text:
User:Egm6321.f2010.team5/hw2p1

=Problem 1 - Verify that the given equation is a non-linear 1st-order ODE=

Given

 * $$\displaystyle M(x,y) + N(x,y)\underbrace{\frac{dy}{dx}}_{y'} = 0$$

Find
Verify that the given equation is an N1-ODE.

Solution
We first recognize that the given equation has 1 independent variable and its derivatives are with respect to that variable; we therefore determine that the given equation is an ordinary differential equation (ODE).

Secondly, the highest order derivative found in the given equation is 1; the given equation is thus a 1st-order ODE.

Linearity
Lastly, we check the linearity of the given equation. A linear equation $$F(y)$$ will satisfy the following two equalities:
 * 1) $$\displaystyle F(u+v) = F(u) + F(v)$$
 * 2) $$\displaystyle F(\alpha u) = \alpha F(u)$$

First condition
Let $$F(y)$$ equal the given equation,
 * $$\displaystyle F(y) := M(x,y) + N(x,y)y' = 0$$

The left side yields,
 * $$\displaystyle F(u+v) = M(x,u+v) + N(x,u+v)(u+v)'$$

The right side yields,
 * $$\displaystyle F(u) + F(v) = M(x,u) + N(x,u)u' + M(x,v) + N(x,v)v'$$

Thus,
 * $$\displaystyle F(u+v) \neq F(u) + F(v)$$

Second condition
The left side yields,
 * $$\displaystyle F(\alpha u) = M(x,\alpha u) + N(x,\alpha u)(\alpha u)'$$

The right side yields,
 * $$\displaystyle \alpha F(u) = \alpha (M(x,u) + N(x,u)u')$$

Thus,
 * $$\displaystyle F(\alpha u) \neq \alpha F(u)$$

The math above proves the non-linearity of the given equation. Thus, from the entire discussion above we can conclude that:

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

The given equation is a non-linear 1st-order ODE.
 * style="width:95%" |
 * style="width:95%" |
 * }

--Egm6321.f2010.team5.riveros 15:34, 21 September 2010 (UTC)

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

title: User:Egm6321.f2010.team5/hw2p1


text:
User:Egm6321.f2010.team5.riveros/hw2p1

=Problem 1 - Verify that the given equation is a non-linear 1st-order ODE=

Given

 * $$\displaystyle M(x,y) + N(x,y)\underbrace{\frac{dy}{dx}}_{y'} = 0$$

Find
Verify that the given equation is an nonlinear 1st-order ordinary differential equation.

Solution
We first recognize that the given equation has 1 independent variable and its derivatives are with respect to that variable; we therefore determine that the given equation is an ordinary differential equation (ODE).

Secondly, the highest order derivative found in the given equation is 1; the given equation is thus a 1st-order ODE.

Linearity
Lastly, we check the linearity of the given equation. A linear equation $$F(y)$$ will satisfy the following two equalities:
 * 1) $$\displaystyle F(u+v) = F(u) + F(v)$$
 * 2) $$\displaystyle F(\alpha u) = \alpha F(u)$$

First condition
Let $$F(y)$$ equal the given equation,
 * $$\displaystyle F(y) := M(x,y) + N(x,y)y' = 0$$

The left side yields,
 * $$\displaystyle F(u+v) = M(x,u+v) + N(x,u+v)(u+v)'$$

The right side yields,
 * $$\displaystyle F(u) + F(v) = M(x,u) + N(x,u)u' + M(x,v) + N(x,v)v'$$

Thus,
 * $$\displaystyle F(u+v) \neq F(u) + F(v)$$

Second condition
The left side yields,
 * $$\displaystyle F(\alpha u) = M(x,\alpha u) + N(x,\alpha u)(\alpha u)'$$

The right side yields,
 * $$\displaystyle \alpha F(u) = \alpha (M(x,u) + N(x,u)u')$$

Thus,
 * $$\displaystyle F(\alpha u) \neq \alpha F(u)$$

The math above proves the non-linearity of the given equation. Thus, from the entire discussion above we can conclude that:

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

The given equation is a non-linear 1st-order ODE.
 * style="width:95%" |
 * style="width:95%" |
 * }

--Egm6321.f2010.team5.riveros 15:34, 21 September 2010 (UTC)

title: User:Egm6321.f2010.team5.riveros/hw2p1


text:
User talk:Marina21~enwikiversity

 Hello Marina21, 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

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Your account will be renamed
Hello,

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

Unfortunately, your account clashes with another account also called Marina21. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Marina21~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:38, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:15, 19 April 2015 (UTC)

title: User talk:Marina21~enwikiversity


text:
Talk:Latin

Directory
Can be found here: Talk:Latin/Directory

Formatting questions
Firstly, thanks for the navigation changes, they are great!

I have some questions for :
 * 1) Can you review what I have done so far and let me know if it is ok? I am very aware that I'm making changes to something I didn't write, so I'd like you to take a look at the copy that I've had to hack around at a bit to remove Duo references and some personal asides.
 * 2) I put the word lists and sentences in bullets at the start, but I have put them into tables later on. I think tables are better, but what do you think?
 * 3) Could you let people know this is happening and see if we can get any help? It'd be better coming from you I think. It's a week of evenings to move the content across I estimate, so it'd be good to get a bit of help. I mentioned it on Reddit but didn't get any bites!

Note: there is a shortcut to link to Wiktionary entries: eg sumo which we can add in the consolidated vocab list for instance.
 * Hi – I have moved all the content across now, in rough shape. There will be a lot of things I have missed though in formatting and links, so I will need to keep going through it for a while doing error corrections. JimKillock (discuss • contribs) 21:27, 2 April 2019 (UTC)

 * Hi - I couldn't be more pleased with all the work you have done to port everything over and organize it. The graphical look is great, especially with the tables and internal links, and there is potential to make this a powerful learning tool. Formatting is a major weakness for me so I truly appreciate the time it took. As my confidence grows I will certainly tidy up the language to make it more appropriate for this forum. My writing style was personal and conversational in the Duo forums, but I envision something a little more restrained here, which you have picked up on. I apologize for being unable to help with the major work you did. At this point I am confident enough to edit the work as it stands now, but not quite ready to add new material or put things into tables. I'm sure time will help with that. Tibi multas gratias ago! --CarpeLanam (discuss • contribs) 16:30, 3 April 2019 (UTC)
 * no problem :) If you do want to add content, just let me know and I can help with the formatting. For now I want to do some more navigation things (pointing people to the practice on Memrise for each lesson). I have also been porting audio from Wiktionary into Latin/Classified Vocabulary List – this is a biggish job but not as bad as moving the lessons over! Once that is done, it may be possible to copy and past some of that into the lessons also. For sentences we will need to find volunteers however, which is a bit harder because we'll need to ask people who are good, I feel, otherwise it will generate complaints. (As an aside we should probably accept by Ecclesiastical and Classical, so long as it's made clear which is which? I personally use Classical but I've no desire to cause arguments over it!) JimKillock (discuss • contribs) 17:59, 3 April 2019 (UTC)
 * I just noticed the audio vocabulary; excellent! At some point I want to add along with the introductory material a Latin alphabet and pronunciation guide, in both classical and ecclesiastical. But I am fine with only classical pronunciation being used in any audio - it's pretty standard. I wish I were better at technical things. I can maybe aspire to recording sentence audio. My accent tends toward a melange of ecclesiastical, Americanized English, and classical pronunciation but I can do classical in a pinch, except for rolling the r's. --CarpeLanam (discuss • contribs) 02:58, 4 April 2019 (UTC)

Directory
Hi, is it possible for us to move the long lesson list off the main Latin page? I tried this and it broke the menu! JimKillock (discuss • contribs) 08:18, 3 April 2019 (UTC)

 * The sidebar menu should continue to work, but it will break the navigation links at the bottom. It would be possible for me to fix that with a bit of effort, but it might take a few days. :Separately, or somewhat related, would it be possible to identify "units" or modules for this content? Research on short-term memory suggests that we should try to break things down into lists of five to nine things, although it is also common to go as high as 15 if we are designing a full semester course. Perhaps something like Basics, Declensions, Nouns, Adjectives, Verbs, etc. would be a way to break down the content into a shorter list. -- Dave Braunschweig (discuss • contribs) 14:26, 3 April 2019 (UTC)
 * yes, I can see that. I think I would take 's advice on that, but I would note that the order is quite fixed, and that the course would probably be at least three or maybe four semesters if taught at school. The content would be dumped on students at Uni in rather less than that but this is much slower paced. I think it is six months for someone going quite fast, and probably more like 12-18 for most people to cover if starting from scratch and not hammering it. Consequently given the breadth grouping the lessons into a few names is likely to make abstract and fairly meaningless groups like "Part One: foundations" etc. JimKillock (discuss • contribs) 19:24, 3 April 2019 (UTC)

Audio files
Hi there, especially : thanks for trying to add some audio. I have found a lot of the basic words on Wikitionary, and brought most but not all of the files into the Latin/Classified Vocabulary List page. I've added the first files onto the Latin/Basics Lesson 1 so you can see how it is done – essentially you use which gives a player thus: The contributor who added these kept a good naming convention, exxcept in a few early cases, so a file is always la-cla-word.ogg or la-ecc-word.ogg if it the ecclesiastical pronunciation. JimKillock (discuss • contribs) 18:59, 4 April 2019 (UTC)

If you are adding audio, if you can keep to similar conventions that would be handy!
 * just to add, tomorrow I will try to add a guide to adding audio, including just adding a long audio file which can edit. It's all probably done via Wiki Commons. JimKillock (discuss • contribs) 22:22, 4 April 2019 (UTC)
 * Here is some further help and guidance: Latin/Contribute Audio JimKillock (discuss • contribs) 09:07, 5 April 2019 (UTC)
 * Update: I have added as many of the audio files as possible from WikiCommons, so we know what is missing from single words. We can also move onto sentences, although it'd be handy for a final check on Macrons. JimKillock (discuss • contribs) 09:04, 15 September 2019 (UTC)

Macrons
Hi there, a quick note to say that it seems inevitable that people will want macrons (editors, eg are adding them sporadically already). I think we should be ok with this, if we do audio it will make more sense with macrons in any case. Hope this is ok. JimKillock (discuss • contribs) 10:03, 14 April 2019 (UTC)

No problem, it is standard in dictionaries and textbooks, it just is quite time-consuming to be consistent and clutters the look a bit. I would not want any testing to require macron use because it tends to discourage new learners, or teach them to fixate on minutia.--CarpeLanam (discuss • contribs) 14:43, 15 April 2019 (UTC)

 * Hi there, I am working on this at last, so it would be great if you can check sometime. I am also adding in links to Wiktionary for each word as I go through the vocab pages (I didn't do this on the first pages but I may go back and do that later). JimKillock (discuss • contribs) 06:30, 1 August 2019 (UTC)

Things we can do to build on the course
Hi all,

Here's a few things I would like to do to build on the course, and things I can't do but could be useful to learners:

 * Macrons: in progress, First pass complete, second pass proofing complete; now ideally needs external checking
 * Audio: I am not sure if my pronunciation is quite up to it but I will have a go without publishing, to see and ask, and also I am asking around to see if we can get external contributors like Luke Ranieri - even a few lines from a few people who are really good could be a good start
 * Anki version: this might be a better alternative for people who find Memrise a bit awkward for sentence practice, and for more customised kinds of excercises.
 * Dialogues: Longer term, we could add short dialongues or short texts for people to read and practice with. These could also be recorded, depending on whether we have people to do it

Hope that helps, --JimKillock (discuss • contribs) 08:27, 7 September 2019 (UTC)

Page on stress needs checking
I added a page on Latin/Pronunciation Stress, but it needs checking. JimKillock (discuss • contribs) 14:04, 8 September 2019 (UTC)

Memrise Links
Memrise moved courses to community-courses.memrise.com, and links to practice are invalid now. Kiyafirs (discuss • contribs) 22:59, 26 April 2024 (UTC)

 * @Kiyafirs thanks, I will work through these and change them to the new locations JimKillock (discuss • contribs) 15:57, 6 May 2024 (UTC)

title: Talk:Latin


text:
Latin/First Declension

Latin nouns are divided into five declensions, plus irregulars. Most first declension nouns are of the feminine gender. Our example word will be: Puella, puellae, f.  girl. The first declension endings are as follows in the singular:

Here is the plural:

Common First Declension Words
First declension nouns are fairly common in Latin, and are essential to learn. To assist with this, here are a few simple nouns in the first declension that will help you in your studies. Note that the order of the forms used for the example nouns are as follows: ''Nominative, Genitive, Gender. Translation.'' (Puella, Puellae, f.  Girl) You will see this convention used in Latin dictionaries and elsewhere in the Wikiversity Latin section. Sometimes, instead of the full genitive form, only the ending will be given, as in "Puella, -ae, f. girl." In the vocabulary sections, English words derived from the latin will be provided where available.

Here are some common nouns of the first declension:
 * agricola, agricolae, m. farmer.  (Agriculture)
 * aqua, aquae, f. water.  (Aquarium)
 * femina, feminae, f. woman (Feminine)
 * insula, insulae, f. island.  (Insular)
 * ira, irae, f. anger.  (Irate)
 * nauta, nautae, m. sailor.  (Nautical)
 * puella, puellae, f. girl.
 * silva, silvae, f. forest.  (Sylvan)
 * via, viae, f. road, way

Peculiarities
Most first declension nouns are feminine. But a few first declension nouns are masculine. This is very unusual, and you can easily remember these nouns using the acronym P.A.I.N. :
 * Poeta, poetae, m. poet; pirata, piratae, m., pirate
 * Agricola, agricolae, m. farmer
 * Incola, incolae, m. inhabitant
 * Nauta, nautae, m. sailor

While noun gender does not affect declension, it does affect the endings of adjectives applied to the noun.

title: Latin/First Declension


text: File:2010 09 21 14 56 48.djvu

title: File:2010 09 21 14 56 48.djvu


text:
Latin/Nouns

Genders
a gender is the way you refer to an object. With people and living things, you refer to it by their sex, however most inanimate objects get their gender and by their nature. in English the two genders are masculine and feminine. however, in Latin, there are three genders masculine, feminine, and neuter. they are usually modified by their suffix or their declension, but there are exceptons

Number
For nouns there are two numbers, singular and plural. singular is used when there 1 of the noun in question and plural at any other time. note that it is different then person, which is for verbs instead of nouns and will be covered later

title: Latin/Nouns


text:
User talk:900674user

 Hello 900674user, 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

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 * Find out about research activities on Wikiversity.

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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 02:06, 22 September 2010 (UTC)

title: User talk:900674user


text:
User talk:Teviko

 Hello Teviko, 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.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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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. --Ottava Rima (talk) 02:07, 22 September 2010 (UTC)

title: User talk:Teviko


text:
Motivation and emotion/Book/2010/Achievement motivation

Overview
The chapter introduces the topic of achievement motivation. First, a definition of achievement motivation will be given. Next, there will be discussion on the main areas of study in achievement motivation, including implicit and self-attributed motives. There will also be some discussion on the measures used, plus some interesting fields of study using achievement motivations, such as entrepreneurs versus managers and gender difference studies. At the conclusion of the chapter the reader will be made aware of the areas of surrounding achievement motivation.

Achievement motivation
Achievement motivation is one type of motivation that helps to determine how and why an individual has behaved in a certain way. It investigates what gives some people “drive” and some don’t. This has lead to research and to the development of achievement motivation theory. The need for achievement has lead many researchers to investigate why and what makes people do and achieve different things. Many wanted to explain the factors involved in high and low achieving personalities. What emerged was not one theory or condition or behaviour, but that achievement motivation is best represented as a multidimensional construct (Ziegler, Schmukle, Egloff & Buhner, 2010). In broad terms achievement motivation represents an important construct in understanding an individual’s motivation to work hard on tasks, provide creative solutions to problems, and assess risk and to control uncertainty (Ziegler et al.).

There have been a number of different theories developed about achievement motivation, such as Murray (1938), who described it as a trait-like characteristic influencing behaviour, specifically in relation to, excelling in a task and attaining a high standard, others theories include, goal theory and expectancy value theory. These all have strong elements of achievement motivation in them. However one model that has influenced a number of these theories has been the model of need for achievement (nAch) developed by David McClelland who along with Atkinson and Murray first defined the theory of “need for achievement”, including its constructs and variables. McClelland considered the need for achievement as fundamental to achievement motivation. McClelland saw the need for achievement (nAch) related to competing with one’s own performance norms. He distinguished between hope for success and fear of failure (Ziegler, Schmukle, Egloff and Buhner, 2010).

McClellend stated that there were a number of variables that influenced a person’s ability to partake in a task, they include the difficulty of a task, the competition (rewards) and what he coined as entrepreneurship (Ziegler, Schmukle, Egloff and Buhner, 2010). These three main influences determined whether a person would take on a task (whether it be school work, a sporting activity or a work situation) and also determined how successful the person would be at completing this task. This would, as Ziegler, Schmukle, Egloff and Buhner (2010)comment depends on the behaviour of the individual, whether their fear of failure/success indicators would move them to complete and succeed in completing the task. This theory was one of the main reasons that achievement motivation was developed, to determine how and why these variables influenced a person’s behaviour.

Need for achievement
Achievement motivation refers to the tendency to set and work hard to meet personal standards and to attain goals within one’s social environment (Ziegler, Schmukle, Egloff and Buhner, 2010). There are many facets of achievement motivation that different researchers have developed ie. Goal theory, achievement behaviour theory and approach/avoidance motivation theory. These theories have all tried to answer questions, such as “What motives people to achieve, to seek success, to do well? The investigation of this question for many started with research into child development and what factors influenced a child’s ability to complete and achieve succeed at different tasks (McClelland, 1965).

This initial research highlighted three main areas of focus social influences (where children develop, what McClelland described as “achievement strivings”), cognitive influences (suggesting that some ways of thinking, such as perceptions of high ability, and high expectations influenced a child’s behaviour), and developmental influences (suggesting that achievement motivation is a struggle between approach and avoidance of achievement situations). (Hart, Stasson, Fulcher and Mahoney, 2001) undefined. The research also highlighted what McClelland, among others found was that there was a struggle between wanting to succeed and the fear of failure. This struggle between attainment of success and avoidance of fear of failure influences whether someone will be oriented towards a high achieving behaviour or low achieving behaviour (Elliot and Church, 1997).

This struggle underlies two competence-relevant motives, the need for achievement and need to avoid failure (Elliot and Church, 1997). Elliot and Church, (1997) also point out that these motives can be considered non conscious and they are posited to energise, select and drive achievement motivation. It is this need for achievement motive that McClelland focused on and developed further. McClelland (1965)summarised that need for Achievement (nAch) influenced behaviour, as it was a nonconscious driver of achievement, similar to an individual’s need to eat (hunger) and to drink. He also predicted that achievement motives predicted long term behaviour trends (McClelland, 1965). McClelland (1965) further identified two major constructs within the need for achievement motive reflecting different aspects of achieving motives, they are implicit motives and self-attributed motives.

These two constructs have emerged from research on the need for achievement as research showed there were two main areas of focus. Motives based on imaginative throughs and those based on self-reported desires or interests. While initial research, such as Murray (1938) assumed that these two motives were part of the one motive type, further research conducted by McClelland and others has helped to identify these as two separate constructs can be measured, tested and assessed to determine a number of different things (Elliot and Church, 1997), such whether an individual suits a certain occupation or whether such things as gender influence a person’s ability to achieve.

Implicit and self-attributed motives, while both aspects of achievement motivation, as McClelland, Koestner and Weinberger, (1989) suggest that the two measures of achievement motives were uncorrelated and their behaviour correlates were different. They also further suggest that these two measures should be distinguished separately in future research. This research lead McClelland to further develop the two constructs, explaining what aspects of achievement motivation they relate to and how to measure them.

Implicit motives
Implicit motives was the original major construct in need for achievement motive, it can be defined as a fantasy function, such as the nonsonscious needs that orient, select and organise behaviour (Pang and Schultheiss, 2005). This definition is similar to that for need for achievement (nAch) which is described as an almost animal like drive in a sense that it energises and selects behaviour (McClelland, Koestner and Weinberger, 1989)and argued that these implicit motives are considered to be a the major predictor of achievement behaviour. However for implicit motives to be considered a predictor there needs to be an incentive or event that triggers a reaction in a certain way. Pang and Schultheiss, (2005) describe this as motive incentives, these are the situations, such as work situations, school situations (ie exams, assignments) or other situations that require individuals to act in certain ways, such as achieving to receive a reward or to behaving to social expectations. These motive incentives play an important role in achievement motivation, Spangler (1992) proposes that if there are no achievement incentives, there is no reason to believe that people who are achievement oriented will not behave differently than those with low achievement levels. This leads to the suggestion that achievement motivation and behaviour is not only influenced by Implicit motives, but also is effected by different situations, or incentives individuals are presented with that it is the interaction between these implicit motives and incentives that activate an individual’s behaviour in an given situation.

These incentives and implicit motives are as McClelland (1985)indicates are developed by the individual and are important, and that implicit motives develop early in life as a result of early experiences to various incentives and are related to physiological processes such as norepinephrine and dopamine which support the proposition of implicit motives being an nonconscious process that directs behaviour (Spangler (1992).

As incentives are strongly related to achievement motives, McClelland, Koestner and Weinberger (1989) further defined them. They found a distinction between they determined as social and activity incentives. Activity incentives are the characteristics of the task, such as difficulty and work context can determine whether and how will an individual performs the tasks, regardless of the reward. Whereas social incentives, while not the activity itself, are the social expectations and/or rewards expected if the task is performed and performed, ie rewards, promotion, etc. These incentives are important as Spangler (1992)submits, achievement behaviour is an interactive effect of implicit motives and achievement incentives, meaning without the incentive there might not be a behavioural reaction.

Activity incentives are strongly related to implicit motiving behaviour. An individual high in implicit motives is influenced by activity incentives and in reinforced by performing the task itself and are a better at predicting behavioural trends over time. Spangler, (1992) reports that these spontaneous behavioural trends are operant in nature as they are nonconscious responses to situations. Research further suggests that implicit motives provide a general orientation and toward certain types of goals, but it is self-attributed motives that reflect social norms, which help to determine more clearly the areas in which these goals are to be accomplished. This has led to more research into the second construct, self-attributed motives.

Self-attributed motives
Self-attributed motives, are different the implicit motives. Self-Attributed motives have been classified as of value achievement (sanAch) or explicit achievement motives and can be described as the value or worth to individuals of specific achievement and are considered to predict responses to immediate and specific situations and choice of behaviour (McClelland and Watson, 1982). They are considered to be a conscious reaction and are part of an individual’s self-concept. They are usually developed later in life as well as from an understanding of social incentives and demands. They are considered more of an immediate reaction to a situation and can help to predict immediate behaviour and choice. (Hart, Stasson, Fulcher and Mahoney, 2001) undefined

As self-attributed motives help determine what immediate direction and choice is made as to where an individual wishes to achieve. Self-attributed motives are considered respondent in nature as it is behaviour that is generated and controlled by the environment and known stimuli(McClelland, Koestner and Weinberger, 1989), highlighting the differences between the two motives. Self-Attributed motives like implicit motives are also influenced by incentives. For self-attributed motives it is the social incentives that have the most influence. Spangler, (1992) states that explicit motives and goals that drive an individual to achieve and to do well in some particular socially accepted domain, such a work situation or excelling in an exam at school.

While these two motives are separate constructs they do not work alone. There is still a strong theme that neither predicts achievement behaviour alone, that both are needed to define achievement motivation. As McClelland, Koestner and Weinberger, (1989) imply that implicit motives which provide the unconscious impulse to achieve and do something well, while the conscious self-attribute motive provides the immediate choice of which actual goals will be achieved (i.e. doing well in a school course or career success), which defines that particular areas in which this impulse will express itself. They further suggest that a measure of both types of motives improved prediction of performance over what either predicts alone. Meaning that implicit motives can show that an individual has a willingness or instinct to strive or do better, but it is the be self-attributed motives that are needed to show which direction that behaviour will be, or which task an individual will perform and why.

Measures of achievement motivations
There is no one measure of achievement motivation. As there are two main types of achievement motives, implicit and self-attributed motives, there are different types of measures for both. As implicit motives was the first motive to be defined the traditional  way to measure achievement motivation is used. The first measures used is the Thematic Appreciation test (TAT), an example of this is a picture story exercise (PSE), while there are a number of different types, such as the IAT and DAMT (Ziegler, Schmukle, Egloff and Buhner, 2010) the measure basically involves 1 to 5 different pictures are shown to participants and they are asked to write a “story” about or relating to that picture. The stories are then scored and assessed by experienced trained scorers using something such as “Writers Manual for Scoring Motive Imagery” (Winter 1994) or Smith, Atkinson and McClelland, who also developed a  scale of scores to assess stories (Pang and Schultheiss, 2005). These scores determine the importance or dominance of different variables, such as achievement, power and affiliation.

Example of PSE/TAT
The PSE, to be reliable has to meet some main crucial points. As set out in (Pang and Schultheiss, 2005) example of PSE exercise http://www.pranjmehta.com/GreenStudy/PSE/PSE1.htm
 * 1) Empirically derived scoring systems or ...
 * 2) Diverse picture stimuli, usually .....

However, there has been criticism of using this method to measures achievement motivation, such as poor test-retest, internal consistency and reliability (Spangler, 1992). Much of this criticism relates to the issues of correlation as (Ziegler, Schmukle, Egloff and Buhner, 2010) remark that fantasy based measures of achievement variables do not satisfy the psychometric requirements of a good measure, whereas questionnaires measures didThis has led to researchers using different measures for the two motives, as they further note that there seems to be a growing consensus that both are distinct motivational systems and need different measures.

As these two constructs are two different and distinct achievement motives, they should and need to be measures differently, as Spangler, (1992) argues that the dispositions measures by the TAT or PSE are implicit motive that are unconscious reactions and as self-attributed motives are not, they would not be measured in the same way. This means that a different measure is needed for self-attributed motives by this measure. Self-report questionnaire have been used to better measure self-atributed motives in a number of studie, such as Pang and Schultheiss (2005), McClelland, Koestner and Weingberger (1989) and Ziegler, Schmukle, Egloff and Mahoney (2001)

Example of self-report questionnaire
Again there are a number of different types including:
 * The achievement motivation scale (revised)example of achievement motivation scale http://www.wwnorton.com/college/psych/psychsci/media/survey.htm
 * Conscientiousness facet achievement striving
 * Personality research form (PRF)

These self-report questionnaires as Spangler(1992) comments, their research supports the distinction made by McClelland and associates that TAT’s and questionnaires appear to be measuring different aspects of personality. TAT measures nonconscious needs for achievement and other motives whereas questionnaires measure more conscious values of achievement. Using the two different measures can improve prediction or performance over using one or the other alone. As McClelland, Koestner and Weingberger (1989) observe implicit motives only predict general behaviour, it cannot give an indication of which area of life in which a person will strive, self-attributed motives and goals are also needs to show the direction in which the individual’s achievement motive will turn. This again is supported by MCClelland and others, such as Pang and Schultheiss (2005), McClelland, Koestner and Weingberger (1989) and Ziegler, Schmukle, Egloff and Mahoney (2001) who argue that the two constructs need to be measured separately, but used together to achieve a better prediction of an individual’s ability to achieve and in what situations they are likely to.

Entrepreneurs vs managers
How does all of this achievement motivation theory help define or determine whether a person is going to be a successful manager or entrepreneur. McClelland (1965) conducted a longitudinal study, looking at graduates 14 years after graduation, and found that significantly more of those originally scoring high in need for achievement where currently in entrepreneurial occupations. This supports the theory that there is a strong relationship between high need for achievement scores and “entrepreneurial” behaviour. What this highlights is that individuals with higher nAch are predisposed to seeking out positions entrepreneurial nature to achieve the satisfaction they are wanting.

These entrepreneurial positions include sales, real estate, owning a small business, consulting, etc and these positions usually involve initiating decisions, individual responsibility and entail more risk. They are commonly related to need for achievement and implicit motives. Whereas positions such as credit/personal manager, office managers positions are more likely to involve high self-control, maintaining systems and managing people are considered positions, associated with low need for achievement or sanAch and high need for power and affiliation (McClelland and Boyatsiz, 1982). Research has shown that individuals seeking these positions do not have a high nAch, and more likely to have a high need for power and are predisposed to positions that provide the ability to influence and manage.

These two types of occupations are quite different, while on the surface one would propose that to achieve success as a manager or entrepreneur an individual would need a high nAch. However McClelland and Boyatcis (1982) argue that while nAch leads to success in small business or sales, where people do most of the work themselves, this motivation should not be associated with managerial success, as people with high nAch are interested in how well they do personally, not in influencing other to do well which is a major focus for managerial success.

McClelland and Boyatzis (1982) study supported their theory as it found that was no significant pattern or relationship between managers, success or personality characteristics and nAch i.e managers with high nAch were no more successful that managers with low nAch, they further suggest that a manager with nPower, nAffiliation were better predictors of success than nAch, in a managerial position as these variables had more influence on an individual’s ability to succeed in a managerial role, than high need for achievement.

Are males and females different when it comes to achievement motivation?
As discussed previously there are differences between managers and entrepreneurs. The next point to consider,is, there gender differences as well. Much of the previous research has been conducted using mainly male participants, so the question, does achievement motivation mainly relate to males or do females also have the ability to be high nach and hence have entrepreneurial behaviour is something to be considered.

When Langon-fox and Relth (1995) investigated this very issue to find out whether female entrepreneurs have the same need for achievement as mean, they firstly point out that that studies of women entrepreneurs are almost non-existent. Their research found that there were three types of female entrepreneurs:
 * 1) need achiever
 * 2) pragmatic entrepreneur
 * 3) the managerial entrepreneur

What they also found was that there the innate need for achievement is essential to those high in need for Achievement and thought to have a general orientation toward certain types of goals. That this general orientation towards achieving goals was no different for females and males (Langan-Fox and Roth, 1995). They also found supporting evidence that female like male entrepreneurs were the need achiever type and had a high need for achievement and fitted into the “need for achievement’ type defined by McClelland. This was highlighted in the first type of entrepreneur they identified. The second was, the pragmatic type, which is a mix of moderate need for Achievement and value Achievement, and the third, the managerial type, was very similar to what McClelland (1965) defined as “managerial”, occupations, with low need for Achievement, high value Achievement and high locus of control and ability to influence/have power.

What this has shown is that there are not many differences between males and females, the difference lays more in the types of definitions/classifications of type of characteristics. This is also support by a study conducted by Pang and Schultheiss, (2005)as part of their research looked at more generally whether there are gender differences and they found that women were higher then men in naffiliation, but no difference in nAchievement scores. They suggest that the gender differences relate more to power and affiliation than need for achievement.


text:
User:Egm6321.f10.team2.oztekin/HW2

=Problem 1: Verify Mdx+Ndy=0 is general N1_ODE=

Given
In general form a N1_ODE can be written as follows:

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$$M\left ( x,y \right )dx+N\left ( x,y \right )dy=0 \displaystyle$$ (Eq 1.1)
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Find
Show Eq1 has a general form.

Solution
We can write Eq as

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$$M\left ( x,y \right )+N\left ( x,y \right )\frac{dy}{dx}=M\left ( x,y \right )+N\left ( x,y \right )y^{'}\displaystyle$$ (Eq 1.2)
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Now let consider about general form as follows :

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$$P\left ( x \right )y{}'+Q\left ( x \right )y=R\left ( x \right )\displaystyle$$ (Eq 1.3)
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You can see from the lecturer notes from the 10-2

If we divide both sides with P(x);

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$$y{}'+\frac{Q(x)}{P(x)}y=\frac{R(x)}{P(x)}\displaystyle$$ (Eq 1.4)
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Then we can say that ;
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$$\bar{N}(x,y)=1\displaystyle$$ (Eq 1.5)
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$$\bar{M}(x,y)=\frac{Q(x)}{P(x)}y-\frac{R(x)}{P(x)}\displaystyle$$ (Eq 1.6) So we concluded that
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$$\bar{M}(x,y)+\bar{N}(x,y)y{}'=0\displaystyle$$ (Eq 1.7)
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Another example for this porblem we can give in book that it says in the page 511 for the Variables Separable kind of differential equations.

It is shown as :

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$$g(y)\frac{dy}{dt}=f(t)\displaystyle$$ (Eq 1.8)
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To solve this ;

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$$\int g(y)\frac{dy}{dt}=\int f(t)\displaystyle$$ (Eq 1.9)
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$$\int g(y)dy=\int f(t)dt\displaystyle$$ (Eq 1.10)
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We can rearrange this

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$$g(y)\frac{dy}{dt}-f\left ( t \right )=0\displaystyle$$ (Eq 1.11)
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And again

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$$\bar{N}(t,y)=g(y),\bar{M}(t,y)=-f(t)\displaystyle$$ (Eq 1.12)
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$$\bar{N}(t,y)\frac{dy}{dt}+\bar{M}(t,y)=0\displaystyle$$ (Eq 1.13)
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An ordinary differential equation of the form

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$$y'+ P(x)y = Q(x)y^n\displaystyle$$ (Eq 1.14)
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is called a Bernoulli equation when n≠1, 0. Bernoulli equations are special because they are nonlinear differential equations with known exact solutions. Dividing by $$y^n$$ yields
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$$\frac{y'}{y^{n}} + \frac{P(x)}{y^{n-1}} = Q(x)\displaystyle$$ (Eq 1.15)
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If we use changing of variables method...
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$$w=\frac{1}{y^{n-1}}\displaystyle$$ (Eq 1.16)
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$$w'=\frac{(1-n)}{y^{n}}y'\displaystyle$$ (Eq 1.17)
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$$\frac{w'}{1-n} + P(x)w = Q(x)\displaystyle$$ (Eq 1.18)
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The substituted equation can be solved using the integrating factor
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$$h(x)= e^{(1-n)\int P(x)dx}\displaystyle$$ (Eq 1.19)
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By showing all of these now we can say that for every type first order nonlinear differansial equations can be written in Eq 1.1 type by some arrangements.

=Problem 2 : Show why Eq 7 in page (7-1) is first order and non linear=

Given
Eq 7 in page 7-1
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$$\left ( 4*x^7+sin(y) \right )+\left ( x^2*y^3 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=0 \displaystyle$$ (Eq 2.1)
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Find
Show why the equation above is first order and non linear

Solution
Since we define our first order differential $$F\left ( x,y,y{}' \right )=0$$ there must be only first degree differential of dependent variable in the function.If we look at the example we can notice that there is only first degree differential of dependent variable multiplied by N(x,y) so we conclude that this is first order differential equation.

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$$F\left ( x,y \right )=\left ( 4*x^7+sin(y) \right )+\left ( x^2*y^3 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=0\displaystyle $$ (Eq 2.2)
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$$F\left ( x,(\alpha u+\beta v) \right )=\left ( 4*x^7+sin(\alpha u+\beta v) \right )+\left ( x^2*(\alpha u+\beta v)^3 \right )(\alpha \frac{\mathrm{d} u}{\mathrm{d} x}+\beta \frac{\mathrm{d} v}{\mathrm{d} x})=0\displaystyle $$ (Eq 2.3)
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$$F\left ( x,\alpha u \right )+F\left ( x,\beta v \right )=\left ( 4*x^7+sin(\alpha u) \right )+\left ( x^2*(\alpha u)^3 \right )\alpha \frac{\mathrm{d} u}{\mathrm{d} x}+\left ( 4*x^7+sin(\beta v) \right )+\left ( x^2*(\beta v)^3 \right )\beta \frac{\mathrm{d} v}{\mathrm{d} x} \displaystyle $$ (Eq 2.4)
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$$F\left ( x,\alpha u \right )+F\left ( x,\beta v \right )\neq F\left ( x,(\alpha u+\beta v) \right )\displaystyle$$ (Eq 2.5)
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=Problem3:Plot Homogeneous Solutions And Show independence=

Given

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$$y_{H}^{1}(x)=x\displaystyle $$ (Eq 3.1)
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$$y_{H}^{2}(x)=\frac{x}{2}*log\left ( \frac{(1+x)}{(1-x)} \right )-1 \displaystyle $$ (Eq 3.2)
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Linear Dependence and Independence
Let us consider defined vector space called V. And x1,x2,x3,....xk are the vectors in this space. E={x1,x2,.....xk} is the cluster of these vectors in this vector space. Now look at the condition of c1*x1+c2*x2+......ck*xk=0.

If the coefficients of x1,x2,...xk which are c1,c2,....ck can not be all together zero then we can say that x1,x2,....xk vectors are linearly dependent to each other. But if the statement c1*x1+c2*x2+.....ck*xk=0 can be valid for only for only where all coefficients are zero which is c1=c2=......ck=0 then we can say that vectors in V space are 'linearly independent'.

Find
Plot $$ y_{H}^{1}(x)$$and$$y_{H}^{2}(x)$$ for visual introduction.

Show that we can not derive $$ y_{H}^{1}(x)$$ from multiplying $$y_{H}^{2}(x)$$ with $$\alpha$$

Solution
IMPORTANT NOTE:For plotting we define the interval of x between -1 to 1 to avoid imaginary parts of $$y_{H}^{2}(x)$$

 Matlab Code:

Since for any X

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$$ y_{H}^{1}(x)-\alpha*y_{H}^{2}(x)=x-\alpha *\frac{x}{2}*log\left ( \frac{(1+x)}{(1-x)} \right )-1\neq 0 \displaystyle $$ (Eq 3.3)
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we can sat that $$ y_{H}^{1}(x) $$ and $$ y_{H}^{2}(x)$$ are independent homogeneous solutions.

=Problem 4=

Given

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$$ \phi (x,y)=x^{2}y^{3/2}+log(x^{3}y^{2}) \displaystyle $$ (Eq 4.1)
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Find
$$ F(x,y,{y}')=\frac{\partial \phi (x,y)}{\partial x} \displaystyle $$, verify that $$ F(x,y,{y}') \displaystyle $$ is an exact non-linear ODE and invent three more non-linear ODE's

Solution
$$ F(x,y,{y}')=\frac{\partial \phi (x,y)}{\partial x}=M(x,y)+N(x,y){y}' \displaystyle $$

where:
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$$ M=\frac{\partial \phi }{\partial y} \displaystyle $$ (Eq 4.2) and
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$$ N=\frac{\partial \phi }{\partial x} \displaystyle $$ (Eq 4.3) Solving for M and N we get:
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$$ M=2xy^{3/2}+\frac{3}{xlog10} \displaystyle $$ (Eq 4.4)
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$$ N=\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10} \displaystyle $$ (Eq 4.5) Plugging into the equation we get:
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$$ F(x,y,{y}')=(2xy^{3/2}+\frac{3}{xlog10})+(\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}){y}' \displaystyle $$ (Eq 4.6) In order for an equation to be an exact N1-ODE the following 2 conditions must be satisified:
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Condition 1
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$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (Eq 4.7) Condition 2
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$$ M_{y}(x,y)=0=N_{x}(x,y) \displaystyle $$ (Eq 4.8)
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For condition 1:
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$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (Eq 4.9) Using $$ M(x,y) \displaystyle $$ and $$ N(x,y) \displaystyle $$ from above we can plug in and we find:
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$$ (2xy^{3/2}+\frac{3}{xlog10})+(\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}){y}'=0 \displaystyle $$ (Eq 4.10) For Condition 2:
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x} \displaystyle $$ (Eq 4.11)
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$$ \frac{\partial M(x,y)}{\partial y}=3xy^{1/2} \displaystyle $$ (Eq 4.12)
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$$ \frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (Eq 4.13) Therefore:
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (Eq 4.14) and $$ F(x,y,{y}') \displaystyle $$ is an exact N1-ODE.
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Three exact N1-ODE's

 Example 1

Let's consider following equation:

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$$\phi \left ( x,y \right )=\frac{x^{2}}{y^{3}}-\frac{1}{y}+c \displaystyle $$ (Eq 4.15)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=\frac{2x}{y^{3}}\displaystyle $$ (Eq 4.16)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=\frac{y^{2}-3x^{2}}{y^{4}}\displaystyle $$ (Eq 4.17)
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Providing second condition of exactness :

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$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=\frac{-6x}{y^{4}}\displaystyle $$ (Eq 4.18)
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Now we can write :

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$$M(x,y)dx+N(x,y)dy=\frac{2x}{y^{3}}dx+\frac{y^{2}-3x^{2}}{y^{4}}dy=0\displaystyle $$ (Eq 4.19)  Example 2
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Let's consider following equation:

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$$\phi \left ( x,y \right )=e^xsiny \displaystyle $$ (Eq 4.20)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=e^xsiny\displaystyle $$ (Eq 4.21)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=-e^xcosy\displaystyle $$ (Eq 4.22)
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Providing second condition of exactness :

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$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=-e^xcosy\displaystyle $$ (Eq 4.23)
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Now we can write :

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$$M(x,y)dx+N(x,y)dy=e^xsinydx+-e^xcosydy=0\displaystyle $$ (Eq 4.24)
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 Example 3

Let's consider following equation:

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$$\phi \left ( x,y \right )=x^{1/2}lny \displaystyle $$ (Eq 4.25)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=\frac{x^{-1/2}lny}{2}\displaystyle $$ (Eq 4.26)
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 * }

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

$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=\frac{x^{1/2}}{y}\displaystyle $$ (Eq 4.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Providing second condition of exactness :

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

$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=\frac{x^{-1/2}}{2y}\displaystyle $$ (Eq 4.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now we can write :

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

$$M(x,y)dx+N(x,y)dy=\frac{x^{-1/2lny}}{2}dx+\frac{x^{1/2}}{y}dy=0\displaystyle $$ (Eq 4.29)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5: Giving example of ODEs whcih cannot satisfy the first condition of exactness =

Given

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

$$ \displaystyle M(x,y)+N(x,y)f(y')=0$$ (Eq 5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

Solution
If we were to create the following parts of such an equation that is described in Eq 5.1, we can get the following functions out of an infinite possibilities pool.

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

M(x,y)=5xy$$

(Eq 5.2)
 *

 * }

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

$$ \displaystyle N(x,y)={{x}^{2}}{{y}^{3}}$$ (Eq 5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle f(y')=10y'{{e}^{y'}}$$ (Eq 5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then plugging in all of the equations into Eq 5.1 will get us the following:

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

$$ \displaystyle 5xy+{{x}^{2}}{{y}^{3}}(10y'{{e}^{y'}})=0$$ (Eq 5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Reorganizing the equation to solve for y’ will yield

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

$$ \displaystyle 5xy+{{x}^{2}}{{y}^{3}}(10y'{{e}^{y'}})=0$$ (Eq 5.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle {{x}^{2}}{{y}^{3}}(10y'{{e}^{y'}})=-5xy$$ (Eq 5.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle y'{{e}^{y'}}=-\frac{5xy}{10{{x}^{2}}{{y}^{3}}}=-\frac{1}{2xy}$$ (Eq 5.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finally taking the natural logarithms of both sides will yield

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

$$ \displaystyle \ln (y')+y'=\ln (-\frac{1}{2xy})$$ (Eq 5.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Therefore, it can be concluded that $$\displaystyle f(y')=10y'{{e}^{y'}}$$ will not yield an exact equation for the given $$\displaystyle M(x,y)$$ & $$\displaystyle N(x,y)$$.

= Problem 6: Verifying the solution of a N1-ODE =

Given

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

$$ \displaystyle y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$, (Eq 6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle M(x,y)+N(x,y)y'=0$$, (Eq 6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle \left\{ \begin{align} & M(x,y)=75{{x}^{4}} \\ & N(x,y)=\cos y \\ \end{align} \right.$$.

Find
Verify that Eq 6.1 satisfies Eq 6.2.

Step 1
When we plug in the given $$\displaystyle M(x,y)$$ & $$\displaystyle N(x,y)$$ values into Eq 6.2, we get:

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

$$ \displaystyle 75{{x}^{4}}+(\cos y)y'=0$$. (Eq 6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Step 2
So now let’s look at Eq 6.1 and rewrite it in a different from such that

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

$$ \displaystyle y(x)={{\sin }^{-1}}(k-15{{x}^{5}})\Rightarrow \sin y=(k-15{{x}^{5}})$$. (Eq 6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Step 3
Taking the derivative of both sides of the new form we get the following:

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

$$ \displaystyle (\sin y)'=(k-15{{x}^{5}})'$$, (Eq 6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle (\cos y)y'=-75{{x}^{4}}$$. (Eq 6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Step 4
Rearranging the result from Step 3 will give us $$\displaystyle 75{{x}^{4}}+(\cos y)y'=0$$; thereby, proof is complete.

= Problem 7: Solving General Non-Homogeneous Linear First Order Ordinary Differential Equations =

Given

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

$$ \displaystyle {y}'+xy=2x+3$$, (Eq 7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle {y}'+\frac{{{a}_{0}}(x)}{{{a}_{1}}(x)}y=\frac{b(x)}{{{a}_{1}}(x)}$$, (Eq 7.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where we assume that for any $$\displaystyle x $$, $$\displaystyle {{a}_{1}}(x)\ne 0 $$.

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

$$ \displaystyle ({{x}^{2}}+1){y}'+xy=2x$$. (Eq 7.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Find
Solutions regarding Eq 7.1 ~ Eq 7.3.

Solution
Before we get started, we should find the method of deriving solutions of non-homogeneous 1st order ODEs.

Method by King et al.
Consider a general non-homogeneous 1st order ODE like:

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

$$ \displaystyle {{a}_{1}}(x){y}'+{{a}_{0}}(x)y=b(x)$$, (Eq 7.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If for any $$\displaystyle x $$, $$\displaystyle {{a}_{1}}(x)\ne 0 $$, then we get another form of $$Eq 7.4 $$,

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

$$ \displaystyle {y}'+P(x)y=Q(x)$$, (Eq 7.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$P(x)=\frac{{{a}_{0}}(x)}{{{a}_{1}}(x)}$$ and $$Q(x)=\frac{b(x)}{{{a}_{1}}(x)}$$.

If we let factor $$\displaystyle \exp \left\{ \int_ – ^{x}{P(s)ds} \right\}$$ times each sides of Eq 7.5, we derive,

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

$$ \displaystyle \exp \left\{ \int_ – ^{x}{P(s)ds} \right\}\frac{dy}{dx}+\exp \left\{ \int_ – ^{x}{P(s)ds} \right\}P(x)y=Q(x)\exp \left\{ \int_ – ^{x}{P(s)ds} \right\}$$. (Eq 7.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can see that the left side of the equation can be simplified, thus,

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

$$ \displaystyle {{h}^{1}}\frac{dy}{dx}+y\frac{d{{h}^{1}}}{dx}=Q{{h}^{1}}$$, (Eq 7.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

By integrating Eq 7.7 we have the solution,

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

$$ \displaystyle y(x)=\frac{1}{{{h}^{1}}(x)}\int_ – ^{x}{{{h}^{1}}(s)Q(s)}ds$$, (Eq 7.8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$\displaystyle {{h}^{^{1}}}(x)=\exp \left\{ \int_ – ^{x}{P(s)ds} \right\} $$.

Method by finding exact total differential
We can also get the solution fully under the idea of using exact total differential to eliminate the $$\displaystyle {{y}'} $$ and utility of line integral. By doing this, Eq 7.5 can be also be written as,

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

$$ \displaystyle \left[ P(x)y-Q(x) \right]dx+dy=0$$. (Eq 7.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If we find a function $$\displaystyle \mu (x,y)$$ whose total differential is same as the left side of Eq 7.9, then we will have,

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

$$ \displaystyle \mu (x,y)=C$$, (Eq 7.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle C$$ is some constant, then we get the implicit or explicit solution from Eq 7.10.

However the prerequisite of existence of such $$\displaystyle \mu (x,y)$$ is not assured, so we need a factor multipled to the Eq 7.9 to make it satisfy the second condition of exactness. Let's call this factor $$\displaystyle {{h}^{2}}$$ and assume it a function only of $$\displaystyle x$$. We have,

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

$$ \displaystyle {{h}^{2}}(x)\left[ P(x)y-Q(x) \right]dx+{{h}^{2}}(x)dy=0$$. (Eq 7.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq 7.11 must satisfy the second condition of exactness, hencce we obtain,

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

$$ \displaystyle \frac{\partial \left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right] \right\}}{\partial y}=\frac{\partial {{h}^{2}}(x)}{\partial x}$$ (Eq 7.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \Rightarrow {{h}^{2}}(x)=\exp \int_ – ^{x}{P(s)}ds$$ (Eq 7.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can see that the assumption "$$\displaystyle {{h}^{2}}(x)$$ is a function only of $$\displaystyle x$$" is automatically satisfied and $$\displaystyle {{h}^{2}}(x)$$ is actually equal to $$\displaystyle {{h}^{1}}(x)$$.

Now we can concentrate on getting the critical important function: $$\displaystyle \mu (x,y)$$.

Theorem 7-1
Let $$\displaystyle I(x,y)$$ and $$\displaystyle J(x,y)$$ be both smooth functions in its domain and such relationship is satisfied:

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

$$ \displaystyle \frac{\partial I}{\partial y}=\frac{\partial J}{\partial x}$$. (Eq 7.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then the total differential of function $$\displaystyle u(x,y)=\int_{({{x}_{0}},{{y}_{0}})}^{(x,y)}{I(x,y)dx+J(x,y)dy}$$ equals $$\displaystyle I(x,y)dx+J(x,y)dy$$ in which the $$\displaystyle u(x,y)$$ is a line integral and $$\displaystyle ({{x}_{0}},{{y}_{0}})$$ is an arbitrary point.

Verification:

To illustrate this theorem, we just need to prove relationships below:

 * $$\displaystyle \frac{\partial u}{\partial x}=I(x,y)$$, $$\displaystyle \frac{\partial u}{\partial y}=J(x,y)$$

Then we have,

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

$$ \displaystyle \frac{\partial u}{\partial x}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{u(x+\Delta x,y)-u(x,y)}{\Delta x}$$ (Eq 7.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The Eq 7.14 satisfied the condition of path independence of line integrals(please see Green's theorem), therefore, line integral $$\displaystyle u(x+\Delta x,y)=\int_{({{x}_{0}},{{y}_{0}})}^{(x+\Delta x,y)}{I(x,y)dx+J(x,y)dy}$$ can be designated onto 2 line segments:

 * $$\displaystyle ({{x}_{0}},{{y}_{0}})\to (x,y)\to (x+\Delta x,y)$$. We obtain,

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

$$ \displaystyle u(x+\Delta x,y)=u(x,y)+\int_{(x,y)}^{(x+\Delta x,y)}{I(x,y)dx+J(x,y)dy}$$, (Eq 7.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \Rightarrow u(x+\Delta x,y)-u(x,y)=\int_{x}^{x+\Delta x}{I(x,y)dx}$$, (Eq 7.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \Rightarrow u(x+\Delta x,y)-u(x,y)=I(x+\omega *\Delta x,y)\Delta x$$, (Eq 7.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle 0\le \omega \le 1$$(please see Intermediate value theorem).

Then we substituting Eq 7.18 back to Eq 7.15, we derive,

 * $$\frac{\partial u}{\partial x}=I(x,y)$$.

Similarly we have,

 * $$\frac{\partial u}{\partial y}=J(x,y)$$.

Thus the whole theorem is proved.

Now we go back to the Eq 6.11. Because Eq 7.12 meets the prerequisite of Eq 7.14, we can apply Theorem 7-1 on Eq 7.11, we have,

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

$$ \displaystyle \begin{align} u(x,y) & =\int_{({{x}_{0}},{{y}_{0}})}^{(x,y)}{\left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right]dx+{{h}^{2}}(x)dy \right\}} \\ & =\int_{({{x}_{0}},{{y}_{0}})}^{(x,{{y}_{0}})}{\left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right] \right\}dx}+\int_{(x,{{y}_{0}})}^{(x,y)}{{{h}^{2}}(x)dy} \end{align}$$, (Eq 7.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle ({{x}_{0}},{{y}_{0}})$$ is an arbitrary point. And if combined with Eq 7.10, we are able to get the solution from the equation below:

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

$$ \displaystyle \int_{({{x}_{0}},{{y}_{0}})}^{(x,{{y}_{0}})}{\left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right] \right\}dx}+\int_{(x,{{y}_{0}})}^{(x,y)}{{{h}^{2}}(x)dy}=C$$, (Eq 7.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle C$$ is some constant.

Solution for Eq 7.1
We use the first method.

From Eq 7.1 we know,

$$\displaystyle P(x)=x$$, $$\displaystyle Q(x)=2x+3$$.

The integrating factor is,

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

$$ \displaystyle \begin{align} h= & \exp \int_ – ^{x}{sds} \\ & ={{e}^{\tfrac{2}}} \end{align}$$. (Eq 7.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

So,

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

$$ \displaystyle \begin{align} y & =\frac{1}{h(x)}\int_ – ^{s}{h(s)Q(s)ds} \\ & ={{e}^{-\frac{2}}}\int_ – ^{x}(2s+3)ds \\ & ={{e}^{-\frac{2}}}\left( 2\int_ – ^{x}{{{e}^{\frac{2}}}d\frac{2}+3\int_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}ds}} \right) \\ & =2+{{e}^{-\frac{2}}}*3\int_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}ds} \\ & =2+\frac{3\sqrt{\pi }}{2}{{e}^{-\frac{2}}}*\left( erfi(x)+C \right) \end{align}$$, (Eq 7.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle C$$ is some constant.

The term $$\displaystyle \int_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}ds}$$ is unable to be described by the combination of elementary functions and their integrals (please see Lists of integrals). But it can be described by imaginary error function(the term $$\displaystyle erfi(x)$$ in Eq 7.22), see also Error Function.

Solution for Eq 7.2
From Eq 7.2 we know,

$$P(x)=\frac{{{a}_{0}}(x)}{{{a}_{1}}(x)}$$ and $$Q(x)=\frac{b(x)}{{{a}_{1}}(x)}$$.

Substituting these two into Eq 7.8, we dereive,

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

$$ \displaystyle y(x)=\frac{1}{\exp \left\{ \int_ – ^{x}{\frac{{{a}_{0}}(s)}{{{a}_{1}}(s)}ds} \right\}}\int_ – ^{x}{\exp \left\{ \int_ – ^{x}{\frac{{{a}_{0}}(s)}{{{a}_{1}}(s)}ds} \right\}\frac{b(s)}{{{a}_{1}}(s)}ds}$$ (Eq 7.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution for Eq 7.3
We use second method.

As $$\displaystyle {{a}_{1}}=({{x}^{2}}+1)\ne 0$$, Eq 7.3 can de written as:

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

$$ \displaystyle {y}'+\frac{x}{{{x}^{2}}+1}y=\frac{2x}{{{x}^{2}}+1}$$ (Eq 7.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The integation factor is,

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

$$ \displaystyle h=\exp \int_ – ^{x}{\frac{s}{{{s}^{2}}+1}ds}={{\left( {{x}^{2}}+1 \right)}^{\frac{1}{2}}}$$ (Eq 7.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Hence Eq 7.24 can be written as,

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

$$ \displaystyle \frac{\left( y-2 \right)x}{\sqrt{{{x}^{2}}+1}}dx+\sqrt{{{x}^{2}}+1}dy=0$$ (Eq 7.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Integrate Eq 7.26, we have,

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

$$ \displaystyle \begin{align} & \int_{(0,0)}^{(x,y)}{\left\{ \frac{\left( y-2 \right)x}{\sqrt{{{x}^{2}}+1}}dx+\sqrt{{{x}^{2}}+1}dy \right\}=0} \\ & \Rightarrow \int_{(0,0)}^{(x,0)}{\frac{\left( y-2 \right)x}{\sqrt{{{x}^{2}}+1}}dx}+\int_{(x,0)}^{(x,y)}{\sqrt{{{x}^{2}}+1}dy}=0 \\ & \Rightarrow -2\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}+1}*y=C \\ & \Rightarrow y=\frac{C}{\sqrt[{}]{{{x}^{2}}+1}}+2 \\ & \\ \end{align}$$, (Eq 7.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

where $$\displaystyle C$$ is some constant.

= Problem 8: Ordinary Differential Equations with Varying Coefficients (ODE_VC) =

Given
Show that the constant, k1, is not necessary when the following expression is evaluated:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=\frac{1}{h\left( x \right)}\int_ – ^{x}{h\left( s \right)b\left( s \right)ds}

$$ (eq 8.1) [1]
 *

 * }

Solution
The function h(x), given in Equation 8.1, is further expressed as:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

h\left( x \right)={{e}^{\int_ – ^{x}{{{a}_{0}}\left( s \right)ds}}}

$$ (eq 8.2) [2]
 *

 * }

By converting the integral using the following principle:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int_ – ^{x}{f\left( s \right)ds}=\int{f\left( x \right)dx}+k $$ (eq 8.3) [2]
 *

 * }

The function h can be expressed as follows:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

h\left( x \right)={{e}^{\left( \int{{{a}_{0}}\left( x \right)dx} \right)+k}}={{e}^{k}}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}} $$ (eq 8.4)
 *

 * }

Setting the constant ek to k yields the following:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

h\left( x \right)={k}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}} $$ (eq 8.5)
 *

 * }

Plugging Equation 8.4 into both instances of h(x) in Equation 8.1 yields the following:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=\frac{1}\int_ – ^{x}{{{k}_{2}}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}}b\left( s \right)ds}=\frac{{e}^{-\int{{{a}_{0}}\left( x \right)dx}}}\int_ – ^{x}{{{e}^{\int{{{a}_{0}}\left( x \right)dx}}}b\left( s \right)ds}

$$ (eq 8.6)
 *

 * }

Let the constant k2=k3k1 which therefore makes k1 unnecessary.

Given
Show that Equation 8.1 agrees with the following:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=A{{y}_{h}}\left( x \right)+{{y}_{p}}\left( x \right)

$$ (eq 8.7) [3]
 *

 * }

Solution
Applying the principle in Equation 7.3 directly into Equation 8.1 yields the following:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=h{{\left( x \right)}^{-1}}\left[ \int{h\left( x \right)b\left( x \right)dx}+k \right]

$$ (eq 8.8)
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=kh{{\left( x \right)}^{-1}}+h{{\left( x \right)}^{-1}}\int{h\left( x \right)b\left( x \right)dx}

$$ (eq 8.9)
 *

 * }

It is shown that the two equations agree by letting A=k and making the following substitutions:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{h}}\left( x \right)=h{{\left( x \right)}^{-1}}

$$ (eq 8.10)
 *

 * }

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)=h{{\left( x \right)}^{-1}}\int{h\left( x \right)b\left( x \right)dx}

$$ (eq 8.11)
 *

 * }

Given
Find yH independently by solving the following equation:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y'+{{a}_{0}}y=0

$$ (eq 8.12)
 *

 * }

Solution
Equation 8.12 can be converted into the following form:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac{dx}+{{a}_{0}}{{y}_{H}}=0

$$ (eq 8.13)
 *

 * }

In solving for yH, it follows that

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{}^{d{{y}_{H}}}\!\!\diagup\!\!{}_\;=-{{a}_{0}}dx

$$
 *

 * }

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int{{}^{d{{y}_{H}}}\!\!\diagup\!\!{}_\;}=-\int{{{a}_{0}}dx} $$
 *

 * }

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\ln \left( {{y}_{H}} \right)=-\int{{{a}_{0}}dx}+k $$
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{H}}=K{{e}^{-\int{{{a}_{0}}dx}}}$$
 *

 * }

= Problem 9: Exact Differential Equations =

Given
Find a N1_ODE that is either exact or can be made exact using the following equation:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\bar{b}\left( x \right)c\left( y \right)y'+a\left( x \right)\bar{c}\left( y \right)=0

$$ (eq 9.1)
 *

 * }

With the following values
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

a\left( x \right)=\sin \left( {{x}^{3}} \right) $$ (eq 9.2)
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

b\left( x \right)=\cos \left( x \right) $$ (eq 9.3)
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

c\left( y \right)={e}^{2y} $$ (eq 9.4)
 *

 * }

Solution
The functions with superscript bars are integrals of the original functions and thus the following equations are derived:

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

b\left( x \right)=\cos \left( x \right)\Rightarrow \bar{b}\left( x \right)=\sin \left( x \right) $$ (eq 9.5)
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

c\left( x \right)={{e}^{2y}}\Rightarrow \bar{c}\left( x \right)={}^{1}\!\!\diagup\!\!{}_{2}\;{{e}^{2y}} $$ (eq 9.6) Inserting all of the values into Equation 9.1 yields the following:
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\underbrace{\sin \left( x \right)\left( {{e}^{2y}} \right)}_{N\left( x,y \right)}y'+\underbrace{\sin \left( {{x}^{3}} \right)\left( {}^{1}\!\!\diagup\!\!{}_{2}\;{{e}^{2y}} \right)}_{M(x,y)}=0 $$ (eq 9.7) Equation 9.7 meets the first condition of exactness which states that:
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

N\left( x,y \right)y'+M\left( x,y \right)=0 $$ (eq 9.8)
 *

 * }

Divide Equation 9.7 by e2y.

Next we can check to see if the equation passes the second condition of exactness in which My must equal Nx
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{M}_{y}}\left( x,y \right)=0 $$ (eq 9.9)
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{N}_{x}}\left( x,y \right)=\cos \left( x \right) $$ (eq 9.10) From the previous equations it is evident that
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{M}_{y}}\left( x,y \right)\ne {{N}_{x}}\left( x,y \right) $$ (eq 9.11) Therefore, in order to make the equation exact, Euler's integrating factor, h(x) must be derived as follows:
 *

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

$$ \displaystyle h\left( x \right)={{e}^{-\int{\frac{1}{N}\left( {{N}_{x}}-{{M}_{y}} \right)dx}}} $$ (eq 9.12)4
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle ={{e}^{-\int{\frac{1}{\sin \left( x \right)}\left( \cos \left( x \right)-0 \right)dx}}} $$
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle ={{e}^{-\ln \left( \sin \left( x \right) \right)}} $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle h\left( x \right)=\frac{1}{\sin \left( x \right)} $$ (eq 9.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

In order to verify that the derived integrating factor is correct it is necessary to multiply Equation 8.8 by h(x) and denote the new values for M and N as follows:
 * {| style="width:100%" border="0"

$$ \displaystyle \underbrace{\left( \frac{1}{\sin \left( x \right)} \right)\left( \sin \left( x \right) \right)}_{\bar{N}=hN\left( x,y \right)}y'+\underbrace{\left( \frac{1}{\sin \left( x \right)} \right)\left( \frac{\sin \left( {{x}^{3}} \right)}{2} \right)}_{\bar{M}=hM\left( x,y \right)}=0 $$ (eq 9.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Again we check to see if the second condition of exactness has been met as follows:

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

$$ \displaystyle {{\bar{M}}_{y}}\left( x,y \right)=0 $$ (eq 9.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle {{\bar{N}}_{x}}\left( x,y \right)=0 $$ (eq 9.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The N1_ODE has been made exact since

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

$$ \displaystyle {{\bar{M}}_{y}}\left( x,y \right)={{\bar{N}}_{x}}\left( x,y \right) $$ (eq 9.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Given
Find the first integral of the N1_ODE used in the previous problem so that

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\phi \left( x,y \right)=k

$$ (eq 9.18)
 *

 * }

Solution
Considering the following equations:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

M\left( x,y \right)={{\phi }_{x}}\left( x,y \right) $$ (9.19)5
 *

 * }

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

N\left( x,y \right)={{\phi }_{y}}\left( x,y \right) $$ (9.20)5
 *

 * }

The following equations can be derived:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int{M\left( x,y \right)dx}+w\left( y \right)=\phi \left( x,y \right) $$ (9.21)6
 *

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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int{N\left( x,y \right)dy}+v\left( x \right)=\phi \left( x,y \right) $$ (9.22)6
 *

 * }

By setting equations 8.21 and 8.22 equal to each other, the functions w(y) and v(x) can be found. From there, equation 8.13 can be solved.
 * {| style="width:100%" border="0"

$$ \displaystyle \int{M}\left( x,y \right)dx+w\left( y \right)=\int{N}\left( x,y \right)dy+v\left( x \right) $$ (eq 9.23)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle \int{\frac{\sin \left( {{x}^{3}} \right)}{2\sin \left( x \right)}dx}+w\left( y \right)=y+v\left( x \right) $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$ \displaystyle \phi \left( x,y \right)=\int{\frac{\sin \left( {{x}^{3}} \right)}{2\sin \left( x \right)}dx}+y=k $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

=Contributions Of Team Members=

Pages
Patrick's Page Zou's Page Patterson's Page

=References=

1Vu-Quoc,L. EGM 6321.F10, 10-3, Equation (6)

2Vu-Quoc,L. EGM 6321.F10, 11-1:11-2

3King, A.C. et al. Differential Equations: Linear, Nonlinear, Ordinary, Partial. Cambridge: Cambridge University Press

4Vu-Quoc,L. EGM 6321.F10, 10-1, Equation (6)&(7)

5Vu-Quoc,L. EGM 6321.F10, 8-2, Equations (1)&(2)

6Zill, D., and Cullen, M. Differential Equations with Boundary-Value Problems. Pacific Grove, CA: Brooks/ Cole,

title: User:Egm6321.f10.team2.oztekin/HW2


text:
User:EGM6321.F10.TEAM1.HW3

Given
For $$ k \ne \ 0\ $$ and $$ v_{x,0}=0\ $$, Eq 1.2 becomes

Find
1) Derive the Equations of motion, (1.1) and (1.2) 2) Verify $$ y(x)\ $$ is a parabola, for the particular case of $$ k=0\ $$ 3.1) Utilizing eq1.3, find $$ v_y(t)\ $$ and $$ y(t)\ $$ for m=constant 3.2) Utilizing eq1.3, find $$ m=m(t)\ $$

Solution
1) Derive the Equations of motion, (1.1) and (1.2) Newton's 1st Law of motion says that a change in a particle's momentum with respect to time is due directly to the net (resultant) force acting on that object. If we assume that the object is moving in an inertial reference frame then we may express Newton's 1st Law mathematically as

What we mean by inertial reference frame is that the coordinate system in which we are observing the motion of the particle is not accelerating; i.e. is moving at the same exact speed as the coordinate system in which we are calculating the motion of the particle. Momentum is defined by the product of both the mass of the particle, and the velocity at which the particle is traveling.

Therefore we may rewrite Newton's 1st Law as If we look closely we see that by stating the change in momentum with respect time is due to the net force, and not just to any force it is implied that a summation of multiple forces acting on the particle is required. Also we must note that because force is a vector quantity, we will be summing vectors which have both a magnitude and direction (this will be important when calculating a force's action on the path and velocity a particle travels). We will assume that the only forces acting on the particle while in motion are gravity and some retarding force from the viscosity of the medium the particle is traveling through. Their summation is expressed as

In cases where we are observing the motion of an object that does not lose a measurable amount of mass during its travel (measurable is a relative term, meaning measurable by only techniques used with in the scale of the experiment), we can assume m to be a constant and pull it out of the time derivative.

Let's consider a case such as a baseball being thrown between two people. Imagine that player A (we pick arbitrarly) throws a baseball to player B. As the baseball flies through the air it will experience the force of gravity acting straight down on it, as well as interactions with the molecules that make up the air the baseball is flying through. This interaction with the air molecules is very complex and therefore we need to make an assumption about it that will allow us to mathematically describe it in a simple way. It turns out that if the speed of the baseball is not changing very quickly during its flight between player A and B, then we can use whats known in physics and engineering as the Power Law. We say that the force opposing the motion of the baseball due to the air molecules colliding with it is directly proportional to the velocity to some n power, and in the opposite direction of its velocity. Using k as a constant to represent the strength of the retarding force we may express it as

Therefore we arrive at the following vector equation to describe the flight of the baseball between player A and B expressed as

From experience we know that a baseball flying between two people will travel both horizontally (or else how would it move from player A to B?) and vertically forming an arc-like path between player A and B. Therefore, we will choose to express this motion in 2 dimensional cartesian coordinates (x,y). Due to the nature of a vector quantity and the fact that both force and velocity are vectors, we must come up with a way of describing the variation of the angels between the force vectors (gravity and retarding) acting on the object and the objects velocity vector during its flight. Due to gravity always acting in a down position we will say that a relationship between the retarding force and velocity will suffice; i.e. gravity has no influence on the x-component of velocity. Therefore, we can say ɑ is the angle between the horizontal direction and the velocity vector of the baseball at any point in time during its flight between player A and B. Using trigonometric arguments to determine the x and y components of the vectors, we arrive at the follow equations describing the flight of the baseball in both the x and y directions.

2) Verify $$ y(x)\ $$ is a parabola, for the particular case of $$ k=0\ $$ First let's attack the x-direction of motion. Setting k = 0 we arrive at the following expression.

Integrating once over x with t gives

Now lets stop and notice something about this expression. By integration we have found the the velocity in the x-direction is constant, i.e. in does not change. From start to finish, the baseball's velocity strictly in the x-direction will not change, and we can see this prediction from the above equation. Now integrating again over the x-direction with time will give us an expression for the instanteous x position of the baseball at any moment during its flight in the form of 2 constants.

Now we must use 2 conditions in which we know about the motion of a the baseball traveling between player A and B to solve for these 2 constants. First we know that at time = 0, the baseball is in the starting position. Therefore, let us take the starting position of the baseball to be the x = 0 position on our x,y plane.

We also know that at the final time of travel T, the baseball has to transverse the entire path between player A and B. Let's call this path length L.

Therefore our final expression for the x-direction of motion is Because we are proving that y(x) is a parabola we know that we must eliminate t between x(t) and y(t). Here is a very good opportunity due to the simplicity of the x(t) expression to use the inverse of x(t) in order to get t(x).

Following the same logic as we did before with the x-direction we start with the following 2nd order differential equation. Integrating over the y-direction with respect to t gives Now, because the velocity in the y-direction is not constant, we must apply a initial condition to it. We state that at t = 0, the velocity in the y-direction will have some generic value of $${{v}_{0}}$$.

Integrate once again over y with respect to t

Now again, to keep the solution to this problem as general as possible, we would like to give the option of inputing different starting heights for the projectile, i.e. in our case allowing for player A to stand on a hill and throw the baseball to player B who is at the bottom of the hill. This is done by specifying the following initial condition.

Now plugging in for the fourth constant we have the equation of motion for the y-direction as the following

Now plugging in x(t) into the above y(t) equation we arrive at the follow expression.

We see that y(x) is of the following form

3.1) Utilizing eq1.3, find $$ v_y(t)\ $$ and $$ y(t)\ $$ for m=constant

Due to the velocity being a function of time, if we take the integrating factor as only a function of time and leave it in its integral form, then technically we have not violated any rules. We can not integrate the integrating factor because we do not have the velocity in an explicit form. Therefore we have the following. Which then will simplify to the following ordinary differential equation. Integrating the left side and leaving the right side in integral form we arrive at the following result for v(t).

Integrating again over time we arrive at the following expression for y(t).

3.2) Utilizing eq1.3, find $$ m=m(t)\ $$

EGM6321.F10.TEAM1.Lang70.185.108.64 07:45, 6 October 2010 (UTC)

Author EGM6321.Langpm

Find
1) Derive eq2.1 and eq 2.2 2.1) Write eq2.1 and eq 2.2 in the form of: 2.2) Find $$ \underline{A}\ $$, $$ \underline{B}\ $$ and $$ \underline{u}\ $$,

Solution
1) Derive equation (2.1) and (2.2)

Equation (2.1) can be derived by linearizing (2.10) for small argument.

Equation (2.2) can be derived by linearizing (2.19) for small argument.

2.1) Write eq2.1 and eq 2.2 in the form of:

2.2) Find $$ \underline{A}\ $$, $$ \underline{B}\ $$ and $$ \underline{u}\ $$,

Author EGM6321.Taeho Kim

Find
Use the integrating factor method to obtain the result of eq3.1 directly

Solution
The given solution is one of a first order linear ODE. The derivative of such an equation has the form:

Where $$\ \tau\, $$ is the constant of integration. We will assume the case of constant coefficients, so this equation becomes:

Equation 3.3 can be rearranged as:

We desire a solution for $$\ x(\tau\,) $$, which may be found using the integration by factors method. First, the function $$\ h(\tau\,) $$ must be found:

Multiplying both sides of equation 3.4 by $$\ h(\tau\,) $$ yields:

This equation may be simplified by recalling that $$\ h(\tau\,)a(\tau\,) = \dot{h}(\tau\,) $$:

Substituting equation 3.5 into equation 3.7 for $$h(\tau\,)$$ yields:

It is now possible to find an expression for x(t) by integrating both sides of this equation with respect to $$\ \tau\, $$:

Performing this integration yields:

By performing some algebra on this equation, x(t) is found:

Author: EGM6321.f10.team1.russo 07:15, 6 October 2010 (UTC)

Find
Expand Eq4.1 in a taylor series

Solution
One definition of the exponential function is:

In addition, the Taylor series of a function about some real or complex point, a, is defined as:

For the specific case of a=0, the series is also called the Maclaurin series which is defined as:

Substituting equation 4.2 into the first part of equation 4.4 yields:

Since the derivative of $${{e}^{x}}$$ is simply $${{e}^{x}}$$ and anything raised to the zeroth power is 1, equation 4.5 can be simplified to.

Finally, equation 4.2 can be used to reintroduce the original equation.

Author Egm6321.f10.team1.allison 13:24, 28 September 2010 (UTC)

Find
Generalize Eq 5.1 to the case of SC_L1_ODE_VC.

Solution
For a L1_ODE_CC we have

and for a L1_ODE_VC we have

Similarly, with

Author

Egm6321.f10.team1.allen 04:30, 6 October 2010 (UTC)

Find
Use Eq6.1, Eq6.2 and Eq6.3 to obtain Eq6.4:

Solution
From Bryson & Ho, 1975, pgs.449-450 Relating Eq(6.3) to Eq (A4.5) from Bryson & Ho, 1975, p.450

The initial condition of the output is from Bryson and Ho Substitute 6.11 into 6.13 and

Author 173.18.203.75 05:36, 6 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Find
Place Eq7.1, Eq7.2 and Eq7.3 in the form of Eq7.4

Solution
Author EGM6321.Taeho Kim

Given
if: then 8.1 is satisfied

Find
Are there any other solutions, without assuming solution given by eq 8.2? Discuss the search for the solution of 8.1.

Solution
Author EGM6321.XXXXXXXXXXXXXXXXXXXX

Given
Exactness condition 1:

Find
Check 2nd condition of exactness using the following:

Solution
First, the various derivatives used must be found:

Now, substituting equations 9.6 and 9.10-9.14 into equation 9.4.

When the terms on the right hand side are canceled with each other and the zero terms dropped, the equation simplifies to:

which can easily be verified by the reflexive property.

Similarly, equations 9.7-9.9 and 9.15 are substituted into 9.5 to yield:

Once again, the zero terms must be dropped and the results is proven by the reflexive property

$$\therefore \begin{align} f_{xp}+pf_{yp}+2f_y=g_{pp} \end{align}$$

Author Egm6321.f10.team1.allison 14:37, 5 October 2010 (UTC)

Given
with cancelled terms removed The first choice for solution assumes thus and since: now 10.4 becomes: solving for $$ \phi\ (x,y,p) $$ moving $$ k_1\ $$ to the other side of the equation and using the equivalency $$ k_2-k_1=k_3\ $$, the solution for choice 1 bcomes:

Find
Finish the argument assuming the following

Solution
Now and using 10.6 and 10.7 and solving for $$ h(x,y)\ $$

solving for $$ \phi\ (x,y,p) $$ moving $$ k_1\ $$ to the other side of the equation and using the equivalency $$ k_2-k_1=k_3\ $$, the solution for choice 2 is:

Author EGM6321.F10.TEAM1.WILKS 03:12, 4 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Signatures
Solved problem 1 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 2 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 3 -- EGM6321.f10.team1.russo 07:14, 6 October 2010 (UTC)

Solved problem 4 -- Egm6321.f10.team1.allison 13:27, 28 September 2010 (UTC)

Solved problem 5 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 6 -- EGM6321.F10.TEAM1.WILKS 06:00, 6 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Solved problem 7 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 8 -- EGM6321.F10.TEAM1.HW3 03:50, 22 September 2010 (UTC)

Solved problem 9 -- Egm6321.f10.team1.allison 14:36, 5 October 2010 (UTC)

Solved problem 10 -- EGM6321.F10.TEAM1.WILKS 03:13, 4 October 2010 (UTC)EGM6321,F10.TEAM1.WILKS

title: User:EGM6321.F10.TEAM1.HW3


text:
User:Issariyaphorn

Hi! My name is Issariyaphorn Boonyawiroch. Also known as Som-o. I was born in May of 7th 1992. I am BBA student and am majoring in International Business. This is my 2nd year now. My hometown is Nakhonsawan. When i was 11 years old, i came to study in Bangkok with my brother and cousins. Later on, i had a chance to go study in India for almost 3 years. Then one year in United States and New Zealand. After i finished the high school, i came back to Bangkok for the university. I applied for MUIC. At first, i got into Pre-College 3. It was really a hard and tough time for me. However, i got into Mahidol University International College.

Thanks

Link Youtube: []

Group Homework: Anusara 5280449, Buranee 5280451, Issariyaphorn 5280452

iPhone CPU& speed: RAM: 600MHz Bus speed : 1GHz Apple A4 Cache size : L1 cache size of 32 KB I-Cache and 32 KB D-Cache, and an L2 cache size of 256 KB Storage(type and amount)  :  Memory (16GB or 32GB flash drive)

iPad CPU & speed : RAM : 256MB Bus speed : 1GHz Apple A4 Cache size : 64 KB of cache L1, 640 KB of cache L2 Storage(type and amount)  : Internal storage (16 GB, 32GB, 64GB Flash)

Blackberry CPU& speed: RAM: 400MHz Bus speed : 100MHz Cache size : 1 MB L2 Storage(type and amount) : external –MicroSD : internal storage 1GB

-	iPhone has many applications, such as making phone call, storages, multitasking and etc. -	Whereas, we can store data in iPad, but we can’t use to make a phone call. -	Blackberry has advanced application which is blackberry messenger, but from our own experience it’s very slow and is always frozen. It has only 1 GB of internal memory which is the lowest.

Conclusion: we think that iPhone is the best because it is advance in applications, speed, and storages.

title: User:Issariyaphorn


text:
User:5280342rut

INTRODUCTION My name is Rut Poolchan, and my nickname is Peach. My birthday is 25 july 1989, and I was 21 years old. I study in Marketing major at MUIC. I study this subject because I am interested about computer, and I see the project that my friends made it last term, so I want to learn about how to use computer and the software of computer.

WELCOME TO MUIC YOUTUBE : http://www.youtube.com/watch?v=897wYVKYbto

SPECS OF MOBILE PHONE iPHONE 3G : CPU : ARM 11 412 MHz, PowerVR MBX-Lite graphics INTERNAL : 128 MB RAM STORAGE : 8 / 16 mb

In my opinion, I think that iPhone is more technological than blackberry because it is mixed up with the technological device that use in iPhone. You can use iPhone in many ways, such as playing games, using the internet, listening to the music, etc. On the other hand, even though Blackberry can also do like the iPhone but not at all. Blackberry's graphic is not clear as much as iPhone do. Because of these reason, I think iPhone is more technological than BlackBerry.

title: User:5280342rut