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User talk:Ura Lee Louis

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User:Stevenarntson/Student Hub/Summer2009

Analytical Writing - Monday Afternoon

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User talk:Oniqua Daymond

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University of Florida/Egm6321/f09.Team2/HW3

Do Not Us This Page.

Go to User:Egm6321.f09.Team2/HW3 for latest information

Homework Assignment #3 - due Wednesday, 10/7, 21:00 UTC

Problem 1
Find $$(m,n)$$ such that eqn. 1 on ([[media:Egm6321.f09.p13-1.png|p.13-1]]) is exact. A first integral is $$ \Phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2 $$ where $$k_1,k_2$$ are constants.

Problem Statement : Given a L2_ODE_VC $$\sqrt{x}y''+2xy'+3y=0$$

Find (m,n) from the integrating factor (xm,yn) that makes the equation exact.

A first integral is $$\phi(x,y,p)=xp+\big(2x^{3/2}\big)y+k_1=k_2$$

 * $$\phi_p=x$$
 * $$\phi_x=p+3x^{1/2}$$
 * $$\phi_y=2x^{3/2}-1$$

$$\phi(x,y,p)=h(x,y)+\int_{}f(x,y,p)dp, f=\phi_p$$

$$\phi(x,y,p)=h(x,y)+\int_{}xdp={h(x,y)+xp}$$

$$g(x,y,p)=\phi_x+\phi_yp=h_x+\phi_x+(h_y+\phi_y)p$$

$$g(x,y,p)=p+3x^{1/2}+(2x^{3/2}-1)p=h_x+x+(h_y+0)p$$

 * $$h_y=2x^{3/2}$$
 * $$h_x=3x^{1/2}-x$$

$$y^n=h_y\Rightarrow lny^n=ln(2x^{3/2})$$  $$n=ln(2x^{3/2})$$ $$x^m=h_x\Rightarrow lnx^m=ln(3x^{1/2}-x)$$  $$m=ln(3x^{1/2}-x)$$

Problem 2
Solve eqn. 2 on ([[media:Egm6321.f09.p13-1.png|p.13-1]]) for $$y(x)$$.

Problem Statement: Given a first integral$$\phi$$ of a L2_ODE_VC, solve for $$y(x)$$.

$$\phi(x,y,p)=xp+(2x^{3\over 2}-1)y+k_1=k_2$$ (1)

where k1 and k2 are const, and $$p=y'$$

Eq. (1) is in the form $$M(x,y)+N(x,y)y'$$ where

$$M(x,y)=(2x^{3\over 2}-1)y$$

$$N(x,y)=x$$

so it satisfies the 1st condition of exactness.

Check if $$M_y=N_x$$ for the 2ndcondition of exactness

$$M_y=2x^{3\over2}-1$$

$$N_x=1$$

$$M_y\ne N_x$$ so we do not satisfy the 2nd condition of exactness.

We must apply the integrating factor method for a L1_ODE_VC.

$$xp+(2x^{3\over 2}-1)y=k_2-k_1$$, divide by x to obtain the form:

$$y'+a_0(x)y=b(x)$$ where:

$$a_o(x)={1\over x}(2x^{3\over 2}-1)=2\sqrt{x}-{1\over x}$$

$$b(x)={k_2-k_1\over x}$$

From our solution of a general non-homogeneous L1_ODE_VC p.[[media:Egm6321.f09.p8-1.png|8-1]]

$$h(x)=exp\int_{}^{x}a_0(s)ds$$

$$h(x)=exp\int_{}^{x}2\sqrt{x}-{1\over x}=exp\bigg({4\over 3}x^{3\over 2}-ln\left|x\right|\bigg)\bigg({{k_2-k_1}\over x}\bigg)=exp{4\over 3}x^{3\over 2}-x$$

From p.[[media:Egm6321.f09.p8-2.png|8-2]] Eq. (4)

$$y(x)={1\over h(x)}\int_{}^{x}h(s)b(s)ds$$

Use the product rule of integration $$\int_{} ab=a\int_{} b-\int_{}a'\int_{}b$$

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-\int_{}^{x}h'(x)\int_{}^{x}b(x)\bigg]$$

In our example $$\int_{}^{x}h'(x)=h(x)$$ so,

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-h(x)\int_{}^{x}b(x)\bigg]$$

$$y(x)=0$$

Problem 3
From ([[media:Egm6321.f09.p13-1.png|p.13-1]]), find the mathematical structure of $$\Phi$$ that yields the above class of ODE.

$$F(x,y,y',y'')={d\phi\over dx}=\phi_x(x,y,p)+\phi_y(x,y,p)p+\phi_p(x,y,p)p'\ where\ p=y'$$

$$\phi=h(x,y)+\int_{}\phi_pdp=h(x,y)+\phi_pp$$

$$\phi_x=h_x+\phi_{px}p$$

$$\phi_y=h_y+\phi_{py}p$$

$$g=\phi_x+\phi_yp=h_x+\phi_{px}p+(h_y+\phi_{py}p)p$$

$$h_y=\phi_y-\phi_{px}$$

Take the integral of $$h_y$$

$$h(x,y)=(\phi_y-\phi_{px})y+k_1$$

Substitute back into the equation for $$\phi$$

$$\phi=(\phi_y-\phi_{px})y+k_1+\phi_pp$$

Rearrange the terms to obtain

 $$\phi(x,y,p)=\phi_pp+(\phi_y-\phi_{px})y+k$$ where,

 * $$P(x)=\phi_p$$
 * $$T(x)=(\phi_y-\phi_{px})y$$
 * $$k=k_1$$

$$\phi(x,y,p)=P(x)p+T(x)y+k$$

Problem 4
From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for the case $$n=1$$ (N1_ODE) $$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$$. Show that $$f_0-\frac{df_1}{dx}=0 \Leftrightarrow\Phi_{xy}=\Phi_{yx}$$. Hint: Use $$f_1=\Phi_y$$. Specifically: 4.1) Find $$f_0$$ in terms of $$\Phi$$ 4.2) Find $$f_1$$ in terms of $$\Phi$$($$f_1=\Phi_y$$) 4.3) Show that $$ f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$$.

Problem Statement: Given a N1_ODE, for the case n=1 $$F(x,y,y')=0\Leftrightarrow {d\phi\over dx}(x,y)$$

Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$ Hint:$$f_1=\phi_y$$

$$F={d\phi\over dx}(x,y^{(0)},....y^{(n-1)}=\phi_x+\phi_y6{(0)}y^{(1)}$$

$$F=\phi_x+\phi_yy'$$

$$f_i:={\partial F\over \partial y^i}$$

4.1
Find $$f_0$$ in terms of $$\phi$$.

$$f_0={\partial F\over \partial y}={\partial (\phi_x+\phi_yy')\over \partial y}=\phi_{xy}$$  $$f_0=\phi_{xy}$$

4.2
Find $$f_1$$ in terms of $$\phi_y$$

$$f_1={\partial F\over \partial y'}={\partial (\phi_x+\phi_yy')\over \partial y'}=\phi_y$$  $$f_1=\phi_y$$

4.3
Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$

$$\phi_{xy}-{d\phi_y\over df_x}=\phi_{xy}-\phi_{yx}=0$$  $$\phi_{xy}=\phi_{yx}$$

Problem 5
From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for the case $$n=2$$ (N2_ODE)  show: 5.1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$ 5.2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$ 5.3) $$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$ 5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

Problem 6
From ([[media:Egm6321.f09.p14-2.png|p.14-2]]), for the Legendre differential equation $$F=(1-x^2)y''-2xy'+n(n+1)y=0$$, 6.1 Verify exactness of this equation using two methods: 6.1a.) ([[media:Egm6321.f09.p10-3.png|p.10-3]]), Equations 4&5. 6.1b.) ([[media:Egm6321.f09.p14-1.png|p.14-1]]), Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

Problem 7
From ([[media:Egm6321.f09.p14-3.png|p.14-3]]), Show that equations 1 and 2, namely 7.1 $$\forall u,v$$ functions of $$x$$, $$L(u+v)=L(u)+L(v)$$. and 7.2 $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\forall$$ functions of $$ x$$. are equivalent to equation 3 on p.3-3.

Problem 8
From ([[media:Egm6321.f09.p15-2.png|p.15-2]]), plot the shape function $$N_{j+1}^{2}(x)$$.

[[Media:Graph1.pdf]]

Problem 9
Problem Statement: From (| p.16-2), show that $$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t\right)$$ $$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t\right)$$

$$y_{xxx}=(d/dx)\bigg[{d/dx\over (d/dx)y}\bigg]=(dt/dx)(d/dt)(dt/dx)(d/dt)(dt/dx)(d/dt)y$$

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$ 'Chain Rule'

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(t_{tt}))$$

$$y_{xxx}=(e^{-t})(d/dt)(-e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

$$y_{xxx}=(e^{-t})(2e^{-2t}(y_t)-e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))$$

$$y_{ttt}=2e-3t(y_t)-e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}9y_{ttt})$$

Factor out $$e^{-3t}$$ and re-arrange terms in ordre of derivative,

$$y_{xxx}=(e^{-3t})(y_{ttt}-3y_{tt}+2y_t)$$

$$y_{xxxx}=(d/dx)(d/dx)(d/dx)(d/dx)y$$

$$y_{xxxx}=(dt/dx)(d/dt)[(dt/dx)(d/dt)]((dt/dx)(d/dt))\langle(dt/dx)(d/dt)y\rangle$$

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(-2e^{-2t}(y_t)+e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(-2e^{-3t}(y_t)+e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}(y_{ttt}))$$

$$y_{xxxx}=(e^{-t})(6e^{-3t}(y_t)-2e^{-3t}(y_{tt})-3e^{-3t}(y_{tt})+e^{-3t}(y_{ttt})+6e^{-3t}(y_{tt})-2e^{-3t}(y_{ttt})-3e^{-3t}(y_{ttt})+e^{-3t}(y_{tttt}))$$

$$y_{xxxx}=6e^{-4t}(y_t)+2e^{-4t}(y_{tt})+3e^{-4t}(y_{tt})-e^{-4t}(y_{ttt})+6e^{-4t}(y_{tt})-2e^{-4t}(y_{ttt})-3e^{-4t}(y_{ttt})+e^{-4t}(y_{tttt}))$$

Factor out $$e^{-4t}$$ and re-arrange terms in order of derivative.

$$y_{xxxx}=(e^{-4t})(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

Problem 10
Problem Statement: From (| p.16-4) Solve equation 1 on p.16-1, $$ x^2y''-2xy'+2y=0 $$ using the method of trial solution $$ y=e^{rx}$$ directly for the boundary conditions $$\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$$ Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

Problem 11
Problem Statement: From (| p.17-4) obtain equation 2 from p.17-3 $$

Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$ using the integrator factor method.

Problem 12
Problem Statement: From (| p.18-1), develop reduction of order method using the following algebraic options

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

$$y(x)=\frac{U(x)}{u_1 (x)}$$

$$y(x)=\frac{u_1 (x)}{U(x)}$$

Problem 13
Problem Statement: From (| p.18-1), Find $$u_{1}(x)$$ and $$ u_{2}(x)$$ of equation 1 on p.18-1 using 2 trial solutions:

$$ y=ax^b$$

$$ y=e^{rx}$$

Compare the two solutions using boundary conditions $$y(0)=1$$ and $$ y(1)=2$$ and compare to the solution by reduction of order method 2. Plot the solutions in Matlab.

Contributing Team Members
Joe Gaddone 16:46, 3 October 2009 (UTC)

Matthew Walker

Egm6321.f09.Team2.sungsik 21:22, 4 November 2009 (UTC)


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User talk:Beao

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User:Sasithorn

Find the specs for the best PC computer

Motherboard:     Gigabyte GA-MA790X-UD4P Processor:         Intel® Core™ 2 Duo Processor Memory:            Intel Core i7 Case:                 Antec Nine Hundred Two.......Cooler Master HAF 932 Black Full Tower Power supply:    Cooler Master eXtreme Power Plus 460W (RS-460-PMSP-A3) Floppy drive:      Supermicro SSERVER 7044A82B TWR/4U-X6DA8-G2 8 SCA 645W BLK Hard drive:         Samsung HD753LJ CD/DVD:             Magnavox ZV450MW8 Keyboard:          Logitech G13 Mouse:               Logitech V450 Nano Monitor:            ViewSonic VX2433wm Sound card:      Asus Xonar Essence STX Video card:        XFX GeForce 8800 GTX Modem:            Belkin N+ Wireless

Homework#2

1) Facebook and Twitter are the same in their kinds of communication. Users can chat and post their pictures on the webpages.

2) Facebook has to worry about the competition with Twitter because Twitter is more public. Facebook users have to add friends otherwise they cannot connect them easily.

title: User:Sasithorn


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User talk:Pashar

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User talk:Seanusmc

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User talk:Riz bell

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User talk:Sasithorn

Homework#2
1) Facebook and Twitter are the same in their kinds of communication. Users can chat and post their pictures on the webpages.  2)  Facebook has to worry about the competition with Twitter because Twitter is more public. Facebook users have to add friends otherwise they cannot connect them easily.

There are many operating systems such as DOS, OS/2, Windows, and Linux

--What operating systems should install on your netbook? --it should be Windows

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User:Sablyth

See my blog: Common Knowledge

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User talk:Francesbell

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 * 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! Jtneill - Talk - c 09:27, 4 October 2009 (UTC)

 Woof! My name is Jack(Russell). I am a dog and a Wikiversity mascot. I am pretty new around Wikiversity. Perhaps we can learn together! Tail wag...

Connectivism glossary
Thanks for your contributions to the connectivism glossary, Frances. Sincerely, James. -- Jtneill - Talk - c 09:27, 4 October 2009 (UTC)

title: User talk:Francesbell


text:
The Tragedy of the Commons

The Tragedy of the Commons takes the prisoner’s dilemma a step closer to how societies interact with their world(s). The concept was introduced in a lecture given by William Foster Lloyd at Oxford University in 1833. Intellectuals like Lloyd and Thomas Malthus (1766–1834) were alarmed at the likely outcome of unrestricted human population growth. In particular they contended that, if unchecked, global human population would increase geometrically (e.g., 2, 4, 8, 16…), whereas food supply would only grow arithmetically. Malthus predicted that food resources per capita would begin to diminish about the middle of the 19th Century. He failed however, to predict the effects of technological advances in agriculture.

In the application of the concept provided here, the world is an isolated island that can expect no assistance from outside. This parable allows us to envisage what may happen in a situation where individuals have free and unlimited access to limited resources.

The goat herder
A broad open pasture, suitable for goat grazing, is the only area of fertile ground on a small, isolated, but inhabited island. Over a period of several centuries, there is a gradual increase in the number of goats and the island families that they support. At one point in time, the carrying capacity of the land is reached, such that the addition of any further goats would lead to overgrazing. Jana, a herdsman, contemplates the situation and, after deliberation, decides to increase his own herd by one goat. Jana rationalises this, in terms of positive and negative elements, in the following way: The extra goat is good for him (and his family), as it increases his income by the proceeds that he will receive after sale of the animal. This can be viewed in broad terms as a ‘+ 1’ for him. Conversely, the effect of overgrazing by just one extra goat would be spread over all the goatherds, resulting in only a very small fraction of a ‘– 1’ being debited against him. The following year, Jana again decides to increase his herd by just one more goat, but this time, unbeknown to him, two of his cousins decide to do the same.

Considerations

 * What are the likely outcomes of Jana’s action?
 * Can Jana’s actions be justified from either a deontological or utilitarian approach?
 * The above scenario is termed the ‘Tragedy of the Commons’. What relevance (if any) does it have to developed nations?
 * Can logic (and history) allow us to predict that technological advances will always allow humans to keep ahead of Malthus’ point of diminishing returns?
 * If he had adhered to contractarian ethics, what would Jana have done in the above exemplar, and what would have been the likely outcomes?

Article taken with permission from: Buckeridge, J.St,J.S., 2008. 4 E's: Ethics, Engineering, Economics and Environment. RMIT University Press, Melbourne, Australia, pp 61-63.

 * Environmental bioethics
 * Animal Rights and Conservation
 * Dimethylmercury or CO2

title: The Tragedy of the Commons


text:
User talk:Janya~enwikiversity

About Me

I am Miss Janya Tummanunkul. I'm a MUIC student. I study BBA (Travel Industry Management). My ID number is easy remembered, which is 5180491. I'm a talkative person, and I believe that I'm one of the most optimistic people in my group, which I considered as my weakness. On my spare, I will sleep, talk, and eat. So there’s no doubt that I look like this.

Homework1

Falcon Northwest Mach V (Intel Core i7-965 Extreme Edition) Windows Vista Ultimate SP1 (64-bit); 3.79GHz (overclocked) Intel Core i7-965 Extreme Edition; 12GB 1,066MHz DDR3 SDRAM; (2) 1GB ATI Radeon HD 4870X2 graphics card: 1TB 7,200rpm Hitachi hard drive; 80GB Intel X-25 solid state hard drive

Homework2

Facebook and twitter are the same because this two blog allow users to post everything they want to share with others in and publish to the public. But facebook may worry about competion from twitter because twitter is much more conversational and filled with observations on the world and many people tend to log into twitter and stay in the site for a longer period. People have made comments in many blogs saying that twitter is more interested; they can scroll for ages reading everyone’s conversation streams and responding where necessary.

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 Janya. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Janya~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 23:39, 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) 06:12, 19 April 2015 (UTC)

title: User talk:Janya~enwikiversity


text:
User:Sallywang1999

left text

'''The specs of the best computer: Motherboard:   Asus Rampage Extreme- motherboard-ATX-IX48-LGA775 Socket Processor:     Inter Core 2 Duo 3 GHz Processor Memory:        Corsair DHX Series 2-Pack 2 GB PC3-10664 DDR3 DIMM Memories Case:          Twelve Hundred Black Tower (ATX, 12) Power supply:  Corsair TX750w Floppy drive:  ECS GF8200A (V 1.0) NVIDIV RAID Hard drive:    West Digital Elements 1 TB USB 2.0 Desktop External Hand Drive WPE 1 UBK 10000 W CD/DVD:        DVD NEXT COPY Keyboard:      BELKIN WIRELESS Mouse:         BELKIN WIRELESS Video card:    GE FORE GTX 260 (1796 MB) Monitor:       View Sonic VX 243WM Sound card:    ASUS Xonar Essence STX Modem:         Sierras Wireless Compass 597

== '''Windows XP is the most widely used operating system in the world with a 63.3% market share, having peaked at 76.1% in January 2007. Palm OS. Was more popular option for handheld device. ='''=

1. How is Facebook and Twitter the same? •	Online social communication tool •	Easy to use them connect friends and family members •	Using them to share thoughts, ideas, and instant messages, for example: 1.	When you post a message in Facebook or Twitter, everybody in your contact list can see it. We don’t send the message to any specific person. 2.	With facebook, we are able to simply make comments about these on my profile page, and if people were interested they were able to see these things.

2.	Why do you think Facebook is worried about competition from Twitter? Personally I have gotten kinds of tired of facebook. I feel as though facebook has many features. It can be very overwhelming for a user who simply wants to communicate with others. But Twitter, while hard to grasp its concept at first glance, does provide a very simple, straightforward method of communicating like we can ask a question and get an instantaneous response.

title: User:Sallywang1999


text:
User:Fang fang

section one, ID: 5080234

title: User:Fang fang


text:
User:Phraeb427/Homework: The Best of the Best

The Best of the Best PC Computer is the.. Yoyotech Fi7epower MLK1610

Memory: 9GB ram

Processor speed: 3.73 ghz

Motherboard: X58 ATX MOTHERBOARD

CD/DVD: Dual-Format DVD/Blu-Ray

Video card: HIS Radeon 4870X2 2GB GDDR5 graphics

Hard drive: 1TB SATAII

Case: HAF 932 ATX CASE

Power Supply: 700W SILENT ATX

Sound card: 5.1 CHANNEL


text:
User talk:Wensel

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 13:54, 4 October 2009 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 13:54, 4 October 2009 (UTC)

title: User talk:Wensel


text:
User:Bonnieblond

20/11/09 HW: HW: 1.Difference of operating systems Operating systems each have their own merits, and all have different good and bad points. The following is a list of operating systems available for the PC*: LINUX - It's UNIX on the PC and other computers. This makes any computer VERY MUCH more powerful. It's the opposite of the "dumbed-down" computing which is a current fad (2001/01). Linux gives YOU the power to do what you want, rather than just assuming you conform to some ordinary notion of a user. Linux makes your PC like a mainframe, and is especially good for programming. In my opinion this is very good. If you want to be in control, this is for you. Microsoft Windows - This is what many salesmen will assume you want, but it's not the only choice. My personal opinion is that it is NOT the best. However, it has the merit of being usable by those who don't really understand computers. My advice is that there are a lot of better options, and you should decide for yourself. But if you really must have it, don't get the latest edition. If you don't like programming and you do like point-and-click systems and moving things about with a mouse, this is for you. BeOS - I have heard this is very good for real-time video and multimedia applications. It's now been bought out by Palm Computing. It lives on in a version called OpenBe. http://open-beos.sourceforge.net/news.php ... was http://www.be.com (domain lost) QNX - Complete real-time operating system available with an Internet exploration browser included. This is the operating system which British Telecom are using (2001/06) in their Multiphones Atheos - A relatively new GUI multi-tasking operating system available on the PC. Pervasive multi-threading. See www.atheos.cx. AtheOS is a free desktop operating system under the GPL license. AtheOS currently (2002) runs on Intel, AMD and other compatible processors and supports the Intel Multi Processor architecture. Gnu Hurd - The long-awaited GNU kernel for Unix. Linspire - Lindows - runs Linux and Windows programs, but time will tell what becomes of it. ReactOS - It's almost exactly like Microsoft Windows, except it's free. (open source). VisOpSys - VISual OPerating SYStem: small, fast, home-brewed, open source OS kernel for x86 Amoeba Operating System - Amoeba - worth a closer look. Plan9 - was http://plan9.bell-labs.com/plan9/ - - another interesting item. HyperOs - allows multiple Windows operating systems on a computer AND allows Linux on multi-boot Symphony OS (was http://www.symphonyos.com/) - "Unity, Beauty, Simplicity" MSDOS - This has the advantage of the command line interpreter, but it's never been as clever as Unix. As regards versions, 5 was better than 4, 6 was better than 5, etc, but the best version was 6.22, and they got worse after that, 7 being not as good as 6.22

2.What operating systems should install on your netbook? Linux is more flexible and configurable than Windows. the Linux netbook operating systems will continue to improve at a much faster rate than the Windows operating systems for netbooks.

16/10/09 HW: 1. How is Facebook and Twitter the same? - They are the websites that people can interact with others and find new friends easily, there are many similarities of facebook and twitter which are comments, share link, status updates, tags pictures, upload photo and video; these websites are more private than some of the website eg. hi5.com

2. Why do you think Facebook is worried about competition from Twitter? - Twitter is allowed people to sign up easier than facebook which means more people will sign up for twitter

HW : The fastest pc is YoYotech Fi7epower MLK1610

PC CASE HAF 932 ATX CASE POWER SUPPLY 700W SILENT ATX MOTHERBOARD X58 ATX MOTHERBOARD PROCESSOR INTEL i7-960 QUAD CORE - O/C @ 3.73Ghz CPU COOLER THERMALRIGHT 120 + FAN MEMORY 9GB DDR3 - 1800Mhz PERFORMANCE MEMORY SOLID STATE DRIVE 80GB SATAII SSD HARD DRIVE 1TB SATAII OPTICAL DRIVER BLU-RAY WRITER GRAPHICS CARD ONE RADEON 4870x2 - 2GB DDR5 GRAPHICS MODE SINGLE CARD SOUND 5.1 CHANNEL OPERATING SYSTEM VISTA HOME PREMINUM 64 BIT - PRE LOADED

title: User:Bonnieblond


text:
User talk:Bzbestzt

My name is Thitipong and nick name is BEST I'm a senior student, but this is my sixth years on Campus,including 2 years in Pre-College. I major in IB because I'm interested in Business especially to be a Shipper. However, I think I always lost performance of speaking and lost the performance of making decision when I'm in the pressure. However, I'm trying to get over those problems in the class.

Best of Best

Motherboard     Gigabyte GA EP45 UD3P

Processor       Intel Dual Core2

Memory          DDR Ram

Case            Antec 900

Power supply    Corsiar HX 1000 Power Supply

Floppy drive    1.44 Floppy Drive

Hard drive      seagate Barracuda 7200.10 750GB Keyboard        auravision eluminX Illuminated Keyboard

Mouse           A4tech X6-20MD

Video card      nvidia gtx295

Monitor         Nagios

CD/DVD          Lite-on BD-ROM drives

Sound card      Creative Soundblaster X-Fi Elite

Video card      nvidia gtx295

Modem           DSL EXT ALCATEL ST516 ADSL SPEEDTOUCH

How is Facebook and Twitter the same?
How is Facebook and Twitter the same? They can be connected to many people and they work like personal home pages.

Why do you think Facebook is worried about competition from Twitter? Because people start using Twitter for tasks more than using facebook. However, I do not think that facebook need to be worried about twitter.

Assignment
How is Facebook and Twitter the same?

They can be connected to many people and they work like personal home pages.

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

Because people start using Twitter for works more than using facebook. People are using facebook to play games or applications more than doing other business. However, I do not think that facebook need to be worried about twitter.

Assignment3
1.Look at different operating system (how many use it) For Window, there are Window XP, 2003 server, ME, 98, Vista, and Seven. For Mac ,MAC OS, OS X, and OS X Snow Leopard.

2.What operating system should you install in Netbook?

We should install Linux in Netbook

title: User talk:Bzbestzt


text:
User:EGM6321.f09.Team3.cgs/HW3

Problem 1. Find m and n such that Equation 1 p 13-1 is exact

Equation 1 p13-1: $$x^{m}y^{n}[\sqrt{x}y''+2xy'+3y]$$

Step 1: Does Eq 1 satisfy Condition 1 for exactness?

Rewrite Eq 1 as follows:

Let p=y’

$$\underbrace{x^{(m+\frac{1}{2})}y^{n}}_{f(x,y,p)}y''+\underbrace{2x^{(m+1)}y^{n}p+3x^{m}y^{(n+1)}}_{g(x,y,p)}$$

$$ f(x,y,p)= x^{(m+\frac{1}{2})}y^{n}$$

$$ g(x,y,p)= 2x^{(m+1)}y^{n}p+3x^{m}y{(n+1)} $$

Eq 1 can be written as f(x,y,p)y”+ g(x,y,p)=0 therefore Condition 1 for exactness has been satisfied.

Step 2: Does Eq 1 satisfy Condition 2 for exactness?

Condition 2 for exactness requires the following:

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

and

$$f_{xp}+ pf_{yp} + 2f_{y} =g_{pp}$$ (Equation 3)

Evaluating each derivative, we obtain:

$$f_{x} = (m+\frac{1}{2})x^{(m-\frac{1}{2})}y^n $$

$$ f_{xx} = (m^2-\frac{1}{4})x^{(m-\frac{3}{2})}y^n $$

$$ f_{xy} = (m+\frac{1}{2})nx^{(m-\frac{1}{2})}y^{n-1} $$

$$ f_{y} =n x^{(m+\frac{1}{2})}y^{n-1} $$

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

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

$$ g_{x}= 2(m+1)x^{m}y^np+3mx^{(m-1)}y^{(n+1)} $$

$$ g_{xp}= 2(m+1)x^{m}y^n $$

$$ g_{y}= 2nx^{(m+1)}y^{(n-1)}p+3(n+1)x^{m}y^{n} $$

$$ g_{yp}= 2nx^{(m+1)}y^{(n-1)} $$

$$ g_{p}= 2x^{(m+1)}y^n $$

$$ g_{pp}= 0 $$

Substituting into Equation 3, we obtain:

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

$$ (m^2-\frac{1}{4})x^{(m-\frac{3}{2})}y^n+2p(m+\frac{1}{2})nx^{(m-\frac{1}{2})}y^{(n-1)}+p^{2} n(n-1) x^{(m+\frac{1}{2})}y^{(n-2)}=2(m+1)x^{m}y^{n}+p^{2}nx^{(m+1)}y^{(n-1)}- 2nx^{(m+1)}y^{(n-1)}p+3(n+1)x^{m}y^{n}$$\\(Equation 4)

Substituting into Equation 4, we obtain:

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

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

In order for Equation 5 to equal 0, n must equal 0 Substitute n= 0 into equation 4 to determine m:

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

$$m=\frac{1}{2} $$

Therefore the integrating factor is

$$h(x,y)= x^{(\frac{1}{2})}y^{0}=x^{(\frac{1}{2})} $$

and the equation that satisfies both exactness conditions is:

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

or

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

Problem 12

Problem: Attempt to develop the reduction of order method 2 using different algebraic operations:

1) $$ y(x)=U(x)\text{±}u_{1}(x)$$

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

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

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

and a known variable $$ u_{1}(x)$$

Problem 12 Part 2

Given the following homogenous equation:

For algebraic operation 1, we will assume that

$$ y(x)=U(x)\text{±}u_{1}(x) $$ (Equation 2)

$$ y'(x)=U'(x)\text{±}u'_{1}(x) $$

$$ y(x)=U(x)\text{±}u''_{1}(x) $$

Substituting these values into Equation 1 has the following result:

$$ [U(x)\text{±}u_{1}(x)]+a_{1}[U'(x)\text{±}u'_{1}(x)]+a_{0}[U(x)\text{±}u_{1}(x)]=0$$

Regrouping the terms by order of U we find that the equation can now be written as:

$$ U(x)+a_{1}U'(x)+a_{0}U(x)\pm\underbrace{[u_{1}(x)+a_{1}u'_{1}(x)+u_{1}(x) }_{g(x,u,u',u'')} =0]$$ (Equation 3)

As $$u_{1}(x)$$ has been defined as an homogeneous function, then $$ g(x,u,u',u'')$$ is also homogeneous and can be set to 0 thus reducing Equation 3 to:

$$ U''(x)+a_{1}U'(x)+a_{0}U(x)$$

There has been no reduction of order, so this algebraic operation is not appropriate to solve the reduction of order problem.

Problem 12 Part 2

Given the algebraic operation

$$ \frac{U}{u_{1}}$$

Determine whether the algebraic operation can be used to reduce the order of Equation 1

$$y= \frac{U}{u_{1}} $$

$$y'= \frac{U'u_{1}-Uu_{1}'}{u_{1}^{2}} $$

Which can be reduced to

$$y'= \frac{U'}{u_{1}}-\frac{Uu_{1}'}{u_{1}^{2}} $$

Let $$g=Uu_{1}' $$

Let $$g'=U'u_{1}' +Uu_{1}''$$

$$y= \frac{U u_{1}- U' u_{1}'}{ u_{1}^{2}}-\frac{g'u_{1}^{2}- g(u_{1}^{2})'}{u_{1}^{4}}$$

$$y= \frac{U}{ u_{1}}- \frac{U' u_{1}'}{ u_{1}^{2}}-\frac{(U'u_{1}'+Uu_{1}'')u_{1}^{2}-Uu_{1}'(u_{1}^{2})'}{u_{1}^{4}}$$

Substituting into Equation 1:

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

$$\frac{U}{ u_{1}}- \frac{U' u_{1}'}{ u_{1}^{2}}-\frac{(U'u_{1}'+Uu_{1})u_{1}^{2}-Uu_{1}'(u_{1}^{2})'}{u_{1}^{4}} +a_{1}\frac{U'}{u_{1}}-\frac{Uu_{1}'}{u_{1}^{2}} )+a_{0}(\frac{U}{u_{1}} )=0$$

Assuming $$u_{1}(x)\neq 0$$ divide equation through by $$u_{1}^{4}$$

$$U u_{1}^{3}-2U' u_{1}' u_{1}^{2}+Uu_{1}u_{1}^{2}-U(u_{1}^{2})'u_{1}' +a_{1}(U' u_{1}^{2}-Uu_{1}^{2}u_{1}')+a_{0}Uu_{1}^{3} =0$$

Regrouping to separate out U, U’ and U’’ results in the following equation:

$$ U u_{1}^{3}+U'(-2u_{1}' u_{1}^{2} +a_{1}u_{1}^{2})+U(u_{1}u_{1}^{2}-(u_{1}^{2})'u_{1}'+ a_{0}u_{1}^{3}$$

Which can’t be reduced.

Problem 12 Part 3

Given the algebraic operation

$$ \frac{u_{1}}{U}$$

Determine whether the algebraic operation can be used to reduce the order of Equation 1

$$y= \frac{U_{1}}{U} $$

$$y'= \frac{u_{1}'U-u_{1}U'}{U^{2}} $$

Which can be reduced to

$$y'= \frac{u_{1}'}{U}-\frac{u_{1}U'}{U^{2}} $$

Let $$g=u_{1}U' $$

Let $$g'=u_{1}'U' u_{1}U''$$

$$y= \frac{u_{1} U- u_{1})' U'}{ U^{2}}-\frac{g'U^{2}- g(U^{2})'}{U^{4}}$$

$$y= \frac{u_{1}}{ U}- \frac{u_{1}' U'}{ U^{2}}-\frac{(u_{1}'U'+u_{1}U'')U^{2}-u_{1}U'(U^{2})'}{U^{4}}$$

Substituting into Equation 1:

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

$$\frac{u_{1}}{ U}- \frac{u_{1}' U'}{ U^{2}}-\frac{(u_{1}'U'+u_{1}U)U^{2}-u_{1}U'(U^{2})'}{U^{4}} +a_{1}\frac{u_{1}'}{U}-\frac{u_{1}U'}{U^{2}} )+a_{0}(\frac{u_{1}}{U} )=0$$

Problem ?? Assuming $$U(x)\neq$$ divide equation through by $$U^{4}$$

$$u_{1} U^{3}-2u_{1}' U' U^{2}+u_{1}UU^{2} -u_{1}(U^{2})'U'+a_{1}(u_{1}' U^{2}-u_{1}U^{2}U')+a_{0}u_{1}U^{3} =0$$

As this equation is non-linear, no reduction of order is possible.

Given

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

Find solution using the trial solution

$$ y=ax^{b} $$

$$ y'=bax^{b-1} $$

$$ y''=b(b-1)ax^{b-2} $$

Substituting into Equation 1

$$ (1-x^{2}) ax^{b-2} -2x bax^{b-1} +2 ax^{b}  =0 $$ $$ (ax^{b-2}) -ax^{b} -2 abx^{b} +2 ax^{b}  =0 $$ $$ (ax^{b-2}) -ax^{b} -2 abx^{b} =0 $$ Assuming $$ ax^{b-2}/ne{0}$$, divide through by $$  ax^{b-2}$$ This results in the following equation:

$$ (x^{-2}) -1-2b =0 $$

Given

$$ (1-x^{2})y''-2xy'+2y=0 $$ (Equation 1) Find solution using the trial solution $$ y=er^{x} $$

$$ y'= re^{rx} $$

$$ y''= r^{2}e^{rx} $$

Substituting into Equation 1

$$ (1-x^{2}) r^{2}e^{rx} -2x re^{rx}  +2 e^{rx}  =0  $$ $$ r^{2}e^{rx}  - r^{2} x^{2} e^{rx}  -2rx e^{rx}  +2 e^{rx}=0 $$ Assuming $$ y=er^{x}/ne{0}$$, divide through by $$  er^{x}$$ This results in the following equation:

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

title: User:EGM6321.f09.Team3.cgs/HW3


text:
User:ChayaninP/Homework 1: The specs of best PC computer.

The specs of best PC computer.

Motherboard    ASUS P6T7 WS SUPERCOMPUTER

Processor      Core i7 975 3.33GHz/1366/FSB 6.4GT/8MB

Memory        G.SKILL DDR3 TRI-CHANNEL 6TD/2000 CL 		9-9-9-24

Case 		Lian-Li PC-A20

Power supply 	ThermalTake 1500W

Hard drive 	G.SKILL SSD 256 GB FALCON SATA2 2.5"

CD/DVD 		DVD-RW SONY 10S

Keyboard 	Razor	KEYBOARD Lycosa

Mouse 		Razor	Mouse Lachesis

Monitor 		DELL 3007WFP

Sound card 	CREATIVE X-FI TITANIUM

Video card 	WINFAST VGA QUADRO FX 5800 PCI-E 16X

Modem          SMC TigerAccess EE Splitter7024/VSP


text:
User talk:AOM4642/questions

On problem 5.3 when looking up the R-values for the air space in the wall it says from 3/4" to 5.5"...since the space in the wall is btw those values we can just use the listed value w/o having to multiply it by its thickness? Similar situation w/ the gypsum board... Aaron Stelck

title: User talk:AOM4642/questions


text:
User talk:Moolek

19/11/09 HW: difference of operating systems

Operating systems each have their own merits, and all have different good and bad points. The following is a list of operating systems available for the PC*:

LINUX - It's UNIX on the PC and other computers. This makes any computer VERY MUCH more powerful. It's the opposite of the "dumbed-down" computing which is a current fad (2001/01). Linux gives YOU the power to do what you want, rather than just assuming you conform to some ordinary notion of a user. Linux makes your PC like a mainframe, and is especially good for programming. In my opinion this is very good. If you want to be in control, this is for you.

Microsoft Windows - This is what many salesmen will assume you want, but it's not the only choice. My personal opinion is that it is NOT the best. However, it has the merit of being usable by those who don't really understand computers. My advice is that there are a lot of better options, and you should decide for yourself. But if you really must have it, don't get the latest edition. If you don't like programming and you do like point-and-click systems and moving things about with a mouse, this is for you.

BeOS - I have heard this is very good for real-time video and multimedia applications. It's now been bought out by Palm Computing. It lives on in a version called OpenBe. http://open-beos.sourceforge.net/news.php ... was http://www.be.com (domain lost)

QNX - Complete real-time operating system available with an Internet exploration browser included. This is the operating system which British Telecom are using (2001/06) in their Multiphones

Atheos - A relatively new GUI multi-tasking operating system available on the PC. Pervasive multi-threading. See www.atheos.cx. AtheOS is a free desktop operating system under the GPL license. AtheOS currently (2002) runs on Intel, AMD and other compatible processors and supports the Intel Multi Processor architecture.

Gnu Hurd - The long-awaited GNU kernel for Unix.

Linspire - Lindows - runs Linux and Windows programs, but time will tell what becomes of it.

ReactOS - It's almost exactly like Microsoft Windows, except it's free. (open source).

VisOpSys - VISual OPerating SYStem: small, fast, home-brewed, open source OS kernel for x86

Amoeba Operating System - Amoeba - worth a closer look.

Plan9 - was http://plan9.bell-labs.com/plan9/ - - another interesting item.

HyperOs - allows multiple Windows operating systems on a computer AND allows Linux on multi-boot

Symphony OS (was http://www.symphonyos.com/) - "Unity, Beauty, Simplicity"

MSDOS - This has the advantage of the command line interpreter, but it's never been as clever as Unix. As regards versions, 5 was better than 4, 6 was better than 5, etc, but the best version was 6.22, and they got worse after that, 7 being not as good as 6.22

What operating systems should install on your netbook?

Linux is more flexible and configurable than Windows. the Linux netbook operating systems will continue to improve at a much faster rate than the Windows operating systems for netbooks.

The fastest personal computer is " YoYotech Fi7epower MLK1610"

16/10/09 HW:

1. How is Facebook and Twitter the same? - There are many similarities of facebook and twitter in term of status updates, tags pictures,comments, links, uploading photos and profiles of users. Both websites are more private because people have to be in the lists of each other. there are lots of information of user that can be inform so others can follow easily. Moreover, people can interact with others and find new friends easily. You can keep contact with the people you want to stay in touch.

2. Why do you think Facebook is worried about competition from Twitter? - Twitter is allowed people to sign up easier than facebook because signing up in facebook you have to inform a lot of detail about yourself. Therefore,it might be the important reason why more people are interested and decide to sign up twitter.

Falcon Northwest Mach Price $8,028 $7,738 Motherboard chipset Intel X58 Nvidia NForce 790i SLI CPU 3.79GHz Intel Core i7-965 Extreme Edition (overclocked) Memory 12GB 1,066MHz DDR3 SDRAM Graphics (2) 1GB ATI Radeon HD 4870 X2 Hard drives 1TB 7,200rpm Hitachi hard drive; 80GB Intel X-25M solid state drive Optical drive 20x dual-layer DVD burner with LightScribe; 4x dual-layer Blu-ray burner 4x dual-layer Blu-ray burner Operating system Windows Vista Ultimate SP1 64-bit

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

title: User talk:Moolek


text:
Swedish Language/Irregular Verbs

Examples
 Det lät som en humla. It sounded like a bumblebee. Hon har burit upp dockan till övervåningen. She has carried the doll to the upper floor. Hon har förlåtit honom. She's forgiven him. Jens la sig i diskussionen. Jens interfered with the discussion. De ville inte vänta längre. They didn't want to wait any longer. Det innebär att du är skyldig mig en massa pengar! That means you owe me a lot of money! Listan bestod av saker som skulle inhandlas. The list consisted of items to be procured. Hon log åt alla vackra män. She smiled at all beautiful men. Vem har tillåtit dig att börja äta? Who has allowed you to start eating? Han har bara legat på soffan och inte gjort någonting. He has just lain on the sofa not doing anything.

Exercises
Please translate into English.  Hon förlät mig för mitt misstag. Lejonet röt när det kom förbi en zebra. Vem har lärt barnen att svära?
Valpen har legat i sin korg hela dagen.
De har farit iväg någonstans.
 Please translate into Swedish.

I flew to London.
He has seen the snow man.
They have found a treasure in the jungle.
Your team won last time but this time we will beat you.
Why are you sneaking around?

title: Swedish Language/Irregular Verbs


text:
User:SvenAERTS

In Belgium/Flanders, there are about 12 people called Sven AERTS. We have a facebook called "Sven AERTS" to gather all and to coordinate one and the other. Some of us have very overlapping profiles. I'm trying to have all Sven AERTS'en to agree to use their FirstnameFAMILYNAME followed by their Birthday. I'm known as SvenAERTS228.

I am working on:

 * Agenda 21 http://en.wikipedia.org/wiki/Agenda_21 -Section IV: Means of Implementation, more specifically education, international institutions and financial mechanisms bringing money of the Low Carbon Economy to the Education Sector.
 * with http://www.coren.be - implementing Agenda 21, EMAS, ISO14000 in the Education sector and managing the European Federation of ECO-School auditors. "Kyoto Protocol" Consulting, how to attract money from the Low Carbon Economy to the Education & Youth Sector.
 * Millennium Development Goal nr2: Universal Education, more specifically: One Laptop Per Child - where I introduced the concept of paying the laptops with CO2e- / Green- and White-Certificates : http://www.olpco2e.org
 * I am the 38th EU recognised Youth Trainer in the EU Salto-Youth Trainers Database: http://www.salto-youth.net/find-a-trainer/38.html
 * Skype: SvenAERTS or +32 2 7 876 7 21 (Europe-Belgium-Brussels time) - SvenAERTS228 at gmail dot com

More ? Why not click through on my blog: http://svenaerts228.blogspot.com/

PS

 * I speak 5 languages and hold minimum 3 degrees.
 * As a "Kyoto Protocol" consultant, my regular consulting involves projects of +10M€ minimum, 6 month engagements with 2000€/month contribution for my overheads and 525€/day for basic consulting. These projects I work on offer generally +100% Return on Investment, with toppers of +1000% with Special Purpose Companies brought in a shell to the stockmarket with exit strategies for the equity providers and business angels.

title: User:SvenAERTS


text:
User:Egm6321.f09.team5.GV/HW3

Problem Statement
Find $$(m,n)$$ such that  $$ (X^{m}Y^{n})\left[\sqrt[]{x}y^{''}+2xy^{'}+3y \right]=0 $$ is exact.

Solution
The first step is to identify what kind of ODE is the given equation. It is identified as a Non-Linear Second-Order ODE (N2.ODE). In order for a second order ODE to be exact the following two conditions must be met:

Condition 1: The ODE must be of the form:

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

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

Condition 2: The following Conditions must be met:

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

$$    \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp}  \qquad $$         Equation 3

By arranging the given equation so it satisfies the first exactness condition (Equation 1) The Following is obtained

$$ F= x^{m}y^{n}\sqrt[]{x}y^{''} + 2x^{m+1}y^{n}P +3x^{m}y^{n+1} $$

Where $$ f(x,y,p) = x^{m}y^{n}\sqrt[]{x} $$             $$g(x,y,p) = 2x^{m+1}y^{n}P +3x^{m}y^{n+1} $$

In order to meet the Second Condition for exactness the following partial derivatives must be identified:

$$ f_{x}=(m+0.5)x^{(m-0.5)}y^{n} \qquad \qquad  g_{y}= 2Pnx^{(m+1)}y^{(n-1)} + 3(n+1)x^{m}y^{n} $$

$$ f_{xx}=(m^{2}-0.25)x^{(m-1.5)}y^{n}  \qquad \qquad g_{yp}= 2nx^{(m+1)}y^{n-1} $$

$$ f_{xy}=(m+0.5)nx^{(m-0.5)}y^{(n-1)}  \qquad \qquad  g_{x}= 2P(m+1)x^{m}y^{m} + 3mx^{(m-1}y^{(n+1)}  $$

$$ f_{y}=nx^{(m+0.5)}y^{(n-1)}          \qquad \qquad \qquad  g_{xp}= 2(m+1)x^{m}y^{n} $$

$$ f_{yy}=n(n-1)x^{(m+0.5)}y^{(n-2)}    \qquad \qquad  g_{p}= 2x^{(m+1)}y^{n} $$

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

By populating equation 3, the following is obtained:

$$ 0 + P(0) + 2nx^{(m+0.5)}y^{(n-1)} = 0 \qquad $$

Therefore $$ n=0 $$

Using the value for n and populating Equation 2, the following is found:

$$ (m^{2}-0.25)x^{(m-1.5)} + 2P(0) + P^{2}(0) = 2(m+1)x^{m} + P(0) - 3x^{m} $$

Rearranging: $$ (m^{2}- 0.25)x^{(m-1.5)} = x^{m}(2m-1)$$

$$ x^{-1.5} = \frac{(2m-1)}{(m^{2}-0.25)} $$

In order for this equation to be satisfied, both sides of the equation must be equal to 0 :

$$ \frac{(2m-1)}{(m^{2}-0.25)} = 0 $$

$$ 2m-1 = 0 $$

$$ m = \frac{1}{2} $$

Problem Statement
$$ \Phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2 $$ where $$k_1,k_2$$ are constants.

Solve for $$y(x)$$ Hint: L1.ODE.VC (int. Factor Method)

Solution
The First step is to test the equation for exactness:

The two conditions for exactness of a L1.ODE.VC are:

The ODE must be of the form:

$$ M(x,y) + N(x,y)y^{'} = 0 $$

The Second Condition is:

$$ M_{y}=N_{x} $$

Applying our equation to the condition test we find that the equation is not exact as follows:

$$ M(x,y)= 2(x^{\frac{3}{2}} - 1)y + K1 - K2 $$

$$ N(x,y)= x $$

$$ M_{y} = (2x^{\frac{3}{2}}-2) $$

$$ N_{x} = 1 $$

In order to find y(x) we must then use Euler's Integration Factor Method as Follows:

An integration factor h is identified and used as follows:

$$hMdx + hNdy=0 \qquad $$ $$h_{x}N - h_{y}M + h(N_{x}-M_{y})=0 \qquad $$ Equation 1

At this point we will consider the particular case when the integration factor is only a function of x.

$$ \ h(x,y)=h(x) $$

With this definition Equation 1, becomes:

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

Which can then be re-arranged as follows:

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

Substituting the known values into this relation we obtain:

$$ \frac{(2-2x^{\frac{3}{2}})}{x} = f(x) $$

Using this expression an integration factor is found as follows:

$$ h(x) = \exp \int_{}^{x}f(s)ds $$

$$ h(x) = \exp (\int_{}^{x}2ds - \int_{}^{x}2s^{\frac{1}{2}}ds $$

$$ h(x) = x^{2} - \exp(\frac{4x^{\frac{3}{2}}}{3}) $$

Using the newly identified integration factor the following can be defined:

The Original Equation can be re-written as:

$$ P + \color{blue}{\frac{2(x^{\frac{3}{2}} - 1)}{x}} y = \color{red}{K_{2} - K_{1}} \qquad $$  $$ \color{blue}{Blue=a_{0}(x)} \qquad  \color{red}{Red=b(x)} $$

A solution for an equation in this form can be identified as follows:

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

By populating the previous equation with our known values the following is obtained:

$$ y(x) = \frac{1}{x^{2}-\exp(\frac{4x^{\frac{3}{2}}}{3})} \int{}^{x} \left[ s^{2} - exp(\frac{4s^{\frac{3}{2}}}{3}) \right] (K_{2}-K_{1})ds $$

$$ y(x) = \frac{K_{2}-K_{1}}{x^{2}-\exp(\frac{4x^{\frac{3}{2}}}{3})} \left[ \frac{x^{3}}{3} - \int{}^{x}exp(\frac{4s^{\frac{3}{2}}}{3}) \right] $$

Problem Statement
Find the mathematical structure of $$\Phi$$ that will yield L2.ODE.VC

Solution
First it is necessary to define the first integral as follows:

$$ F(x,y,y^{'},y^{''}) = \frac{d \phi(x,y,p)}{dx} $$

$$ F(x,y,y^{'},y^{''}) = \phi_{x} + \phi_{y}P + \phi_{p}P^{'} \qquad \qquad \qquad Where \quad P=y^{'}     $$      Equation 1

The next step is to define a 2nd Order ODE as follows:

$$ F = W(x)y^{''} + Q(x)y^{'} + R(x)y $$ Equation 2

Comparing Equation 2 with Equation 1 the following is identified:

$$ \phi_{p}=W(x) $$

$$ \phi_{y}=Q(x) $$

$$ \phi_{x}=R(x)y(x) $$

The next step is to integrate as follows:

$$ \phi= \int{}^{}W(x)dp = W(x)P + C_{1} $$

$$ \phi= \int{}^{}Q(x)dy = Q(x)y + C_{2} $$

$$ \phi= \int{}^{}R(x)y(x)dx = R(x)y(x) + C_{3} $$

Adding all the definition for Phi the following is obtained:

$$ \phi= W(x)P + T(x)y + K \qquad \qquad \qquad Where \quad T(x) = \left[R(x)+Q(x)\right]y+ (C_{2}+C_{3}) \qquad and \quad K = C_{1}+C_{2}+C_{3} \qquad \qquad$$  Equation 3

In order to generate additional exact L2.ODE.VC, the first integral, $$\phi$$, must be of the form outlined in Equation 3.

Problem Statement
Problem 4 From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for the case $$n=1$$ (N1_ODE) $$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$$. Show that $$f_0-\frac{df_1}{dx}=0 \Leftrightarrow\Phi_{xy}=\Phi_{yx}$$. Hint: Use $$f_1=\Phi_y$$. Specifically: 4.1) Find $$f_0$$ in terms of $$\Phi$$ 4.2) Find $$f_1$$ in terms of $$\Phi$$($$f_1=\Phi_y$$) 4.3) Show that $$ f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$$.

Problem Statement
Problem 5 From ([[media:Egm6321.f09.p13-3.png|p.13-3]]), for the case $$n=2$$ (N2_ODE)  show: 5.1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$ 5.2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$ 5.3) $$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$ 5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

Problem Statement
Problem 6 From ([[media:Egm6321.f09.p14-2.png|p.14-2]]), for the Legendre differential equation $$F=(1-x^2)y''-2xy'+n(n+1)y=0$$, 6.1 Verify exactness of this equation using two methods: 6.1a.) ([[media:Egm6321.f09.p10-3.png|p.10-3]]), Equations 4&5. 6.1b.) ([[media:Egm6321.f09.p14-1.png|p.14-1]]), Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

Problem Statement
Problem 7 From ([[media:Egm6321.f09.p14-3.png|p.14-3]]), Show that equations 1 and 2, namely 7.1 $$\forall u,v$$ functions of $$x$$, $$L(u+v)=L(u)+L(v)$$. and 7.2 $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\forall$$ functions of $$ x$$. are equivalent to equation 3 on p.3-3.

Problem Statement
Problem 8 From ([[media:Egm6321.f09.p15-2.png|p.15-2]]), plot the shape function $$N_{j+1}^{2}(x)$$.

Problem Statement
Problem 9 From (| p.16-2), show that $$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t\right)$$ $$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t\right)$$

Problem Statement
Problem 10 From (| p.16-4) Solve equation 1 on p.16-1, $$ x^2y''-2xy'+2y=0 $$ using the method of trial solution $$ y=e^{rx}$$ directly for the boundary conditions $$\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$$ Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

Problem Statement
Prolbem 11 From (| p.17-4) obtain equation 2 from p.17-3

$$

Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$ using the integrator factor method.

Problem Statement
Develop reductions of order method 2 using different algebraic Operations:

1) $$ y(x) = U(x)\pm u_{1}(x) $$ 2) $$ y(x) = U(x) / u_{1}(x) $$ 3) $$ y(x) = u_{1}(x) / U(x) $$

Objective: Proof you will not obtain a solution with a missing dependent variable.

Problem Statement
Find $$ u_{1}(x)$$ and $$ u_{2}(x) $$ of

$$ (1-x^{2})y^{''} - 2xy^{'} + 2y = 0 $$

using trial solutions:

1) $$ y=ax^{b} $$ where a,b are coefficients to be determined.

2) $$ y=e^{rx} $$

Compare the two solutions using the boundary conditions:

$$ y(0)=1 $$ $$ y(1)=2 $$

and also compare to the solutions by reduction of order method 2. Plot Solutions using Matlab

title: User:Egm6321.f09.team5.GV/HW3


text:
User:Egm6321.f09.team5/HW3

Problem Statement
Find $$(m,n)$$ such that  $$ (X^{m}Y^{n})\left[\sqrt[]{x}y^{''}+2xy^{'}+3y \right]=0 $$ is exact.

Solution
The first step is to identify what kind of ODE is the given equation. It is identified as a Non-Linear Second-Order ODE (N2.ODE). In order for a second order ODE to be exact the following two conditions must be met:

Condition 1: The ODE must be of the form:

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

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

 You are mixing upper and lowercase letters for $$f,g,F,G$$ in your partial expressions and exactness criteria, which is generally treacherous ground. Some of your expressions below mix upper/lowercase in the same equation. Recall that on p.13-2 we defined $$P(x)= \Phi_x$$ and $$p=y'$$. You want to be consistent in your notation. --Egm6321.f09.TA 04:24, 15 October 2009 (UTC)

Condition 2: The following Conditions must be met:

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

$$    \ F_{xp} + pF_{yp} + 2F_{y} = G_{pp}  \qquad $$         Equation 3

By arranging the given equation so it satisfies the first exactness condition (Equation 1) The Following is obtained

$$ F= x^{m}y^{n}\sqrt[]{x}y^{''} + 2x^{m+1}y^{n}P +3x^{m}y^{n+1} $$

Where $$ f(x,y,p) = x^{m}y^{n}\sqrt[]{x} $$             $$g(x,y,p) = 2x^{m+1}y^{n}P +3x^{m}y^{n+1} $$

In order to meet the Second Condition for exactness the following partial derivatives must be identified:

$$ f_{x}=(m+0.5)x^{(m-0.5)}y^{n} \qquad \qquad  g_{y}= 2Pnx^{(m+1)}y^{(n-1)} + 3(n+1)x^{m}y^{n} $$

$$ f_{xx}=(m^{2}-0.25)x^{(m-1.5)}y^{n}  \qquad \qquad g_{yp}= 2nx^{(m+1)}y^{n-1} $$

$$ f_{xy}=(m+0.5)nx^{(m-0.5)}y^{(n-1)}  \qquad \qquad  g_{x}= 2P(m+1)x^{m}y^{m} + 3mx^{(m-1}y^{(n+1)} $$

$$ f_{y}=nx^{(m+0.5)}y^{(n-1)}          \qquad \qquad \qquad  g_{xp}= 2(m+1)x^{m}y^{n} $$

$$ f_{yy}=n(n-1)x^{(m+0.5)}y^{(n-2)}    \qquad \qquad  g_{p}= 2x^{(m+1)}y^{n} $$

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

By populating equation 3, the following is obtained:

$$ 0 + P(0) + 2nx^{(m+0.5)}y^{(n-1)} = 0 \qquad $$

Therefore $$ n=0 $$

Using the value for n and populating Equation 2, the following is found:

$$ (m^{2}-0.25)x^{(m-1.5)} + 2P(0) + P^{2}(0) = 2(m+1)x^{m} + P(0) - 3x^{m} $$

Rearranging: $$ (m^{2}- 0.25)x^{(m-1.5)} = x^{m}(2m-1)$$

$$ x^{-1.5} = \frac{(2m-1)}{(m^{2}-0.25)} $$

In order for this equation to be satisfied, both sides of the equation must be equal to 0 :

$$ \frac{(2m-1)}{(m^{2}-0.25)} = 0 $$

$$ \ 2m-1 = 0 $$

$$ m = \frac{1}{2} $$

The exact ODE is:

$$(X^{0.5}Y^{0})\left[\sqrt[]{x}y^{''}+2xy^{'}+3y \right]=0 $$

 You can leave out $$y^0$$ terms in the expression for the exact ODE. --Egm6321.f09.TA 04:24, 15 October 2009 (UTC)

Problem Statement
$$ \Phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2 $$ where $$k_1,k_2$$ are constants.

Solve for $$y(x)$$ Hint: L1.ODE.VC (int. Factor Method)

Solution
The First step is to test the equation for exactness:

The two conditions for exactness of a L1.ODE.VC are:

The ODE must be of the form:

$$ M(x,y) + N(x,y)y^{'} = 0 $$

The Second Condition is:

$$ M_{y}=N_{x} $$

Applying our equation to the condition test we find that the equation is not exact as follows:

$$ M(x,y)= 2(x^{\frac{3}{2}} - 1)y + K1 - K2 $$

$$ N(x,y)= x $$

$$ M_{y} = (2x^{\frac{3}{2}}-2) $$

$$ N_{x} = 1 $$

In order to find y(x) we must then use Euler's Integration Factor Method as Follows:

An integration factor h is identified and used as follows:

$$hMdx + hNdy=0 \qquad $$

$$h_{x}N - h_{y}M + h(N_{x}-M_{y})=0 \qquad $$ Equation 1

At this point we will consider the particular case when the integration factor is only a function of x.

$$ \ h(x,y)=h(x) $$

With this definition Equation 1, becomes:

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

Which can then be re-arranged as follows:

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

Substituting the known values into this relation we obtain:

$$ \frac{(2-2x^{\frac{3}{2}})}{x} = f(x) $$

Using this expression an integration factor is found as follows:

$$ h(x) = \exp \int_{}^{x}f(s)ds $$

$$ h(x) = \exp (\int_{}^{x}2ds - \int_{}^{x}2s^{\frac{1}{2}}ds $$

$$ h(x) = x^{2} - \exp(\frac{4x^{\frac{3}{2}}}{3}) $$

This integrating factor is not correct. Egm6321.f09.TA 01:27, 28 October 2009 (UTC)

Using the newly identified integration factor the following can be defined:

The Original Equation can be re-written as:

$$ P + \color{blue}{\frac{2(x^{\frac{3}{2}} - 1)}{x}} y = \color{red}{K_{2} - K_{1}} \qquad $$  $$ \color{blue}{Blue=a_{0}(x)} \qquad  \color{red}{Red=b(x)} $$

A solution for an equation in this form can be identified as follows:

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

By populating the previous equation with our known values the following is obtained:

$$ y(x) = \frac{1}{x^{2}-\exp(\frac{4x^{\frac{3}{2}}}{3})} \int{}^{x} \left[ s^{2} - exp(\frac{4s^{\frac{3}{2}}}{3}) \right] (K_{2}-K_{1})ds $$

$$ y(x) = \frac{K_{2}-K_{1}}{x^{2}-\exp(\frac{4x^{\frac{3}{2}}}{3})} \left[ \frac{x^{3}}{3} - \int{}^{x}exp(\frac{4s^{\frac{3}{2}}}{3}) \right] $$

Your $$h(x)$$ produces an incorrect result. If you don't verify exactness (as a double check on the integrating factor) then you have to be extra careful when using the result. Egm6321.f09.TA 01:27, 28 October 2009 (UTC)

Problem Statement
Find the mathematical structure of $$\Phi$$ that will yield L2.ODE.VC

Solution
First it is necessary to define the first integral as follows:

$$ F(x,y,y^{'},y^{''}) = \frac{d \phi(x,y,p)}{dx} $$

$$ F(x,y,y^{'},y^{''}) = \phi_{x} + \phi_{y}P + \phi_{p}P^{'} \qquad \qquad \qquad Where \quad P=y^{'}     $$      Equation 1

Why did you change $$p$$ to $$ P $$ ? Egm6321.f09.TA 02:07, 28 October 2009 (UTC)

The next step is to define a 2nd Order ODE as follows:

$$ F = W(x)y^{''} + Q(x)y^{'} + R(x)y $$ Equation 2

Comparing Equation 2 with Equation 1 the following is identified:

$$ \phi_{p}=W(x) $$

$$ \phi_{y}=Q(x) $$

$$ \phi_{x}=R(x)y(x) $$

Why are you introducing $$ W(x)$$ into your notation? Egm6321.f09.TA 02:07, 28 October 2009 (UTC)

The next step is to integrate as follows:

$$ \phi= \int{}^{}W(x)dp = W(x)P + C_{1} $$

$$ \phi= \int{}^{}Q(x)dy = Q(x)y + C_{2} $$

$$ \phi= \int{}^{}R(x)y(x)dx = R(x)y(x) + C_{3} $$

Adding all the definition for Phi the following is obtained:

$$ \phi= W(x)P + T(x)y + K \qquad \qquad \qquad Where \quad T(x) = \left[R(x)+Q(x)\right]y+ (C_{2}+C_{3}) \qquad and \quad K = C_{1}+C_{2}+C_{3} \qquad \qquad$$  Equation 3

In order to generate additional exact L2.ODE.VC, the first integral, $$\phi$$, must be of the form outlined in Equation 3.

This does not quite match the expected form. Egm6321.f09.TA 02:07, 28 October 2009 (UTC)

Problem Statement
Problem 4 From, for the case $$n=1$$ (N1_ODE)

$$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$$.

Show that $$f_0-\frac{df_1}{dx}=0 \Leftrightarrow\Phi_{xy}=\Phi_{yx}$$.

Hint: Use $$f_1=\Phi_y$$.

Specifically: 4.1) Find $$f_0$$ in terms of $$\Phi$$ 4.2) Find $$f_1$$ in terms of $$\Phi$$($$f_1=\Phi_y$$) 4.3) Show that $$ f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$$.

Solution
Note: $$F=\frac{d\Phi_}{dx} =\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

For n=1, $$ F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}$$

$$f_{i}=\frac{dF}{dy^{(i)}}$$

4.1)

$$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy}$$

$$f_0=\Phi_{xy}$$  $$f_0=\frac{\partial}{\partial y}\frac{d\phi}{dx}=\phi_{xy}+\phi_{yy}y'$$. You are missing terms. The differential with respect to y should be a partial. --Egm6321.f09.TA 14:29, 15 October 2009 (UTC)

4.2}

$$f_1=\frac{dF}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy'}$$

$$f_1=\Phi_{y}$$

4.3)

$$ f_0-\frac{df_1}{dx}=0$$

$$ \Phi_{xy}-\frac{d\Phi_y}{dx}=0$$

$$ \Phi_{xy}-\Phi_{yx} = 0$$

 $$ \frac{d\phi_y}{dx}=\phi_{yx}+\phi_{yy}y'$$. --Egm6321.f09.TA 14:29, 15 October 2009 (UTC) $$ \Phi_{xy} = \Phi_{yx}$$

Problem Statement
Problem 5 From, for the case $$n=2$$ (N2_ODE)  show: 5.1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$ 5.2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$ 5.3) $$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$ 5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

Solution
Note: $$F=\frac{d\Phi_}{dx} =\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

For n=2, $$ F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+\Phi_{y^{(1)}}y^{(2)}$$

$$f_{i}=\frac{dF}{dy^{(i)}}$$

 $$f_i=\frac{\partial F}{\partial y^(i)}$$ -> partial wrt to y, not total derivative. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.1)

$$f_2=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y))}{dy''}=\Phi_{y'}$$

$$f_1=\frac{dF}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y))}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy'}+\frac{d(\Phi_{y'}y)}{dy'}=\Phi_y+\frac{d(\frac{d}{dx}(\Phi_{y'}y'))}{dy'}=\Phi_y+\frac{d(\Phi_{y'})}{dx}$$

$$f_1=\frac{df_2}{dx}+\Phi_{y}$$

 This expression for $$f_1$$ is not correct. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.2}

$$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y)}{dy}=\frac{d\Phi_y}{dx}+\frac{d(\Phi_yy')}{dy}+\frac{d(\Phi_yy)}{dy'}=\frac{d\Phi_{y}}{dx}$$

$$f_0=\frac{d\Phi_{y}}{dx}$$

 Your partial differentiations are incorrect. The correct expression was obtained by complimentary errors. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.3)

$$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

$$\frac{d\Phi_{y}}{dx}-\frac{d(\frac{df_2}{dx}+\Phi_{y})}{dx}+\frac{d^2f_2}{dx^2}=0$$

$$\frac{d\Phi_{y}}{dx}-\frac{d^2\Phi_{y'}}{dx^2}-\frac{d\Phi_{y}}{dx}+\frac{d^2\Phi_{y'}}{dx^2}=0$$

$$0=0$$

 I am not sure how you arrived at this result. It is not clear what you have done. --Egm6321.f09.TA 01:48, 16 October 2009 (UTC)

5.4)

Equation (4) from 10-2:

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

Equation (5) from 10-2:

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

$$F(x,y,y',y)=\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y$$

$$F(x,y,p,q)=\Phi_{x}+\Phi_{y}p+\Phi_{p}q=(\Phi_{p})q+(\Phi_x+\Phi_{y}p)=fq+g$$

Find $$f_0, f_1, f_2$$ using $$F(x,y,p,q)$$

$$f_0=\frac{d(fq+g)}{dy}=f_yq+g_y$$

$$f_1=\frac{d(fq+g)}{dp}=f_pq+g_p$$

$$f_2=\frac{d(fq+g)}{dq}=f$$

Plug into $$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

$$(f_yq+g_y)-((f_{px}+f_{py}p+f_{pp}q)q+f_pq'+(g_{px}+g_{py}p+g_{pp}q))+\frac{d(f_x+f_yp+f_pq)}{dx}=$$

$$(f_yq+g_y)-((f_{px}+f_{py}p+f_{pp}q)q+f_pq'+(g_{px}+g_{py}p+g_{pp}q))+(f_{xx}+f_{xy}p+f_{xp}q)+((f_{yx}+f_{yy}p+f_{yp}q)p+f_yq)+((f_{px}+f_{py}p+f_{pp}q)q+f_pq')$$

Which can be simplified down to:

$$f_{xx}+2pf_{xy}+p^2f_{yy}+(f_{xp}+pf_{yp}+2f_y-g_{pp})q-(g_{xp}+pg_{yp}-g_y)=0$$

The previous equation is valid if the following equations are satisfied:

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

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

 good work. --Egm6321.f09.TA 01:51, 16 October 2009 (UTC)

Problem Statement
Problem 6 From, for the Legendre differential equation $$F=(1-x^2)y''-2xy'+n(n+1)y=0$$, 6.1 Verify exactness of this equation using two methods: 6.1a.), Equations 4&5. 6.1b.) , Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

Solution
6.1a)

Equation (4) from 10-2:

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

Equation (5) from 10-2:

$$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

The Legendre differential equation has the form:

$$F=f(x,y,p)y''+g(x,y,p)$$

so:

$$f(x,y,p)=(1-x^2)$$

and

$$g(x,y,p)=(-2xy'+n(n+1)y)$$

To use equation (4) and (5), we must find all the partial derivatives:

$$f_x=-2x$$

$$f_{xx}=-2$$

$$f_{xy}=0$$

$$f_{xp}=0$$

$$f_y=0$$

$$f_{yy}=0$$

$$f_{yp}=0$$

$$g_x=-2p$$

$$g_{xp}=-2$$

$$g_y=n(n+1)$$

$$g_{yp}=0$$

$$g_p=-2x$$

$$g_{pp}=0$$

Plugging these derivatives into equations (4) and (5) we get:

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y \Rightarrow -2+0+0=-2+0-n(n+1)$$

Which is not exact, (unless n equals 0 or -1).

$$f_{xp}+pf_{yp}+2f_y=g_{pp} \Rightarrow 0-0+0=0$$

Which is exact.

We can't show exactness because equation (4) fail to show exactness.

6.1b)

For an alternate method of finding exactness we can use equation (5) from page 14-1:

$$ f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

$$f_0=\frac{dF}{dy}=n(n+1)$$

$$f_1=\frac{dF}{dy'}=-2x$$

$$f_2=\frac{dF}{dy''}=(1-x^2)$$

Plugging these into equation (5) we get:

$$n(n+1)-\frac{d(-2x)}{dx}+\frac{d^2(1-x^2)}{dx^2}=0$$

$$n(n+1)+2-2=0$$

$$n(n+1)=0$$

The exactness condition still isn't satisfied, but it could be if n were to equal 0 or -1.

6.2)

Make $$F = (1-x^2)y''-2xy'+n(n+1)y=0$$ an exact equation by using an integrating factor of the form $$x^my^n$$

$$ x^my^n((1-x^2)y''-2xy'+n(n+1)y) = 0 $$

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y $$          Equation 1

$$f_{xp}+pf_{yp}+2f_y=g_{pp} $$                            Equation 2

Using Equations 1 and 2 we check the for exactness of the function F multiplied by the integrating factor. Equation 2 yields:

$$ 0+0+2(nx^my^{n-1}-n^{m+2}y{n-1}=0 $$

Therefore n = 0.

Populating Equation 1 with n=0 yields.

$$ m(m-1)x^{m-2}-(m+2)(m+1)x^m=-2(m+1)x^m $$

Which can be rearranged to produce:

$$ x^{-2}= \frac{m(m+1)}{m(m-1)} $$

Both sides of the equation must equal zero, therefore

$$m=-1 $$

The exact ODE is:

$$ x^{-1}((1-x^2)y''-2xy'+n(n+1)y)=0 $$

Problem Statement
Problem 7 From, Show that equations 1 and 2, namely 7.1 $$\forall u,v$$ functions of $$x$$, $$L(u+v)=L(u)+L(v)$$. and 7.2 $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\forall$$ functions of $$ x$$. are equivalent to equation 3 on p.3-3.

Solution
Use:

$$L(u+v)=L(u)+L(v)$$    and     $$L(\lambda u)=\lambda L(u)$$

to show that:

$$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$$

Using the additive property we can show that:

$$L(\alpha u+\beta v)=L(\alpha u)+L(\beta v)$$

If α and β are equal to 1, this would give us the additive equation:

$$L(u+v)=L(u)+L(v)$$

Using the multiplicative property we can state that:

$$L(\alpha u)= \alpha L(u)$$

and

$$L(\beta v)= \beta L(v)$$

If both α and β are equal to λ, the multiplicative property is apparent as both equations take the form of the multiplicative equation:

$$L(\lambda u)=\lambda L(u)$$

Putting all of this together, the additive and multiplicative properties show that:

$$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$$

 good explanation, although you do not need to choose values for $$\alpha,\beta$$. --Egm6321.f09.TA 04:05, 16 October 2009 (UTC)

Problem Statement
Problem 8 From, plot the shape function $$N_{j+1}^{2}(x)$$.

Solution

correct. Egm6321.f09.TA 03:07, 28 October 2009 (UTC)

Problem Statement
Problem 9 From (| p.16-2), show that $$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t\right)$$ $$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t\right)$$

Solution
$$\frac{d^3y}{dx^3}=(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt})y)$$

$$\ y_{xxx} = (\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))e^{-t}y_t$$

$$\ = e^{-t}(\frac{d}{dt}(-e^{-2t}y_t+e^{-2t}y_{tt})$$

$$\ = e^{-t}(2e^{-2t}y_t-e^{-2t}-2e^{-2t}y_{tt}+e^{-2t}y_{ttt})$$

$$\ y_{xxx} = e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$$

$$\frac{d^4y}{dx^4}=(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt}))(\frac{dt}{dx}(\frac{d}{dt})y)$$

$$ y_{xxxx} = (\frac{dt}{dx}(\frac{d}{dt})y_{xxx})$$

$$ = (\frac{dt}{dx}(\frac{d}{dt})e^{-3t}(y_{ttt}-3y_{tt}+2y_t))$$

$$ = (\frac{dt}{dx})(-3e^{-3t}(y_{ttt}-3y_{tt}+2y_t)+e^{-3t}(y_{tttt}-3y_{ttt}+2y_{tt})$$

$$ y_{xxxx} = e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

It is not clear from your expressions which terms your differential operators are acting upon. You need some parenthesis or brackets (or alternate expressions) to clarify this. Egm6321.f09.TA 03:26, 28 October 2009 (UTC)

Problem Statement
Solve: $$ x^2y''-2xy'+2y=0 $$ Equation 1

using the method of trial solution $$ y=e^{rx} \quad and \quad y=x^{r}$$ directly for the boundary conditions $$\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$$ Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

Solution
Given that the solution to this ODE is of the form:

$$ y(x) = C_{1}x^{r_{1}} + C_{2}x^{r_{2}} $$ Equation 2

PART 1

Since the trial solution is:

$$ y=x^{r} $$

The following can be determined:

$$ y^{'}=rx^{r-1} $$

$$ y^{''}=r(r+1)x^{r-2} $$

Using these results and using them in Equation 1, the following is obtained:

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

Simplify to find:

$$ r^{2} - 3r + 2= 0 $$

This previous result is the characteristic equation of the ODE. The next step is to find the roots "r".

$$ r_{1}= 2 \qquad r_{2}=1 $$

With these roots and comparing to Equation 2, the following is the particular solution:

$$ y(x) = C_{1}x^{2} + C_{2}x^{1} $$

The constants (C1 and C2) can be then found using the known values of the function y(x):

since: $$ y(1)= 3 \quad and \quad y(2)=4 $$

$$ y(1)= 3 = C_{1} + C_{2} $$  Equation 3.

$$ y(2)= 4 = 4C_{1}+ 2C_{2} $$ Equation 4.

Solving Equations 3 and 4 simultaneously it yields:

$$ C_{1} = -1 \qquad and \qquad C_{2} = 4 $$

Then the solution is of the form:

$$ y(x)= -x^{2} + 4x $$   Equation 5

The Plot of the Solution is shown in Figure 1.

very good. Egm6321.f09.TA 03:51, 28 October 2009 (UTC)

PART 2

Since the trial solution is:

$$ y= \exp^{rx} $$

The following can be determined:

$$ y^{'}=re^{rx} $$

$$ y^{''}=r^{2}e^{rx} $$

Using these results and using them in Equation 1, the following is obtained:

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

Simplify to find:

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

The previous equation cannot be solved without additional information therefore it is assumed to be a bad guess for a solution.

By Using the values for the roots as found in the previous part, the Equation becomes of the form:

$$ y(x)= C_{1}\exp^{2x} + C_{2}\exp^{x} $$

Using the known values for y(x) the constants are found as follows:

$$ C_{1}= -.1204 \qquad C_{2}= 1.4309 $$

It then follows that the solution is of the form:

$$ Y(x)= -.1204\exp^{2x} + 1.4309\exp^{x} $$

The plot of the solution is shown in Figure 2:

Problem Statement
Prolbem 11 From (| p.17-4) obtain equation 2 from p.17-3

$$Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$

using the integrator factor method and equation (1) from page 17-3.

$$u_1(x)Z'+(a_1(x)u_1(x)+2u_1'(x))Z=0$$

Solution
$$u_1(x)Z'+(a_1(x)u_1(x)+2u_1'(x))Z=0$$

Divide by u_1(x) to get:

$$Z'+(a_1(x)+\frac{2u_1'(x)}{u_1{x}})Z=0$$

Multiply by an integrating factor h(x):

$$h(x)Z'+h(x)(a_1(x)+\frac{2u_1'(x)}{u_1{x}})Z=0$$

$$(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=-\frac{Z'}{Z}$$

We can say that:

$$\frac{d(h(x)Z(x))}{dx}=h(x)Z'(x)+h'(x)Z(x)=0$$

thus

$$\frac{h'(x)}{h(x)}=-\frac{Z'(x)}{Z(x)}$$

After some algebraic manipulation of equation (1) multiplied by h(x):

$$(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=-\frac{Z'}{Z}$$

$$(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=\frac{h(x)'}{h(x)}$$

Integrating both sides we get:

$$\int(a_1(x)+\frac{2u_1'(x)}{u_1{x}})=\int\frac{h(x)'}{h(x)}$$

$$\int^x(a_1(s)ds)+ln(u_1)^2+d=1+ln(h(x))+e$$

We can combine the constants d, e, and 1 into another constant b:

$$\int^x(a_1(s)ds)+ln(u_1)^2+b=ln(h(x))$$

$$h(x)=exp(\int^x(a_1(s)ds)+ln(u_1)^2+b)$$

We've already stated that:

$$h(x)Z'(x)+h(x)'Z(x)=0$$

which is the same as:

$$(h(x)Z(x))'=0$$

Integrating both side we get:

$$h(x)Z(x)=k$$

Plugging in h(x) into this we get:

$$exp(\int^x(a_1(s)ds)+ln(u_1)^2+b)Z(x)=k$$

solving for Z(x) we get:

$$Z(x)=\frac{k}{exp(\int^x(a_1(s)ds)+ln(u_1)^2+b)}=\frac{c}{exp(\int^x(a_1(s)ds)+ln(u_1)^2)}=\frac{c}{u_1^2exp(\int^x(a_1(s)ds)}$$

Which can be expressed as:

$$Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$

very good. long but straight forward. Egm6321.f09.TA 04:08, 28 October 2009 (UTC)

Problem Statement
Develop reductions of order method 2 using different algebraic Operations:

1) $$ y(x) = U(x)\pm u_{1}(x) $$ 2) $$ y(x) = U(x) / u_{1}(x) $$ 3) $$ y(x) = u_{1}(x) / U(x) $$

Objective: Proof you will not obtain a solution with a missing dependent variable.

Solution
The form of a homogeneous L2.ODE.VC is as follows:

$$ y^{''} + a_{1}y^{'} + a_{0}y = 0 $$   Equation 1.

1a)

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

$$ y^{'}= U^{'} + u_{1}^{'} $$

$$ y^{} = U^{} + u_{1}^{''} $$

Substituting into Equation 1

$$ 0 = U^{} + u_{1}^{} + a_{1}(U^{'} + u_{1}^{'}) + a_{0}(U + u_{1}) $$

Rearranging yields:

$$ \color{blue}{U^{} + a_{1}U^{'} + a_{0}(U)} + \color{red}{u_{1}^{} + a_{1}u_{1}^{'} + a_{0}(u_{1})} =0 $$

Realizing that the Blue and the Red Part of the previous equations are homogeneous solutions we realize that we cannot reach a form of Equation 1, with a missing dependent Variable.

1b)

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

$$ y^{'}= U^{'} - u_{1}^{'} $$

$$ y^{} = U^{} - u_{1}^{''} $$

Substituting into Equation 1

$$ 0 = U^{} - u_{1}^{} + a_{1}(U^{'} - u_{1}^{'}) + a_{0}(U - u_{1}) $$

Rearranging yields:

$$ \color{blue}{(U^{} + a_{1}U^{'} + a_{0}(U))} - \color{red}{(u_{1}^{} + a_{1}u_{1}^{'} + a_{0}(u_{1}))} =0 $$

Realizing that the Blue and the Red Part of the previous equations are homogeneous solutions we realize that we cannot reach a form of Equation 1, with a missing dependent Variable.

2)

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

$$ y^{'}= \frac{u_{1}U^{'} - Uu_{1}^{'}}{u_{1}^{2}} $$

$$ y^{} = \frac{u_{1}U^{} - U^{'}u_{1}^{'}}{u_{1}^{2}} - \left[ U(\frac{u_{1}^2u_{1}^{''} - u_{1}^{'}(u_{1}^{2})^{'}}{u_{1}^{4}} + \frac {u_{1}U^{'}}{u_{1}^{2}} \right]$$

Substituting into Equation 1:

$$ \frac{u_{1}U^{} - U^{'}u_{1}^{'}}{u_{1}^{2}} - \left[ U(\frac{u_{1}^2u_{1}^{} - u_{1}^{'}(u_{1}^{2})^{'}}{u_{1}^{4}} + \frac {u_{1}U^{'}}{u_{1}^{2}} \right] + \frac{a_{1}U^{'}}{u_{1}} - a_{1}\frac{Uu_{1}^{'}}{u_{1}^{2}} + a_{0} \left[ \frac{Uu_{1}^{'}}{u_{1}^{2}} - \frac{Uu_{1}^{'}}{u_{1}^{2}} \right] $$

From the previous expression it can be seen that an expression with a missing dependent variable is not obtained.

3)

This expression is similar to part 2 and the conclusion is that an expression with a missing dependent variable cannot be obtained.

very good. Egm6321.f09.TA 04:17, 28 October 2009 (UTC)

Problem Statement
Find $$ u_{1}(x)$$ and $$ u_{2}(x) $$ of

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

using trial solutions:

1) $$ y=ax^{b} $$ where a,b are coefficients to be determined.

2) $$ y=e^{rx} $$

Compare the two solutions using the boundary conditions:

$$ y(0)=1 $$ $$ y(1)=2 $$

and also compare to the solutions by reduction of order method 2. Plot Solutions using Matlab

Solution
A solution to the 2nd Order ODE would be of the following form:

$$ y(x) = K_{1}U_{1}(x) + K_{2}U_{2}(x) $$  Equation 2

PART 1

A trial solution will be used as follows:

$$ y = ax^{b} $$

"a" and "b" are coefficients to be determined using initial conditions.

using the trial solution the following is identified:

$$ y^{'}(x) = abx^{b-1}  $$

$$ y^{''}(x) = ab(b-1)x^{b-2} = (ab^{2} - ab)x^{b-2} $$

Replacing the newly defined equations into equation 1, the following is obtained:

$$ (1-x^{2}) \left[ (ab^{2} -ab)x^{b-2}) \right] -2x \left[abx^{b-1} \right] + 2 \left[ax^{b}\right] = 0 $$

Dividing by $$ x^{b} $$:

$$ (1-x^{2})(ab^{2}-ab)x^{-2} - 2x \left[ abx^{-1} \right] + 2a = 0 $$

Rearranging:

$$ (1-x^{2})(a)(b^{2}-b)x^{-2} - 2xabx^{-1} + 2a = 0 $$

Dividing by "a":

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

From here we can then identify the following:

$$ U_{1}(x) = (x^{-2}-1) \qquad \qquad K_{1}=(b^{2}-b) $$

$$ U_{2}(x) = (1) \qquad \qquad \qquad K_{2}=(-2b+2) $$

$$ y(x)= (b^{2}-b)(x^{-2}-1) + (-2b+2)(1) $$

The next step is to identify the coefficient "b". If for both solutions known the value is computed to be the same then the trial solution is determined to be a solution to the ODE.

Since

$$ y(0)=1 $$

Applying this to Equation 2 with our identified values:

$$ y(0) = 1 = (0^{-2} -1)(b^{2}-b) + (-2b +2) $$

The coefficient b cannot be identified as the first part of the equation yields to an undefined solution.

When

$$ y(1) = 2 $$

$$ y(1) = 2 = (1^{-2} -1)(b^{2}-b) + (-2b +2) $$

$$ 2 = -2b + 2 $$

$$ b=0 $$

The Coefficient "b" is then identified. It is concluded that this trial solution is not a solution to the ODE.

PART 2

A trial solution will be used as follows:

$$ y = \exp^{rx} $$

"r" is a coefficient which represents the roots of the ODE's Characteristic Equation.

using the trial solution the following is identified:

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

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

Replacing the newly defined equations into equation 1, the following is obtained:

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

Dividing by $$ \exp^{rx} $$

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

Rearranging:

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

From here we can then identify the following:

$$ U_{1}(x) = (1-x^{2}) \qquad \qquad K_{1}=(r^{2}) $$

$$ U_{2}(x) = (xr+1) \qquad \qquad K_{2}=(-2) $$

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

The next step is to identify the coefficient "r". If for both solutions known the value is computed to be the same then the trial solution is determined to be a solution to the ODE.

Since

$$ y(0)=1 $$

Applying this to Equation 2 with our identified values:

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

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

$$ r = \pm 3 $$

When

$$ y(1) = 2 $$

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

$$ 2 = -2(r+1) $$

$$ r= -2 $$

Since the solutions for "r" do not match then the Trial Solution is not a solution to the ODE.

You should be able to make one of these trials work. Egm6321.f09.TA 05:43, 28 October 2009 (UTC)

Contributing Authors
--Egm6321.f09.team5.GV 04:40, 7 October 2009 (UTC)

--Egm6321.f09.team5.risher 16:39, 7 October 2009 (UTC)

--Egm6321.f09.team5.bear 18:15, 7 October 2009 (UTC)

title: User:Egm6321.f09.team5/HW3