text:
User:E.herbert/E-portfolio

Introduction to Motivation and Emotion
Motivation and emotion is a topic I was keen to dive into this semester, hoping to decipher big questions such as 'why do people do what they do'? and 'why do people want what they want?’ These questions may be simple for the individual participating in the acts to understand, yet how do we funnel the strengths and weaknesses of motivation to benefit individuals on a larger scale? In discovering these questions, in the future, I hope to not only motivate myself towards beneficial goals such as, healthy eating and exercise but also share my discoveries with others. Motivation = Energy + Direction What does this equation mean? Firstly to begin to explain motivation we must recognise the functioning components of energy and direction, and apply them to behaviour. The lecture this week identified motivation to be processes that give behaviour energy and direction. Energy is created through strong, intense and persistent behaviours and forms motivation by combining with direction, behaviours aimed toward achieving a particular purpose or goal. The four sources for motivation (where it gains its energy and direction, strength and purpose) are needs, cognitions, emotions and external events. We can measure motivation and its level of robustness through how it is expressed. These being measurements of behaviour, engagement, brain and physiology activations and self-report.

The true depth of a person's behaviour cannot be explained without the questions addressing the core problems of motivation. Such core problems present themselves through the following questions: These questions can be addressed through behavioural expressions of motivation known as behavioural engagement which consists of; attention, effort, latency, persistence, choice, probability of response, facial expressions and bodily gestures. Behavioural engagement is just one of the four inter-related aspects of engagement, with the three others being emotional engagement, cognitive engagement and voice.
 * Whats starts behaviour?
 * How is behaviour sustained over time?
 * Why is behaviour directed towards some ends but away from others?
 * Why does behaviour change its direction?
 * Why does behaviour stop?

Motivation can come in waves and is not always directed towards a positive goal, so what keeps us motivated? What motivates the individual towards activities which result in a negative outcomes, such as self-sabotage? I hope to take what i know about human motivation and emotion and apply it to practical problems to propose solutions and/or interventions. Hopefully throughout the semester some of these questions can be addressed and possibly answers found. I am looking forward to completing my own textbook chapter as well being able to read other participants textbooks chapters and e-portfolios. I am keen to experience the learning benefits of each of us addressing different aspects of motivation and emotion with our inputs being available to everyone online, making it possible to gain a wider perspective of the subject and be active in our own learning.

How to use Wikiversity
Starting my Wikiversity adventure was a daunting thought, I have never used any form of online bloging system nor attempted to navigate or java script any webpage or social site. Looking at the unit outline for Motivation and Emotion before the start of semester was stress provoking, yet i was excited to jump into the assignment tasks being the first participant to sign up as Wikiversity user for this unit. The tasks show creative thought about extending student learning past lectures and tutorials through utilizing the internet, allowing online interaction between ourselves as participants and all fellow Wikiversity users. Lecture two could not have come sooner, as I was already underway with formatting the beginning of my E-portfolio and in desperate need of any help that could be offered. Since the lecture I have been practicing my Wikiversity editing skills such as editing and saving, watching pages, fiddling with settings and attempting tasks such as headings and tables. I have even gone as far as searching how to edit wiki sites through the internet and encourage others to utilize pages such within Wikiversity such as How to edit a page which offers a great Cheat sheet. Since the start of this adventure I have been grateful for the easy editing buttons found at the top of editing boxes which allow me to use Bold, italics and underline. If all attempts made by me to edit freely fail I shall revert back to using these simple yet extremely helpful methods of beautifying my page.

 The first tutorial of Motivation and Emotion was, in simple terms, fun. There is nothing more daunting than walking into the first tutorial of the semester and not really knowing others in the class. I never would have thought that arranging the class in order of thumb-size would be a great ice-breaker and set the mood for a positive environment for learning and participation. Showing that everyone in the tutorial was similar in the following categories allowed participants to identify with one another whilst creating new friends; Small groups were formed for future collaborative peer review, which would see peers contributing to one another's textbook chapters and E-portfolio. Forming groups allowed for discussion sounding the definitions of Motivation and Emotion, the two concepts which we will be exploring with great detail this semester. A word which was commonly used in definitions put forward for motivation was driving force, as everyone showed belief that motivation is a internal or external driving force which encourages an individual or group to act which enforced text book definitions learnt from earlier lectures;
 * how long they have been studying at the University of Canberra,
 * favourite fast food,
 * voting preference this past election, and
 * how they were feeling emotionally that day.

Motivation is a "process that give behaviour energy and direction", (Reeve, 2009).

It is "needs or desires that energise and direct behaviour", (Gerrig et al., 2008).

Emotion was a harder concept for the tutorial group to put into words, everyone can describe an emotion but when it comes to defining the overall meaning of the word it becomes difficult. A commonality between group definitions was the term state of mind, as groups flagged emotion as being a state of mind which is determined by our thoughts and feelings.

After discussion of what we as a group believed defined these words, we explored areas which fell under their umbrellas such as intrinsic and extrinsic motivation in the workplace, self-control, adultery and the art of hiding/masking emotions. I am mostly interested in the drive to partake in regular/daily exercise, and methods in which to increase people's motivation to exercise. How do we tunnel the energy of those who love and participate in exercise regularly to those who don't?

'The best motivation is self-motivation. The guy says, “I wish someone would come by and turn me on.” What if they don’t show up? You’ve got to have a better plan for your life.' – [| Jim Rohn]

References:

Reeve, J. (2009). Understanding motivation and emotion (5th ed.). Hoboken, NJ: Wiley.

Gerrig, R. J., Zimbardo, P. G., Campbell, A. J., Cumming, S. R., & Wilkes, F. J. (2008). Psychology and Life (Australian edition). Sydney: Pearson Education Australia

Brain & Physiological needs
In week three I found out the motived emotional brain consists of three parts; the thinking brain which focuses on the intellectual and cognitive functions, the motivated brain and the emotional brain. These three brain functions break down behaviours by asking; what is the task the individual is completing? Do they want to take part in the task and what is their mood while preforming the task? Such insights allow critical thinking into the motivation behind behaviours thus creating tools to pinpoint sources of motivation.

A few of the brain structure's which are associated with motivational and emotional states are;
 * Hypothalamus: Deals with positive feelings surrounding drinking, eating and mating
 * Medial Forebrain Bundle: pleasure and reinforcement
 * Orbitofrontal Cortext: Making choices and learning the value of those choices
 * Septal area: Deals with positive feelings surrounding sociability and sexuality

Four motivationally relevant neurotransmitter pathways which allow these parts of the brain to communicate are dopamine, serotonin, norepinephrine and endorphins.

Physiological needs of an individual provide the motivation to gain the essentials to sustain life and well-being. Motivation provides the kick start to fulfill basic psychological needs before any damage is done, pushing the individual to act. These physiological needs have been broken up into three main areas, physiological needs, psychological needs and social needs. Abraham Maslow provides a model showing these areas in order of importance to survival, breaking down each motivational need into sections of most important to least important.

Maslow's hierarchy of needs

Maslow's hierarchy of needs (1943,1954) is psychological theory developed by Abraham Maslow which lists human physiological and psychological needs within a hierarchy that consists of five levels. Labeled the most important to the least is physiological, safety, love/belonging, esteem and self-actualization. Physiological needs are placed at the bottom of the pyramid and are inherent within the workings of human biological systems, these are the most basic physical requirements such as the need for sleep, food, water and warmth. As the pyramid progresses the motivational needs become less physiological and more psychological and social. Showing an individual's motivation for love, friendship and intimacy being more important than those found higher up on the scale such as the need for self-esteem and feelings of confidence and achievement.

Henry Murray (1893 - 1988) was another to look at needs in regards to motivation. Breaking up needs into two groups, viscerogenic needs and psychogenic needs, Murray looked at needs in a more modest structure to Maslow's. Viscerogenic needs were recognised by Murray as being things that were biologically based and psychogenic needs were derived from our biological needs or inherit of our psychological nature. Murray looked into needs from a personality approach, stating the secondary needs drive our personality (Interactive and Digital Media Institute, 2008).
 * Viscerogenic needs:
 * food, water, air, sex avoidance of pain
 * Psychogenic needs:
 * achievement, recognition, acquisition
 * dominance, aggression, autonomy
 * affiliation, rejection
 * nurturance, play, cognizance

References:

[http://www.silentblade.com/presentations/BBostan-2008-2.pdf Interactive and Digital Media Institute. (2008). Player Motivations: A Psychological Perspective. "National University of Singapore", Accessed September 25, 2010 from Google Scholar.]

[http://psg-fl.com/downloads/Koltko-Rivera%202006%20(RGP)%20Maslow%20Self-Transcendence.pdf Koltko-Rivera, M. (2006). Rediscovering the later version of Maslow's hierarchy of needs: Self-transcendence and opportunities for theory, research, and unification. Review of General Psychology, 10(4), 302-317, Accessed September 9, 2010 from PsycARTICLES, EBSCOhost.]

Maslow, A. H. (1969). The farther reaches of human nature. Journal of Transpersonal Psychology, 1(1), 1–9.

Reeve, J. (2009). Understanding motivation and emotion (5th ed.). Hoboken, NJ: Wiley.

Psychological & Social needs
This week the readings included chapter 6 and 7 of Reeve (2009), covering psychological needs and social needs.

Psychological needs, as described in the lecture this week, are an inherent source of motivation that generates the desire to interact with the environment so as to advance personal growth, social development, and psychological well-being.

Self-determination Theory
The self-determination theory, as described by Reeve, approaches different types of motivation focusing on all intensities of self-determination from amotivation to intrinsic motivation. Edward Deci and Richard Ryan the initiators of this theory describe the base of this theory to be one surrounding human motivation, development and wellness with focus on the type of motivation rather than the level. Behaviour is self-determined (antonomous) when our interests play apart in our decision-making process.
 * “Self-determination theory examines people's life goals or aspirations, showing differential relations of intrinsic versus extrinsic life goals to performance and psychological health”, (Deci and Ryan, 2008).

Four essential ways of supporting autonomy (self-determination)
1. Nurtures inner motivational resources

Antonomy-supportive motivators include finding ways to allow others to express interests, preferences, and competences.

Controlling motivators include relying on extrinsic motivators.

2. Relies on Informational Language

Autonomy-supportive motivators include treating poor performance and inappropriate behaviour as motivational problems.

Controlling motivators include using a communication style that is rigid and pressuring.

3. Promotes valuing

Autonomy-supportive motivators involves communicating the value of engaging in uninteresting tasks.

Controlling motivators involves not communicating the importance "Do it because i told you too!"

4. Acknowledges and accepts negative feedback

Autonomy-supportive motivators includes working with others to solve underlying causes of negative affects and resistance. Controlling motivators includes ignoring other peoples expressions of negative affect and resistance.

Tutorial two allowed its participants to share and discuss the topics and plans of their assessment for Motivation and Emotion. Group work led to class discussion in relation to what was needed within a text book chapter contents page and how many sub-topics would be appropriate. It was established that contents pages are helpful in planning, researching and overall structure of the finished product but are not necessarily apart of the first stages of planning. Participants in the tutorial group are scattered through all different stages of planning, as some have completed their contents pages and begun heavy research and others are still stumped on which topic to choose. My approach to this task has begun as i have chosen my topic from the text book table of contents and started the research process. The topic i have chosen is Arousal. I have decided the best method for me is to begin researching before i make any decisions on sub-topics.

Maslow's hierarchy of needs was a starting point in this tutorial for looking at conceptualisations of needs. I have discussed the hierarchy within week 3 of my e-portfolio and have further investigated the five conceptualisations of needs making reference to Maslow's later beliefs of an added conceptualisation, totaling six. In order of most important needs the conceptualisations are: physiological, safety, love/belonging, esteem, self-actualization and self-transcendence.

Brain structure and its functions in motivation were approached in this tutorial, allowing participants to identify specific neural structures and their motivational function. Looking at the physiological aspects of motivation not only facilitated our learning but allowed us to look at the topics we had chosen to approach in the text book chapter assessment item, thus showing us further ways to research and approach our topics from a physiological perspective. These specific neural structures were identified through a cross section of the brain showing the anatomic position of the key brain structures involved in motivation and emotion. The Hypothalamus was identified as being associated with pleasurable feelings surrounding feeding, drinking and mating. The Medial Forebrain Bundle has part in pleasure and reinforcement, with the Orbitofronal Cortex allows learning the incentive value of events and making choices. Another area which is also known for its role in making choices is the Anterior Cingulate Cortex, it is also responsible for mood and volition. The Septal area is known as the pleasure centre associated with sociability and sexuality and the Cerebral Cortex (frontal lobes) have a role in making plans, setting goals and formulating intentions.

Neurotransmitters like neural structures. play an important role motivation through the form of key hormones.
 * Dopamine generates the good feelings associated with rewards,
 * Serotonin has an influence over mood and emotion,
 * Norepinephrin is responsible for regulating arousal and alertness - an interesting piece of information i found, that will help me in the research process of my text book chapter, and
 * Endorphins allow the generations of good feelings which counter negative feelings such as pain, anxiety and fear.

Close attention was paid to the physiological and social needs attributed to motivation and emotion in the tutorial this week with notes on revision of previous lectures given in the form of a contents page. Specific key points were autonomy, competence, relatedness, acquired needs, achievement, power, affiliation and intimacy. The two topics, physiological and social needs, were discussed looking at the social contexts that support psychological needs and methods to integrate classic and contemporary approaches to achieve motivation.

References:

[http://www.psicologia-uniroma4.it/LS/organizzazione/materiale/cap-49-3-182%5B1%5D.pdf Deci, E. L. & Ryan, R. M. (2008). Self-Determination Theory: A Macrotheory of Human Motivation, Development, and Health, Canadian Psychology, 49, 182-185. doi: 10.1037/a0012801]

Reeve, J. (2009). Understanding motivation and emotion (5th ed.). Hoboken, NJ: Wiley.

Intrinsic & Extrinsic Motivation & Goal settting
Intrinsic motivation, in my understanding is an inherent desire towards something that you are interested in with no influence from an external force. Intrinsic motivation causes you to do things because the reward is something within yourself. The origins of intrinsic motivation are looked at within our unit textbook and this weeks lecture. The origins come from our psychological need satisfaction and its three components, autonomy, competence and relatedness. These three components approach to need for support from environment and relationships. There are specific performance and psychological benefits to be found within intrinsic motivation these being; Learning the difference between intrinsic motivation and extrinsic motivation is important, I believe it will give people the tools to swap extrinsic motivation with intrinsic motivation and move towards a greater feeling of self, subjective vitality and self-esteem. Through my understanding from lectures and the textbook I define extrinsic motivation through behaviours that you undertake because of some external event, behaviour that is contingent upon the external response you’ll get back such as money. Three things that are recognised by Reeve (2009) that externally regulate motivation are incentives, consequences and rewards. An extrinsic reward can service positive emotions within individuals and facilitate certain behaviours because of its ability to give individuals an opportunity for personal gain. Yet rewards may or may not act as a reinforcer as some people may not value the reward being offered, the reward will only serve as a motivator to someone who wants it. Individuals who chase awards for their behaviour can experience experience unexpected, unintended, and adverse effects on their intrinsic motivation, high-quality learning and autonomous self-regulation. Sometimes rewards that motivate engagement behaviour within and activity can cause unexpected primary effects such as interferer with the quality and processes of learning, undermining the person’s intrinsic motivation, and interferers with the capacity for autonomous self-regulation. Adding to a person's motivation reduces the intrinsic motivation, therefore having the opposite affect from the desired. Using a reward to engage someone in activity is a popular method and seems tempting to use but when you take into account the side effects, such as affects on psychological emotions and traumatic baggage it just is not worth it. Overuse of external motivators leads to just a behaviour, not a cognition or emotion which becomes reliant on external contingencies for motivation. Like rewards punishments are also popular to use yet also supporting undesirable side effects such as; There does appear to be a lot of negatives associated with extrinsic motivation yet for small tasks it can be extremely useful making otherwise boring and uninteresting tasks seem suddenly worth pursuing. Reeve (2009) shares four reasons not to use extrinsic motivation even in the most un-motivating circumstances
 * Persistence – The amount of persistence a person has with a task can be attributed to the their level on intrinsic motivation. The higher the level of their intrinsic motivation the greater their persistence.
 * Creativity – Great levels of creativity come from how the individual approaches the task, if they are intrinsically motivated (experience interest, enjoyment and satisfaction from the work itself) they tend to show a higher level of creativeness.
 * Conceptual Understanding/ High-Quality Learning – Allows in a conceptual way with flexible thinking and active information processing.
 * Optimal Functioning and Well-Being – People who experience intrinsic motivation are said to benefit from greater self-actualization, subjective vitality and self-esteem. Experiencing less anxiety and depression.
 * Incentives - environmental events that attract or repel a person toward or away from initiating a particular course of action
 * Consequences – recognised through positive and negative reinforcers of behaviour and punishments.
 * Rewards – An offering from one person given to another person in exchange from his or her service or achievement.
 * Negative emotionality such as crying, screaming and feeling afraid,
 * Impaired relationship between the punisher and punishee, and
 * Negative modeling for the punishee showing them how to cope with undesirable behaviour in others.
 * Extrinsic motivators still undermine the quality of performance and interfere with the process of learning,
 * Using rewards distracts attention away from asking the hard question of why another person is being asked to do an uninteresting task in the first place,
 * There are better ways to encourage participation than extrinsic bribery, and
 * Extrinsic motivators still undermine the individual’s long-term capacity for autonomous self-regualtion.

Cognitive evaluation theory
The cognitive evaluation theory predicts the effects of an extrinsic event on motivation. The theory takes the two of the psychological needs I wrote about earlier in this week’s refection, autonomy and competence and looks at how they might be effected. External events have two functions being control behaviour and inform competence, the cognitive evaluation theory states that which ever function is more salient determines how the external event will affect intrinsic and extrinsic motivation. Not all extrinsic motivation is understanding of intrinsic motivation and example of this being if it were an informational function. Events that promote greater perceived competence will enhance intrinsic motivation.

The cognitive evaluation theory gives the three types of motivation, amotivation, extrinsic motivation, and intrinsic motivation their own regulatory styles.

 * Amotivation: Non-Regulation
 * Extrinsic motivation: External Regulation, Introjected Regulation, Identified Regulation and Integrated Regulation
 * Intrinsic motivation: Intrinsic Regulation

References:

Ryan, R. M. and Deci, E. L. (2000). Intrinsic and extrinsic motivations: Classic deﬁnitions and new directions. Contemporary Educational Psychology, 25, 54–67.

Patall, E. A., Cooper, H., & Robinson, J. C. (2008). The Effects of Choice on Intrinsic Motivation and Related Outcomes: A Meta-Analysis of Research Findings. Psychological Bulletin, 134, 270-300. doi: 10.1037/0033-2909.134.2.270

Reeve, J. (2009). Understanding motivation and emotion (5th ed.). Hoboken, NJ: Wiley.

Control beliefs & The self
Motivation to exercise personal control comes from initial assumptions and understandings such as the desire to control the environment in order to make positive outcomes more likely than negative outcomes, and exercising personal control regularly strengthens a person’s belief that they have the power to influence outcomes. The extent to which a person’s will exercise personal control depends on their own beliefs and expectations of being able to do so.
 * Types of expectancy:

Efficacy Expectations: The expectation of being able to perform behaviours needed to cope effectively with the situation. Asking “Can I do it?”

Outcome Expectation: The expectation that behaviour will produce positive outcomes, or prevent negative ones. Asking “Will what I do work?”

Self-efficacy is a belief by an individual in their ability to succeed in particular situations. Such beliefs control how and individual approaches tasks, challenges, and their own goals. People who have high self-efficacy are more likely to see a hard task as something to be mastered, whereas someone with low self-efficacy will look at the same task and think it is something to be avoided. Sources of self-efficacy which contribute towards self-efficacy include their personal behaviour history, vicarious experience (modelling), verbal persuasion (pep talk), and physiological activity. Sources of self-efficacy combined with effects of self-efficacy are responsible for the extent of self-efficacy experienced by an individual. Effects of self-efficacy include choice, effort and persistence, thinking and decision making, and emotional reactions such as stress and anxiety.

Empowerment If an individual possess the knowledge, skills and beliefs that allow them to exert control over their lives they are believed to be empowered. Positive self-efficacy encompasses personal empowerment which can be taught through master modelling programs.

Mastery beliefs are the perceived control one has over attaining desirable outcome and preventing aversive ones. The extent to which they perceive control will determine the strength of the positive outcomes. “To what extent do you believe your actions control the environment?”

 * Learned helplessness

When person beliefs the outcomes in their life are uncontrollable they develop learned helplessness – outcomes that happen to me are uncontrollable and cannot be changed through my behaviour. Three components of learned helplessness are;

Contingency – the objective relationship between a person’s behaviour and the environment’s outcomes

Cognition – subjective personal control beliefs, biases, attributions, and expectancies

Behaviour – listless, demoralised coping behaviour

Intrinsic and extrinsic motivation is a topic this semester which I have found really interesting. In this week tutorial we were asked to look out our own motivations for going to university. The answers my group and I came up with were popular across the board, with the four groups within the tutorial expressing the same motivations.

The question which were asked were, '''Why are you at University? Why do students go to university?'''

BRAINSTORM

My group responded with:

To expand knowledge/ learning/ skills

To further our careers/ gain skills for a career

Rejected the alternative of getting a job straight after school

Social influences

Two popular motivators which we didn't discuss were:

Social opportunities - to meet people, make and explore friendships, enjoy social environment

Altrism - to become better able to help people, help society, help the planet

These responses made me ask the questions; are popular motivations the result of social pressures or modeling? Do we do it because it is expected of us? Do we don't know any different? I believe it is really interesting to acknowledge what motivates people and compare it to your own motivations.

Functionalist perspective to motivation:

A good match between motivations and outcomes leads to satisfaction and retention (or intention to continue), whereas a poor match between motivations and outcomes leads to low satisfaction and risk of drop-out.

1. Our motivations can be multiple and complex.

'''2. The match between our motivations and outcomes is theorised to predict satisfaction adn satisfaction is theorised to predict our likelihood to continue. '''

BRAINSTORM

Group brainstorm on self-constructs!

What is a self construct?? Our group definition was: Positive or negative dimensions to how you see yourself.

Esteem - Worth

Confidence - Relates to belief in your ability to successfully do a task efficiently.

Identity - What kind of identity do you have in society, groups and memberships

Efficacy - Link to how much effort you'll put into a task

Concept - set of descriptors

Belief - Relates to many self-constructs

During the tute we also did a few surveys; L.E.Q. - I and another testing our optimism. The optimism survey was extremely difficult to work out the results, it was kinda frustrating. Yet I love to know the results to my surveys, especially if they are psychology surveys as they are very very interesting. So i persevered to find out I am moderately pessimistic. Overall it was a great tute.

Week seven
There was no lecture this week, and it was not my week to attend tutorials.

Week eight
There were no lectures or tutorials this week due to the University of Canberra's mid semester break.

Nature of emotion
The five perennial questions surrounding the nature of emotion, according to Reeve (2009) are,

 * 1. What is emotion? (definition needs to be established and the relationship between emotion and motivation worked out)

 * 2. What causes emotion? (Involves discussing the affects of biology vs. cognition)

 * 3. How many emotions are there? (Biological perspective, cognitive perspective, and basic emotions)

 * 4. What good are the emotions? (Coping functions, social functions)

 * 5. What is the difference between emotion and mood? (Everyday mood, positive affect)

 Question 1.

What is an emotion? An emotion is made up of,

Feelings - Subjective experience, Phenomenological awareness, and cognition

Bodily Arousal - Physiological activation, Bodily preparation for action, and Motor responses

Sense of purpose - Goal-directed motivational state, and Function aspect

Social-Expressive - Social communication, facial expression, and vocal expresion

 * Emotions are one type of motive which energises and directs behaviour.

 * Emotions serve as an ongoing "readout" to indicate how well poor how poorly personal adaptation is going.

Question 2.

What causes emotions?

Significant situational events -> Cognitive process / Biological process -> which turn into feelings, sense or purpose, bodily arousal, or social-expressive.

 * Biology perspective - Biology lies at the causal core of emotion.

 * Cognitive perspective - Cognitive activity is a necessary prerequisite to emotion.

Question 3.

How many emotions are there?

 * Biology perspective - Places an emphasises on the primary emotions such as anger, and fear.

 * Cognitive perspective - Acknowledges the importance of the primary emotions, but stresses the complex emotions..

Question 4.

What good are the emotions?

Emotions have both coping and social functions.

Coping functions include destruction, reproduction, reunion, affiliation, rejection, and exploration.

Social functions of emotions can be, communicating our feeling to others, influencing how others interact with us, invite and facilitate social interaction, and create, maintain, and dissolve relationships.

Question 5.

What is the difference between emotion and mood?

Emotions emerge from significant life situations and appraisals of their significance to our well-being. Whereas moods emerge from ill-defined processes.

 This week's tutorial was based on emotion. Because I got mixed up with what weeks the tutorials would be held after the break I missed my tutorial (Tutorial one), and instead attended the following weeks tutorial on the same topic (Tutorial 3). This fault on my part turned out quite well as it allowed me to meet others within the unit and work with groups of different people.

This weeks tutorial had alot less content involved than previous weeks, but the tasks were more focused and reliant on critical thinking. I think this tutorial was one of the best as it got groups thinking and collaborating ideas, instead of thinking like textbook robots.

The first task surrounded specific emotional words, and deciding wether or not each word was an emotion. We discussed, agreed and organised the emotional words into an organised model of emotions (into underlying clusters). There were four groups within our tutorial that completed the task, with each group having similar clusters, yet differing in the number of emotional words which they believed were true emotions.

 * A photo was taken of my groups organised model of emotions. Click on the image below for a larger image.

After, we completed and scored the 20-item PANAS-Trait. A mood survey (mood tool), which assesses the degree of positive and negative affect. The scale consists of a number of words that describe different feelings and emotions, we were asked to look at the emotions and respond by indicating to what extent we feel these emotional in general (how we feel on average). My own positive affect was 33, and negative affect, 17.

We finished the tutorial by discussing wikiversity, and how to upload and embed images. Which is a task that I always find helpful, as others ask questions that I would have never thought of, which in turn helps me to improve my own wikiversity pages.

Biological aspects of emotion
Includes autonomic nervous system, endocrine system, neural brain circuits, rate of neural feedback, and facial feedback.

James-Lange theory The James-Lange theory recognises that emotional experience is a way of making sense of bodily changes. The theory has two hypothesis, these being;

1. The body reacts uniquely to different emotion-stimulating events, and

2. The body does not react to non- emotion-stimulating events.

 * Stimulus → Emotion → Bodily reaction

 * or

 * Stimulus → Bodily reaction → Emotion

It is important to understand that each theory has criticisms, for one theory rarely explains every aspect of the issue. Three criticisms which have been identified are:

1. The body reactions were part of a general fight-flight response that did not vary between emotions

2. Emotions are experienced more quickly than physiological reactions

3. Physiological arousal augments rather than causes emotion. Its role is small, supplemental and relatively unimportant.

Contemporary perspective recognises emotions recruit biological and physiological support to enable adapative behaviours such as fighting, fleeing, and nurturing. The contemporary perspective also acknowledges distinct physiological differences (e.g., Heart rate and Skin temperature) are evident for some emotions (e.g., anger, fear, sadness, and disgust).

Cognitive aspects of emotion
Includes appraisals, knowledge, attributions, socialisation history, and cultural identities.

Cognitive aspects of emotion help to recognise that without an antecedent cognitive appraisal of the event, emotions do not occur, and that the appraisal, not the event itself, causes the emotion.

Emotion knowledge

1. We learn to distinguish finer shades of emotion as we develop (distinctions are stored cognitively).

2. An individual's emotion knowledge is the number of emotions s/he can distinguish.

3. Emotion knowledge partially underlies the rationale for teaching emotional intelligence.

Socio-cultural aspects of emotion
Includes socialisation history and cultural identities.

1. Appraisal contributes to a cognitive understanding of emotion

2. The sociocultural context one lives in contributes to a cultural understanding of emotion.

3. Social interaction contributes to a social understanding of emotion

The socio-cultural aspects of emotion in include mimicry, feedback, contagion, emotional socialisation, and managing emotions.

Personality, motivation and emotion
Three motivational principals
 * Happiness
 * Arousal
 * Control

Personality traits as causes of behaviour and motivation
 * Traits cause people to react differently to different situations
 * Traits also cause people to approach and avoid different situations
 * Traits determine how people react to situations, e.g., positively or negatively
 * Traits determine the choice of situations and the altering of situations, e.g., approach, avoid, or modify situation

The big 5 personality traits

 * Neuroticism
 * Extraversion
 * Openness to Experience
 * Agreeableness
 * Conscientiousness

Each of these is measured using 6 facets (traits), these traits are shown in the table below.

The tutorial started out with a chapter review quiz, which include topics happiness and personality, arousal, and control. I did surprisingly well considering I'm not good at any quiz when placed on the spot, which gives credit to how this unit has been run and presented.

The second task we completed was the sensation seeking scale. Sensation Seeking is a construct developed by Marvin Zuckerman in 1971 that has gradually become a personality trait (Roberti, 2004). Through engaging in risky behaviours a sensation seeker’s main objective is to raise levels of internal arousal, with the trait being defined by a willingness of an individual to take risks to fulfil a desire for new stimulation.

Sensation Seeking traits can be measured using a self-report questionnaire developed by Zuckerman, the Sensation Seeking Scale (SSS). The scale measures the presence of four dimensions of sensation seeking in the participant, these being, thrill and adventure seeking, experience seeking, disinhibition, and boredom susceptibility (Roberti, 2004; Carducci, 2009, p. 347). The Zuckerman Sensation Seeking Scale gives a number value to each answer to give an overall accumulative score for each of the four dimensions. The scores then rank the behavioural tendencies of the participant, with low scores indicating low sensation seekers (LSS) and higher scores indicating high sensation seekers (HSS) (Walker & Broughton, 2009, p 54).

Individuals who score high on thrill and adventure (HSS) may participate in behaviours such as fast driving, whereas individuals who score low (LSS) may tend to drive to the speed limit. Experience-seeking individuals may seek arousal in trying unusual foods, such as insects. Disinhibited thrill seekers tend to seek heightened arousal from risking social behaviours like shoplifting, and boredom susceptibility thrill seekers do their best to avoid falling into routines (Carducci, 2009, p. 347; Walker & Broughton, 2009, p 54).

On completing the Sensation Seeking Scale i scored a 3 for boredom susceptibility (BS), 4 for disinhibition (D), 7 for experiencing seeking (ES), and 5 for thrill and adventure seeking (TAS). Most of the results from students in the class were slightly lower or slightly higher than mine. It would be interesting to look at why we were all relatively similar. Does it have something to do with our choice of degree? Do certain sensation seekers seek out different careers? Can differences in sensation seeking traits help predict what university degree people would choose to study?

The tutorial finished with a quick review on assessment items for the unit, as people asked questions and gained knowledge on how to best complete the tasks.

References:

Carducci, B. J. (2009). The Psychology of Personality: Viewpoints, Research, and Applications. Maldon, MA: Wiley-Blackwell

Baechle, T. R., & Earle, R. W. (2008). Essential of Strength Training and Conditioning. Winsor, ON: National Strength and Conditioning Association

Bradley, M. M., Miccoli, L., Escrig, M., & Lang, P. J. (2008). The pupil as a measure of emotional arousal and autonomic activation. Psychophysiology, 45, 602-607. doi:10.1111/j.1469-8986.2008.00654.x

Moodie, C., & Finnigan, F. (2005). A comparison of the autonomic arousal of frequent, infrequent and non-gamblers while playing fruit machines. Addiction, 100, 51-59. doi:10.1111/j.1360-0443.2005.00942.x.

Nagai, Y., Goldstein, l. H., Critchley, H. D., & Fenwick, P. B. (2004). Influence of sympathetic autonomic arousal on cortical arousal: implications for a therapeutic behavioural intervention in epilepsy. Epilepsy research, 58, 185-193. doi:10.1016/j.eplepsyres.2004.02.004

Pfaff, D. (2006). Brain arousal and information theory: neural and genetic mechanisms. Cambridge, MA: Harvard University Press.

Roberti, J. W. (2004). A review of behavioural and biological correlates of sensation seeking. Journal of Research in Personality, 38, 256-279. doi:10.1016/S0092-6566(03)00067-9

Walker, L., & Broughton, P. (2009). Motorcycling and leisure: understanding the recreational PTW rider. Surrey, England: Ashgate Publishing Company

Yerkes, R. M., & Dodson, J. D. (1908). The relation of strength of stimulus to rapidity of habit formation. Journal of Comparative Neurology and Psychology, 18, 459-482. doi: 10.1002/cne.920180503

Unconscious motivation
As I understand it psychodynamic is the study of unconscious psychological processes, the lecture mentioned examples of these to be defense mechanisms and depression.

The Unconscious
Much of mental life is unconcious

There are three contemporary views on the unconscious;

 * 1) Freudian unconscious (Automatically appraises the environment)
 * 2) Adaptive unconscious (sets goals, makes judgements, and initiates action)
 * 3) Implicit motivation (Automatically attend to emotionally linked environmental events)

Psychodynamics
Mental processes operate in parallel with one another.

Illustration of psychodynamics
 * 1) Repression - The process of forgetting information and an experience by ways that are unconscious, unintentional, and automatic. Repression is the ego’s counterforce to the id’s demanding desires.
 * 2) Suppression - The process of removing a thought from attention by ways that are conscious, intentional, and deliberate.

{| cellpadding="10" cellspacing="5" style=": right; width: 100%; background-color: Inherent; margin-left: auto; margin-right: auto"
 * style="width: 40%; background-color: Plum; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 10px;" |

 Further Investigation

Within my text book chapter I spoke about Drive theory. Drive theory was introduced into this weeks lecture. I thought I would share with you some independent research I have done on Drive theory.

Drive Theory
In 1943, Clark Hull proposed a liner relationship between arousal and performance (Movahedi, et. al., 2007) that suggests the more aroused an athlete is the higher quality of his performance (Weinberg & Gould, 2007). The theory surrounds the belief that increased arousal will increase performance of the dominant response, whether it be learned or instinctive (Movahedi, et. al., 2007).

In 1966 Spence and Spence took Hull’s theory and proposed a slightly adapted view, concentrating on the presence of stressors and overarousal. Spence and Spence (1966) referred to a learned response as habit strength, and arousal as drive in their explanatory equation that showed the outcome of a performance as dependent on three variables, incentive value, the strength of arousal, and an individual’s learned responses (McMorris, 2004, p. 246).

 Performance = Habit strength x Drive x Incentive value

Drive theory is concentrated on the experience of learning new skills to complete a task, and the tendency to react instinctively when put under pressure (Weinberg & Gould, 2007). The influence of skill level on arousal and performance suggests that in the early cognitive stage of learning whilst completing an unfamiliar task under circumstances of increased arousal, an individual will act instinctively to the situation resulting in a less then moderate performance. If the same task were to be completed in the latter autonomous stage of learning with the same level of increased arousal, the level of performance is likely to be improved. This is because the instinctive response has been replaced with the correct learned response, and actions performed in heightened arousal are more likely to reflect previous experiences of the task. Learnt skills and behaviours, with practice, become habitual (Movahedi, et. al., 2007).

For more information on Drive theory visit my textbook chapter Motivation and Arousal

 * }

References:

McMorris, T. (2004). Acquisition and performance of sports skills. England: John Wiley & Sons

Movahedi, A., Sheikh, M., Bagherzadeh, F., Hemayattalab, R., & Ashayeri, H. (2007). A Practice-Specificity-Based Model of Arousal for Achieving Peak Performance. Journal of Motor Behavior, 39, 457–462. doi: 10.3200/JMBR.39.6.457-462

Weinberg, R. S., & Gould, D., (2007). Foundations of sport and exercise psychology. United States: Human Kinetics

Growth motivation & positive psychology
Positive psychology looks at people's mental health and the quality of their laves, and seeks to build people's strength and competencies. Positive psychology asks "What could be?"

Looking towards the future and being optimistic is something everyone would like to do, but sometimes we can catch our selfs doing the complete opposite. I believe mastering the art of positive psychology has a very important role within society as peoples mental health can always use a good, positive boost.

Positive psychology has three illustrative personal strengths, these being: Definitions:
 * 1) Optimism
 * 2) Meaning, and
 * 3) Eudamonic Well-being

Optimism - A positive attitude or a good mood that is associated with what one expects to unfold in his or her immediate and long-term future. Related to better psychological & physical health, more health-promoting behaviors, greater persistence, and more effective problem solving.

Meaning - A sense of purpose, internalised values, and high efficacy are the motivational means to cultivate meaning in life. The act of creating meaning helps prevent future sickness.

Eudaimonic Well-being - Eudaimonic well-being is self-realisation. Relatedness satisfaction and pursuit of self-endorsed goals forecast Eudaimonic well-being

Criticisms of positive psychology include pollyanna optimism, unscientific concepts, and unknown origins of inner guides.

Growth can be encouraged by making grows choices, being honest, situationally positioning yourself for peak experiences, giving up defensiveness, and being open to experience.

Actualising tendency - Innate, a continual presence that quietly guides the individual toward genetically determined potentials. Motivates the individual to want to undertake new and challenging experiences.

Organismic Valuation Process - Innate capability for judging whether a specific experience promotes or reverses growth. Provides the interpretive information needed for deciding whether the new undertaking is growth-promoting or not.

Summary and conclusion
As the tutorial rolls around every second week I find myself sitting in class wishing this unit had an exam component. It is not because I have great talent in completing exams (quite the opposite actually) but because I feel that I have learnt and benefited so much from the structure of this unit. Recording in my e-portfolio each week has been a fantastic learning tool and incentive to attend each tutorial and lecture throughout the semester, if only all lecturers could see the benefits of such assessment and replace not so appealing and tedious assessment items (group-assignments and oral presentations to be blunt) with items that facilitate in the task of learning through encouraging attendance and participation.

title: User:E.herbert/E-portfolio


text:
User:Egm6321.f10.team4/HW2

= Problem 1 - Linearity test of ODEs  =

From Meeting 7, p 7-1.

Given

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

$$\displaystyle \ {\rm{M}}\left( {x,y} \right) + {\rm{N}}\left( {x,y} \right)\frac = 0 $$ $$
 * $$\displaystyle (Eq. 1-1)
 * }
 * }

Find
Verify the given general form of N1-ODE-VC is non-linear.

Solution
Let (Eq.1-1) be a function of H(y)
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(y)= {\rm{M}}\left( {x,y} \right) + {\rm{N}}\left( {x,y} \right)\frac $$
 * }
 * }

For the given general equation to be linear, it has to meet both of the following conditions: H is a dummy function and alpha and beta are arbitrary constants.

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

$$\displaystyle H(\alpha )=\alpha H $$ $$
 * $$\displaystyle (Eq. 1-2)
 * }
 * }

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

$$\displaystyle H(\alpha+\beta)=H(\alpha)+H(\beta) $$ $$
 * $$\displaystyle (Eq. 1-3)
 * }
 * }

By the first condition of linearity,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha )= {\rm{M}}\left( {x,\alpha y} \right) + {\rm{N}}\left( {x,\alpha y} \right)(\alpha y)' $$

 * }
 * }

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

$$\displaystyle \alpha H = \alpha[{\rm{M}}\left( {x,y} \right) + {\rm{N}}\left( {x,y} \right)y'] $$ Thus,
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha ) \neq \alpha H' $$
 * }
 * }

By the second condition of linearity,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle H(\alpha+\beta)={\rm{M}}\left( {x,\alpha + \beta} \right) + {\rm{N}}\left( {x, \alpha + \beta} \right) \left(\alpha + \beta \right)' $$
 * }
 * }

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

$$\displaystyle H(\alpha)+H(\beta)={\rm{M}}\left( {x,\alpha} \right) + {\rm{N}}\left( {x, \alpha} \right)(\alpha)'+{\rm{M}}\left( {x,\beta} \right) + {\rm{N}}\left( {x, \beta} \right)(\beta)' $$
 * }
 * }

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

$$\displaystyle H(\alpha+\beta) \neq H(\alpha)+H(\beta) $$
 * }
 * }

Since the given equation satisfies neither of the linearity conditions, the given general N1-ODE-VC is proven to be non-linear.

Egm6321.f10.team4.Yoon 00:49, 16 September 2010 (EDT) - Primary Author & Editor

= Problem 2 - Order of a differential equation and test of linearity =

From Meeting 7, p. 7-1.

Given

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

$$ \left[ {4x^1 + \sin (y)} \right] + \left( {x^2 y^3 } \right)y' = 0$$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

Find
Verify the given equation is an example of a N1-ODE.

Solution
The order of any equation can be easily determined just by inspection. In the case of Eq. (2-1), the highest-order derivative is $$y'$$ and it is of order one. Thus, Eq. (2-1) is a first order differential equation.

To verify that is it a nonlinear equation, the linearity test from the previous problem (Problem 1) should be applied. Let Eq. (2-1) be a function of H(y)
 * {| style="width:100%" border="0" align="left"

$$H(y)= \left( {4x^1 + \sin (y)} \right) + \left( {x^2 y^3 } \right)y' = 0$$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

Let us now apply Eq. (1-2). The left-hand side will take the following form:

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

H(\alpha y) = 4x^7 + \sin (\alpha y) + (x^2 (\alpha y)^3 )y'$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

while the right-hand side will take this form:

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

\alpha H(y) = \alpha \left( {4x^7 + \sin (y)} \right) + \alpha (x^2 y^3 )y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-4)
 * }
 * }

Comparing Eq. (2-3) and Eq. (2-4), it is clear that Eq. (2-1) is nonlinear.

Egm6321.f10.team4.petralanda.n 21:00, 18 September 2010 (UTC) - Author

= Problem 3 - Graphical interpretations of nonlinear variables  =

From Meeting 8, p. 8-1.

Given
General form of a homogeneous Non-linear second order differential equation(N2_ODE_VC) :

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

$$\,{a}_{2} \left( x\right)\ddot{y}+{a}_{1} \left(x \right) \dot{y}+{a}_{0} \left( x\right)y=f\left(x \right)$$
 * $$\,(Eq.3.1)$$
 * }

where,
 * {|style="width:100%" border="0"

$${y}^{1}_{H}\left(x \right)=x$$
 * $$\,(Eq.3.2)$$
 * }

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

$${y}^{2}_{H}\left(x \right)= \frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1$$
 * $$\,(Eq.3.3)$$
 * }

Find
Plot $$ y_H ^1 (x) $$ and $$ y_H ^2 (x) $$ and show $$y_H ^1 (x) \ne \alpha y_H ^2 (x)$$.

Where $$\forall x \in \real $$  and   $$\forall \alpha  \in \real$$

Hint: Find $$\hat x$$ such that $$ y^1 _H \left( {\hat x} \right) - \alpha y^2 _H \left( {\hat x} \right) \ne 0$$

Solution
Assume that,

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

$$ \,{y^1_{H}} \left( x\right) = \alpha{y^2_{H}} \left( x\right) $$
 * $$\,(Eq.3.4)$$
 * }

by our assumption, $$\displaystyle \alpha$$ must be linear so that $$\displaystyle \alpha$$ can satisfy the following (eq 3.5) for all $$\displaystyle x$$ in the $$\real$$eal domain

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

$$\,\alpha= \frac{{y^1_{H}}\left( x\right)}{{y^2_{H}} \left( x\right)}= \frac{x}{\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1}$$

 * $$\,(Eq.3.5)$$
 * }

However, $$\displaystyle \alpha$$ is proven not to be linear by figure.2, our Assumption (Eq. 3.3) is false.

Thus,
 * {|style="width:100%" border="0"

$$ \,{y^1_{H}} \left( x\right) \neq \alpha{y^2_{H}} \left( x\right) $$
 * $$\,(Eq.3.6)$$
 * }

Egm6321.f10.team4.Yoon 00:49, 16 September 2010 (EDT) - Primary Author & Editor

68.17.105.125 02:47, 22 September 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 4 - Exactness of N1-ODEs =

From Meeting 8, p. 8-2.

Given

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

$$ \varphi (x,y) = x^2 y^{3/2} + \log (x^3 y^2 ) = k$$ $$
 * $$\displaystyle (Eq. 4-1)
 * }
 * }

Find
Part A

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

$$F\left( {x,y,y'} \right) = \frac $$ $$
 * $$\displaystyle (Eq. 4-2)
 * }
 * }

Part B

Verify that $$ \displaystyle F $$ is an exact N1-ODE.

Part C

Create an additional three examples.

Solution
Part A

The total derivative given in Eq. (4-2) can be expanded to take the following form:

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

$$\frac = \frac + \frac \frac $$ $$
 *
$$\displaystyle (Eq. 4-3)
 * }
 * }

Evaluating the first term in (Eq. 4-3) will yield the following:

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

$$ \frac{\partial } \left( {x^2 y^{3/2} + \log (x^3 y^2 )} \right) = 2xy^{3/2}  + \frac{3} $$ $$
 *
$$\displaystyle (Eq. 4-4)
 * }
 * }

Evaluating the second term in (Eq. 4-3) will yield the following:

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

$$ \frac{\partial } \left( {x^2 y^{3/2} + \log (x^3 y^2 )} \right) = \frac{3} {2}x^2 y^{1/2} + \frac{2} $$ $$
 *
$$\displaystyle (Eq. 4-5)
 * }
 * }

Combining all together we see that there is in fact a function $$F(x,y,y')$$

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

$$ F(x,y,y') = \left( {2xy^{3/2} + \frac{3} } \right) + \left( {\frac{3} {2}x^2 y^{1/2}  + \frac{2} } \right)\frac = 0$$ $$
 *
$$\displaystyle (Eq. 4-6)
 * }
 * }

Part B

In order for (Eq. 4-6) to be exact it needs to satisfy the following two conditions:

Condition 1:

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

$$ F(x,y,y') = M(x,y) + N(x,y)\frac = 0 $$ $$
 *
$$\displaystyle (Eq. 4-7)
 * }
 * }

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

$$M(x,y) = \left( {2xy^{3/2} + \frac{3} } \right)$$ and  $$N(x,y) = \left( {\frac{3} {2}x^2 y^{1/2}  + \frac{2} } \right)$$
 * }

Then, the first condition is satisfied.

Condition 2:

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

$$ \frac = \frac

$$ $$
 *
$$\displaystyle (Eq. 4-8)
 * }
 * }

If we compute the left side of (Eq. 4-8):
 * {| style="width:100%" border="0" align="left"

$$\frac{\partial } \left( {2xy^{3/2} + \frac{3} } \right) = 3xy^{1/2} $$ $$
 *
$$\displaystyle (Eq. 4-9)
 * }
 * }

and if we compute the right side:

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

$$ \frac{\partial } \left( {\frac{3} {2}x^2 y^{1/2} + \frac{2} } \right) = 3xy^{1/2} $$ $$
 *
$$\displaystyle (Eq. 4-10)
 * }
 * }

Comparing (Eq. 4-9) and (Eq. 4-10) it is clear that the second exactness condition is also satisfied. Therefore $$ \displaystyle F(x,y,y') $$ is exact.

Part C

Example 1

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

$$F(x,y,y') = \frac {2}\sin (x) - y\cos (x)=0$$ $$
 *
$$\displaystyle (Eq. 4-11)
 * }
 * }

Example 2

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

$$ F(x,y,y') = e^{xy} \left( {\frac{1} {x} + \frac{1} {y}y'} \right)=0 $$ $$
 *
$$\displaystyle (Eq. 4-12)
 * }
 * }

Example 3

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

$$ F(x,y,y') = \frac + \frac y'=0$$ $$
 *
$$\displaystyle (Eq. 4-13)
 * }
 * }

Note: to generate the three examples of Part C, an arbitrary function was assigned to (Eq. 4-8). It was then integrated with respect to x and y to create M and N. Egm6321.f10.team4.petralanda.n 23:11, 18 September 2010 (UTC) - Author

68.17.105.125 02:49, 22 September 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 5 - Finding an ODE that is not exact  =

From [http://upload.wikimedia.org/wikiversity/en/1/15/2010_09_09_14_48_41.djvu Mtg. 9], p. 9-2

Given

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

$$ \displaystyle M(x,y) + N(x,y)f(y') = 0 $$ $$
 *
$$\displaystyle (Eq. 5-1)
 * }
 * }

Find

 * $$\displaystyle f(y')$$ such that there is no analytical solution to $$f(y') = - \frac{M}{N}$$
 * (i.e., such that the N1-ODE cannot be exact).

Solution
Two conditions must be met in order for an equation of the form
 * $$\displaystyle F(x,y,y')=0$$

to be exact:

1) $$\displaystyle F$$ must be in the form:

2) From Eq. (5-2), the partial derivative of $$\displaystyle M(x,y)$$ with respect to $$\displaystyle y$$ must equal the partial derivative of $$\displaystyle N(x,y)$$ with respect to $$\displaystyle x$$:

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

M_y(x,y) = N_x(x,y) $$ $$
 *
$$\displaystyle (Eq. 5-3)

For this solution, we need to "invent" some function $$ \displaystyle f(y') $$....

First, suppose {| style="width:100%" border="0" align="left"
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle

f(y')=5y'+7 $$ $$
 *
$$\displaystyle (Eq. 5-4)

Then: Eq. (5-1) becomes:
 * $$\displaystyle M(x,y) + N(x,y)(5y'+7) = 0$$
 * $$\displaystyle \underbrace{M(x,y) + 7N(x,y)}_{M'(x,y)}+\underbrace{5N(x,y)}_{N'(x,y)}y'= 0$$

This reduces to:
 * $$\displaystyle M'(x,y) + N'(x,y)y' = 0$$

Thus the form of Eq (5-2) is maintained (and we must find another function that will not have an analytical solution).

 * Now, suppose that

{| style="width:100%" border="0" align="left"
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * $$\displaystyle

f(y') =\frac{e^{y'}}{sin^3(y')} $$ $$
 *
$$\displaystyle (Eq. 5-5)

Then: Eq. (5-1) becomes:
 * $$\displaystyle M(x,y) + N(x,y)\frac{e^{y'}}{sin^3(y')} = 0$$
 * $$\displaystyle \frac{e^{y'}}{sin^3(y')} = -\frac{M(x,y)}{N(x,y)} $$
 * $$\displaystyle e^{y'} = -\frac{M(x,y)}{N(x,y)}sin^3(y') $$
 * $$\displaystyle y' = -ln \left[\frac{M(x,y)}{N(x,y)}sin^3(y') \right] = ln \left[ N(x,y) \right] -ln \left[ M(x,y) sin^3(y') \right]$$

Rearrange:
 * $$ \displaystyle -ln \left[N(x,y) \right] + ln \left[ M(x,y) sin^3(y') \right]+y'=0$$
 * $$ \displaystyle \underbrace{-ln \left[N(x,y) \right]}_{M'(x,y)} + \underbrace{ln \left[ M(x,y)\right]+ln \left[ sin^3(y') \right]+y'}_{ \neq N'(x,y)y'}=0$$

The equation cannot be arranged in the form of Eq. (5-2), thus Eq (5-5) is an example of a function $$\displaystyle f(y') $$, for which there exists no analytical solution when used in Eq (5-1).

 * Note: There is no need to test the second condition of exactness, Eq (5-3), since we are not given the functions $$\displaystyle M(x,y) $$ and $$\displaystyle N(x,y) $$.

128.227.34.126 14:55, 22 September 2010 (UTC) - Author

= Problem 6 - Verification of the first condition of exactness =

From Meeting 9, p. 9-3.

Given

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

$$\displaystyle y(x) = \sin ^{ - 1} (k - 15x^5 ) $$ $$
 *
$$\displaystyle (Eq. 6-1)
 * }
 * }

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

$$\displaystyle M(x,y)=75x^4 $$ $$
 *
$$\displaystyle (Eq. 6-2)
 * }
 * }

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

$$\displaystyle N(x,y)=cos(y) $$ $$
 *
$$\displaystyle (Eq. 6-3)
 * }
 * }

Find
Verify equation 6-1 satisfies the first condition of exactness,

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

$$\displaystyle M\left(x,y\right)+N(x,y)y'=0 $$ $$
 *
$$\displaystyle (Eq. 6-4)
 * }
 * }

Solution
Start off by assigning a dummy variable $$\displaystyle u$$

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

$$\displaystyle u = \sin{y} = \sin \left[ \sin ^{ - 1} \left(k - 15x^5 \right) \right] = k - 15x^5 $$
 * }

and

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

$$\displaystyle \frac {du}{dx} = - 75x^4 $$
 * }

Now taking the derivative of $$\displaystyle y$$ and applying the chain rule,

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

$$\displaystyle y' = \frac = \frac \frac = \frac (- 75x^4) $$
 * }

Substituting $$\displaystyle u = \sin{y}$$ back into the equation yields

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

$$\displaystyle y' = \frac $$
 * }

or

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

$$\displaystyle y' = \frac = \frac $$ $$
 *
$$\displaystyle (Eq. 6-5)
 * }
 * }

Substituting Eq.s (6-2), (6-3), and (6-5) into Eq. (6-4) yields

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

$$\displaystyle M\left(x,y\right)+N(x,y)y'=75x^4+\cos{y} \left( \frac \right) $$
 * }

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

$$\displaystyle =75x^4- 75x^4 = 0 $$
 * }

Thus equation 6-1 meets the first condition of exactness

Egm6321.f10.team4.Auerbach 02:53, 21 September 2010 (UTC) - Author

= Problem 7 - Solving nonhomogenous L1-ODE-VCs =

From Meeting 10 p. 10-3, 10-4, and Meeting 11, p. 11-1.

Given

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

$$
 * $$\displaystyle a_1 (x)y' + a_0 (x)y = b(x)$$
 *
$$\displaystyle (Eq. 7-1)
 *
$$\displaystyle (Eq. 7-1)
 * }
 * }

Find
Part A

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

|$$\displaystyle a_0 (x)=x $$ $$
 * $$\displaystyle a_1 (x)=1 $$
 * $$\displaystyle a_1 (x)=1 $$
 * $$\displaystyle b(x)=2x+3$$
 *
$$\displaystyle (Eq. 7-2)
 * }
 * }

Solve for $$\displaystyle y $$ in Eq. (7-1).

Part B

Assume

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

 * $$\displaystyle a_1 (x) \ne 0$$ $$ \forall {\text{ }}x$$
 * }
 * }
 * }

Then, (Eq. 7-1) becomes

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

y' + \underbrace {\frac }_{P(x)}y = \underbrace {\frac }_{Q(x)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7-3)
 * }
 * }

Find a solution for $$\displaystyle y(x) $$ in terms of $$\displaystyle a_0 {\text{, }}a_1 {\text{,}} $$ and $$\displaystyle b. $$

Part C

Give an example where

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

a_1 (x) = x^2 + 1{\text{ }}a_0 (x) = x{\text{ and }}b(x) = 2x$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7-4)
 * }
 * }

Solution
Part A

Substituting Eq. (7-2) into (7-1),
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle 1y' + xy = 2x + 3$$
 *
$$\displaystyle (Eq. 7-5)
 *
$$\displaystyle (Eq. 7-5)
 * }
 * }

Eq. (7-5) can be solved by multiplying both sides of the equation with an integration factor:

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

$$
 * $$\displaystyle h(x)(1y' + xy) = h(x)(2x + 3) $$
 *
$$\displaystyle (Eq. 7-6)
 *
$$\displaystyle (Eq. 7-6)
 * }
 * }

where

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

 * $$\displaystyle h(x)=\exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt \right]$$
 * }
 * }
 * }

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

 * $$\displaystyle =\exp \left[\int^{x} \frac{t}{1} dt \right]$$
 * }
 * }
 * }

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

 * $$\displaystyle =\exp \left[\frac{x^2}{2}\right]$$
 * }
 * }
 * }

Rewriting 7-6,

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

 * $$\displaystyle \exp \left[\frac{x^2}{2}\right]y' + \exp \left[\frac{x^2}{2}\right]xy = \exp \left[\frac{x^2}{2}\right](2x + 3) $$
 * }
 * }
 * }

The left side of the equation is equivalent to

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

 * $$\displaystyle \frac{d}{dx}(hy) = hy' + h'y$$
 * }
 * }
 * }

so,

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

 * $$\displaystyle \frac{d}{dx}(hy) = h(2x + 3) $$
 * }
 * }
 * }

integrating both sides

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

 * $$\displaystyle hy = \int^{x} h(s)(2s + 3) ds$$
 * }
 * }
 * }

Dividing both sides of the equation by h,

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

 * $$\displaystyle y = h^{-1} \int^{x} h(s)(2s + 3) ds$$
 * }
 * }
 * }

substituting

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

 * $$\displaystyle y = \exp \left[\frac{-x^2}{2}\right] \int^{x} \exp \left[\frac{s^2}{2}\right](2s + 3) ds$$
 * }
 * }
 * }

completing the integration,

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

 * $$\displaystyle y=\exp \left[\frac{-x^2}{2}\right](2 \exp \left[\frac{x^2}{2}\right]+3 \sqrt{\frac{\pi}{2}} erfi(\frac{x}{\sqrt {2}})+c)$$
 * }
 * }
 * }

expanding,

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

 * $$\displaystyle y=2+3\exp \left[\frac{-x^2}{2}\right] \sqrt{\frac{\pi}{2}} erfi(\frac{x}{\sqrt {2}})+c \exp \left[\frac{-x^2}{2}\right]$$
 * }
 * }
 * }

Part B

starting with equation 7-1,
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle a_1 (x)y' + a_0 (x)y = b(x)$$
 *
$$\displaystyle (Eq. 7-7)
 *
$$\displaystyle (Eq. 7-7)
 * }
 * }

Since we have assumed,
 * {| style="width:100%" border="0" align="left"

 * $$\displaystyle a_1 (x) \ne 0$$ $$ \forall {\text{ }}x$$
 * }
 * }
 * }

equation 7-7 can be rewritten as

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

$$
 * $$\displaystyle y' + \frac{a_0 (x)}{a_1 (x)}y = \frac{b(x)}{a_1 (x)}$$
 *
$$\displaystyle (Eq. 7-8)
 *
$$\displaystyle (Eq. 7-8)
 * }
 * }

Equation 7-8 can be solved by multiplying both sides of the equation with an integration factor:

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

$$
 * $$\displaystyle h(x)(y' + \frac{a_0 (x)}{a_1 (x)}y) = h(x)\frac{b(x)}{a_1 (x)} $$
 *
$$\displaystyle (Eq. 7-9)
 *
$$\displaystyle (Eq. 7-9)
 * }
 * }

where

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

 * $$\displaystyle h(x)=\exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]$$
 * }
 * }
 * }

Rewriting 7-9,

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

 * $$\displaystyle \exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]y' + \exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{a_0 (x)}{a_1 (x)}y = \exp \left[\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{b(x)}{a_1 (x)} $$
 * }
 * }
 * }

The right side of the equation is equivalent to

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

 * $$\displaystyle \frac{d}{dx}(hy) = hy' + h'y$$
 * }
 * }
 * }

so,

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

 * $$\displaystyle \frac{d}{dx}(hy) = h\frac{b(x)}{a_1 (x)} $$
 * }
 * }
 * }

integrating both sides

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

 * $$\displaystyle hy = \int^{x} h(s)\frac{b(s)}{a_1 (s)} ds$$
 * }
 * }
 * }

Dividing both sides of the equation by h,

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

 * $$\displaystyle y = h^{-1} \int^{x} h(s)\frac{b(s)}{a_1 (s)} ds$$
 * }
 * }
 * }

substituting

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

 * $$\displaystyle y = \exp \left[-\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right] \int^{x} \exp \left[\int^{s} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{b(s)}{a_1 (s)} ds$$
 * }
 * }
 * }

completing the integration,

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

$$
 * $$\displaystyle y = \exp \left[-\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right] \int^{x} \exp \left[\int^{s} \frac{a_{0}(t)}{a_{1}(t)}dt\right]\frac{b(s)}{a_1 (s)} ds + c \exp \left[-\int^{x} \frac{a_{0}(t)}{a_{1}(t)}dt\right]$$
 *
$$\displaystyle (Eq. 7-10)
 *
$$\displaystyle (Eq. 7-10)
 * }
 * }

where c is a constant of integration

Part C

Substituting equation 7-4 into equation 7-10 yields

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

 * $$\displaystyle y = \exp \left[-\int^{x} \frac{t}{t^2 + 1}dt\right] \int^{x} \exp \left[\int^{s} \frac{t}{t^2  + 1}dt\right]\frac{2s}{s^2  + 1} ds + c \exp \left[-\int^{x} \frac{t}{t^2  + 1}dt\right]$$
 * }
 * }
 * }

simplifying,

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

 * $$\displaystyle y = \frac{1}{\sqrt{x^2 + 1}} \int^{x} \sqrt{s^2  + 1} \frac{2s}{s^2  + 1} ds + \frac{c}{\sqrt{x^2  + 1}}$$
 * }
 * }
 * }

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

 * $$\displaystyle = \frac{1}{\sqrt{x^2  + 1}} \int^{x}  \frac{2s}{\sqrt{s^2  + 1}} ds + \frac{c}{\sqrt{x^2  + 1}}$$
 * }
 * }
 * }

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

 * $$\displaystyle = \frac{1}{\sqrt{x^2  + 1}} 2 \sqrt{x^2  + 1}+ \frac{c}{\sqrt{x^2  + 1}}$$
 * }
 * }
 * }

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

 * $$\displaystyle y = 2 + \frac{c}{\sqrt{x^2 + 1}}$$
 * }
 * }
 * }

Egm6321.f10.team4.Auerbach 10:55, 22 September 2010 (UTC) - Author

= Solutions Generated by Use of the Integrating Factor Method =

From Meeting 11, p. 11-2.

Given

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

$$\displaystyle \ y\left( x \right) = \frac{1}\int^x h\left( s \right)b\left( s \right)ds = \frac{1}\int h\left( x \right)b\left( x \right)dx+k_2 $$ $$ where
 *
$$\displaystyle (Eq. 8-1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \ h(x) = e^{\int^x a_0\left( s \right)b\left( s \right)ds} = e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} $$ $$ where $$\displaystyle k_1 $$ & $$\displaystyle k_2 $$ are integration constants. Also, $$\displaystyle h(x) $$ is a particular case of $$\displaystyle h(x,y) $$, from the equation:
 *
$$\displaystyle (Eq. 8-2)
 * }
 * }

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

$$\displaystyle \ h(x,y) \left[M(x,y)+N(x,y)y' \right] = 0 $$ $$
 *
$$\displaystyle (Eq. 8-3)
 * }
 * }

in which case $$\displaystyle h $$ is not a function of $$\displaystyle y $$ (and thus $$\displaystyle \frac{\partial h}{\partial y}= h_y=0 $$).

Find
Part A

Show that $$\displaystyle k_1 $$ is an extraneous constant, when solving by the integrating factor method.

Part B

Show that Eq. (8-1) agrees with King p. 512, where

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

$$
 * $$\displaystyle y(x) = Ay_H (x) + y_P (x)$$
 *
$$\displaystyle (Eq. 8-4)
 *
$$\displaystyle (Eq. 8-4)
 * }
 * }

Part C

Find $$ \displaystyle y_H (x)$$ independently.

Hint: solve for

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

 * $$\displaystyle y'+a_0y=0$$
 * }
 * }
 * }

Solution
Part A

In using the Integrating Factor Method (IFM) to find some function $$ h(x,y) $$, such that our solution can be made exact, the condition

 * $$\frac

= \frac $$

must be true, where $$\overline M_y = h_yM + hM_y$$ and $$\overline N_x = h_x + hN_x$$

Part A

In the Euler Integrating Factor Method, distributing $$ \displaystyle h(x,y) $$ in Eq. (8-3) gives:

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

$$\displaystyle \ \underbrace{h(x,y)M(x,y)}_{\overline{M}(x,y)}+\underbrace{h(x,y)N(x,y)}_{\overline{N}(x,y)}y'=0 $$ $$
 *
$$\displaystyle (Eq. 8-5)
 * }
 * }

The function $$\displaystyle h(x) $$ is used to make Eq. (8-3) exact, such that it satisfies the second condition of exactness, from Lecture 8, p. 8-4:

 * $$ \displaystyle \overline{M}_y(x,y)=\overline{N}_x(x,y) $$

Thus, from Eq. (8-5), we have:

 * $$ \displaystyle \not{h_y}M+hM_y= h_xN+hN_x $$
 * (Recall: $$ \displaystyle h_y=0 $$)
 * $$ \displaystyle h_xN+ h \left(N_x-M_y \right )=0 $$

Substitute $$ \displaystyle h_x $$ from Eq. (8-2):

 * $$ \displaystyle \frac{\partial}{\partial x} \left[e^{\int a_0\left( x \right)b\left( x \right)dx+k_1}\right] N+e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} \left(N_x-M_y \right )=0 $$

 * $$ \displaystyle a_0(x) e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} N+e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} \left(N_x-M_y \right )=0 $$

Factor out $$ \displaystyle e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} $$:

 * $$ \displaystyle e^{\int a_0\left( x \right)b\left( x \right)dx+k_1} \left[a_0(x)+N_x-M_y \right]=0 $$

 * $$ \displaystyle a_0(x)+N_x-M_y =0 $$

 * $$ \displaystyle $$

which we have rewritten

 * $$ \displaystyle y' + a_0 (x) y = b(x)$$

We have further substituted the term $$ \displaystyle h(x) $$ to represent our integrating factor, where King does not. By a simple substitution of terms, it can be shown that our results are consistent with King.

 * $$ \displaystyle h(x) = \exp^{\int^{x} P(t) dt} $$

 * $$\displaystyle b(x) = Q(x) $$

 * $$ \displaystyle a_0 (x) = P(x) $$

Our solution $$\displaystyle y(x) = \frac \int^{x} h(s) b(s) ds $$, becomes

 * $$ \displaystyle y(x) = e^{ - \int^{x} P(s) ds } \int^{x} h(s) b(s) ds $$

This is identical to the solution presented on page 512 in King [1]:

 * $$ \displaystyle y_p = \exp \left[ - \int^{x} P(t) dt \right] \int^{x} \underbrace{Q(s)}_{b(s)} \underbrace{\exp \left[ \int^{s} P(t) dt \right]}_{h(s)} ds$$

Part C

Our integrating factor is given by Eq. (8-2):

 * $$ \displaystyle h(x) = e^{\int^{x} a_0(s) ds} $$

To find the homogenous differential equation, start with the given equation:

 * $$ \displaystyle y' + a_0 y = 0 $$

Multiplying both sides by $$ \displaystyle h $$, yields

 * $$ \displaystyle h y' + a_0 h y = 0 $$

But the derivate of $$ \displaystyle h $$ with respect to $$ \displaystyle x $$ is:
 * $$ \displaystyle \frac{dh}{dx} = h' = \frac{d}{dx} e^{\int^{x} a_0(s) ds} = a_0 e^{\int^{x} a_0(s) ds } = a_0 h $$

Therefore we can write:

 * $$ \displaystyle h y' + h' y = 0 $$
 * $$ \displaystyle \frac{d}{dx} \left[ h(x) y(x) \right] = 0 $$
 * $$ \displaystyle h y' + h' y = 0 $$

By integrating both sides:

 * $$ \displaystyle \int^{x} (hy)' dx= \int^{x} 0 dx$$

 * $$ \displaystyle h(x) y_H(x) = A $$

where $$ \displaystyle A $$ is a constant of integration and $$ \displaystyle y_H(x) $$ is the homogenous solution of $$ \displaystyle y(x) $$.

Therefore,

 * $$ \displaystyle y_H(x) = \frac $$

 * $$ \displaystyle y_H(x) = {A}{e^{- \int^{x} a_0(s) ds}}$$

References:

68.17.105.125 05:52, 22 September 2010 (UTC)Egm6321.f10.team4.osentowski - Editor Part A, Author Part B and C 128.227.34.126 14:54, 22 September 2010 (UTC) - Author Part A, Editor Parts B and C

= Problem 9 - Making a N1-ODE exact =

From Meeting 12, p. 12-2.

Given
$$\displaystyle a(x) = \sin (x^3 )$$ $$\displaystyle b(x) = \cos (x) $$ $$\displaystyle c(y) = e^{2y} $$

Find
Part A

Find a N1-ODE that is either exact or can be made exact by using

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

$$
 * $$\displaystyle \bar b(x)c(y)y' + a(x)\bar c(y) = 0$$
 *
$$\displaystyle (Eq. 9-1)
 *
$$\displaystyle (Eq. 9-1)
 * }
 * }

Part B

Find the first integral

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

$$
 * $$\displaystyle \phi (x,y) = k$$
 *
$$\displaystyle (Eq. 9-2)
 *
$$\displaystyle (Eq. 9-2)
 * }
 * }

Solution
Part A

Lets first compute

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

$$
 * $$\displaystyle\bar b(x) = \int\limits_{}^x {b(s)ds} $$
 *
$$\displaystyle (Eq. 9-5)
 *
$$\displaystyle (Eq. 9-5)
 * }
 * }

Which will yield the following

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

$$
 * $$\displaystyle\bar b(x) = \int\limits_{}^x {\cos (s)ds} = \sin (x)$$
 *
$$\displaystyle (Eq. 9-6)
 *
$$\displaystyle (Eq. 9-6)
 * }
 * }

Doing the same for c

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

\bar c(y) = \int\limits_{}^y {c(s)ds} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-7)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

\bar c(y) = \int\limits_{}^y {e^{2s} ds} = \frac{1} {2}e^{2y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-8)
 * }
 * }

Combining all into Eq. (9-1)

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

\underbrace {\sin (x)e^{2y} }_Ny' + \underbrace {\sin (x^3 )\frac{1} {2}e^{2y} }_M = 0$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-9)
 * }
 * }

One can see that Eq. (9-9) clearly satisfies the first condition of exactness. Lets now examine if the second condition shown in (Eq 4-2) is satisfied.

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

\frac = \frac{\partial } \left( {\sin (x^3 )\frac{1} {2}e^{2y} } \right) = \sin (x^3 )e^{2y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-10)
 * }
 * }

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

\frac = \frac{\partial } \left( {\sin (x)\frac{1} {2}e^{2y} } \right) = \cos (x)e^{2y} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-11)
 * }
 * }

Comparing the above two equations it is clear that Eq. (9-9) will have to be made exact by implementing Euler's Integrating Factor Method. However, before doing so, let's assume $$e^{2y} \ne 0$$ and divide it through to get

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

{2}}_M = 0$$ $$
 * $$\displaystyle\underbrace {\sin (x)}_Ny' + \underbrace {\sin (x^3 )\frac{1}
 * $$\displaystyle\underbrace {\sin (x)}_Ny' + \underbrace {\sin (x^3 )\frac{1}
 *
$$\displaystyle (Eq. 9-12)
 * }
 * }

Multiply everything by an integration factor h.

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

$$
 * $$\displaystyle\underbrace {hN}_{\bar N}y' + \underbrace {hM}_{\bar M} = 0$$
 *
$$\displaystyle (Eq. 9-13)
 *
$$\displaystyle (Eq. 9-13)
 * }
 * }

The above (Eq. (9-13)) still satisfies the first condition of exactness, but we are going to force the second one:

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

= \frac $$ $$
 * $$\displaystyle \frac
 * $$\displaystyle \frac
 *
$$\displaystyle (Eq. 9-14)
 * }
 * }

and expanding:

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

h_x N + hN_x = h_y M + hM_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-15)
 * }
 * }

By inspection of Eq. (9-12), we see that both $$\displaystyle N $$ and $$\displaystyle M $$ are only functions of $$\displaystyle x $$. Therefore, we will assume that $$\displaystyle h $$ is also a function of $$\displaystyle x $$ only to get the following:

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

\frac {h} = \frac{1} {N}\left( {M_y - N_x } \right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-16)
 * }
 * }

Integrating Eq. (9-16)

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

\ln \left| {h(x)} \right| = \int {\frac{1} (0 - \cos (x))dx = - \ln \left| {\sin (x)} \right|} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-17)
 * }
 * }

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

h(x) = e^{ - \ln \left| {\sin (x)} \right|} = \frac{1} = \frac{1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-18)
 * }
 * }

The exact N1-ODE will be

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

y' + \frac = 0$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-19)
 * }
 * }

Part B

We know that

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

= \underbrace {\frac }_M + \underbrace {\frac }_N\underbrace {\frac }_{y'} $$ $$
 * $$\displaystyle F(x,y,y') = \frac
 * $$\displaystyle F(x,y,y') = \frac
 *
$$\displaystyle (Eq. 9-20)
 * }
 * }

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

= M(x,y) $$ $$
 * $$\displaystyle \frac
 * $$\displaystyle \frac
 *
$$\displaystyle (Eq. 9-21)
 * }
 * }

and

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

= N(x,y) $$ $$
 * $$\displaystyle \frac
 * $$\displaystyle \frac
 *
$$\displaystyle (Eq. 9-22)
 * }
 * }

Integrating Eq. (9-21) with respect to $$\displaystyle x $$ and Eq. (9-22) with respect to $$\displaystyle y $$ we will get the following:

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

$$ $$
 * $$\displaystyle \int {\partial \phi } = \int {1dy = y + c_1 }
 * $$\displaystyle \int {\partial \phi } = \int {1dy = y + c_1 }
 *
$$\displaystyle (Eq. 9-23)
 * }
 * }

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

\int {\partial \phi } = \int {\frac dx = \frac{1} {2}} \int {\frac dx + c_2 } $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-24)
 * }
 * }

Combining all together:

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

$$\displaystyle \phi (x,y) = \frac{1} {2}\int {\frac dx + c_2 } + y + c_1  = \frac{1} {2}\int {\frac dx + } y + c$$ $$
 *
$$\displaystyle (Eq. 9-25)
 * }
 * }

Egm6321.f10.team4.petralanda.n 03:05, 20 September 2010 (UTC) - Author

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

title: User:Egm6321.f10.team4/HW2


text:
User:EGM6321.F10.TEAM1.HW2

Given
Eq5) from pg 6-3:

Find
Verify the equation given is a N1_ODE

Solution
The largest number of a differentiation is 1. (1.1) equation is 1st order. The differential equation consists of one dependent variable and its derivatives. (1.1) equation is Ordinary Differential Equation. To be a linear differential equation, (1.2) equation must satisfy superposition (1.3).

(1.2) equation does not satisfy superposition. In general (1.1) equation is N1-ODE. In a particular case this equation is L1-ODE.

Example :

$$\therefore $$ (1.12) equation is Linear.

Author EGM6321.f10.team1.Kim.TH

Given
Eq.7 from pg 7-1:

Find
Verify this equation is N1_ODE

Solution
The largest number of a differentiation is 1. (2.1) equation is 1st order. The differential equation consists of one dependent variable and its derivatives. (2.1) equation is Ordinary Differential Equation. To be a linear differential equation, (2.2) equation must satisfy superposition (2.3).

(2.2) equation does not satisfy superposition. (2.1) equation is N1-ODE.

Author EGM6321.f10.team1.Kim.TH

Given
and

Find
1) Show plot of $$ y^1_H(x) $$ and $$ y^2_H(x) $$ and 2) show that there is no $$ \alpha\ $$ such that $$ y^1_H(x) = \ \alpha\ y^2_H(x) $$ for any x

Solution
Plot of $$ y^1_H(x) $$ and $$ y^2_H(x) $$

To prove that there is no $$ \alpha\ $$ such that $$ y^1_H(x) = \ \alpha\ y^2_H(x) $$ for any x The following equation was used:

$$ y^1_H(\tilde{x}) - \alpha\ y^2_H(\tilde{x}) \ne \ 0 $$

This equation would only be satisfied for $$ \alpha = 0 $$.

Then, using $$ \tilde{x}=5 $$ and the newly defined $$ \alpha = 0$$ yields

$$5-0 \ \left( \frac{5}{2}\log \left( \frac{1+5}{1-5} \right)-1 \right)= \ 0$$ which reduces to $$5= 0\ $$ within the realm of complex numbers, but $$5 \ne \ 0\ $$

$$\therefore$$ there is no $$ \alpha\ $$ such that $$ y^1_H(x) = \ \alpha\ y^2_H(x) $$ for any x

Author EGM6321.F10.TEAM1.WILKS

Given

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

$$ \varnothing(x,y)= x^2y^\frac{3}{2}+log(x^3y^2)=k$$ (4.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Find
1)$$ F(x,y,y')=\frac{d\varnothing}{dx}(x,y)$$ 2) Verify that F is exact N1_ODE 3) Invent three more N1_ODE's

Solution
This problem asks for F(x,y,y') and for the verification that F is exactly N1_ODE. In other words, verify that there exists a function $$ \phi(x,y)\ $$ such that:
 * {| style="width:100%" border="0"

$$ F = \frac{d\phi}{dx}$$ (4.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

In other terms, verify that:

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

$$F(x,y,y') = \frac{d}{dx}\phi(x,y) = \frac{d}{dx}k$$ (4.3, 4.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Solving for the partial derivatives in equations 2 and 3 yields:

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

$$F(x,y,y') = \frac{d\phi}{dx}(x,y) + \frac{d\phi}{dx}(x,y)\frac{dy}{dx} = M(x,y) + N(x,y)\frac{dy}{dx} = 0$$ (4.5, 4.6 and 4.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Where:

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

$$M(x,y) = \frac{d\phi}{dx}(x,y) = (2)(y^{3/2})x + \frac{3}{x}$$ (4.8, 4.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$N(x,y) = \frac{d\phi}{dx}(x,y) = \frac{3}{2}(x^2)(y^{1/2})x + \frac{2}{y}$$ (4.10, 4.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

1) and 2) Substituting equations 4.9 and 4.11 into equation 4.5 yields the final answer for F(x,y,y') (4.12) and also verifies that it is exactly N1_ODE (4.13)

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

$$F(x,y,y') = ((2)(y^{3/2})x + \frac{3}{x}) + (\frac{3}{2}(x^2)(y^{1/2})x + \frac{2}{y})y' = 0$$ (4.12, 4.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

3) Below are 3 examples of N1_ODEs

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

$$ F(x,y,y') = 22x^2y + y^5y'\ $$ (4.15)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ F(x,y,y') = sin(x^{7/2}) - \frac{4}{3}y^6y'\ $$ (4.16)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ F(x,y,y') = x^8 + y^4y'\ $$ (4.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author: User:EGM6321.f10.team1.russo

Find
Find a particular function for f(y') such that there is no analytical solution to the given equation. (i.e. find f(y') such that the N1_ODE cannot be exact)

Solution
Begin with an equation in the form of eq.5.1 Then one possible definition of f'(y) would be:

in which case eq. 5.1 would become:

Although, this definition of f(y') appears to be a simple polynomial, the roots of y' cannot be solved analytically. This is, in general, true of polynomials of order higher than four as is discussed here:

Author

Bryan Allison Egm6321.f10.team1.allison 20:26, 21 September 2010 (UTC)

Given
where k is a constant

Find
Verify eq6.1 is a valid solution to the eq6.2

Solution
Start with eq6.2

then by differentiating both sides with respect to x and solving for y'(x) we get:

which can then be substituted into eq6.1 to yield: which is easily reduced to the trivial equation $$ 0=0\ $$

$$\therefore$$ eq6.1 is a valid solution to the eq6.2

Author Bryan Allison Egm6321.f10.team1.allison 20:25, 21 September 2010 (UTC)

Find
1) Find Solution in terms of $$ a_0\ $$, $$ a_1 \ $$ and $$ b \ $$  for $$ y(x)\ $$ Assuming

2) Find Solution in terms of $$ a_0\ $$, $$ a_1 \ $$ and $$ b \ $$  for $$ y(x)\ $$ Assuming $$ a_1(x)\ne \ 0\ $$ for all x 3) Find Solution in terms of $$ a_0\ $$ , $$ a_1 \ $$  and $$ b \ $$  for $$ y(x)\ $$ Assuming

Solution
1) and 2) Substitute equations 7.2, 7.3 and 7.4 into 7.1 to get:

The integrating factor is shown and solved for below.

Multiply both sides by the integrating factor yields

Integration then yields

Divide both sides by $$ \exp(\frac{1}{2}x^2) $$ and

3) Substitute equations 7.5, 7.6 and 7.7 into 7.1 to get: Divide through by $$ (x^2+1)\ $$ to obtain

The integrating factor is shown and solved for below.

Multiply both sides by the integrating factor yields

Integration then yields

Divide both sides by $$ (x^2+1)^\frac{1}{2} $$ gives the final result:

Author

User:Egm6321.f10.team1.allen

Given
Eq6, p.10-3 $$ y(x)=\frac{1}{h(x)} \int_{0}^{x} h(s)b(s)\, ds $$ and Eq1, p.10-3 $$ h(x)=e^{ \int_{0}^{x} a_0(s)\, ds } $$

Find
$$ \alpha\ $$ Show that integration factor, $$k_1\ $$, is unnecessary $$ \beta\ $$ Show that Eq6, p.10-3 agrees with King et.al. p.512 i.e. show that $$ y(x)=Ay_h(x)+y_p(x)\ $$ $$ \gamma\ $$ Find $$ y_h(x)\ $$ independently. i.e. solve $$ y'+a_0y=0\ $$

Solution
$$ \alpha\ $$) Using instead of

gives an integrating factor of

instead of Multiplying both sides by the integrating factor results in the following:

$$\begin{align} \therefore e^{K_1} \end{align}$$ cancels out of the equation 8.6 rendering it unnecessary.

$$ \beta$$) Using:

is derived from:

8.8 is comparable to:

Which is derived from:

As shown in K. et al p. 512.

Combining above statements and eq. 8.13:

$$ \gamma $$) Since $$ e^x\ $$ is repeating after integration, it is ideal for this solution.

Egm6321.f10.team1.allen 03:47, 22 September 2010 (UTC)

Given
Eq1, p.12-1

and

Find
1) Find an N1_ODE that is exact or that can be made to be exact 2) Find $$ \phi\ (x,y)=k \ $$

Solution
First $$\overline{b}(x)$$ and $$ \overline{c}(x)$$ must be found.

Then, substituting into (9.1)

It can be shown that $${{e}^{2y}}\ne 0,\forall y\in \mathbb{R}$$ so these terms can be divided out without reducing the generality of the problem.

Next, check for exactness:

and

$$ \therefore $$ the equation is not exact as shown. Since both M and N are both functions of x only it is safe to assume that the integrating factor, h, must be a function of x.

so that equation (9.1) can then become the exact equation:

This exactness of this equation can be proven by:

$$ \therefore y'+\frac{\sin ({{x}^{3}})}{2\sin (x)}=0 $$ is an exact equation.

The first integral $$\phi$$ can then be found through the following process. Rewrite 9.1 as:

where

and

so N can be integrated with respect to y to find $$ \phi $$

differentiate again with respect to x and set equal to M to solve f'(x)

Then integrating f'(x) one last time to yield $$ \phi $$

$$ \phi =y+\int{f'(x)dx}$$

$$ \therefore \phi =y+\int{\frac{\sin ({{x}^{3}})}{2\sin (x)}dx}$$

which has no analytical solution, but could be solved numerically given boundary conditions.

Author

Egm6321.f10.team1.allison 15:28, 22 September 2010 (UTC)

Signatures
Solved problem 1 -- Taeho Kim (TK)

Solved problem 2 -- Taeho Kim (TK)

Solved problem 3 -- EGM6321.F10.TEAM1.WILKS 05:38, 20 September 2010 (UTC)User:EGM6321,F10.TEAM1.WILKS|David Wilks (DW)

Solved problem 4 -- EGM6321.f10.team1.russo 17:10, 22 September 2010 (UTC)Charles Russo (CR)

Solved problem 5 -- Egm6321.f10.team1.allison 20:24, 21 September 2010 (UTC)Bryan Allison (BA)

Solved problem 6 -- Egm6321.f10.team1.allison 20:24, 21 September 2010 (UTC)Bryan Allison (BA)

Solved problem 7 -- Egm6321.f10.team1.allen 18:24, 22 September 2010 (UTC) Sean Allen (SA)

Solved problem 8 -- Egm6321.f10.team1.allen 18:24, 22 September 2010 (UTC) Sean Allen (SA)

Solved problem 9 -- Egm6321.f10.team1.allison 15:27, 22 September 2010 (UTC)Bryan Allison (BA)

title: User:EGM6321.F10.TEAM1.HW2


text:
User:Minaalah

Example of each Business Models

Virtual storefront - Examples are sell books online. Sell your books the easy way on websites, in emails and more santu.com/supereasy-selling. And Get eBay Paid - Get SMS. SMS to Phones on eBay item Sold Out Or PayPal Stores Payment Received www.sms4ebay.com

Information broker - An example of an information broker in contemporary fiction would be DC Comics' superheroine, the Oracle, Edward G. Robinson's character Sol in the film Soylent Green, or Orihara Izaya in the anime Durarara.

Transaction Broker - Examples are monster.com and expedia.com

Online marketplace - Examples are eCRATER and one4sale.

Content provider - Examples are kapook and google.

Online service provider - Example of an internet service provider, email provider, news provider (press), entertainment provider (music, movies)

Virtual community - Examples are the well and H2G2(Hitchhiker's Guide to the Galaxy).

Portal - Examples are Yahoo, AOL and MSN.

Members 1.MR.PANIT   5202500009 2.MS.THANANYA 5202500021

title: User:Minaalah


text:
User:Minijedi

Example of each Business Models

Virtual storefront - Examples are sell books online. Sell your books the easy way on websites, in emails and more santu.com/supereasy-selling. And Get eBay Paid - Get SMS. SMS to Phones on eBay item Sold Out Or PayPal Stores Payment Received www.sms4ebay.com

Information broker - An example of an information broker in contemporary fiction would be DC Comics' superheroine, the Oracle, Edward G. Robinson's character Sol in the film Soylent Green, or Orihara Izaya in the anime Durarara.

Transaction Broker - Examples are monster.com and expedia.com

Online marketplace - Examples are eCRATER and one4sale.

Content provider - Examples are kapook and google.

Online service provider - Example of an internet service provider, email provider, news provider (press), entertainment provider (music, movies)

Virtual community - Examples are the well and H2G2(Hitchhiker's Guide to the Galaxy).

Portal - Examples are Yahoo, AOL and MSN.

Members 1.MR.PANIT   5202500009 2.MS.THANANYA 5202500021

title: User:Minijedi


text:
UC Women's Group Mentoring Program/This is a sub page

Contact details
Name of storyteller: Gaynor Evans

Name of person recording story: Mark Spain

Location: The Education Centre, 5A22

Date of recording: 14 July 2010

How did you become involved in UC Women’s Group Mentoring Program (WGMP)?
I applied and my participation was approved and fully supported by my supervisor.

1. What changes have you noticed since you have finished the program?

 * 1) Increased knowledge of the university and who to ask for things. I had been working here for six years but the program helped greatly. Knowledge was reinforced by speakers from the uni coming to talk eg time management and work-life integration.
 * 2) Better self organisation skills.
 * 3) Increased confidence and comfort in my existing knowledge and skills.
 * 4) Greater ability to take on work challenges. I took on more a more front and centre leadership role by training a little group/pool of four casual staff members in our systems and procedures each year.
 * 5) I did online courses through ALIA and Uni of Sheffield - one on supervision and the other on quality assurance. I worked with an online buddy from another institution. UC paid for one course and I paid for the other. I wouldn’t have stepped up to these opportunities but for the confidence I gained and became aware of from the WGMP.
 * 6) I built a broader network across the university.
 * 7) I consolidated two part time jobs into one full time job and built more satisfaction into my work.

2. What has been the MOST significant change?
My most significant change was ceasing to work in two part time jobs and applying for and getting a different full time job.

Beginning (situation before the change) One of the part time jobs was 14hrs per week and growing in demands. It needed two people full time to do it justice. I enjoyed this job but it generated a lot of stress. This job was structured in such a way that there was never enough time to accomplish all the tasks.

I was able to talk with my mentoring group about the stress of this job and brainstorm different ways to tackle it. My supervisor was good but had no idea of the impact of the stress of the extra work that just kept coming in.

There was no sense of job satisfaction.

Middle (what happened)

A full time position at a higher level came up in the same area as my other part time job. I applied for it and won it. It was sad to relinquish the other job, because the subject matter was fascinating, but the change was better for me. This was a big decision.

In the previous job I felt like I never had a chance to finish tasks. Through discussions on the WGMP I realised that my work preferences are to be organised and to finish tasks, and look at what I have achieved and obtain a sense of satisfaction.

End (situation after)

This job change has been very good for me. It was a great benefit to me to work with my mentoring group to deconstruct what was going on and come up with some great solutions to work with. The learning insight for me was to break things down into manageable chunks and to use my improved organisation skills to help.

3. Why was this change significant for you?
It did involve a lot of thought and consideration and it’s now a whole new job for me. I’m doing a job that gives me a great sense of satisfaction. I’m now more in tune with the way I work. I like to see myself achieve things in a timeframe I can manage.

I was able to brainstorm with my WGMP buddies and follow through on the outcomes.

Whole new job! = Great significance

from Jess Dart	www.clearhorizons.com.au


text:
User talk:Lobraco1

 Hello Lobraco1, 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! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:17, 16 September 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:17, 16 September 2010 (UTC)

What is the copyright of File:Intersecting great circles.png?
Thenub314 17:01, 20 October 2011 (UTC)

title: User talk:Lobraco1