:Egm6321.f11.team1.zheng.zx/personal place/HW5

= R*5.3 Find the expressions for X(x)in terms of $$\cos Kx,\ \sin Kx,\ \cosh Kx,\ \sinh Kx$$ =

Given
$$ \displaystyle X^{(4)}-K^4X=0 $$

$$ \displaystyle K:= \frac{\omega^2m}{EI} $$

Assume:

$$ \displaystyle X_{(x)}=e^{(rx)} $$

characteristic equation:

$$ \displaystyle r^4=K^4 $$

then:

$$ \displaystyle r_1=K,\ r_2=-K,\ r_3=iK,\ r_4=-iK,\ i:=\sqrt{-1} $$

Find
the expressions for $$X_{(x)}$$ in terms of:

$$ \displaystyle \cos Kx, \ \sin Kx,\ \cosh Kx,\ \sinh Kx $$

Solution

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

\displaystyle X_{(x)}=c_1e^{Kx}+c_2e^{-Kx}+c_3e^{iKx}+c_4e^{-iKx} $$ $$
 * $$\displaystyle (Equation\;5.3.1 )
 * }
 * }

$$ \displaystyle \begin{align} A_1: &=c_1e^{Kx}+c_2e^{-Kx}\ \in \mathbb{R} \\ &= c_1(\frac{e^{Kx}+e^{-Kx}}{2}+\frac{e^{Kx}-e^{-Kx}}{2}) + c_2(\frac{e^{Kx}+e^{-Kx}}{2}-\frac{e^{Kx}-e^{-Kx}}{2})\\ &= c_1(\cosh Kx+ \sinh Kx) + c_2(\cosh Kx- \sinh Kx)\\ &= (c_1+c_2)\cosh Kx + (c_1-c_2)\sinh Kx \end{align} $$

We can rewrite it to:
 * {| style="width:100%" border="0" align="left"

\displaystyle A_1=b_1\cosh Kx + b_2\sinh Kx $$ $$
 * $$\displaystyle (Equation\;5.3.2 )
 * }
 * }

$$ \displaystyle \begin{align} A_2: &=c_3e^{iKx}+c_4e^{-iKx} \\ &=c_3(\cos Kx +i\sin Kx)+c_4(\cos Kx -i\sin Kx) \\ &= (c_3+c_4)\cos Kx + (c_3-c_4)i\sin Kx \ \in \mathbb{R} \end{align} $$

We can rewrite it to:
 * {| style="width:100%" border="0" align="left"

\displaystyle A_2=b_3\cos Kx- b_4\sin Kx $$ $$
 * $$\displaystyle (Equation\;5.3.3 )
 * }
 * }

So, add up the two terms from equation 5.3.2 and equation 5.3.3:
 * {| style="width:100%" border="0" align="left"

\displaystyle X_{x}=A_1+A_2=b_1\cosh Kx + b_2\sinh Kx + b_3\cos Kx- b_4\sin Kx $$ $$
 * $$\displaystyle (Equation\;5.3.4 )
 * }
 * }

Where:

$$ \displaystyle b_1=c_1+c_2,\ b_2=c_1-c_2,\ b_3=c_3+c_4,\ b_4=-i(c_3-c_4) $$

= R5.6 Show that the trial solution in Method 2 is equivalent to the combined trial solutions in Method 1 =

Given
Show the equivalence of two methods using Euler L3-ODE-VC as an example:
 * {| style="width:100%" border="0" align="left"

\displaystyle a_3x^3y^{(3)}+a_2x^2y^{(2)}+a_1xy'+a_0y=0,\ y'=y^{(1)},\ y=y^{(0)} $$ $$
 * $$\displaystyle (Equation\;5.6.1)
 * }
 * }

Find
That the trial solution in Method 2 is equivalent to the combined trial solutions in Method 1.

Solution
1.Method I Stage 1,

$$ \displaystyle x=e^t $$

and we have derived in class that:
 * {| style="width:100%" border="0" align="left"

\displaystyle y'=e^{-t}y_t $$ $$
 * $$\displaystyle (Equation\; 5.6.2)
 * }
 * }

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

\displaystyle y^{(2)}=e^{-2t}(y_{tt}-y_t) $$ $$
 * $$\displaystyle (Equation\; 5.6.3)
 * }
 * }

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

\displaystyle y^{(3)}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t) $$ $$
 * $$\displaystyle (Equation\; 5.6.4)
 * }
 * }

plug equations 2,3,4 into 1, we obtain:
 * {| style="width:100%" border="0" align="left"

\displaystyle a_3e^{3t}\cdot e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+a_2e^{2t}\cdot e^{-2t}(y_{tt}-y_t)+a_1e^t\cdot e^{-t}y_t+a_0y=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.5)
 * }
 * }

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

\displaystyle a_3y_{ttt}+(a_2-3a_3)y_{tt}+(2a_3-a_2+a_1)y_t+a_0y=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.6)
 * }
 * }

Stage 2,

$$ \displaystyle y=e^{rt},\ r=constant $$

substitute into equation 5.6.6, we get:
 * {| style="width:100%" border="0" align="left"

\displaystyle a_3r^3e^{rt}+(a_2-3a_3)r^2e^{rt}+(a_1-a_2+2a_3)re^{rt}+a_0e^{rt}=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.7)
 * }
 * }

Finally:

2.Method II

$$ \displaystyle y=x^r,\ r=constant $$

substitute into equation 1:
 * {| style="width:100%" border="0" align="left"

\displaystyle a_3x^3r(r-1)(r-2)x^{r-3}+a_2x^2r(r-1)x^{r-2}+a_1xrx^{r-1}+a_0x^r=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.8)
 * }
 * }

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

\displaystyle a_3r(r-1)(r-2)x^r+a_2r(r-1)x^r+a_1rx^r+a_0x^r=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.9)
 * }
 * }

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

\displaystyle a_3(r^3-3r^2+2r)+a_2(r^2-r)+a_1r+a_0=0 $$ $$
 * $$\displaystyle (Equation\; 5.6.10)
 * }
 * }

And, finally:

So, using whatever method I or method II, we can derive the exact same characteristic equation

thus, they are just the same.

= References =


text: File:Motivation and Emotion - Lecture 10a - Unconscious motivation 2011 6slidesperpage.pdf


text: File:Motivation and Emotion - Lecture 10a - Unconscious motivation 2011 3slidesperpage.pdf


text: File:FifaPresentationReal.ogg

title: File:FifaPresentationReal.ogg


text:
User talk:Malia02

ok i want to be brainy

title: User talk:Malia02


text: File:Motivation and Emotion - Lecture 10b - How to make a multimedia recording 2011 3slidesperpage.pdf


text:
User:Malia02

hi my real name is lesieli and my last name is tuitavuki.My favriote colurs are green, pink ,blue ,and purple

title: User:Malia02


text: File:Motivation and Emotion - Lecture 10b - How to make a multimedia recording 2011 6slidesperpage.pdf


text:
User:Egm6321.f11.team1.steele.m/Report5

Homework 5 Problems Assigned to Manuel Steele (in pairs)

R5.6 Question from meeting 31, page 31-6: Given the trial solution (5) from p. 30-4 in Method 2 $$y = x^{r},\ r=\text{constant}$$ $$\displaystyle (Equation\;5.6.1)$$

Find that the trial solution is equivalent to the combined trial solutions (3)-(4) p.30-4 in Method 1.

$$x=e^t$$ $$\displaystyle (Equation\;5.6.2)$$ $$y=e^{rt},\ r=\text{constant}$$ $$\displaystyle (Equation\;5.6.3)$$

Solution
Equations 5.6.2 and 5.6.3 are used in method 1 to solve the Euler Ln-ODE-VC below. Equation 5.6.1 represents method 2 for solving the same equation.

$$a_nx^ny^{(n)} + a_{n-1}x^{(n-1)}y^{(n-1)} + \cdots + a_1xy' + a_0y = 0$$ $$\displaystyle (Equation\;5.6.4)$$ Note that $$y' = y^{(1)}$$ and $$y = y^{(0)}$$.

'''Work on a solution is still in progress as of 2:00 AM CST (11/1/11). Team members can update as needed.'''

R5.8 Question from meeting 32, page 32-2: Given: recall p. 12-2, R*2.17 $$y' + a_0(x)y = 0$$ $$\displaystyle (Equation\;5.8.1)$$

Find: use the same idea of variation of constants (parameters) to find the particular solution $$Y_P(x)$$ after knowing the homogeneous solution $$Y_H(x)$$, i.e., let y(x) = A(x)$$y_H(x)$$, with A(x) being the unknown to be found.

From lecture notes [[media:pea1.f11.mtg32.djvu|Mtg 32]]

Solution
The key is to start with equation (3) on p. 11-3 below. $$\underbrace{\color{blue}{1}}_{\color{blue}{a_1(x)}} \cdot \underbrace{y'}_{\color{blue}{y^{(1)}}} + \underbrace{\frac{Q(x)}{P(x)}}_{\color{blue}{a_0(x)}}y^{\color{blue}{(0)}}=\underbrace{\frac{R(x)}{P(x)}}_{\color{blue}{b(x)}}$$ $$\displaystyle (Equation\;5.8.2)$$ The equation is an L1-ODE-VC. Note that the non-homogeneous equation above utilizes the following integration factor for the solution of y. But we must find the homogenous counterpart to this solution

$$y(x) = \frac{1}{h(x)} \left[\int^x h(s)b(s)ds + k_2 \right]$$ $$\displaystyle (Equation\;5.8.4)$$

The homogeneous solution of the L1-ODE-VC was derived in R2.17 by team 1. The solution summary is repeated for clarity. Note that team 1 has already submitted this solution for R2.17 back in Report 2. The steps are from a simple cross-reference used to define the homogeneous solution of 5.8.2  $$\displaystyle \frac{y'}{y} = -{a}_{0} $$ $$\displaystyle (Eq 2.17.1) $$

Integration of both sides of this equation gives

 $$ \ln \left( y\right)= -\int^{x} {a}_{0} \left( s\right)ds + C$$
$$\displaystyle (Eq 2.17.2)$$

Solving for $$y$$ gives

$$ y_H= \exp\left[ -\int^{x} {a}_{0} \left( s\right)ds + C \right]= A\exp\left[ -\int^{x} {a}_{0} \left( s\right)ds\right]$$

Now, the homogenous solution is known from 2.17. The next step is to find the complete solution for the homogenous equation- $$y(x) = A(x)y_H(x)$$. The technique demonstrated in the lecture is to substitute the chosen form of y(x) into the original Euler equation to find the complete solution below. This is a classic application of the variation of parameters: (1) find homogenous solution, (2) find complete solution of Euler form - the L1-ODE-VC, and (3) find the particular solution by using (1) and (2).

$$y' + a_0(x)y = b(x) = 0$$
$$\displaystyle (Equation\;5.8.4)$$ The derivative simplifies the Euler equation. First, it must be derived.
$$y = A(x)y_H$$
$$y' = A_xy_H + Ay_{Hx}$$ The values can be substituted into the L1-ODE-VC to give the form below.
$$A_xy_H + Ay_{Hx} + a_0(x)Ay_H = 0$$ The equation can be rerranged to give a simple form for integration by natural logarithms and subsequent exponentiation for the complete solution of the L1-ODE-VC.
$$A_xy_H = -A(y_{Hx} + a_0(x)y_H$$
$$\frac{A_x}{A} = -\frac{y_{Hx} + a_0(x)y_H}{y_H} = -\frac{y_{Hx}}{y_H} + a_0(x)$$ The rearranged equation can be integrated by standard application of natural logarithms.
$$\int^x \frac{A_x}{A} = -\int^x \frac{y_{Hx}}{y_H} + \int^x a_0(s)ds$$

Use the identity for logarithms: $$\ln \frac{a}{b} = \ln{a} - \ln{b}$$, which gives $$\ln \frac{A(x)}{y_H} = \int^x a_0(s)ds$$. The equation then gives the next output form.
$$A(x) = y_H{\exp \int^x a_0(s)ds } = {y_H}^2$$
$$\displaystyle (Equation\;5.8.5)$$ The final complete solution is y(x) = A(x)$$y_H$$, which gives
$$y(x) = {y_H}^3$$
$$\displaystyle (Equation\;5.8.6)$$ The particular solution is simply found by substration the homogenous solution from the complete solution.
$$y_p = y(x) - y_H = {y_H}^3 - y_H$$
$$\displaystyle (Equation\;5.8.7)$$ References for R5.6 and R*5.8 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 11) [[media:pea1.f11.mtg11.djvu|Mtg 11]] 13 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 12) [[media:pea1.f11.mtg12.djvu|Mtg 12]] 15 Sep 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 30) [[media:pea1.f11.mtg30.djvu|Mtg 30]] 25 Oct 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 31) [[media:pea1.f11.mtg31.djvu|Mtg 31]] 25 Oct 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 32) [[media:pea1.f11.mtg32.djvu|Mtg 32]] 27 Oct 2011.

title: User:Egm6321.f11.team1.steele.m/Report5