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Ejump/Action Research/Action Research Project Guide

Course title: Action Research Project 2.0 Credits: 7.5 ECTS

Code: Dnr: #####

Responsible department: Department of Interactive Media and Learning (IML)

Subject: Educational Science

Level: Advanced level

Field of study: ICT, Media and Learning

Grades: Pass with distinction/Pass with Merits/Pass/Fail

Connection to study programme: Optional course for Masters Programme in Educational Practice Prerequisites

Participants need to have studied for 180 ECTS at the Basic Level of Higher Education in Education, Computer Science or closely related field OR in any subject area with relevant experience of at least 2 years duration in teaching, training, digital media production or consultancy Note: Course participants will need a computer with Internet access. Computer access, video conference facilities etc. are available at the local learning centre if you live in Sweden. For a complete list of learning centres see: www.larcentra.se/karta.htm For those who live in Umeå and the surrounding area the facilities of Umeå University are available. This course has been designed for participants who have completed New Technologies of e-Learning 2.0 or equivalent. Study Period Spring Term 2009, weeks 11-22 Course Location: Internet

Language of instruction: English

Selection: Selection will take place of there are more applicants than available places

Forms of Study: 25% tempo, Distance (Internet)

The course is offered as net-based distance education through an online learning environment that will offer opportunities to explore a range of social software tools as an integrated part of the learning environment. It will make use of e-meeting software, social software and digital portfolio applications and approaches.

Course Leader/Course Team Brian Hudson, Umeå University

Alison Hudson, Umeå University

Albert Sangra, University of Catalonia

Satu Nurmela, University of Turku

Oliver Popov, Iskra Popova, Sylwia Rudnik, CEENet

Sue Harrison, University of East London

Alfredo Soeiro, University of Porto

Jüri Lõssenko, Ene Tammeoru, EITF

Plus other staff members of the e-Jump 2.0 project

Approval: Date of approval ####

Course Content

This course will focus on putting ideas in action through conducting an action research project that involves the application of resources and tools which utilise Web 2.0 and social media applications by integrating them into the design and development of teaching situations, pedagogical activities and learning environments. Web 2.0 enables users to create content in such a way that allows others to both read and write to a web environment. Social media is a feature of Web 2.0 involving tools that are used to communicate in different settings such as one-to many (blog or podcast) and many-to-many (wiki). A major feature of social media is that it enables people to connect together, providing a space in which they can interact and share ideas, experiences and knowledge. ICT supported collaborative learning can be develop to encourage active and inquiry- based approaches to learning. Educational action research is seen as providing a framework for thinking systematically about what happens in teaching situations and learning environments, implementing action for change and monitoring and evaluating the effects of the action with a view to continuing the development for improvement.

Approach to Learning and Teaching

This module will be structured around the process of conducting an action research project that will involve the identification of the key developmental goals and research questions, research methodology and methods, addressing of research ethics, review of relevant literature and an activity plan. It will involve the implementation of an action research project for conducting the educational development, reflection on action and the critical evaluation of its impact on student learning outcomes. There will be a number of moments during the course through which participants will share their ideas and provide peer feedback to others though the social networking community environment. Each participant will have the support of a study supervisor organised by the local institution to provide one to one tuition and support.

Course Learning Outcomes

On successful completion of this course of study participants should be able to:
 * Define relevant developmental goals and appropriate research questions to underpin the design of an educational action research project
 * Defend a position in relation to educational action research, research methodology and methods with reference to relevant literature
 * Critically analyse a range of theories about the nature of action research in order to describe an informed position as an action researcher
 * Address all associated ethical implications from the outset of the study
 * Synthesise the findings from a review of current relevant literature in order to refine developmental goals and associated research questions
 * Implement an action research project for conducting the educational development, reflect on action and critically evaluate its impact on student learning outcomes
 * Communicate the outcomes of the study through the use of a variety of media whilst demonstrating a high standard of organisation, coherence and clarity of academic writing with accurate citation

Approach to Assessment

Each participant will present a commentary in the form of a self assessment of how he or she has been able to meet each of the learning outcomes. In addition this will be supported by an outline of how s/he has participated in all the module activities. The submission of work for final examination will be through an Action Research Report of approximately 3000 words. Best papers will be selected for further refinement and publication in the e-Jump 2.0 Compendium of Research (2009) Course

Literature:

Core reading

Kanuka, Heather and Anderson, Terry (2007) Ethical Issues in Qualitative E-Learning Research, International Journal of Qualitative Methods 6 (2) June 2007 [WWW document] URL http://www.ualberta.ca/~iiqm/backissues/6_2/kanuka.htm (Visited 19 Oct 08)

Kemmis, Stephen (1993) Action research and social movement: a challenge for policy research, Education Policy Analysis Archives, 1, 1. [WWW document] URL http://epaa.asu.edu/epaa/v1n1.html (Visited 19 Oct 08)

Walker, Melanie (2001) Higher education, critical professionalism and educational action research, Debates in Higher Education, University College London. [WWW document] URL http://www.ucl.ac.uk/cishe/seminars/dhe_papers/MWpaper.doc (Visited 19 Oct 08)

Teacher’s Guide to Action Research, The Northeast Florida Science, Technology, and Mathematics Center for Education [WWW document] URL http://www.nefstem.org/teacher_guide/intro/index.htm (Visited 19 Oct 08).

Recommended web resources

Jean McNiff, Action Research for Professional Development [WWW document] URL http://www.jeanmcniff.com/booklet1.html#ite (Visited 19 Oct 08)

John Elliott on Action Research [WWW document] URL http://www.viddler.com/explore/dominiklukes/videos/10/ (Visited 19 Oct 08)

Recommended reading

Carr, Wilfred and Kemmis, Stephen (1986) Becoming Critical: Education, Knowledge and Action Research. London: Falmer Press.

Elliott, John (1991) Action Research for Educational Change. Milton Keynes: Open University Press.

Day, Christopher; Elliott, John; Somekh, Bridget & Winter, Richard (2002) Theory and Practice in Action Research: some international perspectives, Oxford: Symposium Books.

Hudson, Brian (2008) A Didactical Design Perspective on Teacher Presence in an International Online Learning Community. In Tidskrift för lärarutbildning och forskning, Faculty Board for Teacher Education, University of Umeå (In press) http://www.educ.umu.se/presentation/publikationer/lof/index.html

Hudson, Brian (2008) Didaktik Design for Technology Supported Learning, In Meinert A.

Meyer, Stephanie Hellekamps and Manfred Prenzel (Eds.) Perspektiven der Didaktik:

Zeitschrift für Erziehungswissenschaft, Vol 9, 139-158.

Hudson, B. (2008) Didactical Design Research for Teaching as a Design Profession. In Hudson and Zgaga (Eds.) Monograph: Teacher Education Policy in Europe: A Voice of Higher Education Institutions. University of Umeå/University of Ljubljana, 345-365.’ http://www.educ.umu.se/presentation/publikationer/

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Ejump/Action Research/AR Planning Assessment Grid

Action Research Project Plan

RESULT OF EXAMINATION: PASS/REFER for further work/FAIL

Name:

Title of Project:

Criteria for assessment: Participants' work will be judged on evidence of:
 * Clarity of research focus
 * Description of the context for the enquiry
 * Justification for the methodological approach and techniques proposed
 * Consideration of the ethical implications
 * Overall standard of the action research design

To pass students will:
 * Identify a significant and manageable action research focus
 * Describe and critically analyse the context for the enquiry with reference to appropriate literature
 * Provide appropriate justification for the methodological approach and techniques proposed
 * Take account of the ethical issues arising specifically from the design
 * Design a proposal of a sufficient standard to provide a sound basis for a worthwhile action research project

Comments from First Examiner: ……………………….

Comments (Moderating Examiner) (if appropriate): …………………………….

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


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Ejump/Action Research/AR Project Assessment Grid

Action Research 2.0 Project Report

RESULT OF EXAMINATION:  DISTINCTION/PASS WITH MERIT/PASS/MODIFICATIONS NEEDED/FAIL

Name: ..................................................................... Title of Project: ........................................................

First Study Supervisor: ............................................. Second Study Supervisor: ......................................... Moderating Tutor: ....................................................

Criteria for assessment

1. Defined a relevant developmental goals and appropriate research questions. 2. Described approach to research/research methodology and methods with reference to relevant literature 3. Addressed all ethical implications of the study at the outset.

4. Conducted a review of the literature relevant to the developmental goals and research questions. 5. Reflected critically on the process of undertaking the study and on its outcomes with particular emphasis given to the impact on student learning outcomes.

6. Demonstrated organisation, coherence, clarity and a high standard of academic writing with accurate citation

For each of the six criteria one of the following evaluations is selected.
 * Criteria met to an exceptionally high standard;
 * Criteria met to a good standard
 * Criteria met but with some weaknesses
 * Criteria partially met with more weaknesses than strengths
 * Criteria not met and need further work

Comments from First Supervisor:

Comments from Second Supervisor:

Comments (Moderating Tutor): Categories of achievement: Pass with distinction/Pass with merit/Pass/Modifications needed/Fail

The category of achievement is obtained using the evaluation above and the following guidelines:

Best papers will be selected for further refinement and publication in the e-Jump 2.0 Compendium of Research (2009).
 * Pass with Distinction when 70% or more of criteria are met to an exceptionally high standard
 * Pass with merit when 70% or more of criteria are met to a good standard
 * Pass when 50% or more of criteria are met but with some weaknesses
 * Modifications needed when less than 50% of the criteria are met and with more weaknesses than strengths
 * Fail when Less than 40% of the criteria are met and with many more weaknesses than strengths

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User:Egm6321.f09.team5.GV/HW2

 * 1) Obtain h(y) for Case 2 from pages 6-3 and 6-4
 * 2) Show that result after integration of application of 8-2 is y= ((x^3)/4) + (c/x), where c is constant.
 * 3) Show that eq. 3 of 9-2 of 1st order linear ODE with Varying Coefficients (L1.ODE.VC) is exact
 * 4) Show that eq. 8 of 9-3 is an exact N1.ODE
 * 5) Show that the 2nd exactness condition is satisfied. Pg 10-3.
 * 6) P. 10-2 Eq. 5, Derive this relation by differentiating eq. (3) p. 10-1 wrt p=y'. i.e. Gpp=....... (2nd relation in exactness condition 2)
 * 7) P. 12-1 Use PHIxy=PHIyx to obtain eq. 4 from Pg. 10-2
 * 8) Show that application on pg. 12-1 is exact as determined by the 2nd exactness condition.
 * 9) Prove that Eq. 1 on pg. 12-2 in not exact. find f & g.

Solution
If the condition for exactness is not satisfied: $$ \frac{\partial M}{\partial y}$$ does not equal to $$ \frac{\partial N}{\partial x} $$ The functions M(x,y) and N(x,y) may be multiplied by another function H(x,y) which is yet unknown. The result of the multiplication is as follows:

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

$$ \ H(x,y)M(x,y) = M* $$

$$ \ H(x,y)N(x,y) = N* $$

$$ \ M*_{y}=h_{y}M+hM_{y} $$

$$ \ N*_{x}=h_{x}N+hN_{x} $$

$$ \ h_{x}N-h_{y}M+h(N_{x}-M_{y})=0 $$

If we set $$ \ (h_{x}N=0$$, we are able to solve for H(x,y) as follows: $$ \ h_{y}M-h(N_{x}-M_{y})=0$$

$$ \frac{h_{y}}{h} = \frac{1}{M}(N_{x}-M_{y})dy $$

If we integrate both side: $$ \int_{0}^{y} \frac{h_{y)}{N} dy $$

Problem Statement
Consider:

$$ \ xyy^{''} + x(y^{'})^{2} + yy^{'} = 0 $$ Given that the 1st exactness condition is satisfied show that the 2nd exactness condition is satisfied

Solution
By the 1st exactness condition the equation can be written as follows:

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

$$ F(x,y,p)= xy $$

$$ G(x,y,p)= x(y^{'})^{2} + yy^{'} $$

It is then needed to identify the following partial derivatives with respect to the different variables:

$$ F_{x}=y  \qquad \qquad  G_{y}= p $$

$$ F_{xx}=0 \qquad \qquad  G_{yp}= 1 $$

$$ F_{xy}=1 \qquad \qquad  G_{x}= (y^{'})^{2} $$

$$ F_{y}=x  \qquad \qquad  G_{xp}= 2y^{'} $$

$$ F_{yy}=0 \qquad \qquad  G_{p}= 2xp + y $$

$$ F_{yp}=0 \qquad \qquad  G_{pp}= 2x $$

$$ F_{xp}=0 \qquad \qquad $$

The second exactness condition is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Eq. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Eq. 2 $$

Filling in for Equations 1 the results are as follows:

$$ \ 0 + 2p + 0 = 2p + p - p $$

$$ 2p=2p $$  '''Eq. 1''' is satisfied.

Filling in for Equations 2 the results are as follows:

$$ \ 0 + 0 + 2x = 2x $$

$$ \ 2x = 2x $$   '''Eq. 2''' is satisfied.

By satisfying Eq. 1 and Eq. 2, the 2nd order ODE is considered to be exact.

Problem Statement
The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

Derive the relation in Equation 1.

Solution
The form for an exact 2nd order ODE is as follows:

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

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

Since $$ ( \phi_{p} )_{y} = ( \phi_{y})_{p} $$

Rearranging the definitions of F and G, the previous equation can be written as:

$$ \ F_{y} = \left[ \frac{1}{p}(G - \phi_{x}) \right]_{p} $$

This can be evaluated to obtain the following:

$$ \ F_{y} = \frac{-1}{p^{2}}(G - \phi_{x}) + \frac{1}{p}(G_{p} - \phi_{xp}) $$

Rearranging:

$$ \ F_{y}p^{2} = - (G - \phi_{x}) + p(G_{p} - \phi_{xp}) $$

$$ \phi_{x} = F_{y}p^{2} + G - p(G_{p} - F_{x}) \qquad \qquad Equation 3. $$

Taking a partial derivative of Equation 3. with respect to p

$$ \phi_{xp} = F_{y}2p + F_{yp}p^{2} + G_{p} - pG_{pp} - G_{p} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = 2pF_{y} + p^{2}F_{yp} - pG_{pp} + pF_{xp} + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + F_{x} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \phi_{xp} = P(2F_{y} + pF_{yp} - G_{pp} + F_{xp}) + \phi_{xp} $$

$$ \ 0 = 2F_{y} + pF_{yp} - G_{pp} + F_{xp} $$

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

The result is indeed the second equation of the 2nd exactness condition, Equation 2.

Problem Statement
The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

Derive the relation of Equation 2

Solution
The form for an exact 2nd order ODE is as follows:

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

From this relation the following can be stated:

$$ \ F(x,y,p) = \phi_{p} $$

$$ \ G(x,y,p) = \phi_{x} + \phi_{y}p $$

Where p = y'

$$ \ (\phi_{x})_{p} = G_{p} - (p \phi_{y})_{p} $$

Since: $$ (\phi_{x})_{p} = (\phi_{p})_{x} $$

$$ \ F_{x}= G_{p} - (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = (p \phi_{y})_{p} $$

$$ \ G_{p} - F_{x} = p \phi_{yp} + \phi_{y} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p \phi_{yp} $$

$$ \ \phi_{y} = G_{p} - F_{x} - p (F_{y}) \qquad Equation 3 $$

Since $$ (\phi_{p})_{y} = F_{y} \qquad $$ By the definition of F.

From the definition of G, the following can be identified:

$$ \phi_{x} = G - p \phi_{y} \qquad Equation 4 $$

Using Eqs. 1 and 2 the following is found:

$$ \phi_{xy} = \phi_{yx} $$

$$ \left[G - p \phi_{y} \right]_{y} = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p (\phi_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - \left[ p (G_{py} - F_{xy} - pF_{yy}) \right] = G_{px} - F_{xx} - p(F_{yx}) $$

$$ \ G_{y} - p(G_{py}) + pF_{xy} + p^{2}F_{yy}) = G_{px} - F_{xx} - p(F_{yx}) $$

Rearranging Yields:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation 5. $$

Equation 5 is equivalent to the Equation 2 from the exactness condition.

Problem Statement
Given:

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

Show that the Nonlinear 2nd-Order ODE is exact

Solution
Exactness Condition 1:

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

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

It is concluded that it satisfies the first exactness condition.

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

The Following partial derivatives are then identified:

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

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

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

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

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

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

$$ F_{yp}=0 \qquad \qquad $$

Applying the results in Eq. 1:

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

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

Eq. 1 is then satisfied.

Applying the results in Eq. 2:

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

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

Eq. 2 is then satisfied.

Since Equation 1 and Equation 2 are both satisfied, then the 2nd exactness condition is satisfied. It is then concluded that the N2.ODE is exact.

Problem Statement
Given

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

Prove that the equation is not exact.

Solution
Exactness Condition 1:

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

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

It is concluded that it satisfies the first exactness condition.

The 2nd exactness condition for a second order ODE is as follows:

$$ \ F_{xx} + 2pF_{xy} + p^{2}F_{yy} = G_{xp} + pG_{yp} - G_{y} \qquad Equation. 1 $$

$$ \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp} \qquad Equation. 2 $$

The following partial derivatives are found:

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

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

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

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

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

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

$$ F_{yp}=0 \qquad \qquad $$

Using these values in Eq. 1:

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

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

Equation 1. is not satisfied therefore the ODE is not exact.

Equation 2. can be populated to find:

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

$$ \ 0 = 0 $$

Equation 2. is satisfied.

In order for the ODE to be exact it must satisfy both Eq. 1 and Eq. 2, since it fails to do this it is concluded that the ODE is not exact.

title: User:Egm6321.f09.team5.GV/HW2


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User:Egm6321.f09.team3.diggs

User:egm6321.f09.team3.diggs

title: User:Egm6321.f09.team3.diggs


text:
User:Egm6321.f09.team3.diggs/HW1

1. Complete Case 2 details (7-1) to obtain expression for h as a function of g(y).
$$h_xN - h_yM + h(N_x - M_y) = 0 $$ may be used to solve for the integration factor h(x,y). Case 1, derived and discussed in class, is if $$ h_yM = 0 $$. Case 2, derived and discussed below, is if $$ h_xN = 0 $$

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

rearrange to get: $$ \frac{h_y}{h} = \frac{1}{M}(N_x - M_y) $$

integrate to get: $$ ln(h) = \int - \frac{1}{M} (N_x - M_y) dy $$

which yields, $$ h(x,y) = exp(- \int \frac{1}{M}(N_x - M_y) dy) $$

The case $$ h_xN = 0 $$ implies $$h_x = 0 $$ which implies h(y) only!

therefore, $$ \frac{-1}{M(x,y)}[\partial {N(x,y)}{x} - \partial{M(x,y)}{y}] = g(y) $$ only

if the equation above is satisfied, then $$ h(y) = exp[\int g(s) ds] $$

2. & 3. Given $$ y'+ \frac{1} {x} y = x^2 $$ where $$ a_0 = \frac{1} {x} $$ and $$ b = x^2 $$, find h(x) and solve for y(x).
Recall, $$ h(x) = exp( \int a_0(s) ds) $$, which yields $$ h(x) = exp(\int \frac{1}{s} ds = exp(ln(x)) = x $$, as expected from class notes.

Additionally, the solution to the ODE is obtained by:

multiplying h(x) to the ODE: $$ hy' + ha_0y = hb $$

which is simplified as: $$ (hy)' = hb = x \cdot x^2 = x^3 $$ (once it is recognized $$ha_0 = h'$$)

integrate both sides to get: $$hy = \frac{x^4}{4} + C$$

and solve for y: $$y(x) = \frac{x^3}{4} + \frac{C}{x}$$, as expected from class notes.

4. Given the form of an exact, linear, 1st order ODE as $$ \bar{b}(x)y' + [a(x)y + k] = 0 $$, where $$a(x) = x^4$$, $$b(x) = x$$, $$\bar{b}(x) = \frac{x^2}{2}$$, and $$k=10$$, show the ODE is exact.
ODE is: $$ \frac{x^2}{2}y' + [x^4y+10] = 0 $$

where $$M(x,y) + N(x,y)\cdot y' = 0$$ is the first condition of exactness, and the second condition of exactness is $$M_y = N_x$$

Here, $$M(x,y) = x^4y + 10$$ and $$N(x,y) = \frac{x^2}{2}$$

$$ M_y = x^4 $$ and $$ N_x = x $$

$$ M_y ~= N_x $$, therefore the equation is not exact.

The ODE is not exact, but can be solved bu integration factors method.

ODE is: $$ y' + 2x^2y = -20/x $$

here, $$ a_0 = 2x^2 $$ and $$ b = -\frac{20}{x} $$

$$ h(x) = exp \int a_0(s) ds = exp \int 2s^2ds = exp[ \frac{2}{3} x^3] $$

Multiply both sides of the ODE by the integrating factor: $$ hy' + ha_0y + hb $$

which is: $$ (yh)' = hb = exp[\frac{2}{3} x^3 (-\frac{20}{x}) $$

$$ hy = \int (\frac{2}{3} s^3)(-\frac{20}{s}) $$

solving for y, FINISH THE INTEGRAL!!

== 5. Given the form of an exact, linear, 1st order ODE as $$ a(x) \bar{c}(y) + \bar{b} (x) c(y) y' = 0 $$, where $$a(x) = 5x^3+2$$, $$b(x) = x^2$$, $$\bar{b}(x) = \frac{x^3}{3}$$, $$ c(y) =y^4 $$ and $$\bar{c}(y) = \frac{y^5}{5}$$, show the ODE is exact. ==

ODE is: $$(5x^3+2) \frac {y^5} {5} + \frac{x^3}{3} y^4 y' = 0 $$

where $$M(x,y) + N(x,y)\cdot y' = 0$$ is the first condition of exactness, and the second condition of exactness is $$M_y = N_x$$

Here, $$M(x,y) = (5x^3+2)\frac{y^5}{5} $$ and $$N(x,y) = \frac{x^3}{3}y^4$$

$$ M_y = 5x^3y^4 + 2y^4 $$ and $$ N_x = \frac{2}{3}x^2y^4 $$

$$ M_y ~= N_x $$, therefore the equation is not exact.

The ODE is not exact, but can be solved using integrating factor method.

The ODE is: $$ y' + (3 + \frac{6}{5x^3})y = 0 $$

where $$ a_0 = 3 + \frac{6}{5x^3} $$ and $$ b = 0 $$

$$ h(x,y) = exp \int a_0(s)ds = exp \int 3 + \frac{6}{5s^3} ds $$

$$ h(x.y) = exp(3x - \frac{3}{5x^2}) $$

Multiply both sides of the ODE by h(x) to solve, of the form:

$$ hy' + ha_0y = 0 $$

which is $$ (hy)' = 0 $$

integrate to get: $$ hy = C $$ where C is a constant

$$ y = \frac{C}{h} = \frac{C}{exp(3x - \frac{3}{5x^2}} $$

The solution is $$ y(x) = \frac{C}{exp[3x - \frac{3}{5x^2}} $$, where C is a constant

6. Is the following ODE exact? $$ xyy'' + x(y')^2 + yy' = 0 $$
The first condition is satisfied, as $$ f(x,y,p) = xy $$ and $$ g(x,y,p) = xp^2 + yp $$.

The second exactness conditions are:

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

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

Find the partial derivatives: $$ f_x = y $$

$$ f_{xx} = 0 $$

$$ f_{xy} = 1 $$

$$ f_{yy} = 0 $$

$$ f_p = f_{py} = f_{px} = f_{py} = f_{pp} = 0 $$

$$ g_x = p^2 $$

$$ g_{xy} = 0 $$

$$ g_{xp} = 2p $$

$$ g_y = p $$

$$ g_{yp} = 1 $$

$$ g_{yy} = g_{yx} = 0 $$

$$ g_p = 2xp + y $$

$$ g_{py} = 1 $$

$$ g_{px} = 2p $$

$$ g_{pp} = 2x $$

(i). $$ 0 + 2p + p^2 \cdot 0 = 2p + p - p $$, satisfied!

(ii). $$ 0 + p \cdot 0 + 2x = 2x $$, satisfied!

Yes, the ODE is exact!

7. Derive $$ f_{xy} + pf_{yp} + 2f_y = g_{pp} $$ (10-2, Eqn 5)
Differentiate $$ g = \phi _x + \phi _y p $$ with respect to p

$$ g_p = \phi _{xp} + \phi _{yp} p + \phi _y $$

Recall, $$ f = \phi _p $$, and $$ g = \phi _x + \phi_y p $$

$$ g_{pp} = \phi _{xpp} + \phi _{ypp}p + \phi _{yp} + \phi _{yp} $$

$$ g_{pp} = (\phi _p) _{xpp} + (\phi _p) _{yp}p + 2 (\phi _p)_y $$

$$ g_{pp} = f_{xp} + f_{yp}p + 2f_y $$, which is the equation above.

8. Use $$ \phi _{xy} = \phi _{yx} $$ to obtain $$ f_{xx} + 2pf_{xy} + p^2 f_{yy} = g_{xp} + pg_{yp} - g{y} $$ (10-2, Eqn 4)
$$ \phi _{xy} = \phi _{yx} $$ is also $$ (\phi _x)_y = (\phi _y)_x $$

$$ \phi_x = g - p (g_p) - f_x) + p^2f_y $$ $$ \phi_y = g_y = f_yp - f_x $$

$$ [\phi_x = g - p (g_p) - f_x) + p^2f_y]_y = (\phi_y = g_y = f_yp - f_x)_x $$

$$ g_y - p(g_{py} - f_{xy}) + p^2f_{yy} = g_{px} - f_{yx}p - f_{xx} $$ which is $$ f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{px} + pg_{yp} - g_y $$, as expected.

9. Show $$ (8x^5y')'' + 2x^2y' + 20x^4(y')^2 + 4xy = 0 $$ is exact by condition 2
$$ f(x,y,p) = 8x^5p $$ $$ g(x,y,p) = 2x^2 + 20x^4p^2 + 4xy $$

Take partial derivatives:

$$ f_x = 40x^4p $$

$$ f_{xx} = 160x^3p $$

$$ f_{xy} = 0 $$

$$ f_{xp} = 40x^4 $$

$$ f_y = f_{yy} = f_{yp} = 0 $$

$$ f_p = 8x^5 $$

$$ f_{pp} = 0 $$

$$ g_x = 4xp + 80x^3p^2 + 4y $$

$$ g_{xx} = 4p + 240x^2p$$

$$ g_{xy} = 4$$

$$ g_{xp} = 4x + 160 x^3 p $$

$$ g_y = 4x $$

$$ g_{yy} = 0 $$

$$ g_{yp} = 0 $$

$$ g_p = 2x^2 + 40x^4p $$

$$ g_{pp} = 40x^4 $$

(i). $$ f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y $$ $$ 160x^3p + 2p \cdot 0 + p^2 \cdot 0 = 4x + 160 x^3p + p \cdot 0 - 4x $$, which is satisfied.

(ii). $$ f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ $$ 40x^4 + p \cdot 0 + 2 \cdot 0 = 40x^4 $$, which is satisfied.

The equation is exact.

10. Verify exactness for $$ \sqrt {x} y'' + 2xy' + 3y = 0 $$
where $$ f(x,y,p) = \sqrt {x} $$ and $$ f(x,y,p) = 2xp + 3y $$

satisfies condition 1 of exactness.

Take the partial derivatives:

$$ f_x = \frac {1} {2 \sqrt{x}} $$

$$ f_{xx} = \frac {-1}{ 4 x^{-3/2}} $$

$$ f_{xy} = f_{xp} = 0 $$

$$ f_y = f_{yp} = f_{yy} = f_p = f_pp = 0 $$

$$ g_x = 2p $$

$$ g_{xx} = g_{xy} = 0 $$

$$ g_{xp} = 2 $$

$$ g_y = 3 $$

$$ g_{yp} = g_{yy} = 0 $$

$$ g_p = 2x $$

$$ g_{pp} = 0 $$

Exactness Conditions

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

$$ \frac {-1}{4x^{-3/2}} + 2p \cdot 0 + p^2 \cdot 0 = 2 + p \cdot 0 - 3 $$, which does not satisfy the condition.

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

$$ 0 + p \cdot 0 + 2 \cdot 0 = 0 $$, which does satisfy the condition.

Since both equations do not satisfy the condition, the equation is not exact.

title: User:Egm6321.f09.team3.diggs/HW1


text:
WS3 2009 - 22 September 2009 Topics

Welcome
Hello all, I hope that everyone is tuned into WS3 2009. I just wanted to say that I hope there are folks out there listening because this is going to be a critical time for us to get the portal up, running, and moving in the right direction. I might be the only person that writes on the topic list but this a record of where we have come from. A year from now I hope that there will be many more.

General Curriculum
Does anyone have any ideas about how we can structrue a basic curriculum for the Social Sciences portal? --JaylanHaley 01:59, 22 September 2009 (UTC)

Administrivia
Does any have any feedback for the Social Sciences Scholar's Council? --JaylanHaley 01:59, 22 September 2009 (UTC)

title: WS3 2009 - 22 September 2009 Topics


text:
User:Pemika 5080693

Hello! everyone my name is pemika. i'm a tim student. this trimester, i decided to register computer essentials, but i'm not good at using computer.i'm quite worried about it. anyway, i'll try my best as i can to understand every lesson and pass this subject with my friends.

FIGHTING!!!! FIGHTING!!!! FIGHTING!!!!

Homework 1 Pemika ID 5080693 Sec. 1

The specs for the best PC computer

INTEL Performance PC

1. CPU : Intel i7-940 2.93 GHz

2. Mainboard : ASUS P6T Deluxe X58+ICH10R

3. Memory : OCZ DDR3 Gold Edition 6GB 1600MHz Triple Channel (2GBx3) 8-8-8-24

4. Hard Disk : Barracuda ES.2 ST3500320NS 500GB SATA-II 32MB NCQ (Enterprise Drive)

5. Graphics Card : XFX GTX285 ZDFF 1024MB DDR3 512

6. Monitor : Samsung LCD 22" T220

7. Optical Drive : ASUS DRW2014L1T 20X, SATA-Interface

Other equipment

1. computer keyboard : .Optimus Maximus

2. Power supply : 600 WATT POWER SUPPLY Intel AMD PC Desktop Computer

3. Mouse : The Logitech Gaming Mouse G500

4. Modem : Modem (Intel 82536 Chipset)

5. Sound card : Creamware- SCOPE home ( LUNA II )

6. Video card : HP ATI Radeon X1300 Video Card

7. CD/DVD : HP/Compaq Pavilion DV1000, DV1100 & DV1300 Dual Format 8X DVD+/-RW Drive

8. Case : Case RaidMax SAGITTA (921WB) (pro) (No Power)

9. Floppy drive : Intel (AXXUSBFLOPPY) (AXXUSBFLOPPY) Internal Floppy Drive

Pemika 5080693 05:01, 3 October 2009 (UTC)

Retrieved from "http://en.wikiversity.org/wiki/User_talk:Pemika_5080693"

1.How is Facebook and Twitter the same?

Facebook and Twitter is quite the same. These two networks are social connection. People are able to know what is happening when the users update their information in Facebook and Twitter. No matter where people are, they just access to the internet and log in these two networks, so they will know what their families, friends, relationships, or celebrities type messages on the page. People can contact together.

2.Why do you think Facebook is worried about competition from Twitter?

I think Facebook is quite worried about competition form Twitter. Twitter’s users can ask questions and get answers easily and quickly. It does not take a long time to communicate with other people, and people can contact with strangers and show their ideas and response to what people post on Twitter by clicking “Follow” at the other users you want. They can keep up with the latest messages, so they can connect with people any time they want. Moreover, Twitter can be used to promote several types of people, products, knowledge and even news with the length of message not more than 140 characters. A few years ago, Twitter becomes more and more worldwide. It is like a public network while Facebook is used by small communities (in their groups such as friends and colleges). It means Twitter makes users know more people. Now, I’m following Obama’s twitter and my favorite actors’ twitter. Last two weeks, I knew Obama got the Nobel Prize on Twitter before reading in the newspaper. In addition, donating money via this network is very useful for helping people. Accordingly, Twitter is very beneficial for people to update any messages and share their powerful ideas with others.

--Pemika 5080693 02:04, 17 October 2009 (UTC)

Which oprating system is the best for you

types of OS that people use in daily life

79.07% use Windows xP

7.91% use window vista

3.41% use mac intel

3.16% use window 2000

3.14% use Mac Os

0.82% use windows 98

2.39% use others

personally, i use window Xp becaue I just use it for work and school, it is easy to use. What operating system shoulf install in my notebook?

the best operating system for me is windows 7 for netbook because it is better than window XP. it perform faster than XP. Applications is better, and it also has a number of new features, such as advances in touch and handwriting recognition. It is worth to use window 7.

I think Hercules eCAFE EC-1000W is interesting.

SPEC

 * Windows 7 Starter operating system with Windows Live® Suite: Messenger, Mail, Photo Gallery, Movie Maker & Media Player
 * 10.1″ high-brightness screen (1024×600)
 * Intel Atom N270 1.6GHz Processor
 * 1GB Ram
 * 250GB HDD Storage + an extra 50GB of online storage
 * Comfortable chiclet keyboard with separated flat keys
 * WiFi N technology
 * 6-Cell 4400 mAh battery
 * Built-in webcam with microphone
 * Connections: 1x VGA, 3x USB, 1x headphones, 1x microphone, 1x LAN RJ45, 1x DC-in 12V, 1x card reader: SD/MMC/SDHC/MS

Pemika 5080693 10:58, 18 November 2009 (UTC)

title: User:Pemika 5080693


text: Educational technology/See also

title: Educational technology/See also


text:
Educational technology/Learning projects

Wikipedia

MindMeister An online collaborative mapping tool.

eBlogger A free online journal.

RubiStar   Create rubrics for your project-based learning activities.

YouTube A video sharing website.

Make Beliefs Comix Create your very own comic strip, choosing from a variety of characters, background colors and dialogue boxes.


text:
User:Egm6321.f09.team3.Kuljevan.Francisco/HW2

=Problem 1=

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

Given
For case #2

$$ h_xN = 0 \! $$

Find
Complete the details on how to obtain h(y) for case #2

Solution
General form

\begin{align} &h_xN - h_yM + (N_x-My) = 0 \qquad \text{investigating the case for} \qquad h_xN = 0 \\ &-h_yM =h(N_x - M_y) \qquad \text{factoring out}\\ &\frac{h_y}{h} = \frac{1}{M}\left(N_x-M_y\right) \qquad \text{integrating both sides gives}\\ &\int\left(\frac{h_y}{h}\right)dy = \int\frac{1}{M}\left(N_x-M_y\right)dy \qquad \text{evaluating the integral}\\ &\ln h =\int\frac{1}{M}\left(N_x-M_y\right)dy \end{align} $$

We know that
 * $$ h_x \underbrace{N}_{\neq 0} = 0 $$

which means that $$h(x,y) = h(y)$$

and that,

 * $$\frac{1}{M(x,y)}\left(N_x(x,y)-M_y(x,y)\right) = -g(y)$$

since

$$\ln h = \underbrace{\int\frac{1}{M}\left(N_x-M_y\right)dy}_{-g(y)}$$

then we arrived to $$h(y)$$

 * $$h(y)=\exp \int g(s)ds$$

 * }

=Problem 2=

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

Given
A non-homogeous L1-ODE-VC

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

Find
Show that

 * $$y(x)=\frac{x^{3}}{4}+\frac{C}{x}$$

Solution
The general form for a non-homogenous L1-ODE-VC is:

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

The one given in the problem can be re-written as:

 * $$\underbrace{1}_{N(x,y)}\dfrac{dy}{dx} + \underbrace{\frac{1}{x}y}_{M(x,y)}=b(x)$$

We know that:

\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ 1(0 - a_0(x)) &= -f(x)\qquad \text{which yields}\\ f(x) &= a_0(x)\\ \end{align} $$

Then making use of integrating factor knowledge:

\begin{align} a_0(x) &= \frac{1}{x}\\ h(x)&=\exp\int^{x}a_0(s)ds\qquad \text{replacing values for}\quad a_0(x) \quad \text{and evaluating the integral}\\ h(x)&=\exp(\ln x)\\ h(x) &= x \end{align} $$

Now we can solve for $$y(x)$$

\begin{align} y(x) &= \frac{1}{h(x)}\int^{x} h(s)b(s)ds \qquad \text{plugging known values in this expression yields}\\ y(x)&=\frac{1}{x}\int^{x}s s^2 ds \qquad \text{analyzing this integral and expanding terms out we end up with our desire result}\\ y(x)&=\frac{x^3}{4} + \frac{C}{x} \end{align} $$
 * }

=Problem 3=

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

Given
A non-homogenous L1_ODE_VC

 * $$ \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$

Find
Show that the above non-homogenous L1_ODE_VC is exact

Solution
For the non-homogenous L1_ODE_VC to be exact it must staisfy the 2 conditions of exactness

\begin{align} N(x,y)y' + M(x,y) &= 0 \\ M_y(x,y) &= N_x(x,y) \end{align} $$

So the first condition is met by:

 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y+10]}_{M(x,y)}= 0 $$

Let's take a look at the 2nd condition:

\begin{align} M_y(x,y) & = x^4 \\ N_x(x,y) & = x \end{align} $$

Clearly $$M_y(x,y) \neq N_x(x,y)$$, so we make use of the integrating factor to see if we can make it exactly integrable.

So we re-write the ODE

 * $$ \underbrace{\frac{1}{2}x^{2}}_{N(x,y)}y'+\underbrace{[x^{4}y]}_{M(x,y)}= -10 $$

Looking at case #1 for the integrating factor we obtain:

\begin{align} \frac{1}{N}\left(N_x - M_y\right) &= -f(x) \qquad \text{replacing values found above}\\ \frac{2}{x^2}\left(x - x^4\right) &= -f(x) \qquad \text{applying knowledge from integrating factor we know}\\ h(x)&=\exp \int^{x}f(s)ds \qquad \text{replacing f(x) into the integral and evaluating it, obtain}\\ h(x) &= \frac{1}{x^2}\exp \left(\frac{2x^3}{3}\right) \end{align} $$

Now we can try the 2nd condition of exactness to see if by using integrating factor we can make this ODE exact

\begin{align} \bar{M}_y &= \bar{N}_x\\ \underbrace{(hM)}_{\bar{M}}dx&=\underbrace{(hN)}_{\bar{N}}dy\qquad \text{calculating the respective values we obtain}\\ \bar{N}_x &=x^2 \exp\left(\frac{2x^3}{3}\right)\\ \bar{M}_y &=x^2 \exp\left(\frac{2x^3}{3}\right) \end{align} $$

Hence the non-homogenous L1_ODE_VC is exact.

 * }

=Problem 4=

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

Given
An aribitrary 1st order ODE:

 * $$ \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)=0$$

Find
Show that the above is non-linear and exact

Solution
So let's determine that the above ODE is non-linear by making use of the $$D(.)\!$$ operator

To be a linear ODE the following must be true:

 * $$D\left(\alpha u + \beta v \right) = D\left(\alpha u \right) + D\left(\beta v \right)$$

Now we will define the $$D(.)\!$$ operator for the above ODE:

 * $$D(.)= \left(\frac{1}{3}x^{3}\right)(\cdot)^{4}\frac{d(\cdot)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\cdot)^{5}\right)$$

Then let's evaluate the linearity condition to see if it's true:

\begin{align} D\left(\alpha u + \beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u + \beta v)^{4}\frac{d}{dx}(\alpha u + \beta v)+(5x^{3}+2)\left(\frac{1}{5}(\alpha u + \beta v)^{5}\right)\\ D\left(\alpha u \right) + D\left(\beta v \right) &= \left(\frac{1}{3}x^{3}\right)(\alpha u)^{4}\frac{d(\alpha u)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\alpha u)^{5}\right) +  \left(\frac{1}{3}x^{3}\right)(\beta v)^{4}\frac{d(\beta v)}{dx}+(5x^{3}+2)\left(\frac{1}{5}(\beta v)^{5}\right) \end{align} $$

Clearly $$D\left(\alpha u + \beta v \right) \neq D\left(\alpha u \right) + D\left(\beta v \right)$$ so the ODE is non-linear

To prove that the ODE is exact, let's first look at the 1st condition of exactness:

 * $$ \underbrace{\left(\frac{1}{3}x^{3}\right)(y^{4})}_{N(x,y)}y'+\underbrace{(5x^{3}+2)\left(\frac{1}{5}y^{5}\right)}_{M(x,y)}=0$$

hence $$N(x,y)y' + M(x,y) = 0\!$$ so 1st condition is met.

Looking at the 2nd condition where $$M_y = N_x\!$$. However, from above we can clearly see that:

 * $$M_y \neq N_x$$

So we will try to find an integrating factor to find if our ODE can be exactly integrable

\begin{align} \left(\frac{1}{3}x^{3}\right)(y^{4})y'+(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0 \qquad \text{let us multiply througout by} \qquad \frac{3}{x^3y^4}\\ y' +\frac{3}{x^3y^4}(5x^{3}+2)\left(\frac{1}{5}y^{5}\right) &=0\qquad \text{re-writing the equation obtain}\\ \underbrace{1}_{N}\cdot y' + \underbrace{\left(3 + \frac{6}{5x^3}\right)}_{a_0}y &=0 \end{align} $$

Now we will make use of integrating factor knowledge:

\begin{align} h(x,y) &= \exp\int^{x}a_0(s)ds\qquad \text{replacing the value of} \quad a_0\\ h(x,y) &= \exp\int^{x}\left(3 + \frac{6}{5s^3}\right)ds\qquad \text{evaluating the integral obtain}\\ h(x,y) &= \exp\left(3x - \frac{3}{5x^2}\right) \end{align} $$

Now we will try to evaluate the 2nd condition of exactness which is:

\begin{align} (hM)_{y}&=(hN)_x \qquad \text{the condition says:} \\ \bar{M}_y &= \bar{N}_x \qquad \text{evaluating the derivatives we obtain}\\ \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} } & = \dfrac{6\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}} x^3} + \dfrac{3\exp(3x)}{\left(\exp^{\frac{1}{x^2}}\right)^{\frac{3}{5}}} \end{align} $$

Hence the ODE given is then, exact N1-ODE-VC
 * }

=Problem 5=

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

Given
The following 2nd order ODE

 * $$(xy)y''+x(y')^{2}+yy'=0\!$$

Find
Show that the second exactness condition is satisfied for the above equation

Solution

\begin{align} \underbrace{(xy)}_{f(x,y,p)}y''+\underbrace{x(y')^{2}+yy'}_{g(x,y,p)}&=0 \qquad \text{we can re-write the ODE}\\ (xy)y''+xp^{2}+yp&=0 \end{align} $$

The second exactness condition is the following

\begin{align} f_{xx} +2pf_{xy} + p^2f_{yy} & = g_{xp} + pg_{yp} - g_{y}\qquad \text{evaluating each derivative, we obtain}\\ 0 + 2p(1) +p^2(0) &= 2p(1) + 0 + p - (0) - p \qquad \text{which gives}\\ 2p & = 2p \end{align} $$

Also, the following must be met

\begin{align} f_{xp} + pf_{yp} + 2f_{y} &=g_{pp} \qquad \text{evaluating each derivative, we obtain}\\ 0 + 0 + 2x &= 2x + 0\qquad \text{which gives}\\ 2x &=2x \end{align} $$

Hence the second exactness condition is met for this particular case
 * }

=Problem 6=

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

Given
The following condition is given

 * $$f_{xp} + pf_{yp} + 2f_{y} =g_{pp} \!$$

Find
Derive the above equation by differentiating

 * $$g(x,y,p) = \phi_x +\phi_yp\!$$

with respect to $$p (y')\!$$

Solution
We know that:

\begin{align} g &= \phi_x +\phi_yp \qquad \text{differentiating once wrt p}\\ g_p &= \phi_{xp} +\phi_y +p\phi_{yp} \qquad \text{differentiating again wrt p}\\ g_{pp} &= \phi_{xpp} +\phi_{yp} +\phi_{yp} + p\phi_{ypp} \qquad \text{taking into parenthesis}\quad \phi_p \quad \text{we obtain}\\ g_{pp} &= (\phi_p)_{xp} +(\phi_p)_y +(\phi_p)_y + p(\phi_p)_{yp} \qquad \text{finally arriving to the given equation because}\quad \phi_p = f\\ g_{pp} &=f_{xp} + pf_{yp} + 2f_{y} \end{align} $$

=Problem 7=
 * }

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

Given
The following condition is given

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

Find
Use $$\phi_{xy}=\phi_{yx}\!$$ to obtain the equation given above

Solution
We know that

\begin{align} \phi_x &= g - p(g_p - f_x) + p^2f_y \qquad \text{and that}\\ \phi_y &= g_p -pf_y - f_x \end{align} $$

So what we will need to do is find $$\phi_{xy}=\phi_{yx} \!$$ respectively and equate them to find the given condition

\begin{align} &\phi_{xy} = g_y - pg_{py} + pf_{xy} + p^2f_{yy} \\ &\phi_{yx} = g_{px} - pf_{yx} - f_{xx} \qquad \text{after re-arranging we obtain the desire given condition}\\ &f_{xx} +2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \end{align} $$

=Problem 8=
 * }

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

Given
The following 2nd order ODE

 * $$(8x^5y')y'' + 2x^2y' +20x^4(y')^2 +4xy = 0\!$$

Find
Verify the second exactness condition on the above ODE

Solution
First let us identify some parameters in the ODE

\begin{align} \underbrace{(8x^5y')}_{f}y'' + \underbrace{2x^2y' +20x^4(y')^2 +4xy}_{g} &= 0 \qquad \text{let us look at the exactness conditions needed to be analyzed}\\ f_{xx} +2pf_{xy} + p^{2}f_{yy} &= g_{xp} + pg_{yp} - g_{y} \qquad \text{evaluating each derivative}\\ 160x^3p +2p(0) + p^2(0) & = 160x^3p +4x +p(0) - 4x \qquad \text{which gives}\\ 160x^3p &= 160x^3p \end{align} $$

Also, the other condition is as follow:

\begin{align} f_{xp} + pf_{yp} + 2f_{y}&=g_{pp} \qquad \text{evaluating each derivative}\\ 40x^4 + p(0) + 2(0) &=40x^4\qquad \text{which gives}\\ 40x^4 &= 40x^4 \end{align} $$

Hence the ODE given is exact!

=Problem 9=
 * }

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

Given
The following 2nd order ODE is given

 * $$\sqrt{x}y''+2xy'+3y=0$$

Find
Verify the exactness of the above ODE

Solution
For the ODE to be exact it must satisfy the two conditions of exactness

 * $$\underbrace{\sqrt{x}}_{f}y''+\underbrace{2xy'+3y}_{g}=0$$

Which means that the 1st condition of exactness is met.

\begin{align} f_{xp} + pf_{yp} + 2f_{y} &=g_{pp} \qquad \text{evaluating each derivative, we obtain}\\ (0)+ p(0) + 2 (0) &= 0 \\ 0 &= 0 \qquad \text{which satisfies the first condition} \end{align} $$

However, the ODE must satisfy both conditions to be exact. So let's evaluate the next condition

\begin{align} f_{xx} +2pf_{xy} + p^2f_{yy} & = g_{xp} + pg_{yp} - g_{y}\qquad \text{evaluating each derivative, we obtain}\\ -\dfrac{1}{4x^{\frac{3}{2}}} + 2p(0) + p^2(0) &= 2 + p(0) - 3 \\ -\dfrac{1}{4x^{\frac{3}{2}}} &\neq -1 \end{align} $$

Since the ODE did not comply with the all of the exactness conditions then the ODE is non-exact
 * }


text:
User:Cutedatchy

Hey there!

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Homework Week 1: My Wikiversity Account is: http://en.wikiversity.org/wiki/User:Cutedatchy

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Homework Week 2: Supercomputer!!! the best computer on the face of the earth

Motherboard: Asus P5K3 Premium Motherboard

Processor: Intel i7 Core 2 Quad 3.06 GHz, core speed

Memory: 2 Ram G.SKILL DDR3 4G (2G x 2) bus 1600

Case: AlienwareTM Aurora ALX with AlienwareTM High-Performance Liquid Cooling

Power supply: GIGABYTE Odin 550W Power Supply (GE-S550A-D1)

Floppy drive: Ultra ULT40366 3.5" Floppy Drive with Multi Card Reader

Hard drive: Seagate Barracuda 7200.11 ST31500341AS 1.5TB 7200 RPM 32MB Cache SATA 3.0Gb/s 3.5" Internal Hard Drive

CD/DVD: SONY Blue Ray ROM BD-ROM BDU-X10S

Keyboard: Logitech Cordless Desktop® MX™ 5500 Revolution (Keyboard)

Mouse: Logitech Cordless Desktop® MX™ 5500 Revolution (Keyboard)

Video card: Nvidia GeForce 9800 GX2

Monitor: ASUS VH236H 23" Widescreen HD LCD Monitor

Sound card: Creative SoundBlaster X-Fi Titanium Fatal 1 ty Pro Series

Modem: USRobotics 56K* V.92 Internal Faxmodem PC

>w<

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Homework Week 3: Twitter

My Twitter Account: http://twitter.com/cutedatchy

Q:

1) How is Facebook and Twitter the same?

2) Why do you think Facebook is worried about competition from Twitter?

A:

1) The similarity between Facebook and Twitter are --

a) you could let other people know what you are doing

b) find people/make new friends

c) Have your own profile page -- where you can customised it e.g. changing background themes, colour layout etc.

d) you can send messages to other people [a bit like a conversation]

2) Well, Twitter is newer and the majority of the clients loves to follow new trends, and you could directly connect the account to your phone BB phone. Another reason might be because the celebrities are also using twitter so all the fans could catch up with their latest updates through the "what are you doing now?" status. You could also post alternative links such as other website that you have e.g. facebook, hi5, myspace etc. so that people could contact other than twitter.

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Homework Week 9: Operating System

Q: What Operating Systems should install on your netbook?

A: Linux; "Ubuntu"

why? because it's

1). Hardware -- Netbook hardware is the perfect match for Linux — because netbook is designed for internet use and they do it well, and it's secure and fast as well :)

2). Security -- Linux-based netbook can travel anywhere you want and you won’t have to worry about picking up viruses or spyware like you would with a Windows-based netbook

3). Interface -- the Eeebuntu Netbook Remix desktop introduces an interface that is perfectly suited for the desktop size offered by netbooks. This interface makes the netbook experience far more efficient than any Windows interface. It’s cleaner, faster, and more user-friendly.

4). Installtion -- Very fast to install programs onto your netbook

5). Cheaper -- Linux is open source, so any software you are installing will be free

6). Speed -- Faster than Windows

7). Support -- Finding support for a Windows-based netbook isn’t as easy as it is for a standard laptop or desktop. The nice thing about the Linux netbook community is that the developers are interested in making sure the OS works well and are quick to listen to their users. So you know if you have a problem with your Linux-based netbook, a quick search on Google will most likely find a simple solution to your problem.

For more information: http://blogs.techrepublic.com.com/10things/?p=945

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title: User:Cutedatchy


text:
User:Tuntacpan

''' Hello World~! - -!! '''

1st Homework

CPU: Intel Intel® Core™2 Quad Processor Q9650 12MB shared L22 3.0 GHz 1333 MHz

Motherboard:DFI LANPARTY DK P55-T3eH9 Intel i5 P55 Chipset ATX DFI Motherboard + AlienwareTM High-Performance Liquid cooling.

Memory: Kingston Part: KTS5090/4G DDR 400 Kit (2GB x 2)

Case Aquarium PC Mod

Power Supply: Antec Signature 850

Flobby drive: Sony MPF920-Z/CU1 Internal 3.5" Floppy Drive

Hard drive: Seagate Barracuda 7200.11 1.5TB

CD/DVD: Lite-On IHAS22406 Internal DVD Writer

Keyboard: diNovo Edge™ (Logitech)

Mouse: Logitech MX Revolution

Video Card: XFX G-force 9800 gtx Video Card

Monitor: HP w2338h 23" Diagonal Full HD Widescreen LCD Monitor

Sound Card: Sound Blaster X-Fi Titanium Driver 2.17.0007

Modem: D-Link Internal 56K Pci Bus Modem DFM-562IS

2nd Homework

1.How is Facebook and Twitter the same?

It is to communicate with your friends and to talk about events in your life. It’s the same that when you post something up and it shows to all of your friends. It also let you post your picture and search for some interesting things.

2.Why do you think Facebook is worried about competition from Twitter? because twitter is a new one that still can upgrade themselves quite quickly and may be because of it is new so many people want to try this out. Also that the system is simplier to use with no add ons things that you have to install in order to use an application.

 3rd Homework

What operating systems should install on your netbook?

Linux

title: User:Tuntacpan


text:
User:Htaikseng

Htaik Seng

title: User:Htaikseng


text:
User:Patrol110

My home wiki and user talk.

title: User:Patrol110


text:
User:Egm6321.f09.Team8/HW2

= Problem #1 =

Given
([[media:Egm6321.f09.p7-1.png|7-1]]): Complete the details of case 2, when $$h_{x}N=0$$ to obtain $$h(y)$$

Solution
[[Media:HW2, Page 1.PNG]] [[Media:HW2, Page 2.PNG]] [[Media:HW2, Page 3.PNG]]

= Problem #2 =

Given
([[media:Egm6321.f09.p8-3.png|8-3]]): Show that the solution of $$ y'+\frac{1}{x}y=x^{2} $$  is  $$y=\frac{x^{3}}{4}+\frac{c}{x}$$.

Solution
[[Media:HW2, Page 3.PNG]] [[Media:HW2, Page 4.PNG]]

= Problem #3 =

Given
([[media:Egm6321.f09.p9-2.png|9-2]]): Show that  the L1_ODE_VC $$ \frac{1}{2}x^{2}y'+[x^{4}y+10]= 0 $$  is exact.

Solution
[[Media:HW2, Page 5.PNG]] [[Media:HW2, Page 6.PNG]]

= Problem #4 =

Given
([[media:Egm6321.f09.p9-3.png|9-3]]): Show that $$ (\frac{1}{3}x^{3})(y^{4})y'+(5x^{3}+2)(\frac{1}{5}y^{5})=0$$ is an exact nonlinear, first order ODE.

Solution
[[Media:HW2, Page 7.PNG]] [[Media:HW2, Page 8.PNG]] [[Media:HW2, Page 9.PNG]]

= Problem #5 =

Given
([[media:Egm6321.f09.p10-3.png|10-3]]): Show that the second exactness condition for $$xyy''+x(y')^{2}+yy'=0$$ is satisfied.

Solution
[[Media:HW2, Page 10.PNG]] [[Media:HW2, Page 11.PNG]]

= Problem #6 =

Given
(11-2): Derive eq. 5 on p.([[media:Egm6321.f09.p10-2.png|10-2]]) by differentiating eq. 3 on p.([[media:Egm6321.f09.p10-1.png|10-1]]) with respect to $$p=y' $$.

Solution
[[Media:HW2, Page 12.PNG]]

= Problem #7 =

Given
([[media:Egm6321.f09.p12-1.png|12-1]]): Use $$\phi_{xy}=\phi_{yx}$$ to obtain eq. 4 on p.([[media:Egm6321.f09.p10-2.png|10-2]]).

Solution
[[Media:HW2, Page 13.PNG]] [[Media:HW2, Page 14.PNG]]

= Problem #8 =

Given
([[media:Egm6321.f09.p12-2.png|12-2]]): Verify exactness condition 2, equations 4&5 on p.([[media:Egm6321.f09.p10-2.png|10-2]]).

Solution
[[Media:HW2, Page 14.PNG]] [[Media:HW2, Page 15.PNG]]

= Problem #9 =

Given
([[media:Egm6321.f09.p12-3.png|12-3]]): Verify the exactness of the ODE $$\sqrt{x}y''+2xy'+3y=0$$.

Solution
[[Media:HW2, Page 16.PNG]] [[Media:HW2, Page 17.PNG]]

title: User:Egm6321.f09.Team8/HW2


text:
User:Egm6321.f09.Team1.vasquez/Homework2

= Problem 7: Second condition of exactness, relation (1) for a N2-ODE =

Given
A general N2-ODE is given as,

 * {| style=";width:70%" border="0" align="center"

F(x,y,y',y'') = 0 $$
 * $$\displaystyle

 * }
 * }

and satisfies exactness condition (1) if it is in the form,

 * {| style=";width:70%" border="0" align="center"

f(x,y,p)y'' + g(x,y,p) = 0 $$
 * $$\displaystyle

 * }
 * }

where $$\displaystyle p:=y' $$

Find
To derive 2nd condition of exactness, relation (1) of a N2-ODE from $$\displaystyle \phi_{xy}=\phi_{yx}$$, which is given by

 * {| style=";width:70%" border="0" align="center"

f_{xx} + 2p f_{xy} + p^2 f_{yy} = g_{xp} + p g_{yp} - g_{y} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (34)

 * }
 * }

Solution
We know that $$\displaystyle \phi_{xy}=\phi_{yx}$$

where,

 * {| style=";width:70%" border="0" align="center"

\phi_{x} = g - p ( g_p - f_x) + p^2 f_y $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (35)

 * }
 * }

 * {| style=";width:70%" border="0" align="center"

\phi_{y} = g_p - f_y p - f_x $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (36)

 * }
 * }

Taking the partial derivative of (35) with respect to $$\displaystyle y$$ and the partial derivative of (36) with respect to $$\displaystyle x$$ gives

 * {| style=";width:70%" border="0" align="center"

\phi_{xy} = g_y - p_y ( g_p - f_x) - p (g_{py} - f_{xy}) + 2 p_y f_y + p^2 f_{yy} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (37)

 * }
 * }

 * {| style=";width:70%" border="0" align="center"

\phi_{yx} = g_{px} - f_{yx} p - f_y p_x - f_{xx} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (38)

 * }
 * }

Since (37) and (38) are equal

 * {| style=";width:70%" border="0" align="center"

g_y - \cancelto{0}{p_y} g_p + \cancelto{0}{p_y} f_x - p g_{py} + p f_{xy} + 2 \cancelto{0}{p_y} f_y + p^2 f_{yy} = g_{px} - f_{yx} p - f_y \cancelto{0}{p_x} - f_{xx} $$
 * $$\displaystyle

 * }
 * }

Re-arranging terms yields

 * {| style=";width:70%" border="0" align="center"

$$\displaystyle f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |

 * }
 * }

= Problem 8: Second condition of exactness, relations (1) and (2) for a N2-ODE =

Given
The following N2-ODE,

 * {| style=";width:70%" border="0" align="center"

(8 x^5y')y''+2x^2y'+20x^4(y')2+4xy=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (39)
 * }
 * }

Find
Verify the second condition of exactness, relations (1) and (2)

Solution
Let $$\displaystyle p=y'$$, then (39) can be written as follows:

 * {| style=";width:70%" border="0" align="center"

(8 x^5p)y''+2x^2p+20x^4(p)2+4xy=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (40)

 * }
 * }

From (40) we know that

 * {| style=";width:70%" border="0" align="center"

f(x,y,p)=8 x^5p $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (41)

 * }
 * }

 * {| style=";width:70%" border="0" align="center"

g(x,y,p)=2x^2p+20x^4(p)2+4xy $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (42)
 * }
 * }

Taking the partial derivatives of (41) and (42) gives

 * {| style=";width:70%" border="0" align="center"

\begin{array}{*{20}l} {f_x = 40x^4 p} & {f_{xx}  = 160x^3 p} & {g_x  = 4xp + 80x^3 p^2  + 4y} & {g_{xp}  = 4x + 160x^3 p}  \\ {} & {f_{xy} = 0} & {g_y  = 4x} & {g_{yp}  = 0}  \\ {} & {f_{xp} = 40x^4 } & {g_p  = 2x^2 p^2  + 40x^4 p} & {g_{pp}  = 40x^4 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }

The first relation of the second exactness condition is given by

 * {| style=";width:70%" border="0" align="center"

f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * $$\displaystyle \longrightarrow (43)
 * }
 * }

Substituting the elements of (43) gives

 * {| style=";width:70%" border="0" align="center"

(160x^3 p) + 2 p (0) + p^2 (0)  = (4x + 160x^3 p) + p (0) - (4x) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,

 * {| style=";width:70%" border="0" align="center"

160x^3 p = \cancel{4x} + 160x^3 p - \cancel{4x} $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the first relation is satisfied

The second relation of the second exactness condition is given by

 * {| style=";width:70%" border="0" align="center"

f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * $$\displaystyle \longrightarrow (44)
 * }
 * }

Substituting the elements of (44) gives

 * {| style=";width:70%" border="0" align="center"

(0) + p(0) + 2(0) = (0) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,

 * {| style=";width:70%" border="0" align="center"

0 = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the second relation is satisfied

Since both relations (43) and (44) are satisfied we conclude that the given N2_ODE is exact.

= Problem 9: Exactness of a N2_ODE =

Given
The following differential equation,

 * {| style=";width:70%" border="0" align="center"

\sqrt{x} y'' + 2xy' + 3y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (45)
 * }
 * }

Find
Verify the exactness of the given ODE

Solution
Let $$\displaystyle p=y'$$, then (45) can be written as follows:

 * {| style=";width:70%" border="0" align="center"

x^{\frac{1}{2}} y'' + 2xp + 3y=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (46)

 * }
 * }

From (46) we verify the first condition of exactness, where

 * {| style=";width:70%" border="0" align="center"

f(x,y,p)=x^{\frac{1}{2}} $$ $$
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (47)

 * }
 * }

 * {| style=";width:70%" border="0" align="center"

g(x,y,p)=2xp + 3y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \longrightarrow (48)
 * }
 * }

Taking the partial derivatives of (47) and (48) gives

 * {| style=";width:70%" border="0" align="center"

\begin{array}{*{20}l} {f_x = \frac{1}{2}x^{-\frac{1}{2}}} & {f_{xx}  = -\frac{1}{4}x^{-\frac{3}{2}}} & {g_x  = 2p} & {g_{xp}  = 2}  \\ {} & {f_{xy} = 0} & {g_y  = 3} & {g_{yp}  = 0}  \\ {} & {f_{xp} = 0 } & {g_p  = 2x} & {g_{pp}  = 0 }  \\ {f_y = 0} & {f_{yy}  = 0} & {} & {}  \\ {} & {f_{yp} = 0} & {} & {}  \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }

The first relation of the second exactness condition is given by

 * {| style=";width:70%" border="0" align="center"

f_{xx} + 2 p f_{xy} + p^2 f_{yy}  = g_{xp} + p g_{yp} - g_y $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * $$\displaystyle \longrightarrow (49)
 * }
 * }

Substituting the elements of (49) gives

 * {| style=";width:70%" border="0" align="center"

(-\frac{1}{4}x^{-\frac{3}{2}}) + 2 p (0) + p^2 (0)  = (2) + p (0) - (3) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,

 * {| style=";width:70%" border="0" align="center"

-\frac{1}{4}x^{-\frac{3}{2}} \ne -1 $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the first relation is not satisfied.

The second relation of the second exactness condition is given by

 * {| style=";width:70%" border="0" align="center"

f_{xp} + pf_{yp} + 2f_y = g_{pp} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * <p style="text-align:right;">$$\displaystyle \longrightarrow (50)
 * }
 * }

Substituting the elements of (50) gives

 * {| style=";width:70%" border="0" align="center"

(0) + p(0) + 2(0) = (0) $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Then,

 * {| style=";width:70%" border="0" align="center"

0 = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * style= |
 * }
 * }

Therefore, the second relation is satisfied.

Since the first relation of the second exactness condition is not satisfied we conclude that the given N2_ODE is not exact.


text:
User talk:T.kullatida

My name is Kullatida Teeraratpol. You are welcome to call me May (No, I wasn’t born in May. I was actually born in August). I just turned 18 but still no driving license. My favorite thing to do is reading. I love the book Great Gatzby and am now enjoy reading the Twilight Saga. My family consists of five people, which are my dad, my mom, my younger and older sisters, and me. Both of my parents are successful gas station owners. My older sister is 2 years older then me. She is doing her third year at Thammasat University. My younger sister is 14 and going to go aboard to the United States next year. We are a very close family. I am doing my first trimester of a second year at MUIC. My major is TIM. I have a service-mind and hope to become a successful hotel owner in the future.

 Homeowrk 1: The Best of the Best computers  By Kullatida T., ID 5180285 (Computer Essential Section 1)

 * Motherboard : GigaByte GA-MA790FXT-UD5P Motherboard
 * Processor : one of the AMD's BE2xx0 or X2 3800+ or Intel's E21x0 or E4x00
 * Memory : OCZ DDR3 Gold Edition 6GB 1600MHz Triple Channel (2GBx3) 8-8-8-24
 * Case : Case RaidMax SAGITTA
 * Power supply : OCZ ProXStream 1000
 * Floppy drive : Sony's HiFD 200 MB drive
 * Hard drive : Western Digital My Book Essential 1TB
 * CD/DVD : HP/Compaq Pavilion DV1000, DV1100 & DV1300 Dual Format 8X DVD+/-RW Drive
 * Keyboard : IBM Model M
 * Mouse : The Logitech MX Revolution
 * Video card : HP ATI Radeon X1300 Video Card
 * Monitor : CineMassive MasterPlex 30D
 * Sound card : Asus Xonar Essence STX
 * Video card : Three EVGA GeForce GTX 285 1024 MB in Three-Way SLI
 * Modem : Dell 56 Kbps External USB Modem for Select Systems

Posted on Oct 3rd at 2.12 P.M.

-

Homework 2: Twitter and Faceoff By Kullatida T., ID 5180285 (Computer Essential Section 1)

1. Both Facebook and Twitter are social-networking websites. The other similarity I saw from both of them is that you can update your status in there. On the Home page, they allow you to see your friends' status. You can post pictures and video on Facebook and also in Twitter. To leave a comment, on facebook you can go to your friends' pages and leave a comment. For Twitter, reply to tweets is the same as to leave a comment.

2. I think the reason for Facebook to be worried about the competition from Twitter is that Facebook requires a person to add and to be accepted to be friend and to view profile of other person while in Twitter you can be friend with anyone by following them without their permission. It's your option whether to accept them or not. But in Facebook, both of the users have to be mutually agreed to be friend and to view each other' profiles. Moreover, Twitter is building a society where it brings the real you. By this, I mean that Twitter users are more open-up than Facebook users. I think they are the reasons why Facebook is worried about the competition from Twitter.

Posted on Oct 16th, 8.07 P.M.

-

Homework 3: Which operating system is best for you? By Kullatida T., ID 5180285 (Computer Essential Section 1)

Netbook is known to have limitd abilities, so the operating it uses is very important. Due to this, I choose Window XP as a operating system for my netbook for three reasons. First, XP is very well supported by many many software company, which capable it to run any programs from old to modern ones. Secondly, it remains stable operating system becuase Window has been in the market for many years. It has a reputation and is very popular. Lastly, I find Window XP is much easier to use than any other operating systems like Ubuntu or MAC OS. All in all, Window XP will be great for my netbook because it provides many features while still remaining easy for the limited-abilities netbook to run.

Posted on November 17th at 9.45 p.m.

--

Retrieved from "http://en.wikiversity.org/wiki/User:T.kullatida"

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

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

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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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! --mikeu talk 14:51, 22 September 2009 (UTC)

title: User talk:T.kullatida


text:
User:Scully.peter

This user is a current medical student in the US who graduated with a Bachelor of Science in Physics with a concentration in Biophysics.

title: User:Scully.peter


text:
Technical writing/SDLC EE

back to Technical Writing Level 1 --Chatsami (discuss • contribs) 00:07, 13 November 2014 (UTC)=The Systems Development Life Cycle=

The Systems Development Life Cycle (SDLC) is a logical process that helps to design good new products and software efficiently. The SDLC system emphasises planning.

The SDLC system was developed because:

 * Before the SDLC many government projects failed.
 * Many projects were delivered to the Sponsor' late ( the Sponsor is the Person or organisation that pays for the project).
 * Many projects cost more than double the original estimate

--Chatsami (discuss • contribs) 00:04, 13 November 2014 (UTC)==The reason why projects failed==

 * Inventors invented what they imagined, not what was needed.
 * Products were designed without market research.
 * The inventor or developer did not know if anyone would buy the product until the end of the process.
 * Inventors and developers did not have a logical systematic method when they started to design and develop ideas.
 * The inventor or developer added features to the product that delayed the product launch (many of these features did not improve the performance or make the product easier to use).

--Chatsami (discuss • contribs) 00:04, 13 November 2014 (UTC)==How the SDLC works== The SDLC is a logical system that checks at every stage whether the project is valid and whether the product will be useful.

The SDLC is a process that repeats several times and is represented by either a circle or a triangle.

Start at the top corner of the triangle. The Sponsor Goals are the first input

Initialisation Phase
A Product user is someone who uses a product or service.

The Initialisation Phase starts when a Potential Product User with money (a Sponsor) identifies a need or wish for something new a (Goal).

This Goal may be a totally new product or a change to an existing product to make it easier to use.

Goals
Sponsors wishes are called Goals.

Sponsors want to:
 * use the latest technology
 * Improve the technology they use
 * reduce the number of staff required to perform a set of tasks
 * communicate
 * feel intelligent
 * be entertained
 * make more money by doing less work
 * have more free time by doing less work

Feasibility study
Many companies do a Feasibility study before the project continues to discover whether the project is possible.

Requirements and Feasibility
The Goals of the Sponsor are translated into Requirements.

A System Analyst will ask the Sponsor questions to define  what the true goals are.

The System Analyst will then discuss the goals with
 * Project Managers
 * Product Managers
 * Developers
 * Subject Matter Experts (SME).

The team of Experts may decide that the project is not possible.

The System Analyst will then discuss this with the Sponsor to discover if the goals can be reduced or modified.

Functions
When the initial Requirements have been discovered, then the team of Experts translates the Requirements into Functions that --Chatsami (discuss • contribs) 00:10, 13 November 2014 (UTC)--Chatsami (discuss • contribs) 00:10, 13 November 2014 (UTC)meet the Sponsor's Goals.

Specification
Specification details all the Functions that must be in the product

Analysts
Analysts must translate the Goals of the Sponsor into  Functions that can be tested.

The SDLC is a process that allows:

to analyse and prioritize the Functions needed to fulfil the Goals of the Product user.
 * Sponsors
 * Analysts
 * Managers
 * Developers

This initial description of Product User Requirements is vital to make the project succeed.

Functions
Functions are system actions caused by Product Users.

Functions must serve the Goals of the Sponsor described in the Requirements.

The Functions of the system lead to the more detailed Specification of the logical and physical system Structure.

Functions describe WHAT must happen, then designers and developers specify HOW the function happens in the Structure'.

Structures
Structures both support and limit the system.

Structures respond to the environment of the Users, as well as the capabilities of the developers.

Structures provide interfaces, data, workflow, integration, safety, and security. Together, they are tested and delivered to the Users as a Product.

The Users interact with the Product. If the User's Goals are met in the Functions of the Product, the process continues allowing the User to discover new Goals and developers to add new Functions and Structures in the system.

Products
Product users achieve these Goals by using Products. The Products must satisfy all the requirements

When these Products need an improvement then we return to the Goals of the product user.

Example 1
The task is to design and produce a container which will not affect performance beyond certain limits
 * 1) The Product users have cars
 * 2) The Product users work for the same company. All the cars are basically the same model.
 * 3) The cars are economical (The cars are modern fuel efficient Skodas)
 * 4) The Product Users wish to drive to Switzerland for a company sponsored holiday.
 * 5) The Product Users wish to add a container above the car to transport skis without loss of performance.

 * define the limits
 * Define the deadline (the date when the project must deliver the finished and working product)
 * define whether the Product User intends to install the product or whether specialists will install

The Analyst must ask the Product User the following questions to define the task:

 * 1) What constitutes a loss of performance. The engine of the car will perform within the manufacturers specifications. The real performance of the car will not become worse. But:
 * 2) * The additional container will introduce drag and drag will decrease the speed of the car and increase the quantity of fuel that the engine uses
 * 3) * The additional weight will decrease the speed of the car and increase the quantity of fuel that the engine uses
 * 4) How many skis must the container carry?. Suggest no more pairs of skis than the number of passengers.
 * 5) What is the total weight of the skis? ( this affects the performance per litre of fuel)
 * 6) What dimensions are the skis ( this affects the dimensions of the container)
 * 7) How can the container be attached to the roof of the car?
 * 8) When must the container be ready?

--Chatsami (discuss • contribs) 00:11, 13 November 2014 (UTC)=End of lesson Test=

 * 1) Why did projects fail before the SDLC? Products were designed without market research.
 * 2) What job does an analyst perform? The analyst translates the Sponsor's goals into testable Functions.
 * 3) With whom should an analyst discuss the initial requirements? The Sponsor.
 * 4) Why do companies run feasibility studies? To determine if the proposed product is capable of being developed into a strong revenue-producing product.

back to Technical Writing Level 1

title: Technical writing/SDLC EE


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User:EGM6321.f09.Team3.cgs

HW 1 [[User:EGM6321.f09.Team3.cgs/HW1]|

HW 2 ]

HW 3 [[User:EGM6321.f09.Team3.cgs/HW3]|

HW 4 ]

HW 5 [[User:EGM6321.f09.Team3.cgs/HW5]|

Misc formulas Formulas]

title: User:EGM6321.f09.Team3.cgs


text:
User talk:Scully.peter

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --JWSchmidt 16:33, 22 September 2009 (UTC)

title: User talk:Scully.peter


text:
User:Wissutak

Hello! My name is Wissuta Khaonuna. My nickname is Woon. and I am Thai :)

 * /Homework No.1/
 * /Homework No.2 Twitter Faceoff/
 * /Homework No.3 Which Operating System is best/

title: User:Wissutak


text:
User:Yourockmyworld

Hello, Visitors.. :D

My name is Pitchpavee Bunyakiat. I'm a sophomore of Mahidol University International College in Thailand. I do love to use the computer, often try to learn some new programs, and waste the time surfing on the Internet.

The Best Specifications for PC Motherboard: Intel DP55KG Motherboard •Support for the Intel Core i7 and Intel Core i5 processors in the LGA-1156 package •Intel P55 Express Chipset •Dual-Channel DDR3-1600+ (OC)/1333/1066MHz memory support Processor: Intel Core i7 975XE Processor •Intel Smart Cache 8MB •Clock Speed 3.33GHz •QPI Speed 6.4GT/second •Memory Type/Speed 800/1066MHz Memory: Kingston Valueram 16GB DDR3 SDRAM Memory Module •DDR PC3500, SDRAM PC100, Rambus PC1066 - 1 GB - 1.33 GHz - PC - DIMM, IC DRAM Card •16GB 1333MHz DDR3 ECC Reg w/Par CL9 DIMM (Kit of 2) DR/ x4 Case: SilverStone Fortress FT01 case •Material: 3.0mm-6.0mm uni-body aluminium outer frame and aluminium body •Drive Bay: External 5.25"x5 Internal 3.5"x7 •Cooling System: Front 1 x180mm intake fan 700 RPM, 18 dBA Rear: 1 x120mm exhaust fan 900RPM, 18dBA Top: 1 x180mm intake fan 700RPM, 18dBA •Front I/O Port: USB2.0x2 •Power Supply: 1x Optional Standard PS/2 (ATX) •Expansion Card: Compatible with expansion card up to 12" •Net Weight: 8.66 kg •Dimension: 211(W)x496mm(H)x494.5mm(D) Power Supply: IBM BladeCenter E 8677 Rack-mountable - Power supply - hot-plug 2000 Watt Floppy Disk: No Longer In Use Hard Drive: SuperTalent 512GB MasterDrive RX Series with RAIDSSD Technology •Storage Capacity: 512GB •Data Intregrity: 10 years •Endurance: 280 years @50GB write-erase/day •Warranty: Lifetime Warranty CD/DVD: LG 8x Internal Blu-ray Disc/Double-Layer DVD RW/CD-RW Drive Keyboard: Optimus Maximus Keyboard •Innovative OLED-based keys; completely customizable key layout; powerful configuration software; built-in SD card slot lets you transport custom layouts. Mouse: Logitech MX Air mouse The Logitech MX Air mouse is powered by advanced laser technology that gives it precise tracking and instant response. It has something called ‘In the air, innovative Freespace motion sensing’ that transforms hand motions into natural, responsive cursor controls. The Logitech MX Air comes with a Micro-receiver and a Recharging station. Video Card: Nvidia eVGA GeForce GTX 295 CO-OP Hydro Copper Video Card - 1792MB DDR3 PCI-Express 2.0 (2) Dual Link DVI Q Monitor: Samsung Syncmaster 2443BWX •Versatile 24-inch LCD Monitor •Resolution is 1920 pixels by 1200 pixels Sound Card: •CI-Express Plug-in card •5.1 channel surround •24-bit sound Modem: NetGear DG834G ADSL Modem Router

Similarities between Facebook and Twitter

1. On Facebook, you can write new status updates. On Twitter, you can create new tweets, and you can also add your location to your tweets or pictures.

2. You will receive notifications" and feeds on Facebook, and you will receive tweets on Twitter.

3. You can post pictures on both Facebook and Twitter. On Facebook, it will post your pictures, but it will Twitpic and tweets with a url to your picture on Twitter.

4. You can reply your friends as comments in Facebook, and you have to reply to tweets in Twitter.

5. You can poke other people on Facebook and you can nudge other people on Twitter.

6. You can send private messages between you and another user as message on Facebook, and as direct message on Twitter.

7. On Facebook, you can tag people who are in the pictures, but you have to type @name to callout a person to tag on Twitter.

8. You can share links on Facebook, and tweet tiny URLs on Twitter.

9. Facebook is private profiles, but Twitter is closed updates approved by user.

10. On Facebook, you can see a shared link and you can share the link, and you have to retweet to show other users a shared link on Twitter.

Why do you think Facebook is worried about competition from Twitter?

Facebook is worried about competition from Twitter because Facebook makes some changes to compete with Twitter. It's a part of marketing strategies on Facebook's part to get one step forward of Twitter. Facebook is beginning to envy of Twitter's following among large companies and celebrities, which suggests that Twitter could defeat Facebook as the leading professional network engine. The site has main changes include breaking down the difference between private profiles and public pages and removing the 5,000 friend limit. Facebook is also changing the news updates to real-time, rather than updating them every 10 minutes or so.

Differences of Operating System'' or OS that most people use it

'Top 27 List Of Operating System'' '''

1. Linux is a free Unix-type operating system originally created by Linus Torvalds with the assistance of developers around the world. Developed under the GNU General Public License, the source code for Linux is freely available to everyone. Click on the link below to find out more about the operating system that is causing a revolution in the world of computers.

2. Mac OS X is the world’s most advanced operating system. Built on a rock-solid UNIX foundation and designed to be simple and intuitive, it’s what makes the Mac innovative, highly secure, compatible, and easy to use. Quite simply, there is nothing else like it.

3. UNIX is a computer operating system originally developed in 1969 by a group of AT&T employees. Unix operating systems are widely used in both servers and workstations.

4. FreeBSD is an advanced operating system for modern server, desktop, and embedded computer platforms. FreeBSD's code base has undergone over thirty years of continuous development, improvement, and optimization. It is developed and maintained by a large team of individuals. FreeBSD provides advanced networking, impressive security features, and world class performance and is used by some of the world's busiest web sites and most pervasive embedded networking and storage devices.

5. ReactOS

6. Solaris

7. OpenBSD

8. AmigaOS

9. NEXTSTEP

10. AtheOS

11. SkyOS

12. AROS

13. brickOS™

14. QNX

15. Darwin

16. OS/2

17. MenuetOS

18. leJOS

19. Plan 9

20. Xbox OS

21. HP-UX

22. ZETA

23. List of Unix flavors

24. new OS

25. Windows XP

26. MachTen

27. Windows Vista

What operating systems should install on your netbook?

Netbook hardware is the perfect match for Linux because it is the perfect networking operating system. It plays well with other OSes, it’s secure, and it’s fast. Using Windows make users are a risk for viruses, but with Linux, users won't risk for viruses or spyware. Linux is open source, so any software you are installing will be free. Moreover, because of the cost of the Linux operating system, the cost of netbooks can continue to fall. If Microsoft were to attempt to use Vista or Windows 7, the cost of your average netbook would not fall. Linux offers more flavors to choose from.The Linux netbook operating systems will continue to improve at a much faster rate than the Windows operating systems for netbooks. It is well known that open source bugs are found and fixed far faster than Windows bugs. This will help improve the Linux netbook OS far faster than anything Microsoft can manage. The nice thing about the Linux netbook community is that the developers are interested in making sure the OS works well and are quick to listen to their users. So you know if you have a problem with your Linux-based netbook, a quick search on Google will most likely find a simple solution to your problem.

title: User:Yourockmyworld


text:
Talk:Ethics/Nonkilling/Linguistics

What?--Brad Rapstars 17:21, 22 September 2009 (UTC)

title: Talk:Ethics/Nonkilling/Linguistics


text:
User:JWSchmidt/Blog/22 September 2009

The Letter from Michael Snow and Jimmy Wales about a "strategic plan" invites us all to "Post ideas on your blog". I was struck by the letter's mention of experts.

The role of experts within Wikimedia is an interesting issue. If you go to Google Knol and search for "Wikipedia" you can find articles such as "Wikipedia Inquisition" by professor Claes Johnson which mentions Wikipedia's "contempt for expertize". You can follow links from professor Johnson's knol to other published sources that discuss the problems that Wikipedia has with experts.

Perhaps the most most famous window into the role of experts at Wikipedia is the relationship between Jimmy Wales and Larry Sanger. Sanger has discussed Wikipedia's "lack of respect for expertise". Rather than make use of people like Sanger who could have long ago facilitated the participation of experts in Wikimedia, experts are routinely and systematically driven away from Wikimedia. Sanger had to go and start another wiki project, Citizendium. At Citizendium there is significant interest in the idea that if you are going to be publishing reliable information on the internet then your wiki editors should be encouraged to work under their real names, not by editing from anonymous user accounts.

It is interesting to explore the kinds of "experts" who are allowed at Wikimedia projects and those who are driven away. A famous case is Essjay, a Wikipedian who claimed to to be a tenured professor with several advanced degrees. In reality, the anonymous editor behind the "Essjay" user account was a community college dropout. I will not repeat all the issues raised by this one case; it is only symptomatic of the Wikimedia approach to providing educational resources, an approach which facilitates falsification of biographical information and other problems. It is enough to point out that the Essjay debacle shows that the "Wikimedia way" of favoring anonymous editing over editing by experts leads to problems. One of the most serious problems is that when non-experts are allowed to take control of an information resource like Wikipedia, then POV-pushers will have an irresistible incentive to aggressively work to keep away the real experts who could challenge and expose the phonies like "Essjay" and errors in wiki content that are easy for experts to spot.

Is Wikiversity infected with "Wikipedia disease" and headed down the path of driving away experts? When Wikiversity started, participants who showed their value by creating and participating in learning projects were made custodians. Participants who showed up wanting to participate in the role playing game of "delete and block" were turned away. Since the hostile takeover of Wikiversity last Summer that situation has changed. Since last summer policy violators and deletionists have been encouraged to become custodians. Rather than being welcomed as a full custodian, User:Leighblackall was "put on hold" for not having made enough use of the "tools" (see Candidates for Custodianship/Leighblackall). Why is it that two years ago, having participated at Wikiversity as a learner and scholar was enough to allow you to become a custodian, but not any more? Wikiversity needs scholars as custodians because custodians are supposed to examine community discussions and decide which arguments support the educational mission of the project. Under the new Ruling Party, you can show that you qualify for custodianship by coming to Wikiversity on a mission aimed at getting another Wikiversity participant banned, deleting the good faith contributions of other wiki participants and violating Wikiversity policy, for example, by secretly using your bot in an unauthorized way (see User:JWSchmidt/Blog/14 November 2008). It should be placed on the Main Page: "Interested in helping out as a custodian? All you have done is create learning resources? Get lost!" That would really encourage learners, scholars and experts to participate here.

How does Wikiversity treat experts? A few interesting "case studies" come to mind such as User:Jon Awbrey and Moulton. I've previously described how Jon Awbrey was treated when he dared to post links to some of his scholarly work on his user page. Both Awbrey and Moulton have been accused of posting "spam" when they have linked to their off-wiki scholarly resources. For such "friendly" treatment of scholars you can be rewarded with Wikiversity custodianship. What does this say to experts who think about participating at Wikiversity?

I've also previously written about Moulton and his treatment by Wikimedia. He tried to fix some some bogus biographies at Wikipedia, "encyclopedia articles" that were POV-pushing hatchet jobs that violated Wikiversity policy. For his trouble he was banned from Wikipedia. Moulton came to Wikiversity with a large amount of expertise and experience with online learning communities. However, rather than make use of his expertise, Moulton was driven away from this community. Why? Because Moulton dared to suggest that it is unethical to allow anonymous wiki editors to publish lies in online educational resources like Wikipedia.

In closing, I want to mention my own experience within Wikimedia. I've worked hard to attract experts to Wikimedia projects. I started the Molecular and Cellular Biology WikiProject as a way to encourage biologists to participate at Wikipedia. Due to the bad reputation of Wikimedia among academics, it is hard to get expert biologists to participate. I've been appalled to watch abusive Wikipedia administrators drive away academics like Johnson, Sanger, Awbrey and Moulton and I've participated in the never-ending struggle against anonymous POV-pushers who daily flood Wikipedia with lies and propaganda. I've watched anti-vandalism tools like blocks and checkuser misused to create a hostile environment for experts, academics and innocent non-vandals. I've been warned by Wikimedia "leaders" to shut up and not raise such problems for discussion and I've been invited to fuck off and leave Wikiversity. I've been threatened with being de-sysopped at Wikipedia for defending the right of experts to participate at that project. I was falsely accused of facilitating an evil "agenda" within Wikiversity and subjected to a bogus "emergency" de-sysop where there was no emergency, only an attempt to dump yet another an academic out of Wikimedia. I have decades of training and experience in academic research. I've had to watch Cite sources be reverted from "official policy" back to "proposed". Why can't a scholarly learning community have a policy about citing sources? Such absurdities abound, yet my efforts to make Wikiversity a place for scholarly research projects has been falsely called "policy manipulation". I've had to watch while my own Wikiversity research project was falsely called an "attack page" and I've watched other good faith contributions to Wikiversity proposed for deletion and ruled unwelcome.

If Jimmy Wales is serious about getting better involvement of experts within Wikimedia then why not start with a house cleaning and a conversion of Wikimedia from a place that is all too often hostile to experts while facilitating abusive by POV-pushers and giving a free ride to abusive administrators who think they have the right to prevent or end discussions is by imposing bogus blocks and bans.

Related Reading

 * List of English-language Knols about Wikipedia

title: User:JWSchmidt/Blog/22 September 2009