text:
User:Norton~enwikiversity/Wikistution/Article VI (Debts, Supremacy, Oaths)

Article VI

All Debts contracted and Engagements entered into, before the Adoption of this Constitution, shall be as valid against the United States under this Constitution, as under the Confederation.

This Constitution, and the Laws of the United States which shall be made in Pursuance thereof; and all Treaties made, or which shall be made, under the Authority of the United States, shall be the supreme Law of the Land; and the Judges in every State shall be bound thereby, any Thing in the Constitution or Laws of any State to the Contrary notwithstanding.

The Senators and Representatives before mentioned, and the Members of the several State Legislatures, and all executive and judicial Officers, both of the United States and of the several States, shall be bound by Oath or Affirmation, to support this Constitution; but no religious Test shall ever be required as a Qualification to any Office or public Trust under the United States.


text:
User:Norton~enwikiversity/Wikistution/Article V (Amendments)

Article V

[The Congress, whenever two thirds of both Houses shall deem it necessary, shall propose Amendments to this Constitution, or, on the Application of the Legislatures of two thirds of the several States, shall call a Convention for proposing Amendments, which, in either Case, shall be valid to all Intents and Purposes, as Part of this Constitution, when ratified by the Legislatures of three fourths of the several States, or by Conventions in three fourths thereof, as the one or the other Mode of Ratification may be proposed by the Congress;] Provided that no Amendment which may be made prior to the Year One thousand eight hundred and eight shall in any Manner affect the first and fourth Clauses in the Ninth Section of the first Article; and that no State, without its Consent, shall be deprived of its equal Suffrage in the Senate.

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


text:
User:Norton~enwikiversity/Wikistution/Preamble

Preamble

We the People of the United States, in Order to form a more perfect Union, establish Justice, insure domestic Tranquility, provide for the common defence, promote the general Welfare, and secure the Blessings of Liberty to ourselves and our Posterity, do ordain and establish this Constitution for the United States of America.


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User talk:SimpleMan~enwikiversity

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User:Egm6321.f10.team6.lee/HW4

= Problem 1. Solve N2_ODE_VC =

Required
1. show eqn(1.1) is exact

2. find $$\begin{align}\phi\end{align}$$

3. solve for $$\begin{align}y(x)\end{align}$$

1st condition of exactness
The 1st condition of exactness can be satisfied, if the equation is written in the form of eqn(1.2)

where,

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

$$\begin{align}g(x,y,p) := {\phi}_{x}+{\phi}_{y}p\end{align}$$

$$\begin{align}f(x,y,p) := {\phi}_{p}(x,y,p)\end{align}$$

let's rearrange the eqn(1.1) in the form of eqn(1.2)

where,

$$\begin{align}g(x,y,p)= \underbrace{({x}^{2}-\mathrm{sin}x)}_{{\phi}_{y}}p+\underbrace{2xy}_{{\phi}_{x}}\end{align}$$

$$\begin{align}f(x,y,p)=\underbrace{\mathrm{cos}x}_{{\phi}_{p}}\end{align}$$

The given equation is satisfied the 1st condition of exactness.

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

 * first relation

 * second relation

As both exactness conditions are satisfied, we can conclude that the eqn(1.1) is exact.

Find $$\begin{align}\phi\end{align}$$
As,

using definition of $$\begin{align}g(x,y,p)\end{align}$$

It should be the same to the $$\begin{align}g(x,y,p)\end{align}$$ in the eqn(1.3)

Assume,

Then,

Let's derivate eqn(1.13) with respect to y to find $$\begin{align}f(y)\end{align}$$

it should be the same to the $$\begin{align}{h}_{y}\end{align}$$ in eqn(1.11)

Then,

let's put it to the eqn(1.8)

by rearranging, we can get,

where, $$\begin{align}k={k}_{2}-{k}_{1}\end{align}$$

we can easily check wheter it is correct or not by find partial derivatives wrt x, y, p

using definition of $$\begin{align}F(x,y,y',y'')\end{align}$$

This is the same equation to the eqn(1.1). So we know that eqn(1.17) is correct.

Solve for $$\begin{align}y(x)\end{align}$$
Now, we have to solve N1_ODE_VC to find y(x). let's rearrange eqn(1.17)

let's find integrating factor i(x) first.

then, we can find y(x).

title: User:Egm6321.f10.team6.lee/HW4


text:
User:Rf~enwikiversity

I'm a student of computer science minoring in economics at the University of Waterloo.

Scholarly Interests
In computer science:
 * Compiler design and optimization
 * 3D graphics, especially massive worlds
 * CPU design and architecture

In economics, I tend to subscribe to the Post Keynesian school of thought:
 * Modern Monetary Theory
 * Steve Keen's money circuit models, and dynamic, disequilibrium, continuous-time models in general
 * Real world behaviour of firms and households
 * Energy economics and policy

Projects

 * Monetary Circuit Model - On hold until Minsky is released.
 * Economics page rewrite (finished)
 * Economics cruft list
 * Detailed outlines of IRL courses I'm taking:
 * Macroeconomics 3
 * Monetary economics
 * Calculus 2
 * Algorithms
 * Computer organization and design

Resources

 * Being educational
 * Broader Questions and a Bigger Toolbox: A problem-centered and student-centered approach to teaching pluralist economics

title: User:Rf~enwikiversity


text:
User talk:Rf~enwikiversity

Colloquium
Thanks for initiating that thread on the colloquium. Very useful. --HappyCamper 18:10, 4 January 2012 (UTC)

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User:EGM6321.f10.team6.cook/hw4

=Solve N2_ODE_VC =

==Given ==

Required

 * 1) Show equation (1.1) is exact
 * 2) Find $$\begin{align}\phi\end{align}$$
 * 3) Solve for $$\begin{align}y(x)\end{align}$$

==Solution   ==

Exactness
To be exact the first and second conditions of exactness must be met.

1st condition of exactness
The 1st condition of exactness is shown to be satisfied if the equation can be written in the form of equation (1.2)

where,

$$\begin{align}&p:=y'\\ &g(x,y,p) := {\phi}_{x}+{\phi}_{y}p\\ &f(x,y,p) := {\phi}_{p}(x,y,p)\end{align}$$

Rearranging equation (1.1) into the form of equation (1.2),

where,

$$\begin{align}g(x,y,p)= ({x}^{2}-\mathrm{sin}x)p+2xy \end{align}$$

$$\begin{align}f(x,y,p)=\mathrm{cos}x\end{align}$$

The given equation thus satisfies the 1st condition of exactness.

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

 * first relation

 * second relation

As both of the exactness conditions are satisfied, we can conclude that the equation 1.1 is exact.

Find $$\begin{align}\phi\end{align}$$
As,

using the definition of $$\begin{align}g(x,y,p)\end{align}$$

Which is equivalent to $$\begin{align}g(x,y,p)\end{align}$$ in equation (1.3)

Assume,

Then,

Taking the derivative of equation (1.13) with respect to $$y$$ to find $$\begin{align}f(y)\end{align}$$

and equating it to $$\begin{align}{h}_{y}\end{align}$$ in equation (1.11)

Then,

Substituting back into equation (1.8)

Consolidating constants,

where, $$\begin{align}k={k}_{2}-{k}_{1}\end{align}$$

The solution can be easily verified by finding the partial derivatives with respect to $$x$$, $$y$$, $$p$$

Using the definition of $$\begin{align}F(x,y,y',y'')\end{align}$$

This is equation (1.1), so we know that equation (1.18) is correct.

Solve for $$\begin{align}y(x)\end{align}$$
To solve the first order nonlinear ordinary differential equation with variable coefficients (N1_ODE_VC) for y(x), equation (1.18) is first rearranged.

Finding the integrating factor $$i(x)$$ first.

then, we can find y(x).

Matlab code for solving this N2-ODE-VC:

Which verifies our solution.

Given
The Bessel Equation,
 * {| style="width:100%" border="0"

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} x^{2} y'' + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.1)
 *
 * }

Required
Verify exactness of Bessel Equation(2.1) using 2 methods; first using equations (1) & (2) from the lecture notes on p. 15_3.

The 2nd condition of exactness

Secondly, using equation (2) from the lecture notes on p. 22_4

where $$\begin{align} f_i = \frac{\partial F}{\partial y^{(i)}} ,i= 0,1,2 \end{align}$$

Finally, if equation (2.1) is not exact, see whether it can be made exact using the Integrating Factor Method with $$\begin{align} h(x,y)=x^{m} y^{n}\end{align}$$; See lecture notes p. 19_1, p.22_4

First Method
From equations (2.1), (2.2)

Substituting equation (2.6) into equation (2.2)

if equation (2.7) is met, equation (2.1) is exact.

From equation (2.3)

Substituting equation (2.8) for equation (2.3)

This satisfies the 2nd condition of exactness.

Second Method

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

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} F(x,y,y',y) = x^{2} y + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.10)
 *
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} &f_0 = \frac{\partial F }{\partial y^{(0)}} = x^2 - {\nu}^2\\ &f_1 = \frac{\partial F }{\partial y^{(1)}} = x\\ &f_2 = \frac{\partial F }{\partial y^{(2)}} = x^2\\ \end{align}$$

(2.11)
 *
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} \frac{\mathrm{d} f_1 }{\mathrm{d} x} =1 {~}{~}{~}{~}{~} \frac{\mathrm{d^2} f_2 }{\mathrm{d} x^2} =2 \end{align}$$

(2.12)
 *
 * }

Therefore,

Which is the same result as the first method; if equation (2.13) is met, equation (2.1) is exact.

Making the Equation Exact
Verifying the exactness of equation (2.1) using the 2nd condition of exactness, equations (2.2), (2.3)

Equation (2.18) substituted into equation (2.3)

To satisfy equation (2.19), n=0.

Substituting equations (2.16), (2.17) into equation (2.2)

Solve the quadratic equation for m,

Thus, when equation (2.22) is met, the equation is exact.

Problem Statement
The Euler-Bernoulli beam equation, given in Lecture 24 Fall 2010, is as follows:

If we assume that $${{u}_{\left( x,t \right)}}={{X}_{\left( x \right)}}{{T}_{\left( t \right)}}$$, we can apply the technique of separation of variables, which yields the following :

If we further assume that T has the form in equation 3.3, we can rewrite 3.2 as equation 3.4.

After rearranging terms and doing some algebra we obtain:

If we apply the method of undetermined coefficients and assume $${{X}_{\left( x \right)}}=\exp \left( bx \right)$$ we obtain the following two sets of roots (p25_2 Lecture 25 ):

It follows that the solution to equation 3.1 is the linear combination of the above roots:

Required
Express equation 3.7 in terms of sines, cosines, hyperbolic sines, and hyperbolic cosines.

Sine and Cosine Terms
We begin by examining the imaginary terms, to which we will apply Euler’s formula in order to obtain sines and cosines. Euler’s formula is given as :

Since sin(-u)=-sin(u) and cos(-u)=cos(u) we can rewrite 3.9 as follows:

Hyperbolic Sine and Cosine Terms
We now consider the real terms. We apply the definition of hyperbolic functions, however, instead of looking at the terms let us look at the hyperbolic functions themselves.

The hyperbolic sine and cosine are defined as:

If we add 3.11 to 3.12 we obtain the following:

If we let a = 1/k, equation 3.13 becomes the positive real root term:

Considering the identities in equation 3.15 and 3.16 we can add the negative sinh and cosh to obtain equation 3.17:

Now consider the following identities:

If we let a=1/k, equation 3.17 becomes the negative real root term:

Combining 3.14 and 3.18 we obtain:

Combining Terms
If we now combine equation 3.10 with 3.19 we obtain the final form for the solution to equation 3.1:

title: User:EGM6321.f10.team6.cook/hw4


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User talk:Juztreboot

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --La comadreja formerly AFriedman RESEARCH (talk) 03:32, 18 October 2010 (UTC)

 Hello , you can call me Jacob Frank. I'd like to be your mascot and welcome you to Wikiversity. Even though I was a plain and untutored man back in 18th-century Europe, that still didn't stop me from contributing my ideas to the world. Neither should it stop you. As a religious leader who developed the doctrine of "purification through transgression," my advice to you is to be bold and to ignore all rules—yes, these are two of Wikiversity's official policies! Happy editing.

title: User talk:Ashishdave54


text:
User:Egm6321.f10.team2.patterson/Homework 4

= Problem 1: Show Exactness of Second Order Linear Ordinary Differential Equations with Varying Coefficients (L2_ODE_VC) =

Given
For the following equation:
 * {| style="width:100%" border="0"

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

F\left(x,y,y',y\right)=\left(cos x \right)y+\left(x^{2}-sin x\right)y'+ 2xy=0

$$ (eq 1.1)
 *
 * }

(1) Show that Eq. 1.1 is exact

(2) Find the first integral,Φ, and

(3) Solve for y(x)

(1) Proof of Exactness
To be exact, the equation must satisfy two conditions.

1st Condition of Exactness
First, the equation must be presented in the following format
 * {| style="width:100%" border="0"

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

F\left(x,y,y',y\right)=g\left(x,y,y'\right)+f\left(x,y,y'\right)y

$$ (eq 1.2) By letting y' = p, the first condition of exactness is satisfied as shown in the following equation:
 *
 * }

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

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

F\left(x,y,p,y\right)=\underbrace{\left(x^{2}-sin x \right)p+ 2xy}_{g\left(x,y,p\right)}+\underbrace{\left(cos x \right)}_{f\left(x,y,p\right)}y

$$

(eq 1.3)
 *
 * }

2nd Condition of Exactness
In order to meet the second condition of exactness, the following equations must be satisfied:

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

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

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

$$

(eq 1.4)
 *
 * }
 * {| style="width:100%" border="0"

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

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

$$

(eq 1.5) Each element of equations 1.4 and 1.5 are calculated to be the following:
 *
 * }

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

$$ \displaystyle f_{xx}=-cos\left(x\right)$$
 * style="width:95%" |
 * style="width:95%" |

$$\displaystyle f_{xy}=f_{yy}=f_{y}=f_{xp}=f_{yp}=0$$

$$\displaystyle g_{xp}=2x-cos\left(x\right)$$

$$\displaystyle g_{y}=2x$$

$$\displaystyle g_{yp}=g_{pp}=0 $$

 *
 * }

Plugging each relevant element into equation 1.4 yields the following:
 * {| style="width:100%" border="0"

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

-cos\left(x\right)+2p\cancel{f_{xy}}^{0}+p^{2}\cancel{f_{yy}}^{0}=2x-cos\left(x\right)+p\cancel{g_{yp}}^{0}-2x $$

$$-cos\left(x\right)=-cos\left(x\right)$$ (eq 1.6)
 *
 * }

Plugging each relevant element into equation 1.5 yields the following:
 * {| style="width:100%" border="0"

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

\cancel{f_{xp}}^{0}+p\cancel{f_{yp}}^{0}+2\cancel{f_{y}}^{0}=\cancel{g_{pp}}^{0} $$

$$ \displaystyle 0=0$$ (eq 1.7) Equations 1.4 and 1.5 have been satisfied and therefore the equation is exact.
 *
 * }

(2) Solving for the First Integral,Φ
The first integral is defined by the following:

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

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

\phi=h\left(x,y\right)+\underbrace{\int \underbrace{f\left(x,y,p\right)}_{cos\left(x\right)}\,dp}_{pcos\left(x\right)}

$$

(eq 1.8)
 *
 * }

From equation 1.8 the following equations can be derived:
 * {| style="width:100%" border="0"

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

\phi_{x}=h_{x}+	\frac{\partial}{\partial x}\left[pcos\left(x\right)\right]=h_{x}-psin\left(x\right)

$$

(eq 1.9)
 *
 * }
 * {| style="width:100%" border="0"

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

\phi_{y}=h_{y}+	\frac{\partial}{\partial y}\left[pcos\left(x\right)\right]=h_{y}

$$

(eq 1.10) Using the definition of g to solve for h, yields the following:
 *
 * }
 * {| style="width:100%" border="0"

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

g\left(x,y,p\right)=\phi_{y}p+\phi_{x}=h_{y}p+\left[h_{x}-psin\left(x\right)\right]

$$

$$g\left(x,y,p\right)=\left[h_{y}-sin\left(x\right)\right]p+h_{x}$$

(eq 1.11) By equating equation 1.11 with the g(x,y,p) shown in equation 1.3, the partial derivatives of h can be solved as shown:
 *
 * }
 * {| style="width:100%" border="0"

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

\left[h_{y}-sin\left(x\right)\right]p+h_{x}=\left[x^{2}-sin\left(x\right)\right]p+2xy $$ (eq 1.12)
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle h_{y}=x^{2}$$ (eq 1.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle h_{x}=2xy$$
 * style="width:95%" |
 * style="width:95%" |

(eq 1.14) Next, equations 1.13 and 1.14 can be integrated separately to solve for h(x,y) as follows:
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle h=\int h_{x}\,dx +w\left(y\right)=x^{2}y+ w\left(y\right)$$
 * style="width:95%" |
 * style="width:95%" |

(eq 1.15)
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle h=\int h_{y}\,dy +v\left(x\right)=x^{2}y+ v\left(x\right)$$
 * style="width:95%" |
 * style="width:95%" |

(eq 1.16)
 *

 * }

Equating equations 1.15 and 1.16 show that
 * {| style="width:100%" border="0"

$$\displaystyle h=x^{2}y$$ (eq 1.17) Therefore, equation 1.8 can be rewritten as:
 * style="width:95%" |
 * style="width:95%" |
 *

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

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

\phi=x^{2}y+pcos\left(x\right)+k_{1}

$$

(eq 1.18)
 *

 * }

(3) Solving for y(x)
When combined with the definition of the first integral:
 * {| style="width:100%" border="0"

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

\phi=k_{2} $$

(eq 1.19)
 *

 * }

Equation 1.18 can be rewritten as:

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

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

\phi=x^{2}y+cos\left(x\right)y'=K $$

Where $$\displaystyle K=k_{2}-k_{1}$$ (eq 1.20) The equation is divided by cos(x) in order to find an integrating factor:
 *

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

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

y'+\frac{x^{2}}{cos\left(x\right)}y=\frac{K}{cos\left(x\right)} $$

(eq 1.21)
 *

 * }

Multiply the entire equation by $$e^{\int \frac{x^{2}}{cos(x)}\,dx}$$ yields:
 * {| style="width:100%" border="0"

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

e^{\int \frac{x^{2}}{cos(x)}\,dx}y'+\frac{x^{2}}{cos\left(x\right)}e^{\int \frac{x^{2}}{cos(x)}\,dx}y=e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{K}{cos\left(x\right)} $$

(eq 1.22) The left hand side of the equation can be condensed to the following:
 *

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

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

(e^{\int \frac{x^{2}}{cos(x)}\,dx}y)'=e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{K}{cos\left(x\right)} $$

(eq 1.23) Integrating both sides with respect to x gives:
 *

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

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

e^{\int \frac{x^{2}}{cos(x)}\,dx}y=K\int e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{1}{cos\left(x\right)}\,dx $$

(eq 1.24) Dividing both sides by $$e^{\int \frac{x^{2}}{cos(x)}\,dx}$$ yields the solution for y:
 *

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

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

y=Ke^{-\int \frac{x^{2}}{cos(x)}\,dx}\int e^{\int \frac{x^{2}}{cos(x)}\,dx}\frac{1}{cos\left(x\right)}\,dx $$

(eq 1.25)
 *

 * }


text:
User talk:Mack paris

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:17, 18 October 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:17, 18 October 2010 (UTC)

title: User talk:Mack paris


text: User:Nicodemuz/ChineseProject

title: User:Nicodemuz/ChineseProject


text:
Category:Theory of infinite nesting of matter

The theory of infinite nesting of matter is the domain of science, studies cosmic systems, containing objects from the smallest particles up to infinitely large metagalactic systems. These objects can not be limited to inanimate nature, but to be living organisms. As an interdisciplinary science, the theory of infinite nesting of matter for describing the subject of it research uses the methods of other sciences, including physics, astronomy, biology, systems science, etc.


text:
User talk:Deepak.ail

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 13:04, 18 October 2010 (UTC)

title: User talk:Deepak.ail


text:
User:Dipaolatommaso

{{robelbox|title=Basic filmmaking student: Dipaolatommaso}}

 * Dipaolatommaso has completed the following assignments in the course on Basic filmmaking (pre-production) for narrative motion pictures.

 * [[Image:Blancovinkjeup.png|16px]] Formatting the Script [[Image:Nuvola apps package wordprocessing.png|28px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Pop Quiz - What is the first shot? [[Image:Crystal Clear app clock.svg|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Creating the thumbnail storyboards [[Image:Crystal Clear app gimp.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Quick Exercise - Over-the-shoulder shot [[Image:Applications-accessories.svg|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Creating the 3D storyboards [[Image:Crystal Clear app Login Manager.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] A simple DVD-Video Editing Workshop [[Image:Crystal Clear device cdwriter unmount.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Learning about camera lenses [[Image:Crystal Clear app camera.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Record the dialog for the animatic [[Image:Crystal Clear app krec.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Intro to narrative filmmaking [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Draw the movie poster for "Star Wars" [[Image:Crystal Clear app kcoloredit.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Understanding dailies [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Create a matte painting [[Image:Crystal Clear app kcoloredit.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Understanding editing [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Pop Quiz - Creating the moods (film scoring) [[Image:Crystal Clear app clock.svg|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Understanding "L-Cuts" [[Image:Clapboard-Small.png|24px]]
 * [[Image:Blancovinkjeemptysmall.png|16px]] Create the animatic [[Image:Clapboard-Small.png|28px]]

{{Robelbox/close}}

title: User:Dipaolatommaso


text:
User:Brylie/LessonPlans

I am creating several lesson plans for the Peer to Peer University course entitled Collaborative Lesson Planning.

The lesson plans can be found at my Music Theory course page.

title: User:Brylie/LessonPlans


text:
User:Arided

Hi, I'm Joe Corneli... more information about me is in my "learning profile".

=Paragogy=
 * Starting with a page about paragogy, a theory for peer-to-peer learning. This has been further developed in  a paper co-written with Charlie Danoff (and subsequently on paragogy.net and peeragogy.org).
 * I'm adding a short lesson plan about implementing paragogy.
 * I've written a short abstract of a research proposal about building a distributed university.

=Other=
 * Along with other participants in an Intro to Computer Supported Collaborative Learning at P2PU, I helped draft a Free Learning Manifesto.
 * In order to keep the page uncluttered, I'm moving my math course stuff to a Math Courses page.

title: User:Arided


text:
User:Arided/Paragogy

Outline of Paragogy
1. Context as decentered center.

For learning design in a peer-to-peer context, understanding the learner's self-concept may be less important than understanding the concept of `shared context in motion'.

2. Meta-learning as a font of knowledge.

We all have a lot to learn about learning.

3. Peers are equal, but different.

The learner mustn't seek only to confirm what they already know, and must therefor confront and make sense of difference as part of the learning experience.

4. Learning is distributed and nonlinear.

''Side-tracking is OK, but dissipation isn't likely to work. Part of paragogy is learning how to work one's way around in a given social field.''

5. Realise your dream and then wake up!

Paragogy is the art of fulfilling motivations when this is possible, then going on to the next thing.

Paragogy compared with Andragogy
I came up with these principles as responses to the "five principles of andragogy" as they were stated by Malcolm Knowles. In super-succinct form, Knowles's andragogical principles are: (1) that adults are self-directed learners, (2) that adult learners bring a wealth of experience to the educational setting, (3) that adults enter educational settings ready to learn, (4) that adults are problem-centered in their learning, and (5) that adults are best motivated by internal factors.

While I don't see anything wrong with the idea of "andragogy", I do think that it is not necessarily the best set of principles to apply in peer-based learning. A useful review and critique that goes beyond the soundbytes above is given by Laurie C. Blondy. With regard to the first andragogical principle, she writes:

 * "Robinson reported that learners did not prefer to be self-directed in their learning because they specifically indicated a desire for clear instructions regarding how to complete assignments as well as precise information from the course facilitator about which assignments would be graded. [...] Cheren stated that while learners may express a desire to be self-directed in their learning, most lack the required understanding of learning necessary to be self-directed and thus need guidance and encouragement in the learning process."

Paragogy and basho
The first paragogical principle stresses the importance of understanding the idea of "shared context in motion" -- but what's that?

The philosophical foundations of this notion, summarized by Abe, describe the way in which events and objects arise from their larger contexts. In other words, the idea of basho can help us think about how a context constrains or supports different types of (inter-)actions, and also about how what we (re-)shape the contexts we find ourselves in.

Nonaka and Toyama take this idea and apply it to knowledge creation. They suggest that knowledge is created as people interact over time in a shared context, in a process that can be broken up into repeated phases they call Socialisation, Externalisation, Combination, and Internalisation (SECI). I prefer to think of these phases in more simple terms, as "what I do", "what we do", "how we do it", and "what it's all about".

My point with the first paragogical principle is that instead of just focusing on how learners see themselves (e.g. as "self-directed" or "dependent" or something else), we should be asking how the learning context shapes what learners are actually able to do. Note that this includes looking at ways in which learners can contribute to shaping their learning context. In a peer-based learning context, the learner is faced with various choices and possible interactions at all levels, ranging from deciding whether or not to do a given assignment; to deciding whether to enroll in a short version of a course or longer version; to choosing to facilitate a course; to becoming involved with co-creating the global features of the learning environment by working with other facilitators, technologists, and activists.

The key aspect of paragogy is that for the most part all of these options are available all the time: the learner isn't expected to "progress" from being a confused course participant to being a facilitator, technologist, or activist. Further, instead of simply saying "so-and-so lacks the required understanding of learning, so I need to help them", a facilitator should keep in mind that there are likely to be contextual features of the learning environment that are "blocking" self-directed learning -- possibly these include features that block the ability of learners to make adjustments on their own behalf, or to ask for help. This doesn't say what to do, exactly, but it invites all participants to come to the learning process with an open mind, not just an attitude that reduces everything to questions of "maturity".

To-Do
Taken from instructor's comments in his Collaborative Lesson Planning journal.
 * The next question would be: how to design a course or "lesson" that embodies paragogic principles. 6 October 2010
 * I still need to make some actual *lesson plan*. I do have some ideas in mind based on my DIY Math experiences and I think I just need to bang them out. I'll plan to post this stuff on Wikiversity (at least the first drafts will go there). 18 October 2010
 * I could add another layer of background later as part of an "assessment" phase, but this seems good for an "introduction" phase. 19 October 2010

Suggestions from Others
 * My first comment would be to make sure you're giving credit to others for their ideas, unless of course, they are all yours. from, Charles Jeffrey Danoff, Wikiversity User: Charles Jeffrey Danoff, comment on this version of this document. (done)
 * Done: I added a couple more sections of background and some key references -- which massively improves the write-up, I feel! by, Joe
 * The practice of Agile software development has a practice similar to what you describe. Collaborators have daily and weekly 'scrums' where they check in and explain what they are working on and where they have had difficulty. One interesting difference is that instead of herding cats, stakeholders are categorized as 'pigs and chickens.' from, Brylie Oxley, Wikiversity User: Brylie (Compare the video game "Angry Birds": “Players take control of a flock of birds that are attempting to retrieve their eggs from a group of evil pigs that have stolen them." -- from Wikipedia's entry on the game, care of the NYT 07/11/2010.)
 * Joe: I would love to see your lesson plan. You have some good ideas. Thanks. Dr. King

cf.

 * Instructor's earlier writing on paragogy.
 * Instructor's critique of a pilot study putting paragogical-esque principles into action with his P2PU course DIY Math, in Crowdsourcing a Personalized Learning Environment for Mathematics, pages 13 - 22.

title: User:Arided/Paragogy


text:
User:Egm6321.f10.team6.Kim.MK/hw4

= Problem 2 Verify exactness of Bessel Equation =

Given

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

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} x^{2} y'' + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.1)
 *
 * }

(a) (1) & (2) lecture note p. 15_3

 * 2nd condition of exactness

(b) (2) lecture note p. 22_4
where $$\begin{align} f_i = \frac{\partial F}{\partial y^{(i)}} ,i= 0,1,2 \end{align}$$

2) If (2.1) is not exact, see whether it can be made exact using Integrating Factor Method with $$\begin{align} h(x,y)=x^{m} y^{n}\end{align}$$

 * See lecture note p. 19_1, p.22_4

(a)
From (2.1), (2.2)

(2.6) substitutes for (2.2)

if $$\begin{align}{\nu}^2 = x^{2} +1 \end{align}$$, (2.1) is exact.

From (2.3)

(2.8) substitutes for (2.3)

This satisfies the 2nd condition of exactness.

(b)

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

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} F(x,y,y',y) = x^{2} y + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.10)
 *
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} &f_0 = \frac{\partial F }{\partial y^{(0)}} = x^2 - {\nu}^2\\ &f_1 = \frac{\partial F }{\partial y^{(1)}} = x\\ &f_2 = \frac{\partial F }{\partial y^{(2)}} = x^2\\ \end{align}$$

(2.11)
 *
 * }

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

 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} \frac{\mathrm{d} f_1 }{\mathrm{d} x} =1 {~}{~}{~}{~}{~} \frac{\mathrm{d^2} f_2 }{\mathrm{d} x^2} =2 \end{align}$$

(2.12)
 *
 * }

Therefore,

As same as (a), if $$\begin{align}{\nu}^2 = x^{2} +1 \end{align}$$, (2.1) is exact.

2)
Verify exactness of (2.1) using 2nd condition of exactness (2.2), (2.3)

(2.18) substitutes for (2.3)

To satisfy (2.19), n=0.

(2.16), (2.17) substitute for (2.2)

Solve the quadratic equation for m,

Thus, when $$\begin{align} m= \frac{-2 {\pm} \sqrt{{\nu}^{2}-x^{2}}}{2}, n=0\end{align}$$ (2.14) is exact.

title: User:Egm6321.f10.team6.Kim.MK/hw4


text:
User:Egm6321.f10.team3.cook/HW4

=Problem 3: Find $$ \chi $$ using trial solution=

Given
Method of trial solution, see King p 513, A5.3. Where $$ r$$ are the roots given by page 25-2 of the lecture notes
 * {| style="width:100%" border="0"

$$ {r}_{1,2} = \pm  \frac{1}{k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (1)
 * }

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

$$ {r}_{3,4} = \pm  \frac{i}{k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (2)
 * }

where $$ i $$ is the imaginary number and $$ k =  \sqrt[4]{ \frac{EI}{m { \omega}^{2}}} $$ given in pg 25-1 of the lecture notes

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

\chi (x) = {e}^{rx} $$ $$
 * $$\displaystyle (3)
 * style= |
 * }

Solution
First we solve $$\displaystyle (3) $$ for the root in $$\displaystyle (2) $$ using the following relationship

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

{e}^{i \theta} = \cos( \theta) + i\sin( \theta) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (4)
 * style= |
 * }

The solution to $$\displaystyle (3) $$ becomes

 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{\chi (x)}_{1,2}= {e}^{\frac{x}{k}}+{e}^{-\frac{x}{k}} = \cos( \frac{x}{k}) + i\sin( \frac{x}{k})+ \frac{1}{\cos( \frac{x}{k}) + i\sin( \frac{x}{k}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (5)
 * style= |
 * }

Next we solve $$\displaystyle (3) $$ for the root in $$\displaystyle (1) $$ using the following relationship

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

{e}^{\theta} +{e}^{-\theta} = 2\cosh \theta $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)
 * style= |
 * }

The solution to $$\displaystyle (3) $$ becomes

 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{\chi (x)}_{3,4}= {e}^{\frac{x}{k}}+{e}^{-\frac{x}{k}} = 2\cosh(\frac{x}{k}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)
 * style= |
 * }

Using superposition, the solution to $$\displaystyle (3) $$ is

 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

\chi (x)= {\chi (x)}_{1,2} + {\chi (x)}_{3,4} = \cos( \frac{x}{k}) + i\sin( \frac{x}{k})+ \frac{1}{\cos( \frac{x}{k}) + i\sin( \frac{x}{k}} + 2\cosh(\frac{x}{k}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (6)
 * style= |
 * }

title: User:Egm6321.f10.team3.cook/HW4


text:
User talk:Arided

<< User:Arided

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If you have time, I'd like your thoughts on ...
a log I have been keeping on my CELTA studies. Feel free to wikignome or whatever else if you feel so inclined. --Charles Jeffrey Danoff 05:31, 6 December 2010 (UTC)

title: User talk:Arided


text:
User:Egm6321.f10.team03/Hwk4

=Problem 1: Solution of a L2-ODE-VC=

Given
The following Linear second order ODE with Varying Coefficients is given:
 * {| style="width:100%" border="0"

$$\,F=2xy+\left( {{x}^{2}}-\sin (x) \right){y}'+\cos (x){{y}'}'=0$$
 * $$\,(Eq. 1.1) $$
 * style= |
 * }

Find
It has to be shown that Eq. 1.1 is exact (or either can be made exact), the function $$\Phi $$ has to be found and the solution for y needs to be calculated.

Exactness condition 1
The first exactness conditon is the one where the form of the given function F in Eq. 1.1 has to match the pattern $$g(x,y,p)+f(x,y,p)y''=0$$. Looking at Eq. 1.1 it can be seen that:
 * {| style="width:100%" border="0"

$$\underbrace{2xy+\left( {{x}^{2}}-\sin (x) \right){y}'}_{g(x,y,p)}+\underbrace{\cos \left( x \right)}_{f(x,y,p)}{{y}'}'=0$$ and thus the first condition of exactness is met.
 * $$\,(Eq. 1.2) $$
 * style= |
 * }

Exactness condition 2.1
As known, the second exactness condition for an L2-ODE has two sub-conditions, the first one says that:
 * {| style="width:100%" border="0"

$${{f}_{xx}}+2p{{f}_{xy}}+{{p}^{2}}{{f}_{yy}}\overset{!}{\mathop{=}}\,{{g}_{xp}}+p{{g}_{yp}}-{{g}_{y}}$$ where in this particular problem the single values are:
 * $$\,(Eq. 1.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{f}_{x}}=-\sin \left( x \right) \\ & {{f}_{xx}}=-\cos \left( x \right) \\ & {{f}_{xy}}={{f}_{xp}}={{f}_{yy}}={{f}_{y}}={{f}_{yp}}=0 \\ \end{align}$$

and
 * $$\,(Eq.set 1.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_{x}}=2y+p\left( 2x-\cos \left( x \right) \right) \\ & {{g}_{y}}=2x \\ & {{g}_{xp}}=2x-\cos \left( x \right) \\ & {{g}_{yp}}={{g}_{pp}}=0 \\ \end{align}$$

Inserting the values of Eq.set 1.4 and 1.5 into Eq. 1.3 the first sub-condition of the second exactness condition becomes:
 * $$\,(Eq.set 1.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=2x-\cos \left( x \right)-2x$$ which is equal to
 * $$\,(Eq. 1.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=-\cos \left( x \right)$$ and thus the first sub-condition is also met.
 * $$\,(Eq. 1.7) $$
 * style= |
 * }

Exactness condition 2.2
The second sub-condition of the second exactness condition for L2-ODE's, says that:
 * {| style="width:100%" border="0"

$${{f}_{xp}}+p{{f}_{yp}}+2{{f}_{y}}\overset{!}{\mathop{=}}\,{{g}_{pp}}$$ inserting the appropriate values from Eq.set 1.4 and 1.5 in to Eq. 1.8 yields to:
 * $$\,(Eq. 1.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,0=0$$ and which clearly states that the left and right side of Eq. 1.8 are equal. '''Thus also the second sub-condition is met and therefore the given function Equation in Eq. 1.1 is exact!'''
 * $$\,(Eq. 1.9) $$
 * style= |
 * }

Find $$\Phi $$
From Eq. (3) of the lecture notes in pp. 15-3, we know that function $$\Phi $$ is composed of the following sum:
 * {| style="width:100%" border="0"

$$\Phi =h(x,y)+\int{f\left( x,y,p \right)}\,dp$$ The second term of Eq. 1.0 can be easily calculated to be:
 * $$\,(Eq. 1.10) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{f\left( x,y,p \right)}\,dp=\cos \left( x \right)p$$ And thus the function $$\Phi$$ can be expressed as:
 * $$\,(Eq. 1.11) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\Phi =h\left( x,y \right)+\cos \left( x \right)p$$ Now, from Eq. (3) of the lecture notes in pp. 15-2, it's known that
 * $$\,(Eq. 1.12) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}$$ which in our specific case is equal to
 * $$\,(Eq. 1.13) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}=2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p$$ If we take the derivatives of $$\Phi$$ in Eq. 1.12 with respect to x and with respect to y, we get:
 * $$\,(Eq. 1.14) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{\Phi }_{x}}={{h}_{x}}-p\cdot \sin \left( x \right)$$
 * $$\,(Eq. 1.15) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{\Phi }_{y}}={{h}_{y}}$$ Inserting Eq. 1.15 and Eq. 1.16 into Eq. 1.13 enables to write Eq. 1.14 as
 * $$\,(Eq. 1.16) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p={{h}_{x}}-p\cdot \sin \left( x \right)+{{h}_{y}}p={{h}_{x}}+\left( {{h}_{y}}-\sin \left( x \right) \right)p$$ and by comparison of the left and right side of the Eq. 1.17, it can be seen that:
 *
$$\,(Eq. 1.17) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{x}}=2xy$$
 *
$$\,(Eq. 1.18) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{y}}={{x}^{2}}$$ The function h(x,y) can be obtained by integrating $${{h}_{x}}$$ with respect to x, that for our particular case yields to:
 *
$$\,(Eq. 1.19) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{{{h}_{x}}\,dx}=h\left( x,y \right)={{x}^{2}}y+{{k}_{1}}(y)$$ If we differentiate Eq. 1.20 with respect to y and compare the result with Eq. 1.19, then the value for $${{k}_{1}}(y)$$ can be found:
 *
$$\,(Eq. 1.20) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{dh\left( x,y \right)}{dy}={{x}^{2}}+{{k}_{1}}^{\prime }(y)$$
 *
$$\,(Eq. 1.21) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{h}_{y}}={{x}^{2}}+{{k}_{1}}^{\prime }(y)\overset{!}{\mathop{=}}\,{{x}^{2}}$$ thus,
 *
$$\,(Eq. 1.22) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{k}_{1}}^{\prime }\left( y \right)=0\Rightarrow {{k}_{1}}\left( y \right)=C$$ With the knowledge of Eq. 1.23 we can rewrite Eq. 1.20 as follows for the function h(x,y):
 *
$$\,(Eq. 1.23) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,h(x,y)={{x}^{2}}y+C$$ and thus the function $$\Phi$$ can be written as being:
 *
$$\,(Eq. 1.24) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\Phi \left( x,y,p \right)={{x}^{2}}y+\cos \left( x \right)p+C={{x}^{2}}y+\cos \left( x \right){y}'+C=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\,(Eq. 1.25) $$
 * style= |
 * }

Solve for y
At this point solving for y is not very difficult anymore, since we only need to integrate p which is the derivative of y. For that, Eq. 1.25 is rewritten as:
 * {| style="width:100%" border="0"

$$p\left( x \right)=\frac{-({{x}^{2}}y+C)}{\cos \left( x \right)}=\frac{-{{x}^{2}}y}{\cos \left( x \right)}-\frac{C}{\cos \left( x \right)}={y}'$$ Integrating Eq. 1.26 with respect to x gives the solution for y:
 *
$$\,(Eq. 1.26) $$
 * style= |
 * }

Contributing Members

 * Solved and posted by Egm6321.f10.team03.sigillo 01:01, 19 October 2010 (UTC)

 * Reviewed by Egm6321.f10.team03 03:02, 19 October 2010 (UTC) Schulze

 * Reviewed by Egm6321.f10.team03 13:19, 19 October 2010 (UTC) Hong SJ

 * Solved, Proof read and edited by Adam Franklin 02:09, 20 October 2010 (UTC)

= Problem 2 - Verify Exactness of Bessel Equation =

Given

 * Bessel Equation is given by
 * {| style="width:100%" border="0"

$$
 * $${{x}^{2}}{{y}^{''}}+x{{y}^{'}}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y=0$$
 *
$$\displaystyle (Eq. 2.1)
 *
$$\displaystyle (Eq. 2.1)
 * }
 * }

 * Where, $$\displaystyle\upsilon \in \mathbb{R}$$

Find

 * $$\displaystyle\left( 1 \right)$$ Verify Exactness of $$\displaystyle Eq.2.1$$ using 2 methods

 *
Method 1

 * Second conditions of Exactness:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * First Relation:$$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}
 * First Relation:$$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}
 *
$$\displaystyle (Eq.2.2)
 * }
 * }

 * and

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

$$ $$
 * Second relation:$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}
 * Second relation:$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}
 *
$$\displaystyle (Eq.2.3)
 * }
 * }

 *
Method 2

 * Another form of second Exactness condition for a $$\displaystyle N2-ODE$$:

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

f_0 - \frac {df_1}{dx} + \frac {d^2f_2}{dx^2}=0 $$ $$.
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq.2.4)
 * }
 * }
 * $$\displaystyle\left( 2 \right)$$ If $$\displaystyle Eq.2.1$$ not Exact, see whether it can be made Exact using Integrating Factor Method with:

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

$$
 * $$h\left( x,y \right)={{x}^{m}}{{y}^{n}}$$
 *
$$\displaystyle (Eq. 2.5)
 *
$$\displaystyle (Eq. 2.5)
 * }
 * }

 * That is,find $$\displaystyle (m,n) $$ such that Eq 2.1 is exact.

Solution

 *
PART 1: Verify Exactness of Bessel Equation

 * The first condition for exactness of $$\displaystyle N2-ODE$$ is that it is in the following form:
 * {| style="width:100%" border="0" align="left"

F(x,y,p) = f(x,y,p)y'' + g(x,y,p)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2.6)
 * }
 * }

 * where

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

p = y' $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2.7)
 * }
 * }

 * Now comparing Eq.2.1 with the form of Eq.2.6,we can see that:

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

$$
 * $$\displaystyle f(x,y,p)={{x}^{2}}$$
 *
$$\displaystyle (Eq.2.8)
 *
$$\displaystyle (Eq.2.8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$
 * $$\displaystyle g(x,y,p)=x{p}+\left( {{x}^{2}}-{{\upsilon }^{2}} \right)y$$
 *
$$\displaystyle (Eq.2.9)
 *
$$\displaystyle (Eq.2.9)
 * }
 * }

 * Hence the first condition for exactness is satisfied.

 * Now verify second condition of Exactness:

 * (a) Using Method 1- Eq.2.2 and Eq.2.3

 * Taking derivative of Eq.2.8 yields:

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

\begin{array}{*{20}l} {f_x = 2x} & {f_{xx}  = 2} & {f_{xy}  = 0} & {f_{xp}  = 0 }  \\ {f_y = 0} & {f_{yy}  = 0} & {f_{yp}  = 0}\\ \end{array} $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }
 * Similarly taking derivative of Eq.2.9 yields:

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

\begin{array}{*{20}l} {g_x = p+2xy} & {g_{xp} = 1} & {{g}_{y}}={{x}^{2}}-{{\upsilon }^{2}} & {g_{yp}  = 0 }  \\ {g_p = x} & {g_{pp}  = 0} \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting the above values to Eq.2.3 (first relation) and Eq.2.4 (second relation):

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

 * $$2+2p\times 0+{{p}^{2}}\times 0=1+p\times 0-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 * }
 * }
 * }

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

$$
 * $$\Rightarrow 2=1-\left( {{x}^{2}}-{{\upsilon }^{2}} \right)$$
 *
$$\displaystyle (Eq.2.10)
 *
$$\displaystyle (Eq.2.10)
 * }
 * }

 * So first relation for second exactness condition is not satisfied.

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

$$
 * $$0+p\times 0+2\times 0=0\Rightarrow 0=0$$, second relation is satisfied
 *
$$\displaystyle (Eq.2.11)
 *
$$\displaystyle (Eq.2.11)
 * }
 * }

 * Since the first relation is not satisfied, the Eq.2.1 is not Exact

 * (b) Using Method 2- Eq.2.4

 * Differentiating $$\displaystyle Eq.2.1$$, partially with respect to $$\displaystyle y,{{y}^{'}},{{y}^{''}}$$,

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

 * $${{f}_{0}}=\frac{\partial F}{\partial y}={{x}^{2}}-{{\upsilon }^{2}}$$
 * }
 * }
 * }

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

 * $${{f}_{1}}=\frac{\partial F}{\partial {{y}^{'}}}=x$$
 * }
 * }
 * }

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

 * $${{f}_{2}}=\frac{\partial F}{\partial {{y}^{''}}}={{x}^{2}}$$
 * }
 * }
 * }

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

 * Now, $$\frac{\partial {{f}_{1}}}{\partial x}=1$$, and $$\frac{\partial {{x}^{2}}}=2$$
 * }
 * }
 * }

 * Substittuting above values in $$\displaystyle Eq.2.4$$ yields,

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

 * $${{x}^{2}}-{{\upsilon }^{2}}-1+2\ne 0\Rightarrow $$not satisfied and hence the Eq.2.1 is not Exact.
 * }
 * }
 * }

 *
PART 2: Make Bessel Equation exact using Integrating Factor Method

 * Multiply $$\displaystyle Eq.2.1$$ by Integrating Factor ($$\displaystyle Eq.2.5$$)$$\Rightarrow $$

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

$$
 * $$\underbrace_{f}{{y}^{''}}+\underbrace{{{x}^{m+1}}{{y}^{n}}{{y}^{'}}+{{x}^{m+2}}{{y}^{n+1}}-{{x}^{m}}{{y}^{n+1}}{{\upsilon }^{2}}}_{g}=0$$
 *
$$\displaystyle (Eq.2.12)
 *
$$\displaystyle (Eq.2.12)
 * }
 * }

 * New $$\displaystyle f$$ and $$\displaystyle g$$ are:

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

$$
 * $$f\left( x,y,p \right)={{x}^{m+2}}{{y}^{n}}$$
 *
$$\displaystyle (Eq.2.13)
 *
$$\displaystyle (Eq.2.13)
 * }
 * }

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

$$
 * $$g\left( x,y,p \right)={{x}^{m+1}}{{y}^{n}}p+{{x}^{m+2}}{{y}^{n+1}}-{{x}^{m}}{{y}^{n+1}}{{\upsilon }^{2}}$$
 *
$$\displaystyle (Eq.2.14)
 *
$$\displaystyle (Eq.2.14)
 * }
 * }

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

Analyzing the second relation of second exactness condition ($$\displaystyle Eq.2.3 $$):$$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}$$, it can be noted that most of the terms drop out since $$\displaystyle f$$ is independent of $$\displaystyle p$$. Consequently all derivatives of $$\displaystyle f$$ with respect to $$\displaystyle p$$ become zero. Moreover $$\displaystyle {{g}_{pp}}=0$$ and hence $$\displaystyle Eq.2.3 $$ is simplified to,
 * }
 * }

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

$$
 * $$\displaystyle 2{{f}_{y}}=0$$
 *
$$\displaystyle (Eq.2.15)
 *
$$\displaystyle (Eq.2.15)
 * }
 * }

 * Taking derivative of $$\displaystyle (Eq.2.13)$$ with repect to $$\displaystyle y$$ and substituting to$$\displaystyle (Eq.2.15)$$ yields,

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

 * $$\displaystyle n{{x}^{m+2}}{{y}^{n-1}}=0$$
 * }
 * }
 * }

 * The above equation is true only when,

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

$$\displaystyle n=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.16)
 * }
 * }

 * As $$\displaystyle n=0 $$, the function $$\displaystyle f$$ and $$\displaystyle g$$ can be rewritten as:

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

$$
 * $$f\left( x,y,p \right)={{x}^{m+2}}$$
 *
$$\displaystyle (Eq.2.17)
 *
$$\displaystyle (Eq.2.17)
 * }
 * }

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

$$
 * $$g\left( x,y,p \right)={{x}^{m+1}}p+{{x}^{m+2}}y-{{x}^{m}}y{{\upsilon }^{2}}$$
 *
$$\displaystyle (Eq.2.18)
 *
$$\displaystyle (Eq.2.18)
 * }
 * }

 * Taking derivative of $$\displaystyle Eq.2.17$$ yields,

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

\begin{array}{*{20}l} {{f}_{x}}=\left( m+2 \right){{x}^{m+1}}, & {{f}_{xx}}=\left( m+2 \right)\left( m+1 \right){{x}^{m}}, & {f_{xy} = 0}, & {f_{xp}  = 0 }  \\ {f_y = 0}, & {f_{yy}  = 0}, & {f_{yp}  = 0}\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Similarly taking derivative of $$\displaystyle Eq.2.18$$ yields,

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

\begin{array}{*{20}l} {{g}_{x}}=\left( m+1 \right){{x}^{m}}p+\left( m+2 \right){{x}^{m+1}}y-m{{x}^{m-1}}y{{\upsilon }^{2}}\\ {{g}_{xp}}=\left( m+1 \right){{x}^{m}}\\ {{g}_{y}}={{x}^{m+2}}-{{x}^{m}}{{\upsilon }^{2}}\\ {g_{yp} = 0} \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Substituting above differentials in $$\displaystyle Eq.2.2$$ gives,

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

 * $$\left( m+2 \right)\left( m+1 \right){{x}^{m}}+0+0=\left( m+1 \right){{x}^{m}}+0-\left( {{x}^{m+2}}-{{x}^{m}}{{\upsilon }^{2}} \right)$$
 * }
 * }
 * }

 * Simplifying,

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

 * $$\left( m+2 \right)\left( m+1 \right)=m+1+{{\upsilon }^{2}}-{{x}^{2}}$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{m}^{2}}+2m+\left( {{x}^{2}}-{{\upsilon }^{2}}+1 \right)=0$$
 * }
 * }
 * }

 * $$\displaystyle m$$ is obtained by solving the above quadratic equation:

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

$$m=-1\pm \sqrt{\left( {{\upsilon }^{2}}-{{x}^{2}} \right)}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.19)
 * }
 * }

 * So the integration factor is obtained by substituting $$\displaystyle Eq. 2.16 $$ and $$\displaystyle Eq. 2.19 $$ in $$\displaystyle Eq. 2.5 $$:

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

$$h={{x}^{-1\pm \sqrt{\left( {{\upsilon }^{2}}-{{x}^{2}} \right)}}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 2.20)
 * }
 * }

 * Thus Bessel Equation can be made exact using the above integrating factor.

Contributing Members

 * Solved and Posted by Egm6321.f10.team3.Sudheesh 01:51, 19 October 2010 (UTC)

 * Solved and Proof read by Adam Franklin 02:50, 20 October 2010 (UTC)

=Problem 3: Solution of Free Vibration Equation for Euler bernoulli Beam using Method of Trial Solution=

Given
Free vibration equation for Euler Bernoulli beam is given by,

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

 * $$\displaystyle -EI\frac{{{\partial }^{4}}u}{\partial {{x}^{4}}}=m\frac{{{\partial }^{2}}u}{\partial {{t}^{2}}}$$
 * style= |
 * }
 * }

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

 * Assuming, $$u(x,t)=\chi \left( x \right)T\left( t \right)$$, and using the separation of variable technique,we obtain the following relation:
 * style= |
 * }
 * }

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

 * $${{k}^{4}}{{\chi }^{\left( 4 \right)}}=\chi $$
 * style= |
 * }
 * }

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

 * Where, $${k}^{4}=\frac{EI}{m{{\omega }^{2}}}$$
 * style= |
 * }
 * }

Now using Method of trial solution(see King p 513, A5.3), that is assuming $$X\left( x \right)={{e}^{rx}}$$,we obtain undetermined coefficient(roots) $$\displaystyle r$$ as given in [[media:2010_10_14_14_56_10.djvu|page 25-2]] of the lecture notes

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

$$ {r}_{1,2} = \pm  \frac{1}{k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (3.1)
 * }

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

$$ {r}_{3,4} = \pm  \frac{i}{k} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (3.2)
 * }

where $$ i $$ is the imaginary number and $$ k =  \sqrt[4]{ \frac{EI}{m { \omega}^{2}}} $$ given in pg 25-1 of the lecture notes

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

\chi (x) = {e}^{rx} $$ $$ in terms of sin, cos, sinh and cosh
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (3.3)
 * style= |
 * }

Solution
First we solve $$\displaystyle (3.3) $$ for the root in $$\displaystyle (3.2) $$ using the following relationships

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

{e}^{i \theta} = \cos( \theta) + i\sin( \theta) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (3.4)
 * style= |
 * }

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

$$
 * $$\displaystyle {{e}^{-i\theta }}=\cos (\theta )-i\sin (\theta )$$
 *
$$\displaystyle (3.5)
 *
$$\displaystyle (3.5)
 * style= |
 * }

The solution to $$\displaystyle (3.3) $$ becomes

Next we solve $$\displaystyle (3.3) $$ for the root in $$\displaystyle (3.1) $$ using the following relationships

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

$$
 * $$\displaystyle {{e}^{\theta }}=\cosh (\theta )+\sinh (\theta )$$
 *
$$\displaystyle (3.7)
 *
$$\displaystyle (3.7)
 * style= |
 * }

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

$$
 * $$\displaystyle {{e}^{-\theta }}=\cosh (\theta )-\sinh (\theta )$$
 *
$$\displaystyle (3.8)
 *
$$\displaystyle (3.8)
 * style= |
 * }

The solution to $$\displaystyle (3.3) $$ becomes

Using superposition, the solution to $$\displaystyle (3.3) $$ is

Contributing Members

 * Solved and proofread by James Roark 11:42, 19 October 2010 (UTC)
 * Posted by Chris Cook --98.71.213.114 22:18, 18 October 2010 (UTC)
 * Solved and proofread by Adam Franklin 02:55, 20 October 2010 (UTC)
 * Solved,proofread Egm6321.f10.team03.sigillo 03:57, 20 October 2010 (UTC)
 * Solved, edited and proofread by Egm6321.f10.team3.Sudheesh 14:56, 20 October 2010 (UTC)

=Contributing Team Members=
 * Egm6321.f10.team3.Sudheesh 14:56, 20 October 2010 (UTC)- Solved and posted problem 2 and proofread problem 3

title: User:Egm6321.f10.team03/Hwk4


text:
User:Egm6321.f10.team03.sigillo/Hwk4

=Problem 1: Solution of a L2-ODE-VC=

Given
The following Linear second order ODE with Varying Coefficients is given:
 * {| style="width:100%" border="0"

$$\,F=2xy+\left( {{x}^{2}}-\sin (x) \right){y}'+\cos (x){{y}'}'=0$$
 * $$\,(Eq. 1.1) $$
 * style= |
 * }

Find
It has to be shown that Eq. 1.1 is exact (or either can be made exact), the function $$\Phi $$ has to be found and the solution for y needs to be calculated.

Exactness condition 1
The first exactness conditon is the one where the form of the given function F in Eq. 1.1 has to match the pattern $$g(x,y,p)+f(x,y,p)y''=0$$. Looking at Eq. 1.1 it can be seen that:
 * {| style="width:100%" border="0"

$$\underbrace{2xy+\left( {{x}^{2}}-\sin (x) \right){y}'}_{g(x,y,p)}+\underbrace{\cos \left( x \right)}_{f(x,y,p)}{{y}'}'=0$$ and thus the first condition of exactness is met.
 * $$\,(Eq. 1.2) $$
 * style= |
 * }

Exactness condition 2.1
As known, the second exactness condition for an L2-ODE has to sub-conditions, the first one says that:
 * {| style="width:100%" border="0"

$${{f}_{xx}}+2p{{f}_{xy}}+{{p}^{2}}{{f}_{yy}}\overset{!}{\mathop{=}}\,{{g}_{xp}}+p{{g}_{yp}}+{{g}_{y}}$$ where in this particular problem the single values are:
 * $$\,(Eq. 1.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{f}_{x}}=-\sin \left( x \right) \\ & {{f}_{xx}}=-\cos \left( x \right) \\ & {{f}_{xy}}={{f}_{xp}}={{f}_{yy}}={{f}_{y}}={{f}_{yp}}=0 \\ \end{align}$$

and
 * $$\,(Eq.set 1.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\begin{align} & {{g}_{x}}=2y+p\left( 2x-\cos \left( x \right) \right) \\ & {{g}_{y}}=2x \\ & {{g}_{xp}}=2x-\cos \left( x \right) \\ & {{g}_{yp}}={{g}_{pp}}=0 \\ \end{align}$$

Inserting the values of Eq.set 1.4 and 1.5 into Eq. 1.3 the first sub-condition of the second exactness condition becomes:
 * $$\,(Eq.set 1.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=2x-\cos \left( x \right)-2x$$ which is equal to
 * $$\,(Eq. 1.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\cos \left( x \right)=-\cos \left( x \right)$$ and thus the first sub-condition is also met.
 * $$\,(Eq. 1.7) $$
 * style= |
 * }

Exactness condition 2.2
The second sub-conditino of the second exactness condition for L2-ODE's, says that:
 * {| style="width:100%" border="0"

$${{f}_{xp}}+p{{f}_{yp}}+2{{f}_{y}}\overset{!}{\mathop{=}}\,{{g}_{pp}}$$ inserting the appropriate values from Eq.set 1.4 and 1.5 in to Eq. 1.8 yields to:
 * $$\,(Eq. 1.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,0=0$$ and which clearly states that the left and right side of Eq. 1.8 are equal. '''Thus also the second sub-condition is met and therefore the given function Equation in Eq. 1.1 is exact!'''
 * $$\,(Eq. 1.9) $$
 * style= |
 * }

Find $$\Phi $$
From Eq. (3) of the lecture notes in pp. 15-3, we know that function $$\Phi $$ is composed of the following sum:
 * {| style="width:100%" border="0"

$$\Phi =h(x,y)+\int{f\left( x,y,p \right)}\,dp$$ The second term of Eq. 1.0 can be easily calculated to be:
 * $$\,(Eq. 1.10) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{f\left( x,y,p \right)}\,dp=\cos \left( x \right)p$$ And thus the function $$\Phi$$ can be expressed as:
 * $$\,(Eq. 1.11) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\Phi =h\left( x,y \right)+\cos \left( x \right)p$$ Now, from Eq. (3) of the lecture notes in pp. 15-2, it's know that
 * $$\,(Eq. 1.12) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}$$ which in our specific case is equal to
 * $$\,(Eq. 1.13) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$g(x,y,p)={{\Phi }_{y}}p+{{\Phi }_{x}}=2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p$$ If we take the derivatives of $$\Phi$$ in Eq. 1.12 with respect to x and with respect to y, we get:
 * $$\,(Eq. 1.14) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{\Phi }_{x}}={{h}_{x}}-p\cdot \sin \left( x \right)$$
 * $$\,(Eq. 1.15) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{\Phi }_{y}}={{h}_{y}}$$ Inserting Eq. 1.15 and Eq. 1.16 into Eq. 1.13 enables to write Eq. 1.14 as
 * $$\,(Eq. 1.16) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$2xy+\left( {{x}^{2}}-\sin \left( x \right) \right)p={{h}_{x}}-p\cdot \sin \left( x \right)+{{h}_{y}}p={{h}_{x}}+\left( {{h}_{y}}-\sin \left( x \right) \right)p$$ and by comparison of the left and right side of the Eq. 1.17, it can be seen that:
 *
$$\,(Eq. 1.17) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{x}}=2xy$$
 *
$$\,(Eq. 1.18) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{y}}={{x}^{2}}$$ The function h(x,y) can also be obtained by integrating $${{h}_{x}}$$ with respect to x, that for our particular case yields to:
 *
$$\,(Eq. 1.19) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{{{h}_{x}}\,dx}=h\left( x,y \right)={{x}^{2}}y+{{k}_{1}}(y)$$ If we differentiate Eq. 1.20 with respect to y and compare the result with Eq. 1.19, then the value for $${{k}_{1}}(y)$$ can be found:
 *
$$\,(Eq. 1.20) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{dh\left( x,y \right)}{dy}={{x}^{2}}+{{k}_{1}}^{\prime }(y)$$
 *
$$\,(Eq. 1.21) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{h}_{y}}={{x}^{2}}+{{k}_{1}}^{\prime }(y)\overset{!}{\mathop{=}}\,{{x}^{2}}$$ thus,
 *
$$\,(Eq. 1.22) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{k}_{1}}^{\prime }\left( y \right)=0\Rightarrow {{k}_{1}}\left( y \right)=C$$ With the knowledge of Eq. 1.23 we can rewrite Eq. 1.20 as follows for the functino h(x,y):
 *
$$\,(Eq. 1.23) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,h(x,y)={{x}^{2}}y$$ and thus the function $$\Phi$$ can be written as being:
 *
$$\,(Eq. 1.24) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\Phi \left( x,y,p \right)={{x}^{2}}y+\cos \left( x \right)p+C={{x}^{2}}y+\cos \left( x \right){y}'+C=0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\,(Eq. 1.25) $$
 * style= |
 * }

Sove for y
At this point solving for y is not very difficult anymore, since we only need to integrate p which is the derivative of y. For that, Eq. 1.25 is rewritten as:
 * {| style="width:100%" border="0"

$$p\left( x \right)=\frac{-({{x}^{2}}y+C)}{\cos \left( x \right)}=\frac{-{{x}^{2}}y}{\cos \left( x \right)}-\frac{C}{\cos \left( x \right)}={y}'$$ Integrating Eq. 1.26 with respect to x gives the solution for y:
 *
$$\,(Eq. 1.26) $$
 * style= |
 * }

Contributing Members
Solved and posted by Egm6321.f10.team03.sigillo 01:01, 19 October 2010 (UTC)

title: User:Egm6321.f10.team03.sigillo/Hwk4


text:
User:Arided/ShortPrecalculus

Here is an outline from http://www.wmueller.com/precalculus/ that seems super useful:

 * 1) What Is Precalculus?
 * 2) Functions and Their Changes
 * 3) Functions as Models
 * 4) Families of Functions
 * 5) Functions Defined by Data
 * 6) New Functions From Old
 * 7) Advanced Numbers & Functions

(Details at the link above.)

title: User:Arided/ShortPrecalculus


text:
User:Arided/ShortCalculus

This is based on the outline at http://www.sosmath.com/calculus/calculus.html -- not necessarily the most reputable source but it's a start.

 * 1) sequences & series
 * 2) limit and continuity
 * 3) differentiation
 * 4) integration
 * 5) techniques of integration
 * 6) taylor polynomials and power series

title: User:Arided/ShortCalculus


text:
User:Arided/ShortAdvancedCalculus

This is based on the outlines at http://tutorial.math.lamar.edu/Classes/CalcII/CalcII.aspx and http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx -- Note that you can view this as two classes:

 * 1) Integration Techniques
 * 2) Applications of Integrals
 * 3) Parametric Equations and Polar Coordinates
 * 4) Sequences and Series (including Fourier Series)
 * 5) Vectors
 * 6) Three Dimensional Space and Very Short Linear Algebra

...

 * 1) Partial Derivatives
 * 2) Applications of Partial Derivatives
 * 3) Multiple Integrals
 * 4) Line Integrals
 * 5) Surface Integrals
 * 6) Very Short Differential Equations

Or as one very busy class :) -- here the 1st week would be a "zeroth" week of review? Or we could leave off the "short differential equations" bit at the end and work a bit more of it in to the weeks just before that.

 * 1) Integration Techniques and Applications of Integrals
 * 2) Parametric Equations and Polar Coordinates
 * 3) Sequences and Series (including Fourier Series)
 * 4) Vectors, Three Dimensional Space and Very Short Linear Algebra
 * 5) Partial Derivatives and Applications of Partial Derivatives
 * 6) Multiple Integrals, Line Integrals, and Surface Integrals
 * 7) Very Short Differential Equations

title: User:Arided/ShortAdvancedCalculus


text:
User:Arided/MathForProgrammers

Maybe this is a case where we would want a couple of courses in a sequence... This comes from "Concrete Mathematics", which doesn't immediately lend itself to being divided up into bunches of 6... So, something to think about. Seems better to add more topics from "programming" and spread it out over 2 sections. E.g. mix in some "intro to functional programming" material, such as the stuff from here: http://www.cl.cam.ac.uk/teaching/Lectures/funprog-jrh-1996/ (we can let people choose their own languages). Right now this is sort of randomly thrown together:

 * 1) Effective Functional Programming
 * 2) Lambda calculus as a programming language
 * 3) Recurrent problems
 * 4) Sums
 * 5) Integer Functions
 * 6) Number Theory

...

 * 1) Binomial Coefficients
 * 2) Special Numbers
 * 3) Generating Functions
 * 4) Discrete Probability
 * 5) Asymptotics
 * 6) Proving programs correct

=See Also= From the talk page: a relevant topic has come up on stackoverflow: http://stackoverflow.com/questions/11743/useful-math-for-programmers

title: User:Arided/MathForProgrammers


text:
User:Arided/ShortStatistics

This outline is care of Timothy Teravainen.

 * 1) probability
 * 2) sampling methods
 * 3) "easy" machine learning things like nearest neighbors (pt 1)
 * 4) "easy" machine learning things pt 2
 * 5) bootstrap-type stuff
 * 6) central limit theorem

=See Also= The talk page for this article, as well as Felipe Ortega's proposal to run a short intro to stats on P2PU using GNU R.

title: User:Arided/ShortStatistics


title: File:JE1.png


text:
Wikiversity talk:Journal of the future

Comments not yet incorporated
New Criterion: Science manuscripts are not just an aggregation of individual figures, and the scientist of the future will make the process of data acquisition and analysis an open process. The journal of the future will show no prejudice when deciding to publish data and processes that have been conducted in the open. --Kubke 22:58, 19 October 2010 (UTC)
 * Rephrased Criterion 6 — does this fit? --Daniel Mietchen (talk) 14:36, 20 October 2010 (UTC)
 * I take it you mean the 'Dynamic' criterion (now 1)? Yes, I like the idea of the inclusion of workflows. I still do wonder though: if I make my research process and data and analysis open all along the way, how will that affect my chances of getting the work into a journal? Will it pass the test of the questions journals ask as to whether what you are submitting constitutes new, unpublished data? I am also getting that question asked to me almost every time I say that I am making my embryology work 'open'. I dont have the answer. --Kubke 10:44, 21 October 2010 (UTC)

Comments already incorporated
"The scientific journal of the future promotes the publication of research data and detailed methodologies, including code, as the basis of any publication." It might be worth considering whether the term should be 'promotes' or 'supports'. By promote I understand 'encourage'; by 'supports' I understand 'provides the solution (e.g., archival) but it can also be interpreted as 'encourage'. Or, should the journal of the future 'require'?--Kubke 06:58, 19 October 2010 (UTC)
 * Done. --Daniel Mietchen (talk) 07:23, 19 October 2010 (UTC)

"It also provides a space for rapid and non-conventional types of publications, e.g. by way of post-publication peer review. " there may be two different things mixed up here. For example PLoS ONE offers post publication peer review through comments, but also requires pre-publication peer-review. So from the 'types' of publications that are non-conventional I would think more of the minimum publishable unit (eg a figure), preliminary data, etc. The pre/post publication peer review to me is a separate issue. Perhaps move the 'eg by way [...] review' to the end of the previous sentence (after '[...] encompassing the University community'. --Kubke 07:06, 19 October 2010 (UTC)
 * Done. --Daniel Mietchen (talk) 07:23, 19 October 2010 (UTC)

"The scientific journal of the future, therefore supports initiatives for the on-wiki development of [...]" again I think there may be ambiguity with the word 'supports'. I interpret it as 'provides solution' whereas it can also be interpreted as 'encourage'. (said Kubke)
 * Done. --Daniel Mietchen (talk) 07:27, 19 October 2010 (UTC)

'online audiences are not homogeneous': scope should not be limited to 'topic' or 'methodology' but language should be accessible outside the field of expertise (plain language while not loosing precision) (said Kubke)
 * Tried it, please check. --Daniel Mietchen (talk) 08:14, 19 October 2010 (UTC)
 * Rephrased, please check --Kubke 08:30, 19 October 2010 (UTC)
 * OK, let's leave it at that. --Daniel Mietchen (talk) 09:16, 19 October 2010 (UTC)

Practical test
Moved here from main page.

The criteria shall be put to a first practical test by being applied to a range of publishing platforms. The scoring method is still evolving. Everyone is welcome to add some more journals. Numbers in parentheses represent an estimate for the relative weight of these criteria (they add up to 100% over all criteria).

Since editing wikitables is not very convenient, I turned this one into a collaboratively editable spreadsheet, and since embedding is not possible here at Wikiversity, I did so in my lab notebook. See also blog commentary.

--Daniel Mietchen (talk) 05:26, 14 March 2011 (UTC)

Redirect to WikiJournal
I think these concepts have been integrated in WikiJournal as far as has been conceivable. As such, I think we should redirect this page to there, with a mention of these proposals at WikiJournal User Group/Milestones, such as: Mikael Häggström (discuss • contribs) 11:57, 14 August 2018 (UTC)
 * 20 October 2010: Core criteria described

title: Wikiversity talk:Journal of the future


text:
User talk:Dipaolatommaso

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:17, 19 October 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:17, 19 October 2010 (UTC)

title: User talk:Dipaolatommaso


text:
User:Kubke

M Fabiana Kubke University of Auckland New Zealand

I am a neuroethologist interested in brain evolution and development.

title: User:Kubke


text:
User:PAC2

See my wikibooks page : b:User:PAC2

title: User:PAC2


text:
User talk:Praewploy/Specs on Mobile Equipment

I PHONE 3Gs

CPU and speed: 600 MHz, 256 MB of RAM

Cache size: L1 cache size of 32 KB I-Cache, 32 KB D-Cache, L2 cache size of 256 KB

Bus speed: 150mhz

Storage: 8GB, 16GB, 32 G

IPHONE 4

CPU: Apple A4 (ARM Cortex-A8)

CPU Speed: 1 GHz

Onboard RAM: 256 MB

Maximum RAM: 256 MB

L1 cache 32 KB Instruction + 32KB Data

L2 cache 640 KB

 * Storage

Flash Drive: 16/32 GB

Memory 512 MB eDRAM

IPAD

CPU and Speed: 1.67 GHz

Bus speed:167 MHz

Cache Size: 512 KB

Storage: 16GB, 32GB, 64GB

Which device is more technologically advance?

According to the information above, IPAD is the most advance device. Both, CPU speed and bus speed, of an IPAD are faster than the two IPHONES. This means an IPAD will basically run and do the processing a lot faster. The IPAD's cache size is also bigger than an IPHONE 3Gs. All three devices has a variety of storage capacity. However, an IPAD has up to 64GB. The users can save and store a large amount of datas, files, musics, movies, photos, games and so on.