text:
User:Egm4313.s12.team13.rlh.r2

Problem Statement
5. Find a general solution. Check by substitution.

$$  \; {y}''+2\pi {y}'+\pi ^{2}y=0 $$

Solution
$$  \; {y}''+2\pi {y}'+\pi ^{2}y= \frac{d^2y}{dx^2}\ + 2 \pi \frac{dy}{dt}\ + \pi ^{2}y=0   $$ (4.0a)

If $$ \; \frac{d}{dx}\ = \lambda $$,

The homogeneous linear ODE with constant coefficients becomes:

$$ \; \lambda ^{2}+2\pi \lambda +\pi ^{2}=0 $$ (4.1a)

By using the quadratic formula, or in this case factoring, we find the solutions of the equation:

Solving for $$ \; \lambda $$,

$$ \; (\lambda +\pi) (\lambda +\pi)=0 $$ (4.2a)

$$ \; \lambda =-\pi, -\pi $$

General Solution:

General solution in the case of distinct real roots:

$$  \; y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x}  $$ (4.3a)

So the general solution becomes:

$$  \; y=c_1e ^ {-\pi x} + c_2xe ^ {-\pi x}= e ^ {-\pi x} (c_1+c_2x) $$ (4.4a)

Check by Substitution:

To check by substitution, you need to find $$  \; y, y', y'' $$ and plug them into the original ODE.

$$  \; y= e ^ {-\pi x} (c_1+c_2x) $$

$$  \; y'= -\pi c_1e ^ {-\pi x} + c_2 (e ^ {-\pi x} -\pi xe ^ {-\pi x})= - \pi e ^{- \pi x} (c_1 + c_2x) + c_2 e^ {- \pi x} $$(4.5a)

$$  \; y''= \pi ^{2} c_1e ^ {- \pi x} - \pi c_2e ^ {-\pi x} -\pi c_2(e ^ {- \pi x} - \pi xe^ {- \pi x} )= \pi ^ {2} e^ {- \pi x} (c_1+ c_2x) -2 \pi c_2 e^ {- \pi x} $$ (4.6a)

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+2\pi {y}'+\pi ^{2}y=0 $$ :

$$  \; [ \pi ^ {2} e^ {- \pi x} (c_1 + c_2x) - 2 \pi c_2e^ {- \pi x}] + 2 \pi [ - \pi e^ {- \pi x} (c_1 + c_2x) + c_2e^ {- \pi x} ] + \pi ^ {2} [ e^ {- \pi x} (c_1+c_2x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

Problem Statement
6. Find the general solution. Check by substitution.

$$ 10y''-32y'+25.6y=0 $$

Solution
The original problem can be re-written as:
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$$  \displaystyle 10\frac{d^2y}{dx^2}-32\frac{dy}{dx}+25.6y=0 $$     (4.0b)
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then let
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$$  \displaystyle \frac{d}{dx}=\lambda $$     (4.1b) By substituting equation 4.1b into 4.0b we get:
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$$  \displaystyle 10\lambda^2-32\lambda+25.6=0 $$     (4.2b) Solve for $$ \; \lambda $$ using the quadratic equation.
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Quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (4.3b) By using equation 4.3b it is shown that $$ \lambda=\frac{8}{5}$$
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Since $$ \; \lambda$$ is a solution to the ODE then
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$$  \displaystyle y=c_1e ^ { \lambda _1 x} + c_2e ^ { \lambda _2 x} $$ (4.4b) is the solution of the ODE with real and distinct roots.
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By plugging the value of $$ \; \lambda $$ into equation 4.4b we come to the general solution.

General Solution $$ y=e^{\frac{8}{5}x}({c_1+c_2 x}) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.

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$$  \displaystyle y'=\frac{8}{5}e^{\frac{8}{5}x}(c_1+c_2 x)+c_2 e^{\frac{8}{5}x} $$     (4.5b)
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$$  \displaystyle y''=\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+\frac{16}{5}c_2 e^{\frac{8}{5}x} $$     (4.6b) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 4.5b, and 4.6b back into the original equation.
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After substitution we get:
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$$  \displaystyle \frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)-\frac{64}{25}e^{\frac{8}{5}x}(c_1+c_2 x)+32c_2e^{\frac{8}{5}x}-32c_2e^{\frac{8}{5}x}=?? $$ This equation clearly equals  zero  which proves our general solution is  correct
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Problem Statement
8. Find the general solution. Check by substitution.

$$ y''+y'+3.25y=0 $$

Solution
The original problem can be re-written as:
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$$  \displaystyle \frac{d^2y}{dx^2}+\frac{dy}{dx}+3.25y=0 $$     (8.0a)
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then let
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$$  \displaystyle \frac{d}{dx}=\lambda $$     (8.1a) By substituting equation 8.1a into 8.0a we get:
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$$  \displaystyle \lambda^2+\lambda+3.25=0 $$     (8.2a) Solve for $$ \; \lambda $$ using the quadratic equation.
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Quadratic equation:
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$$  \displaystyle \lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (8.3a) By using equation 8.3a it is shown that $$ \lambda=\frac{-1\pm i \sqrt{12}}{2}$$
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Since $$ \; \lambda$$ is a solution to the ODE then
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$$  \displaystyle y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.4a) is the solution of the ODE with real and distinct roots.
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General Solution $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$

Checking by Substitution To check by substitution, you need to find $$ \; y, y', y''$$ and plug them into the original ODE.

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$$  \displaystyle y'=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)+e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x) $$     (8.5a)
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$$  \displaystyle y''=\frac{1}{4}e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}c_1 \sin\sqrt{12}x+\sqrt{12}c_2 \cos\sqrt{12}x)$$
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$$+\frac{1}{2}e^{-\frac{1}{2}x}(-12c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$     (8.6a) Since we already know what $$ \; y $$ is equal to from our general solution we are ready to substitute the general solution, equation 8.5a, and 8.6a back into the original equation.
 *

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After substitution we get:
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$$  \displaystyle y''+y'+3.25y=0 $$ So the solution to the ODE is: $$ y=e^{-\frac{1}{2}x}(c_1 \cos\sqrt{12}x+c_2 \sin\sqrt{12}x) $$
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title: User:Egm4313.s12.team13.rlh.r2


text:
University of Florida/Egm4313/s12.team11.perez.gp/R2.3

Problem Statement (K 2011 p.59 pb. 3)
Find a general solution. Check your answer by substitution.

Given
$$ y''+6y'+8.96y=0 \! $$

Solution
We can write the above differential equation in the following form:

$$ \frac{d^2y}{dx^2}+6\frac{dy}{dx}+8.96y=0 $$

Let $$ \frac{d}{dx} = \lambda. $$

The characteristic equation of the given DE is

$$ \lambda^2+6\lambda+8.96=0 \! $$

Now, in order to solve for $$ \lambda $$, we can use the quadratic formula:

$$ \lambda=\frac{-(6)\pm \sqrt{(6)^2-4(1)(8.96)}}{2(1)} $$

$$ \lambda=\frac{-6\pm \sqrt{36-35.84}}{2} $$

$$ \lambda=\frac{-6\pm \sqrt{0.16}}{2} $$

$$ \lambda=\frac{-6\pm 0.4}{2} $$

Therefore, we have:

$$ \lambda_1=\frac{-6 + 0.4}{2}=\frac{-5.8}{2}=-2.8 $$

and

$$ \lambda_1=\frac{-6 - 0.4}{2}=\frac{-6.4}{2}=-3.2 $$

Thus, we have found that the general solution of the DE is actually:

$$ y=c_1 e^{-2.8x} + c_2 e^{-3.2x} \! $$

Check:

To check if $$ y $$ is indeed the solution of the given DE, we can differentiate the

what we found to be the general solution.

$$ \frac{dy}{dx}=y'=-2.8x c_1 e^{-2.8x} + -3.2x c_2 e^{-3.2x} \! $$

$$ \frac{d^2y}{dx^2}=y''=7.84x c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} \! $$

Substituting the values of $$ y, y' $$ and $$ y'' $$ in the given equation, we get:

$$ 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}+6(-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x})+ 8.96(c_1e^{-2.8x}+c_2e^{-3.2x}) = 0 \! $$

$$ 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x}-16.8c_1e^{-2.8x}-19.2c_2e^{-3.2x}+ 8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x} = 0 \! $$

and thus:

$$ 0 \equiv 0. \! $$

Therefore, the solution of the given DE is in fact:

$$ y=c_1 e^{-2.8x} + c_2 e^{-3.2x} \! $$

Problem Statement (K 2011 p.59 pb. 4)
Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given
$$ {y}''+ 4{y}'+(\pi^2+4)y = 0\! $$

Solution
The characteristic equation of this ODE is therefore:

$$\lambda^2 + a\lambda + b = \lambda^2 + 4\lambda  + (\pi^2+4 ) = 0\!$$

Evaluating the discriminant:

$$ a^2 - 4b = 4^2 - 4(\pi^2 + 4) = -4\pi^2 < 0\!$$

Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:

$$ y = e^{-ax/2}(Acos(\omega x) + Bsin(\omega x))\!$$

Where: $$ \omega = \sqrt{b - \frac{1}{4}a^2} = \sqrt{\pi^2 + 4 - \frac{1}{4}(4)^2} = \sqrt{\pi^2} = \pi\!$$

And finally we find the general homogenous solution:

$$ y = e^{-2x}(Acos(\pi x )+ Bsin(\pi x))\!$$

Check:

We found that:

$$ y = e^{-2x}(Acos(\pi x )+ Bsin(\pi x))\!$$

Differentiating $$ y\!$$ to obtain $${y}'\!$$ and $${y}''\!$$ respectively:

$$ {y}' = -2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + e^{-2x}(\pi Acos(\pi x)-\pi Bsin(\pi x))\!$$ $$ {y}' = -2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + \pi e^{-2x}( Acos(\pi x)- Bsin(\pi x))\!$$

$$ {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x))- 2 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ {y}'' = 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

Substituting these equations into the original ODE yields: $$ 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ + 4(-2e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + \pi e^{-2x}( Acos(\pi x)- Bsin(\pi x))) + (\pi^2 + 4)(e^{-2x}(Acos(\pi x )+ Bsin(\pi x))) = 0\!$$

$$ 4e^{-2x}(Acos(\pi x)+Bsin(\pi x)) -4 \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) -\pi^2 e^{-2x}(Acos(\pi x)+Bsin(\pi x))\!$$

$$ -8e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + 4\pi e^{-2x}( Acos(\pi x)- Bsin(\pi x)) + \pi^2 e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) + 4e^{-2x}(Acos(\pi x )+ Bsin(\pi x)) = 0\!$$ $$ (4-8+4)e^{-2x}(Acos(\pi x)+Bsin(\pi x)) + (-4+4) \pi e^{-2x}(Acos(\pi x) - Bsin(\pi x)) + (\pi^2-\pi^2) e^{-2x}(Acos(\pi x)+Bsin(\pi x)) = 0\!$$

$$ 0 \equiv 0 \!$$ Therefore, the solution is correct.

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


text:
User:Egm4313.s12.team13.Neisha.R2

Problem R 2.5B
'''pg.59 #17

Problem Statement:

17. Find an ODE for the given solutions.

$$ e ^{- \sqrt{5} x} $$, $$ e ^{- \sqrt{5} x} $$

using $$ \; {y}''+a {y}'+b y=0 $$ for the given basis.

Solution:

Case: Real double root ODE

where $$ \; \lambda = -a/2 $$

and the general solution is $$ \; y= e ^ {-ax} (c_1+ c_2x)  $$

So $$   \; e ^ { \lambda x} =  e ^ {- \sqrt {5} x}  $$

$$  \; e ^ {-ax/2} =  e ^ {- \sqrt {5} x}  $$

Solving for a,

$$  \;  -ax/2= - \sqrt {5} x  $$

$$  \; a= 2 \sqrt {5}  $$

By plugging into the determinant $$ \; a^2-4b=0 $$, Solve for b.

$$  \; {2 \sqrt {5}} -4b=0  $$

$$  \; 4(5)=4b $$

$$  \; b=5  $$

By plugging in a and b, the ODE is:

$$  \; {y}''+ 2 \sqrt {5}{y}'+ 5y=0 $$

title: User:Egm4313.s12.team13.Neisha.R2


text:
User:Peteymills

Climate scientist.

My credentials

My first article: On the negation of modal verbs

title: User:Peteymills


text:
User talk:Peteymills

 Hello Peteymills, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --HappyCamper 06:51, 5 February 2012 (UTC)

title: User talk:Peteymills


text:
Talk:TESOL/Question tags

This article is hilarious. Keep up the good work. Peteymills 06:53, 5 February 2012 (UTC)

title: Talk:TESOL/Question tags


text:
On the negation of modal verbs

Germans use the phrase "must not" in the opposite sense of English speakers to mean "need not." Thinking about this, I realized that there is an implicit "or" in any statement involving modal verbs and the sense of the negative depends upon which part of the logical proposition is negated.

For example, I would translate the phrase:

"You must do A"

into a logical proposition as follows:

^A -> P

where P is some form of punishment. Or:

A or P

We could negate the phrase either by negating the whole thing:

^A and ^P

Or we could negate only one part:

A or ^P = P -> A (e.g., "You (must not) play in the street.")

^A or P = A -> P (e.g. "You must (not play in the street.)")

The first example would seem to be how the Germans use the phrase since whether we do A or not, we will not get punished for it, while the third form is more in line with how we use the phrase, that is, A implies punishment. The second example seems rather more ambiguous and in fact inverts the construct: now, getting punished implies that we have done A.

It has rather deep implications, since all of ethics, law and morality is related to the use of modal verbs. Can we use this idea to justify breaking the Ten Commandments?

I translate,

"I shall go to the store,"

as:

F (uture) -> S (going to the store) = ^F or S

The sixth commandment becomes:

You shall (not kill.) = F -> ^ K = ^F or ^K = ^(F and K)

You (shall not) kill. = ^F -> K = F or K

or perhaps,

^(F -> K) = F and ^K

The second says that if there's a future, there may or may not be killing while the first and third say what we want them to say: if F is true, K must be false.

title: On the negation of modal verbs


text:
Magic glasses

Imagine wearing a pair of glasses that change all of the colours, for instance, into their compliments like a photographic negative. This type of experiment is well known: your brain will re-interpret all the colours so that you see green as red and vice-versa.

Things become more interesting, however, if there is a constant shifting of the colours. Now your brain must come up with a "meta-concept" to label the colours by identifying objects that more permanently keep their colours. For example, there are two objects in the room (at the time of first writing) that would allow me to identify the colour red with almost 100% certainty: a Coke bottle and a bouguet of fake roses, unless the roses are yellow and Coca-Cola has decided to rebrand their bottles in the same colour, both highly unlikely eventualities.

It gives one pause for thought: how much do we rely on reference points to calibrate our colour-sense, which is deceptively constant, despite widely varying lighting conditions? With the magic glasses on, the mind is called upon to create this constant colour-sense, even though the colours of objects are constantly shifting at the same time as the environment. It would be an interesting way to mess with someone's mind: yellow Coke bottle labels, orange leaves, black roses, etc.

The magic glasses would be easy to build: a pair of web-cams, virtual reality glasses and a palm-top computer would be all you need.

title: Magic glasses


text:
Template:Countdown clock

   000 Days to  Stolen from http://wikimania2012.wikimedia.org front page.

title: Template:Countdown clock


text:
User talk:Shashwtendra

sir ,kindly help me in java object,

title: User talk:Shashwtendra


text:
User:Egm4313.s12.team13.Neisha.R2.1

Problem R 2.8B
'''pg.59 #15

Problem Statement:

15. Find a general solution. Check by substitution.

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y=0 $$

Solution:

$$  \; {y}''+0.54 {y}'+(0.0729+ \pi )y= \frac{d^2y}{dx^2}\ + 0.54 \frac{dy}{dt}\ + (0.00729+ \pi )y=0   $$

If  $$ \; \frac{d}{dx}\ = \lambda $$,

$$ \; \lambda ^{2}+0.54 \lambda +(0.0729+ \pi)=0 $$

Solving for $$ \; \lambda $$ by using the quadratic formula,

$$ \lambda_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\,\!$$

$$ \lambda_{1} = -0.27 -{ \sqrt{ \pi } i} $$ $$ \lambda_{2} = -0.27 +{ \sqrt{ \pi } i} $$

In the standard form: $$ \; \lambda=a + bi $$.

General Solution:

The general homogeneous solution is:

$$ y= c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\ \!$$

When plugging in values for a and b, the general solution becomes:

$$ y= c_1 e^{-0.27 x} \cos \sqrt{ \pi }x + c_2 e^{-0.27 x} \sin \sqrt{ \pi }x\ \!$$

So $$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

Check by Substitution:

Solving for derivatives:

$$ y=e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

$$ \; y'=e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x) \ \!$$

$$ \; y''=e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x) \ \!$$

Plugging in  $$   \; y, y', y $$ into $$   \; {y}+0.54 {y}'+(0.0729+ \pi )y=0 $$:

$$ \; [e^{-0.27 x}(- \pi c_1 \cos \sqrt{ \pi }x - \pi c_2 \sin \sqrt{ \pi }x)+ 0.0729e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)-0.54e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x + \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)]$$

$$ \; +0.54 [e^{-0.27 x}(- \sqrt{ \pi } c_1 \sin \sqrt{ \pi }x +  \sqrt{ \pi } c_2 \cos \sqrt{ \pi }x)-0.27e^{-0.27x}( c_1 \cos \sqrt { \pi }x + c_2 \sin \sqrt{ \pi }x)]+(0.0729+ \pi )[e^{-0.27 x} ( c_1 \cos \sqrt{ \pi }x + c_2 \sin \sqrt{ \pi }x)]=0 $$

When plugging in, the general solution is confirmed because it equals zero.

title: User:Egm4313.s12.team13.Neisha.R2.1


text:
User:Egm4313.s12.team13.rossmeissl/R2work

Problem Statement
Using the spring-dashpot-mass system shown below (from Fig p1.4) and following similar characteristics of (3)p5.5 but using the double real root $$\displaystyle \lambda = -3 $$, find values for the parameters k,c,m.

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$$  \displaystyle {(\lambda -5)}^{2} = {\lambda }^{2}-10\lambda +25 = 0 $$     ((3)p5.5)
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Solution
Rewriting ((3)p5.5) with the double real root $$\displaystyle \lambda = -3 $$ gives the new characteristic equation

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$$  \displaystyle {(\lambda -3)}^{2} = {\lambda }^{2}-6\lambda +9 = 0 $$     (6.0)
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From kinematics:

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$$  \displaystyle y = {y}_{k}+{y}_{c} $$     (6.1)
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From kinetics:

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$$  \displaystyle my''+{f}_{1} = f(t) $$     (6.2)
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$$  \displaystyle {f}_{1} = {f}_{k} = {f}_{c} $$     (6.3)
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Relationships:

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$$  \displaystyle {f}_{k} = k{y}_{k} $$     (6.4)
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 *
 * }

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

$$  \displaystyle {f}_{c} = c{y'}_{c} $$     (6.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From equation (6.1) take the second derivative

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

$$  \displaystyle y = {y}_{k} = {y''}_{c} $$     (6.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From equations (6.3), (6.4) & (6.5)

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

$$  \displaystyle {f}_{k} = {f}_{c} \Rightarrow k{y}_{k} = c{y'}_{c} \Rightarrow {y'}_{c} = \frac{k}{c}{y}_{k} $$     (6.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From equations (6.6) & (6.7)

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

$$  \displaystyle y = {y}_{k}+({y'}_{c})' \Rightarrow {y''}_{k}+(\frac{k}{c}{y}_{k})' $$     (6.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From equations (6.2), (6.3) & (6.8)

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

$$  \displaystyle m({y''}_{k}+\frac{k}{c}{y'}_{k})+k{y}_{k} = f(t) $$     (6.9)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From the characteristic equation (6.0), the corresponding homogeneous L2-ODE-CC is

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

$$  \displaystyle y''+6y'+9y = 0 $$     (6.10)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Now take equation (6.9) and divide through by m

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

$$  \displaystyle {y''}_{k}+\frac{k}{cm}{y'}_{k}+\frac{k}{m}{y}_{k} = \frac{f(t)}{m} = 0 $$     (6.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Comparing to equation (6.10) gives

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

$$  \displaystyle \frac{k}{cm} = 6 $$     (6.12) &
 * style="width:95%" |
 * style="width:95%" |
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \frac{k}{m} = 9 $$     (6.13)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

There are an infinite number of solutions to these two equations, one of these is

$$   \displaystyle k = 9\;\;m = 1\;\;c = \frac{3}{2}$$

Section 1 Lecture Notes

Section 5 Lecture Notes


text:
User talk:LemonLime05

 Hello LemonLime05, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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title: User talk:LemonLime05


text:
User talk:Dcalloway

 Hello Dcalloway, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Katarighe (Talk · Contributions · E-mail) 20:12, 5 February 2012 (UTC)

title: User talk:Dcalloway


text: Template:Global sysop

title: Template:Global sysop


text: Template:Global rollbacker

title: Template:Global rollbacker


text: Template:User VRT

title: Template:User VRT


text:
User:Nihatch

School
I’m pursuing my M.S. in Instructional Systems Technology (IST) at Indiana University. I have completed 27/36 credits and hope to complete the program in May 2013. I'm currently taking R626 Instructional Strategies and Tactics with Dr. Peter Honebein. As part of this course, I will contribute an article to Wikiversity. I am brainstorming topics right now and once decided, I will update this page.

Work
I work for Dynamics Research Corporation—a management consulting firm, based in Arlington, VA. As part of their Learning and Organizational Development group, I design/develop learning solutions for various Federal clients including FDIC, USAID, FEMA, the U.S. Department of Agriculture, and the Office of Naval Intelligence. I have experience in Web-based training and instructor-led training development, group facilitation, and program/training evaluation. Currently, I am about to start supporting a large change management project for a federal client.

Interests
I’m passionate about learning, design, and technology and how the three disciplines can be used to solve real-world problems. In addition, I enjoy reading, traveling, exploring, hiking, rock climbing, yoga, swimming, and playing the ukulele.

title: User:Nihatch


text:
User:Kalki

I am most active on Wikimedia wikis at the english Wikiquote project, where I have been a primary contributor since 2003. Here is a link to my user page there. ~ ♞☤☮♌Kalki·†·⚓⊙☳☶⚡ >

Ω

title: User:Kalki


text:
University of Florida/Egm4313/s12.team11.imponenti/R2.9

Problem Statement
Find and plot the solution for the L2-ODE-CC corresponding to

$$\lambda^2 + 4 \lambda + 13\!$$

with $$r(x)=0\!$$

and initial conditions $$y(0)=1\!$$, $$y'(0)=0\!$$

In another figure, superimpose 3 figs.:(a)this fig. (b) the fig. in R2.6 p.5-6, and (c) the fig. in R2.1 p.3-7

Quadratic Equation
$$\lambda = \frac{-b \pm \sqrt(b^2-4ac)}{2a}\!$$ with $$a=1, b=4,  c=13\!$$

$$\lambda=\frac{-4 \pm \sqrt(4^2-4*1*13)}{2*1}=-2 \pm 3i\!$$

$$\lambda=-2 \pm 3i\!$$

Homogeneous Solution
The solution to a L2-ODE-CC with two complex roots is given by

$$y(x)=e^{-\frac{a}{2}x}[Acos(\omega x)+Bsin(\omega x)]\!$$

where $$\lambda=-\frac{a}{2} \pm \omega i=-2 \pm 3i\!$$

$$y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!$$

Solving for A and B
first initial condition $$y(0)=1\!$$

$$y(x)=e^{-2x}[Acos(3x)+Bsin(3x)]\!$$

$$y(0)=e^{-2*0}[Acos(3*0)+Bsin(3*0)]=1\!$$

$$A=1\!$$

second initial condition $$y'(0)=0\!$$

$$y'(x)=\frac{d}{dx}y(x)=\frac{d}{dx}e^{-2x}[cos(3x)+Bsin(3x)]\!$$

$$y'(x)=e^{-2x}[(-2B-3)sin(3x)+(3B-2)cos(3x)]\!$$

$$y'(0)=e^{-2*0}[(-2B-3)sin(3*0)+(3B-2)cos(3*0)]\!$$

$$0=3B-2\!$$

$$B=\frac{2}{3}\!$$

so the solution to our L2-ODE-CC is                       $$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!$$

Solution to R2.6
After solving for the constants $$\frac{k}{c}$$ and $$\frac{k}{m}$$ we have the following homogeneous equation

$$y''(x)+6y'(x)+9y=0\!$$

Characteristic Equation and Roots
$$\lambda^2 + 6\lambda + 9=0\!$$

$$(\lambda+3)(\lambda+3)=0\!$$

We have a real double root $$\lambda=-3\!$$

Homogeneous Solution
We know the homogeneous solution to a L2-ODE-CC with a double real root to be

$$y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}\!$$

Assuming object starts from rest

$$y(0)=1\!$$, $$y'(0)=0\!$$

Plugging in $$\lambda$$ and applying our first initial condition

$$y(0)=c_1e^{-3*0}+c_2*0*e^{-3*0}=1\!$$

$$c_1=1\!$$

Taking the derivative and applying our second condition

$$y'(x)=\frac{d}{dx}y(x)=\frac{d}{dx}e^{-3x}+c_2xe^{-3x}\!$$

$$y'(x)=-3e^{-3x}+c_2e^{-3x}-3c_2xe^{-3x}\!$$

$$y'(0)=-3e^{-3*0}+c_2e^{-3*0}-3c_2*0*e^{-3*0}=0\!$$

$$c_2=3\!$$

Giving us the final solution $$y(x)=e^{-3x}+3xe^{-3x}\!$$

Solution to this Equation
$$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!$$

Superimposed Graph
Our solution: $$y(x)=e^{-2x}[cos(3x)+\frac{2}{3}sin(3x)]\!$$ shown in blue

Equation for fig. in R2.1 p.3-7: $$y(x)=\frac{5}{4}e^{-2x}-\frac{1}{4}e^{5x}\!$$ shown in red

Equation for fig. in R2.6 p.5-6:$$y(x)=e^{-3x}+3xe^{-3x}\!$$ shown in green

Egm4313.s12.team11.imponenti 03:38, 8 February 2012 (UTC)


text:
User:Super Quantum immortal/Ibus English-chinese table 1000

(fix the formating latter)

This is part of Chinese learning technology(User:Super Quantum immortal)

These are the 1000 most frequent English works.

click edit, then copy TODO add chinese char.

Table
s	們	2000000 you	你	1222421 I	我	1052546 to	至	823661 the	一	770161 a	一	563578 and	而	480214 that	彼	413389 it	它	388320 of	的	332038 me	我	312326 what	何	285826 is	是	282222 in	在	266544 this	這	249860 know	知	241548 Im	我	230304 for	為	216535 no	不	212463 have	有	210523 my	我	210429 dont	做	206663 just	剛	196810 not	不	195058 do	做	194183 be	是	191823 on	上	191646 your	你	188060 was	是	184170 we	我	181863 its	其	176919 with	同	169586 so	然	169088 but	但	159109 all	都	158281 well	良	155201 are	是	152650 he	他	150009 oh	哦	148780 about	左右	142750 right	右	137218 youre	你	135738 get	得	126849 here	此地	124368 out	出	121867 going	去	121435 like	同	121194 yeah	是	121148 if	倘	120553 her	她	116265 she	她	110122 can	可	107652 up	上	103826 want	要	102822 think	思	102403 thats	那	102146 now	現	100909 go	去	100905 him	他	96536 at	在	93433 how	何	87908 got	有	86085 there	在	85662 one	一	84057 did	作	79053 why	何	78554 see	見	75968 come	來	72807 good	好	72798 they	他	69192 really	真	68311 as	如	67977 would	將	67718 look	看	67262 when	何時	65117 time	時	64891 will	將	64811 okay	好	63509 back	後	63115 cant	能	62602 mean		61504 tell	訴	61352 Ill	將	61169 from	由	59972 hey	嘿	58699 were	為	58127 hes	他	57551 could	可	57285 didnt	做	56732 yes	是	55828 his	他	55502 been	是	55235 or	或	55062 something	某物	54736 who	誰	53799 because	因	53566 some	些	53265 had	有	52561 then		52197 say	說	50213 ok	好	49968 take		49868 an		49832 way		49794 us		49492 little		47885 make		46154 need		46040 gonna		45124 never		44728 were		43491 too		43360 love		42994 shes		42795 Ive		42160 sure		40757 them		40366 more		40239 over		40001 our		39896 sorry		39854 where		39330 whats		39305 let		38858 thing		38121 am		37814 maybe		37653 down		37644 man		37355 has		37145 uh		36435 very		36140 by		35500 theres		35242 should		34829 anything		34635 said		34568 much		34270 any		34203 life		34192 even		33895 off		33887 please		33628 doing		32831 thank		31728 give		31073 only		30599 thought		30211 help		29825 two		29788 talk		28834 people		28517 god		28502 still		28308 wait		27920 into		27354 find		26757 nothing		26592 again		26472 things		26305 lets		26132 doesnt		25926 call		25880 told		25594 great		25435 before		25329 better		25282 ever		25127 night		24900 than		24815 away		24543 first		24509 believe		24210 other		23936 feel		23850 everything		23714 work		23683 youve		23483 fine		22993 home		22901 after		22182 last		21965 these		21929 day		21860 keep		21842 does		21731 put		21617 around		21591 stop		21341 theyre		21309 Id		21221 guy		21126 long		21086 isnt		20796 always		20581 listen		20142 wanted		20074 Mr		20053 guys		19953 huh		19871 those		19849 big		19520 lot		19421 happened		19282 thanks		19233 wont		19208 trying		19173 kind		19031 wrong		19025 through		18903 talking		18875 made		18872 new		18860 being		18666 guess		18473 hi		18285 care		18042 bad		17636 mom		17514 remember		17266 getting		17199 well		17077 together		17074 dad		17003 leave		16978 mother		16855 place		16796 understand		16724 wouldnt		16615 actually		16585 hear		16523 baby		16514 nice		16275 father		16233 else		16214 stay		16167 done		16010 wasnt		16007 their		15907 course		15866 might		15774 mind		15364 every		15350 enough		15322 try		15235 hell		15191 came		15142 someone		15035 youll		14958 own		14849 family		14765 whole		14623 another		14541 house		14458 jack		14324 yourself		14291 idea		14203 ask		14194 best		14094 must		14089 coming		14082 old		14041 looking		13843 woman		13734 hello		13655 which		13652 years		13526 room		13467 money		13456 left		13428 knew		13357 tonight		13305 real		13161 son		13131 hope		13093 name		13044 same		13024 went		12892 um		12883 hmm		12819 happy		12788 pretty		12746 saw		12707 girl		12641 sir		12557 show		12392 friend		12245 already		12233 saying		12204 may		12199 next		12147 three		12088 job		12047 problem		11931 minute		11903 found		11875 world		11754 thinking		11721 havent		11699 heard		11692 honey		11659 matter		11578 myself		11559 couldnt		11550 exactly		11502 having		11481 ah		11347 probably		11332 happen		11273 weve		11200 hurt		11145 boy		11050 both		11035 while		11006 dead		11002 gotta		11000 alone		10813 since		10757 excuse		10747 start		10704 kill		10676 hard		10608 youd		10577 today		10561 car		10485 ready		10461 until		10432 without		10421 whatever		10363 wants		10269 hold		10254 wanna		10250 yet		10225 seen		10192 deal		10143 took		10083 once		9931 gone		9877 called		9871 morning		9823 supposed		9818 friends		9814 head		9794 stuff		9709 most		9699 used		9623 worry		9620 second		9605 part		9518 live		9507 truth		9470 school		9443 face		9441 forget		9322 true		9296 business		9295 each		9277 cause		9277 soon		9261 knows		9217 few		9164 telling		9129 wife		9125 whos		9109 use		9096 chance		9081 run		9076 move		9073 anyone		9050 person		8997 bye		8969 j		8942 somebody		8923 dr		8917 heart		8824 such		8813 miss		8806 married		8798 point		8707 later		8636 making		8623 meet		8608 anyway		8517 many		8504 phone		8424 reason		8395 damn		8367 lost		8348 looks		8337 bring		8320 case		8287 turn		8264 wish		8262 tomorrow		8254 kids		8247 trust		8187 check		8185 change		8176 end		8171 late		8164 anymore		8160 five		8137 least		8119 town		8099 arent		8065 ha		8033 working		8016 year		8013 makes		8001 taking		7996 means		7980 brother		7952 play		7883 hate		7839 ago		7836 says		7821 beautiful		7813 gave		7759 fact		7654 crazy		7650 party		7633 sit		7625 open		7598 afraid		7587 between		7585 important		7582 rest		7417 fun		7404 kid		7380 word		7355 watch		7310 glad		7277 everyone		7214 days		7213 sister		7188 minutes		7185 everybody		7163 bit		7158 couple		7150 whoa		7081 either		7075 mrs		7064 feeling		7058 daughter		7057 wow		7036 gets		7023 asked		7009 under		6999 break		6941 promise		6926 door		6907 set		6878 close		6877 hand		6872 easy		6848 question		6841 doctor		6841 tried		6830 far		6728 walk		6619 needs		6555 trouble		6510 mine		6503 though		6502 times		6475 different		6464 killed		6390 hospital		6378 anybody		6376 sam		6354 alright		6348 wedding		6334 shut		6294 able		6289 die		6280 perfect		6276 police		6260 stand		6257 comes		6256 hit		6248 story		6231 ya		6212 mm		6192 waiting		6132 dinner		6099 against		6064 funny		6054 husband		6038 almost		6037 stupid		6020 pay		6016 answer		6016 four		6014 office		5955 cool		5954 eyes		5950 news		5936 child		5913 shouldnt		5874 half		5819 side		5813 yours		5805 moment		5770 sleep		5747 read		5743 wheres		5711 started		5707 young		5680 men		5678 sounds		5665 sonny		5635 lucky		5631 pick		5610 sometimes		5596 em		5590 bed		5589 also		5583 date		5581 line		5564 plan		5557 hours		5503 lose		5459 fire		5452 free		5433 hands		5428 serious		5390 leo		5390 shit		5388 behind		5388 inside		5379 high		5377 ahead		5364 week		5356 wonderful		5341 t		5329 fight		5326 past		5315 cut		5305 quite		5304 number		5289 hell		5262 sick		5256 s		5255 itll		5217 game		5212 eat		5202 nobody		5195 goes		5194 death		5165 along		5147 save		5144 seems		5138 finally		5118 lives		5095 worried		5066 upset		5064 Theresa		5056 Carly		5055 Ethan		5033 met		5032 book		5027 brought		5010 seem		5001 sort		5000 safe		4999 living		4999 children		4944 werent		4896 leaving		4889 front		4868 shot		4866 loved		4851 asking		4851 running		4842 clear		4836 figure		4785 hot		4777 felt		4748 six		4737 parents		4725 drink		4706 absolutely		4704 hows		4652 daddy		4641 sweet		4628 alive		4628 Paul		4616 sense		4597 meant		4596 happens		4589 David		4589 special		4573 bet		4548 blood		4542 aint		4541 kidding		4539 lie		4536 full		4533 meeting		4532 dear		4516 coffee		4499 seeing		4495 sound		4488 fault		4484 water		4469 fuck		4453 ten		4448 women		4441 john		4413 welcome		4373 buy		4345 months		4333 hour		4325 speak		4322 lady		4301 jen		4301 thinks		4298 Christmas		4284 body		4271 order		4270 outside		4260 hang		4243 possible		4228 worse		4218 company		4209 mistake		4185 ooh		4184 handle		4181 spend		4175 c		4166 totally		4157 giving		4143 control		4134 heres		4130 marriage		4129 realize		4114 d		4108 power		4094 president		4086 unless		4078 sex		4049 girls		4048 send		4045 needed		4043 o		4030 taken		3997 died		3996 scared		3991 picture		3984 talked		3975 jake		3962 al		3958 ass		3947 hundred		3942 changed		3929 completely		3919 explain		3913 playing		3899 certainly		3895 sign		3890 boys		3888 relationship		3880 Michael		3878 loves		3876 fucking		3843 hair		3839 lying		3837 choice		3828 anywhere		3821 secret		3810 future		3798 weird		3797 luck		3792 shell		3780 max		3773 Luis		3773 turned		3770 known		3757 touch		3755 kiss		3750 crane		3744 questions		3741 obviously		3741 wonder		3735 pain		3720 calling		3714 somewhere		3707 throw		3697 straight		3695 grace		3693 cold		3690 white		3665 fast		3661 Natalie		3660 words		3647 r		3636 food		3632 none		3615 drive		3607 feelings		3603 theyll		3593 worked		3577 marry		3564 light		3554 test		3547 drop		3546 cannot		3537 frank		3525 sent		3524 city		3515 dream		3471 protect		3465 twenty		3463 class		3458 lucy		3450 surprise		3439 its		3439 sweetheart		3434 forever		3432 poor		3431 looked		3431 mad		3407 except		3406 gun		3396 yknow		3395 dance		3394 takes		3386 appreciate		3379 especially		3376 situation		3359 besides		3355 weeks		3350 pull		3340 himself		3331 hasnt		3318 act		3307 worth		3303 Sheridan		3299 amazing		3294 top		3283 given		3283 expect		3277 ben		3262 rather		3241 Julian		3240 involved		3240 swear		3239 piece		3239 busy		3229 law		3225 decided		3224 black		3217 joey		3216 happening		3210 movie		3209 wed		3205 catch		3194 antonio		3188 country		3187 less		3173 perhaps		3165 step		3161 fall		3157 watching		3152 kept		3152 darling		3146 dog		3141 ms		3131 win		3126 air		3126 honor		3106 personal		3103 moving		3102 till		3091 admit		3091 problems		3090 murder		3089 strong		3088 hed		3087 evil		3078 definitely		3068 feels		3067 information		3063 honest		3063 eye		3063 broke		3044 missed		3043 longer		3043 dollars		3042 tired		3029 Jason		3026 George		3013 evening		2998 human		2986 starting		2984 Ross		2981 red		2976 entire		2976 trip		2975 Brooke		2969 e		2964 club		2963 Niles		2956 suppose		2952 calm		2948 imagine		2946 Todd		2943 fair		2941 caught		2933 b		2928 blame		2923 street		2920 sitting		2920 favor		2920 apartment		2920 court		2919 terrible		2914 clean		2914 tony		2913 learn		2912 Alison		2910 Rick		2909 works		2903 Rose		2890 Frasier		2887 relax		2877 york		2873 million		2873 charity		2865 accident		2860 wake		2857 prove		2852 Danny		2848 smart		2847 message		2846 missing		2845 forgot		2845 small		2843 interested		2838 table		2835 nbsp		2824 become		2816 craig		2810 mouth		2807 pregnant		2805 middle		2802 billy		2797 ring		2786 careful		2785 shall		2779 dude		2775 team		2773 ride		2768 figured		2766 wear		2765 shoot		2759 stick		2753 ray		2752 follow		2752 bo		2751 angry		2747 instead		2744 buddy		2743 write		2741 stopped		2719 early		2719 angel		2701 nick		2699 ran		2692 war		2690 standing		2690 forgive		2690 jail		2684 wearing		2678 Miguel		2678 ladies		2670 kinda		2662 lunch		2660 Cristian		2659 eight		2655 Greenlee		2654 gotten		2640 hoping		2639 phoebe		2638 thousand		2633 ridge		2633 music		2631 luke		2614 paper		2607 tough		2606 tape		2605 Emily		2595 state		2592 count		2589 college		2589 boyfriend		2589 proud		2586 agree		2584 birthday		2580 bill		2572 seven		2569 theyve		2566 Timmy		2565 history		2564 share		2563 offer		2563 hurry		2559 ow		2557 feet		2555 wondering		2549 simple		2549 decision		2548 building		2546 ones		2539 finish		2535 voice		2530 herself		2530 Chris		2529 wouldve		2523 list		2518 Kay		2517 mess		2516 deserve		2516 evidence		2514 cute		2504 Jerry		2503 dress		2501 Richard		2499 interesting		2497 Jesus		2487 James		2478 hotel		2463 enjoy		2463 Ryan		2461 Lindsay		2461 quiet		2454 concerned		2452 road		2448 eve		2445 staying		2443 short		2440 m		2440 beat		2440 sweetie		2439 mention		2434 clothes		2434 finished		2423 fell		2419 neither		2417 mmm		2414 fix		2414 victor		2412 respect		2410 spent		2406 prison		2401 attention		2397 holding		2394 calls		2389 near		2386 surprised		2381 bar		2376 beth		2375 pass		2374 keeping		2371 gift		2371 hadnt		2369 putting		2366 dark		2361 self		2358 owe		2354 using		2345 nora		2345 ice		2345 helping		2342 bitch		2341 normal		2339 aunt		2339 lawyer		2338 apart		2336 certain		2333 plans		2327 Jax		2326 girlfriend		2325 floor		2325 whether		2316 everythings		2316 present		2313 earth		2312 private		2311 Jessica		2310 box		2304 Dawson		2303 cover		2298 judge		2294 upstairs		2293 Alexis		2287 Shawn		2281 sake		2281 mommy		2281 possibly		2280 worst		2276


text: File:Plotr2 9.jpg

title: File:Plotr2 9.jpg


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Video journalism/Script a documentary

To write a high quality documentary script, you retain intellectual property, while building on and developing the complex issues of representation and attribution learnt in the earlier assignments based in news and current affairs.

Continue to develop the project and topic you started as a news story, earlier in this learning Schedule. Do so in terms of this assignment and the readings in the Topics, the Internet resources and the script example template in the Resources (Script Example, Cinéma Vérité / Direct Cinema, 'Delinquent Angel' Documentary). While the formatting has gone astray in some places as it was converted to a PDF (I will fix this), the script is a good example as to how to set out a film script that might be used for seeking funds, or providing an editor as they sit down to go through all the material you have acquired to build a film.

Cite a range of sources to support your script and remember that a documentary is a complex narrative that takes the line usually, of an argument, as a scholarly thesis might. So write the theoretical argument into the script, through the intended interviewee/s, their monologue, in dialogue, by way of voice over narration or actor driven dialogue. You could use any of the characters that may emerge in your materials, both in video and print.

Ultimately, you are required to provide a discussion, a through line, of the video material, the reference papers and any background readings that you have brought to the film project. This material now must be reworked into a coherent a script based narrative.

You should couple references with other references, insights or experiences, and discuss what is behind the production in terms of style and genre within documentary film — make this transparent in the script.

All this should be done using interviewees and events — real or fictional — of the documentary form. Keep the response simple, dont get bogged down in some of our reference materials apparent complexity or precociousness.

Resources

How to write a documentary script

Documentary script templates

In search of the dragon's tale - Example documentary script

Triumph of the will - Example propaganda documentary script

Screenwriters Utopia

How to write a script for documentary

title: Video journalism/Script a documentary


text:
Literature/1979/Stafford Beer

Excerpts

 * "The measure of complexity is called VARIETY. VARIETY is defined as the number of possible states of whatever it is whose complexity we want to measure." (p, 32)

title: Literature/1979/Stafford Beer


text:
University of Florida/Egm4313/s12.team11.imponenti/R2.2

Problem Statement
Find and plot the solution for the homogeneous L2-ODE-CC
 * $$y''(x) - 10y'(x) + 25y(x)=0\!$$

with initial conditions $$y(0)=1\!$$ ,and  $$y'(0)=0\!$$

Characteristic Equation
$$\lambda^2-10\lambda+25=0\!$$

$$(\lambda-5)(\lambda-5)=0\!$$

$$\lambda=5\!$$

Homogeneous Solution
The solution to a L2-ODE-CC with real double root is given by $$y(x)=c_1e^{\lambda x}+c_2xe^{\lambda x}\!$$

First initial condition

$$y(0)=1\!$$

$$y(0)=c_1e^{5*0}+c_2*0*e^{5*0}=1\!$$

$$c_1=1\!$$

Second initial condition

$$y'(0)=0\!$$

$$\frac{d}{dx}y(x)=y'(x)=5e^{5x}+c_2e^{5x}(5x+1)\!$$

$$y'(0)=5e^{5*0}+c_2e^{5*0}(5*0+1)=0\!$$

$$5+c_2=0\!$$

$$c_2=-5\!$$

The solution to our L2-ODE-CC is

$$y(x)=e^{5x}(1-5x)\!$$

Plot
$$y(x)=e^{5x}(1-5x)\!$$

Egm4313.s12.team11.imponenti 00:30, 8 February 2012 (UTC)


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File:Graph 1 problem 2.9.1.jpg

Licensing
for school project

title: File:Graph 1 problem 2.9.1.jpg


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File:P2 2 3.png

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for school project

title: File:P2 2 3.png


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User:Dblackal/Books/Video Journalism

 * Video journalism
 * Video journalism/Identify a news story
 * Video journalism/Produce a news story
 * Video journalism/Develop a current affairs story
 * Video journalism/Script a documentary
 * Video journalism/A Picture's Worth
 * Video journalism/The Camera's Eye
 * Video journalism/Constructing Vision
 * Video journalism/Telling Stories
 * Video journalism/Editing Grammar
 * Video journalism/Ethics of the Gaze

title: User:Dblackal/Books/Video Journalism


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File:4313team12r11-21.jpg

Licensing
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File:4313team12r11-22.jpg

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File:4313team12r11-23.jpg

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File:4313team12r11-24.jpg

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title: File:4313team12r11-24.jpg


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File:4313team12r11-25.jpg

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title: File:4313team12r11-25.jpg


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File:4313team12r11-26.jpg

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for school project

title: File:4313team12r11-26.jpg


title: File:4313team12r11-27.jpg


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File:4313team12r11-28.jpg

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for school project

title: File:4313team12r11-28.jpg


title: File:4313team12r11-31.jpg


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File:4313team12r11-32.jpg

Licensing
for school project

title: File:4313team12r11-32.jpg


text:
User:MrFidaAliEngr/Nikah

Under most stringent yet most careful religious rules, regulations and laws, nikah is essential for the continuity of humanity. The holy Quran and other religious books must be consulted for details under guidance of pious religious scholars in order to benefit from nikah marriage system.

title: User:MrFidaAliEngr/Nikah


text:
User:Dineshkumar Ponnusamy

Please visit me in English Wiki.

Alternately, if you wish to contact me, leave a new message here

ta:பயனர்:Dineshkumar_Ponnusamy

title: User:Dineshkumar Ponnusamy


text:
User:Saif ansari

Mohammad Saifuddin Ansari (born 26 February 1987), often credited as Saif Ansari, is an ITC Infotech India Ltd Employee, IT Professional and Film Literal. Saif has Studied in St Thomas Boys School, Kolkata and did his graduation from HKBK Degree College, Bangalore. He started his IT career from ITC Infotech India Ltd as PLM Developer.

title: User:Saif ansari


title: File:Butterfly.SW.1.A.Idea.20120206.pdf


text: File:Butterfly.SW.1.A.Idea.20120206.2.pdf


text:
Wikiversity:Request custodian action/Archive/14

Templates borked
notice to Jtneill. It looks like at least one or more of the template imports from Wikipedia that Jtneill made yesterday broke the collapse template and possibly more. Once I realized the possible extent of this, I decided to bring this here and not just work on this alone. This might require custodial tools to fix, and it also might require expertise that I don't have. Enjoy! (I give evidence for the broken template on User talk:Jtneill, the place linked.) --Abd 18:00, 24 November 2011 (UTC)
 * Fixed the collapse template problem with . That might not be optimal, and there might be other problems. But at least collapse works now. You know, I just gotta collapse sometimes. --Abd 18:28, 24 November 2011 (UTC)
 * Thanks, Abd, for alerting and fixing. Sorry about that. Yes, looks like the WV version must be working slightly differently from the WP version, even though I had imported updates to both collapse top and collapse bottom from WP. It would be nice if they worked the same way to also prevent this from happening in future? -- Jtneill - Talk - c 20:50, 24 November 2011 (UTC)
 * The collapse template uses Template:Navbox, and the WV version is different from the Wikipedia version. See Template:Navbox-test, the differences can be seen by comparing the two versions in History. That, or some other included template, directly or indirectly, might be the problem. Handle with care! Changes to oft-used templates should be tested first, with dummy template names (as I did with Template:Collapse bottom-test before making the change. If I had time, I'd probably be able to figure out what's happening, I do see some clues. But I don't have time, and there are others active with more template experience than I. --Abd 23:51, 24 November 2011 (UTC)
 * I think templates do tend to differ across projects. My guess is jtneill checked "include all templates" when importing a page, which imported templates Wikiversity already had along with it. I think Wikipedia templates especially tend to have a lot of dependencies that are eventually simplified at other projects because they don't have the people who know how to make sense of it to maintain the templates as-is. -- dark lama  01:54, 25 November 2011 (UTC)
 * Just passing, and in the remote chance that this is at all helpful, I based the Template:cot and Template:cob in my MW installation originally on the WV code (because it was simpler than WP). So if you've lost track of the code, looking at or  and their histories might help "fill in any gaps". I don't think this version actually uses Template:Navbox (see ), unless I'm missing it, but it does use the CSS classes "navbox" and "NavHead" etc from MediaWiki:Common.css/Collapse.css (although even that doesn't seem essential to functionality - I don't have all that CSS on that subpage, just some stuff for div.NavHead and a.NavToggle in my common.css) -  It also relies on the Javascript at MediaWiki:Navigation.js (which is edited on my installation to give a slightly different "collapse button" wording/style) Begoon 04:14, 25 November 2011 (UTC)

Continued incivility by Ottava Rima, after warning, block request
Ottava Rima's edit summary: (Undo revision 837811 by Abd (talk) Policy need consensus first, any claim something without clear consensus is a policy is, by definition, vandalism)

On the talk page, Ottava had brought up a series of irrelevancies and incivilities:
 * "The page is policy, it's long been so labelled" This page, even though it was labelled such by a banned user, was never made policy and never officially consented to by the community. The RfCs above were about that and still didn't pass. We have a process, and things don't happen because you someone claims they are true. Abd was banned on Meta for such magical rethinking of policy and concepts of what is true or not, and he lost his bid for adminship on such behavior. The community even agreed before to topic ban him from discussions such as this even though the admin refused to enforce it. Ottava Rima (talk) 19:17, 26 November 2011 (UTC)

 * This was a gratuitous diversion from the topic under civil discussion, the issue of custodians and email, and a possible change to the policy back to what had originally been written. The statement of mine that Ottava quoted as "magical thinking" was simply the truth. The page was labelled policy, it represented, accurately, actual practice (except where Ottava violated it!), and the designation had stood for four years. Ottava apparently believes that policy is only created through some explicit and very formal process, and he's argued tendentiously elsewhere about this, railing against administrators and stewards because they did something not explicitly allowed in formal policy.

 * JWSchmidt added the policy label in 2007. He's not banned, and even if he were, that would be irrelevant.
 * The page had been linked as a policy in March 2007, by Mu301, and the extensive subsequent editing of the policy page, while leaving "policy" in place, shows that it was broadly accepted.
 * The status was questioned in 2008 by Darklama This may have been a purely technical concern. It ultimately went nowhere and minor issues about the text were eventually addressed with consensus. In 2010, the policy tag was changed back to proposed policy, which lasted about thirty minutes. There is no major dispute over the content, there are only some proposed tweaks.

 * Thenub314, apparently unfamiliar with the history, seems to have believed Ottava's misrepresentations, and made the change again to "proposal." Because this has been discussed before, because the page does reflect long-standing actual practice (and prior criticism of the policy was mostly about small details, I reverted, with "See Talk," and then Ottava reverted with the "vandalism" claim in the summary of the edit linked above. The only major disagreements with the policy were the attempts by Ottava to remove inactive custodians and gut mentored custodianship, this year, through the RfCs he mentions, and those were explicitly attempts to amend the policy, not referenda on approving it.
 * The second half of Ottava's post quoted above is a personal attack, completely irrelevant to the issue being discussed. as well as highly misleading.
 * After reverting my restoration of the long-standing policy template, Ottava again attacked me for "clear vandalism" on the Talk page, suggesting I be blocked, long-term, for disruption.
 * See Ottava's block log here. (From the earliest block, you can see where his enmity for me arose. His extreme reactions, including unblocking himself and attempting to hide the block log entries, eventually led to his desysopping.)
 * Having avoided possible sanctions from a pending Community Review by ceasing his editing, Ottava eventually returned to do two things: welcome users, and argue tendentiously, disruptively, and uncivilly, mostly to attack me and anyone who expresses support for me, particularly if new. He has been warned many times. See his most recent block, which was over his focus on "on 'taking down' someone who has historically been [his] rival."
 * He sometimes stops in response to warnings, but never admits error, see this strong warning, where his final comment is, essentially, "I'm right." He continued to repeat, on-wiki, the uncivil (and wrong) claim he had made. He could have been blocked for that alone.
 * Given extensive prior warnings combined with continued blatant and unnecessary incivility, please block. If he continues editing here, a Community Review will be necessary. Were he here to improve resources, it might be worth it. He's not. --Abd 03:53, 27 November 2011 (UTC)
 * Given my own involvement I will personally abstain from acting. But I do want to comment that I found none of Ottava's recent comments any less civil then comments you've made about me, wikilayering, blocking for political reasons, violating policy etc.  In follow up you asked me to view the statements you made as either your opinion or fact, and you could equally well view these comments with detachment you seemed to expect from me.  Got it? Thenub314 04:20, 27 November 2011 (UTC)
 * No, I don't get it. Above, I cite specific statements, specific warnings, including a claim that I should be blocked for "vandalism," including continued attack on a newbie after warning, and you have cited nothing clear. I'm not the issue here. If I've been uncivil to you, or anyone, I should be warned, preferably by someone neutral, and blocked if I improperly continue, per WV:Blocking policy, and that has nothing to do with Ottava. If I'm a monster, it doesn't make his behavior acceptable.--Abd 05:05, 27 November 2011 (UTC)

Suggestion of editing restrictions for Abd
I think this would pretty much sort our current issues with Abd's editing. -- Simone 14:30, 27 November 2011 (UTC) Comments:
 * I'd rather we imposed specific sanctions on Abd to prevent him disrupting Wikiversity rather than going for the nuclear option. In particular:
 * Abd should be topic banned from editing policies. If he wants a policy to be changed, he should request that other editors make his proposed changes on the talk page of the policy.
 * There should be a total interaction ban between Ottava Rima and Abd.
 * Any breach of the above editing restrictions should be dealt with by a 24h block.
 * Abd should be reminded that "walls of text" can be considered disruptive. He should attempt to limit the length of his comments to a few lines.
 * -- Simone 14:30, 27 November 2011 (UTC)
 * Support - This was proposed before and passed, but never enforced. If it was enforced long ago 90% of the problems of this community right now probably would have never turned into problems. The interaction ban was added later - it wouldn't work - Abd has attempted to change many important policies and other matters which affect everyone in the community. I have been here for 3 years and working in every single area here. I have a connection to all of those pages long before Abd, and Abd's only actions have been in attempts to change policy or the encouraging of banned users to continue the actions that got them banned. Ottava Rima (talk) 14:40, 27 November 2011 (UTC)
 * Oppose. Big surprise? Evidence behind this suggestion? And is RCA a page to be used to decide a ban? Normally, that would take a Community Review, this is a page to request custodian action, i.e., things that can properly be done ad hoc, such as a short block for incivility. It is not a community discussion page, though certain people tend to use it that way. I've even done it myself, fallen into discussion here.... Uh, where did I attempt to "change" policy other than through seeking consensus? --Abd 19:15, 27 November 2011 (UTC)

Discussion
Well I see Abd and Ottava Rima both think editing privileges need to be revoked or restricted again. I suggest both be blocked on the grounds that Wikiversity's mission is best achieved when participants aim for mutual understanding and focus on solutions, and concerns like this are best resolved through discussion with the other person first. I suggest both of them remain blocked until demonstrated they mutually understand each other, and they come up with a solution for working together that they both agree with, instead of having custodians or the community get involved. -- dark lama  16:19, 27 November 2011 (UTC)
 * ✅. This seems to be the most reasonable suggestion made on this page. --SB_Johnny talk 20:54, 27 November 2011 (UTC)
 * Darklama, please address the point of the request here. Evidence was provided of total disregard of warnings, with the kind of monomaniacal attack that SB_Johnny mentioned. I've not harassed Ottava; the example that was considered harassment, in SBJ's earlier comment, was a suggestion to him that he be more careful about welcoming users, to look at the contributions. I was surprised to see his response, and I didn't repeat that. Ottava is not here to work on resources, nor to improve this community, that's become obvious. He is here for revenge. I asked for a block, only after he attacked, once again, with no attention from custodians who had warned him, suggesting I should be long-term blocked, that's cited above. This request did not come out of the blue, and it's actually brief compared to what could be shown. I'm suggesting that a custodian might avoid the need for a Community Review, that's all, if the custodian sees the matter as sufficiently obvious. Otherwise, there was a CR closed because Ottava had stopped editing. Should it be re-opened? If I've been disruptive, I should be warned and sanctioned if necessary. But, I'd hope, this would be based on *actual evidence*, not just loud claims. --Abd 19:22, 27 November 2011 (UTC)
 * Really? So we should allow people to edit war on a policy to claim a false consensus? How does that make any sense? Thenub already pointed out that Abd was wrong, and SB Johnny pointed that out before. That is two admin who previously told Abd to stop, which makes my pointing out that Abd is acting inappropriately the correct course of action. Your statement is, by definition, wheeling against that. Why have you been doing that quite a bit lately? - Admin have already taken action/made a statement then you act as if they haven't. That is really, really odd. Then there is your premise that Abd is capable of working anywhere here, which has been demonstrated constantly not to be true. I have actually provided content and did most of the organization and cleanup around here. Abd has verified that he is just here to disrupt, and even his welcoming was a political game that disappeared as soon as his adminship. Ottava Rima (talk) 16:27, 27 November 2011 (UTC)
 * Great everyone is satisfied with how custodians responded, why is any further action requested or supported here then? -- dark lama  16:51, 27 November 2011 (UTC)
 * ...What? No one is satisfied by Abd's actions. He repeatedly over many months insisted on making a page that was never consented to by the community as a policy to be labelled as a policy, and continues his walls of text and "it is a policy because I determined it so" approach. He reported me for being "incivil" because I said labeling something as something it clearly is not is vandalism. On every other Wiki, labeling a page as a "policy" when it is not a policy is vandalism, reverted, and the person blocked. Why this is suddenly not true on Wikiversity is baffling. Ottava Rima (talk) 16:55, 27 November 2011 (UTC)
 * Policies started on Wikipedia with no votes. Some started here with no voting. "repeatedly over many months" refers to, I think, two edits separated by six months, and the first one obviously stood -- and it was discussed. Yes, Ottava was uncivil in calling "vandalism" an edit that supported the status quo on a page that had been labelled policy for four years. Even if Ottava were correct, such an edit could not be vandalism, and that kind of polemic is not to be tolerated. Blocked for reverting to the status quo? Why? Because it's "wrong," i.e, because Ottava says it's wrong? This isn't over the content position, and it's not about his revert itself. It's over the incivility, and that set of comments was just one very immediate example. Are more needed? There is enough reported there to justify some sanction, I have no doubt about that. Ottava has always justified his incivility by insisting that he's right. --Abd 19:31, 27 November 2011 (UTC)
 * "Status quo" does not mean community consensus nor not vandalism. There are plenty of pages on Wikipedia that say "penis" on them for over a year and yet it is removed as vandalism. There are certain things that are never acceptable no matter how long they have been present. Declaring something a policy without any community consensus to do that is one of those things. Ottava Rima (talk) 20:06, 27 November 2011 (UTC)

 * A note - as pointed out on IRC that I might not have been clear - when I mentioned "wheel warring", I was referring to 2 admin who previously determined that the custodianship "policy" was not yet a policy and reverted Abd before. By saying Darklama would be "wheel warring" was if he was to think that Abd wasn't committing an act of vandalism by labeling a page as a policy when it was never approved by the community as that. I do not know if Darklama actually will go that way, but it is an implication in the original post about blocking me. Ottava Rima (talk) 17:03, 27 November 2011 (UTC)

Block request
Perhaps a block for user again? He already has a rather long block log - his last block was 1 month. Trijnstel 17:16, 28 November 2011 (UTC)
 * FYI, I noticed this IP is an open proxy, therefore I globally blocked it for 1 year. Trijnstel 17:20, 28 November 2011 (UTC)
 * Okay. I've blocked it locally for three months. Thanks for your assistance, Trijnstel! Mathonius 17:25, 28 November 2011 (UTC)

Navigation Popups Gadget
Hi.
 * The version of the Navigation Popups Gadget (in preferences) installed here is very old (2007) and doesn't work properly for me - diffs/history don't pop up when hovered. (Google Chrome, at least).
 * I tested a more recent copy which I've been using in my MediaWiki installation, tested using User:Begoon/vector.css and User:Begoon/vector.js, (it's the only thing in those 2 pages at the moment) and it seems to be working fine here.
 * (The version in my user js also has a fix to stop it annoyingly popping up on the edit toolbar - so is not identical to any WP version.)
 * To fix it for everyone, you would need to update MediaWiki:Gadget-popups.js and MediaWiki:Gadget-popups.css - which I can't do because its in the MW namespace, and only sysops can edit that namespace. Begoon  -  talk  13:39, 29 November 2011 (UTC)
 * I can make those edits, but I'd like to wait a little before doing so. What's the simplest snippet of code we can use to fix the problem? Your .js files look substantially different...I'm not sure if we should substitute the entire piece of code over to Wikiversity until we've tested it more. --HappyCamper 16:28, 29 November 2011 (UTC)
 * OK, well looks like another custodian is being bold and updating things from Wikipedia right now. Cheers. --HappyCamper 16:31, 29 November 2011 (UTC)
 * I have updated to the latest version at Wikipedia. diffs/history work when hovered now (Firefox, at least). -- dark lama  18:35, 29 November 2011 (UTC)
 * Wow - thanks for doing that so quickly - it works great as far as I can see - cool. The "annoyance" I mentioned above was that the Navpop will "pop" over the links like "Special Characters" and some custom toolbars in the edit toolbar, while you are editing a page, with a useless pop up link. This can hide the buttons below if you're clumsy like me. The diff for where I fixed that in my copy is - which basically turns off PopUps in edit mode. It's not much of an issue, though, so if you prefer to keep the base code in line with WP, that would probably make more sense for code management - I can always hack that behaviour in my user js - it's really a user preference anyway. Thanks again.  Begoon  -  talk  00:21, 30 November 2011 (UTC)

 * I would prefer to make the code more manageable by reducing its complexity, 240KB is rather large for one Gadget. -- dark lama  01:05, 30 November 2011 (UTC)
 * :-) I can understand that... It is a very complex gadget though, when you consider how much it provides. It works perfectly now, and very useful. (the edit form hack doesn't work in my user js, but it's a very minor thing - I'm living with it - not worth any extra effort for a tiny niggle. I guess the problem with simplifying the gadget base code (aside from the effort in doing so) would be that once you "fork" it from the WP code, you need to maintain the "fork" as an extra task, when it inevitably needs updating one day because a new "skin" becomes default, or the base MW code necessitates changes to the gadget. (or copy the new "WP" version and "simplify" all over again...:-)) Begoon  -  talk  01:29, 30 November 2011 (UTC)

Page: The Beatles
This is linkspam, needs deleting. No Beatles songs have been released by copyright holders for free online streaming/download.
 * Refer: for similar incident.

thanks Begoon  -  talk  07:39, 1 December 2011 (UTC)
 * Nuked. --HappyCamper 10:15, 1 December 2011 (UTC)
 * Thanks :) Begoon  -  talk  14:13, 1 December 2011 (UTC)

How to Meditate
Dear Sir/Madam

I am currently developing the article 'How to Meditate'. I expect to spend a little time on it and what I am hoping is it might enjoy some protection from editing by others. Meditation is a specialist topic subject to a great deal of conflicting opinions. I have been practising meditation for about 45 years and I would like to use Wikiversity as a suitable platform to present the topic in a non sectarian and logical way. I am quite agreeable if people wish to add their thoughts and experience to my offering but that they do not have the capacity to alter and edit my text. Meditation is a difficult thing and greatly misunderstood. I am hoping to clarify things, as much as I am able. My fear is there are plenty of individuals who would take my efforts as their opportunity to promote their own 'brand'. I am seeking to produce an article which rises above such.

Please look at what I have uploaded so far and let me know.

Sincerely

Marcus Zerbini (The preceding unsigned comment was added by Marcus Zerbini (talk • contribs) .)

 * Thanks for the note and good luck with your work. Unfortunately there is not way to protect pages so that just you can edit.  As a technical matter it is not possible to do what your asking with the software as it is.  Collaboration is at the heart of the project, so it is unclear that it would be in the projects interests to set such a precedent.  This is a rather a moot point, as we uanble as it is. Thenub314 22:53, 2 December 2011 (UTC)

Request certification of NEW accounts
Hi all custodians,

In Spring 2012, I will be teaching 2 courses: One undergraduate course with possibly well above 100 students, and one advanced graduate course EGM 6322 Spring 2012 with a much smaller number of students.

I appreciate if a custodian could certify the new accounts relatively quickly enough so students can begin to use their accounts and to upload files.

In particular, I created the account EGM 6322 Spring 2012 more than 2 weeks ago, back on 1 Dec 2011, but it still has not been certified. Past experience indicated that it took only a few days (2,3 days, at most 4 days) for a new account to be certified by a custodian.

I will create the course wiki for the undergraduate course EGM 4313 Spring 2012 soon.

Thanks,

Egm6321.f11 12:26, 15 December 2011 (UTC)

I just checked the class roll for the undergraduate course EGM 4313 Spring 2012: It currently has 163 students registered.

Egm6321.f11 12:37, 15 December 2011 (UTC)

 * I could possibly help here, but I'm not aware of any sort of special certification procedure for your accounts. What was done in the past for you? --HappyCamper 19:51, 15 December 2011 (UTC)

Thanks for certifying EGM 4313 Spring 2012 quickly. could you also certify EGM 6322 Spring 2012, which has been opened for more than 2 weeks already... ?  thanks. i called what you just did "certification", but maybe you use a different terminology for the same action. Egm4313.s12 22:32, 15 December 2011 (UTC)
 * Ah, OK I see. I think I know what is going on. What you mean by certification is just our standard method for greeting new users on Wikiversity. We'll definitely continue to welcome your students on Wikiversity whenever we see them. That message, however, doesn't have any bearing on whether the account can upload files. Whenever a new account is made on Wikiversity I think there is a grace period of 4 days and 10 edits before a user can upload files. The captcha should go away after that point. --HappyCamper 04:58, 16 December 2011 (UTC)

ok, great. now i understand the process. before i associated the greeting message with the ability to upload after 4 days. but your explanation clarifies the process. i now have a better explanation to students. Egm4313.s12 11:36, 16 December 2011 (UTC)

cs:Wikiverzita:Nástěnka správců fr:Wikiversité:Requêtes aux bibliothécaires pt:Wikiversidade:Pedidos a administradores

Deletion request
Please delete User:Onlineservice (it's blatant spam) and perhaps block the user. Thanks! Trijnstel 00:14, 20 December 2011 (UTC)
 * One more: User:Hygeia. Trijnstel 00:16, 20 December 2011 (UTC)
 * Thanks. --HappyCamper 01:42, 20 December 2011 (UTC)

And much more speedy deletion requests (everything is spam):

Trijnstel 19:48, 20 December 2011 (UTC)
 * BMI Calculator: Helps You Achieve The Perfect Shape
 * User:Bmiindexchartwomen
 * Identifying The BMI Index: How It Assists To Keep You In Good Shape
 * Psychic Reading- How to Perform Psychic Readings
 * Breakdown of www.psychicreviewonline.com
 * User:Thebestpsychicreviewonline
 * A Virtual Software For Projecting Based On Online Psychic Ratings
 * Find Your Current Psychic Studying Using Dependable Testimonials
 * How To Get The Particular Reliability Of The Online Psychic Viewer?
 * Discovering Legit Online Psychic Viewers
 * Thanks Trijnstel. --HappyCamper 20:30, 20 December 2011 (UTC)

Unblock requests
User talk:Abd and User talk:Ottava Rima have outstanding unblock requests. Custodian input could be helpful in resolving these requests. User talk:Poetlister is another outstanding unblock request. Please feel free to discuss here or on the respective talk pages. Sincerely, James -- Jtneill - Talk - c 01:13, 21 December 2011 (UTC)
 * Poetlister would be, if unblocked, still not able to edit as his account is globally locked (along with all other socks and per consensus on Meta). Poetlister1 is not locked though. Trijnstel 16:02, 21 December 2011 (UTC)

Deletion requests
Hi. Please delete Mooreandrewpage and User:Mooreandrew2411 because of crosswiki spam (the account is already globally locked). Trijnstel 14:24, 28 December 2011 (UTC)
 * ✅, thanks for the report. Mathonius 15:31, 28 December 2011 (UTC)

MediaWiki:Uploadtext
There's a typo in this system message. The <tt> before "in your learning resource" should be </tt>. Adambro 19:47, 28 December 2011 (UTC)
 * done, thanks. --SB_Johnny talk 20:17, 28 December 2011 (UTC)

Template:Pp-meta
On Template:Pp-meta, please replace Category:Wikipedia pages with incorrect protection templates with Category:Wikiversity pages with incorrect protection templates so that the correct category is used. Thanks. Adambro 16:06, 29 December 2011 (UTC)
 * OK I think it is done. --HappyCamper 19:22, 29 December 2011 (UTC)

A custodian may wish to consider deleting the two pages this user has created and it would probably be a good idea to keep an eye on their contributions. I understand that this is someone who has been creating hoax articles and inserting obviously false information into articles on Wikipedia and other sister projects including Wikispecies. There is more info here. Whilst Wikiversity has a broad scope, hoax articles here would seem similarly problematic as elsewhere. Adambro 16:27, 1 January 2012 (UTC)

 * I deleted those pages. Wikiversity's broad scope is not used to house everything under the sun. Not a perfect heuristic, but to get a sense of what this wiki is intented to nurture, I generally start with Featured and go from there. --HappyCamper 17:12, 1 January 2012 (UTC)


text:
User:Egm4313.s12.team11.arrieta/Report 2

Problem Statement

 * Develop the McLaurin Series (Taylor Series at t=0) for $$ e^t, \cos t , \sin t \!$$

Solution

 * $$ e^t= \sum_{n=0}^{\infty }\frac{t^n}{n!} \!$$

 * $$=\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+...\frac{t^n}{n!} \!$$

$$e^t=1+t+\frac{t^2}{4}+\frac{t^3}{6}+...\frac{t^n}{n!}\!$$

 * $$ \cos t= \sum_{k=0}^{\infty }\frac{(-1)^kt^{2k}}{(2k)!} \!$$

 * $$=\frac{(-1)^0t^{2(0)}}{(2(0))!}+\frac{(-1)^1t^{2(1)}}{(2(1))!}+\frac{(-1)^2t^{2(2)}}{(2(2))!}+\frac{(-1)^3t^{2(3)}}{(2(3))!}+...\frac{(-1)^kt^{2k}}{(2k)!} \!$$

$$\cos t=1-\frac{t^2}{2}+\frac{t^4}{24}+\frac{t^6}{720}...\frac{(-1)t^{2k}}{(2k)!} \!$$

 * $$ \sin t= \sum_{k=0}^{\infty }\frac{(-1)^kt^{2k+1}}{(2k+1)!} \!$$

 * $$=\frac{(-1)^0t^{2(0)+1}}{(2(0)+1)!}+\frac{(-1)^1t^{2(1)+1}}{(2(1)+1)!}+\frac{(-1)^2t^{2(2)+1}}{(2(2)+1)!}+\frac{(-1)^3t^{2(3)+1}}{(2(3)+1)!}+...\frac{(-1)^kt^{2k+1}}{(2k+1)!} \!$$

$$\sin t=t-\frac{t^3}{6}+\frac{t^5}{120}-\frac{t^7}{5040}+...\frac{(-1)^kt^{2k+1}}{(2k+1)!} \!$$

--Egm4313.s12.team11.arrieta 17:07, 6 February 2012 (UTC)


text:
User:Egm4313.s12.team3.pliskow/R2work/Problem8PartAVerify

Verification of General Solution: Report 2, Problem 8, Part A

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

$$\displaystyle y''+y'+3.25y=0 $$     (8.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Let's verify that the solution to the above ODE is indeed what was determined:

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

$$\displaystyle y=e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Before verifying that (1) satisfies the original ODE (8.1), we must calculate the first derivative:

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

$$\displaystyle y'=e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y'=e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B] $$    (3) And the second derivative:
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y''=e^{-\frac{x}{2}}[-3Acos(\sqrt{3}x)-3Bsin(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\sqrt{3}Asin(\sqrt{3}x)+\sqrt{3}Bcos(\sqrt{3}x)]+\frac{1}{4}e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y''=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B] $$    (5) With the derivatives calculated, we can check to see if it satisfies the original ODE.
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\frac{\sqrt{3}}{2}B-\frac{\sqrt{3}}{2}B+\frac{1}{4}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\frac{\sqrt{3}}{2}A+\frac{\sqrt{3}}{2}A+\frac{1}{4}B]+ e^{-\frac{x}{2}}cos(\sqrt{3}x)[\sqrt{3}B-\frac{1}{2}A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-\sqrt{3}A-\frac{1}{2}B]+3.25e^{-\frac{x}{2}}[Acos(\sqrt{3}x)+Bsin(\sqrt{3}x)] $$    (6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A-\sqrt{3}B+\frac{1}{4}A+\sqrt{3}B-\frac{1}{2}A+3.25A]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[-3B+\sqrt{3}A+\frac{1}{4}B-\sqrt{3}A-\frac{1}{2}B+3.25B]
 * style="width:95%" |
 * style="width:95%" |

$$    (7)
 *
 * }

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

$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[-3A+\frac{1}{4}A-\frac{1}{2}A+3.25A-\sqrt{3}B+\sqrt{3}B]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[\sqrt{3}A-\sqrt{3}A+\frac{1}{4}B-\frac{1}{2}B+3.25B-3B] $$    (8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y''+y'+3.25y=e^{-\frac{x}{2}}cos(\sqrt{3}x)[0]+e^{-\frac{x}{2}}sin(\sqrt{3}x)[0] $$    (8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$\displaystyle y''+y'+3.25y=0 $$    (8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

Since (8) matches the form of the original ODE, we can conclude that the general solution does indeed satisfy the ODE.
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }


text:
User:Kyexz

well I'm 19 years old my birthday 26 feb born on the 92 join wikiversity for 2 reasons trying to look for a little bit more of wise dome and the meaning of life hah

title: User:Kyexz


text:
User:EGM4313.s12.team6.davis/R2

R2.3
This problem consists of finding a general solution to two linear, second order differential equations with constant coefficients. The problems are taken from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics and can be found as problems 3 and 4 of problem set 2.2 on page 59.

Given:
$$y''+6y'+8.96y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let   $$y=e^{\lambda x}$$

Therefore:

 * $$y'=\lambda e^{\lambda x}$$
 * and
 * $$y''=\lambda^2 e^{\lambda x}$$

So,

 * $$y''+6y'+8.96y=0$$

can be written as

 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=6$$   and   $$b=8.96$$

Since the exponential $$e^{\lambda x}$$ can never equal zero,

 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :

 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values:

 * $$\lambda =\frac{-(6)\pm \sqrt[]{(6)^2-4(8.96)}}{2}$$

Which evaluates to:

 * $$\lambda =-3\pm \frac{\sqrt[]{0.16}}{2}$$

 * or

 * $$\lambda =-3\pm 0.2$$

Therefore, we have:

 * $$\lambda_{1}=-2.8$$

and
 * $$\lambda_{2}=-3.2$$

So,
 * $$y_{1}=e^{-2.8x}$$

and
 * $$y_{2}=e^{-3.2x}$$

Superimposing the two solutions gives the homogeneous solution:

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

$$y=C_{1}e^{-2.8X}+C_{2}e^{-3.2x}$$
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

Given:
$$y''+6y'+(\pi^2+4)y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let   $$y=e^{\lambda x}$$

Therefore:

 * $$y'=\lambda e^{\lambda x}$$
 * and
 * $$y''=\lambda^2 e^{\lambda x}$$

So,

 * $$y''+6y'+(\pi^2+4)y=0$$

can be written as

 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=6$$   and   $$b=\pi^2+4$$

Since the exponential $$e^{\lambda x}$$ can never equal zero,

 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :

 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values:

 * $$\lambda =\frac{-(6)\pm \sqrt[]{(6)^2-4(\pi^2+4)}}{2}$$

Which evaluates to:

 * $$\lambda =-3\pm \frac{\sqrt[]{-19.478}}{2}$$

 * or

 * $$\lambda =-3\pm 2.207i$$

Since the discriminant is less than zero we let $$\omega=2.207$$ and the homogeneous solution will be of the form:

 * $$y=e^{\frac{-ax}{2}}(Acos(\omega x)+Bsin(\omega x))$$

Substituting values we have the homogeneous solution:

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

$$y=e^(Acos(2.21 x)+Bsin(2.21 x))$$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

R2.4
This problem consists of finding a general solution to two linear, second order differential equations with constant coefficients. The problems are taken from the tenth edition of Erwin Kreyszig's Advanced Engineering Mathematics and can be found as problems 5 and 6 of problem set 2.2 on page 59.

Given:
$$y''+2\pi y'+\pi^2y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
Use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let   $$y=e^{\lambda x}$$

Therefore:

 * $$y'=\lambda e^{\lambda x}$$
 * and
 * $$y''=\lambda^2 e^{\lambda x}$$

So,

 * $$y''+2\pi y'+\pi^2y=0$$

can be written as

 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=2\pi$$ and $$b=\pi^2$$

Since the exponential  $$e^{\lambda x}$$   can never equal zero,

 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :

 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values we can see that the discriminant is zero:

 * $$\lambda =\frac{-2\pi\pm \sqrt[]{4\pi^2-4\pi^2}}{2}$$

Therefore, only one solution can be found from this equation:

 * $$y_{1}=e^{-\pi x}$$

For the second solution reduction of order must be used:

Let:

 * $$y_{2}=uy_{1}$$

Then,

 * $$y_{2}'=u'y_{1}+y_{1}'u$$

and
 * $$y_{2}=uy_{1}+2u'y_{1}'+y_{1}''u$$

Substituting these into the original equation gives:

 * $$(uy_{1}+2u'y_{1}'+y_{1}u)+a(u'y_{1}+y_{1}'u)+ buy_{1}=0$$

Which simplifies to:
 * $$uy_{1}'+u'(2y_{1}'+ay_{1})+u(y_{1}+ay_{1}'+by_{1})=0$$

Since,

 * $$2y_{1}'=-ae^{\frac{-a}{2}x}=-ay_{1}$$

and
 * $$y_{1}''+ay_{1}'+by_{1}=0$$

What remains is:

 * $$u''y_{1}=0$$

or
 * $$u''=0$$

Therefore,

 * $$u=C_{1}x+C_{2}$$

Now we let  $$C_{1}=1$$   and   $$C_{2}=0$$   so that:

 * $$u=x$$

Now,

 * $$y_{2}=xy_{1}=xe^{-\pi x}$$

Using  $$y_{1}$$   and   $$y_{2}$$   we have the general solution:

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

$$y=(C_{1}+C_{2}x)e^{-\pi x}$$
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

Given:
$$10y''-32y'+25.6y=0$$

Find:
The homogeneous solution to the differential equation.

Solution:
First we put the equation into standard form by dividing through by 10 so that the equation becomes:

 * $$y''-3.2y'+2.56y=0$$

Now we can use the method of undetermined coefficients, also known as the method of trial solution, to solve the differential equation.

Let   $$y=e^{\lambda x}$$

Therefore:

 * $$y'=\lambda e^{\lambda x}$$
 * and
 * $$y''=\lambda^2 e^{\lambda x}$$

So,

 * $$y''-3.2y'+2.56y=0$$

can be written as

 * $$e^{\lambda x}(\lambda^2+a\lambda+b)=0$$  where   $$a=-3.2$$ and $$b=2.56$$

Since the exponential  $$e^{\lambda x}$$   can never equal zero,

 * $$\lambda^2+a\lambda+b=0$$

This equation can be solved for lambda using the quadratic equation :

 * $$\lambda =\frac{-a\pm \sqrt[]{a^2-4b}}{2}$$

Substituting values:

 * $$\lambda =\frac{-(-3.2)\pm \sqrt[]{(-3.2)^2-4(2.56)}}{2}$$

or
 * $$\lambda =\frac{3.2\pm \sqrt[]{0}}{2}$$

Which evaluates to:
 * $$\lambda = 1.6$$

Therefore, only one solution can be found from this equation:

 * $$y_{1}=e^{1.6x}$$

For the second solution reduction of order must be used:

Let:

 * $$y_{2}=uy_{1}$$

Then,

 * $$y_{2}'=u'y_{1}+y_{1}'u$$

and
 * $$y_{2}=uy_{1}+2u'y_{1}'+y_{1}''u$$

Substituting these into the original equation gives:

 * $$(uy_{1}+2u'y_{1}'+y_{1}u)+a(u'y_{1}+y_{1}'u)+ buy_{1}=0$$

Which simplifies to:
 * $$uy_{1}'+u'(2y_{1}'+ay_{1})+u(y_{1}+ay_{1}'+by_{1})=0$$

Since,

 * $$2y_{1}'=-ae^{\frac{-a}{2}x}=-ay_{1}$$

and
 * $$y_{1}''+ay_{1}'+by_{1}=0$$

What remains is:

 * $$u''y_{1}=0$$

or
 * $$u''=0$$

Therefore,

 * $$u=C_{1}x+C_{2}$$

Now we let  $$C_{1}=1$$   and   $$C_{2}=0$$   so that:

 * $$u=x$$

Now,

 * $$y_{2}=xy_{1}=xe^{1.6x}$$

Using  $$y_{1}$$   and   $$y_{2}$$   we have the general solution:

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

$$y=(C_{1}+C_{2}x)e^{1.6x}$$
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 *
 * }
 * }

title: User:EGM4313.s12.team6.davis/R2


text:
User:Egm4313.s12.team8.colocar/R2.5

Problem
Kreyszig 2011 p.59 pbs.16-17

FIND AN ODE $$ \displaystyle y'' + ay' + by = 0 $$    (1)   for the given basis.

16. $$ \displaystyle e^{2.6x}, e^{-4.3x} $$

17. $$ \displaystyle e^{- \sqrt{5}x}, xe^{- \sqrt{5}x} $$

Solution
 16.

Note: Given $$ \displaystyle y = e^{\lambda x} $$ as a solution of the second-order homogeneous linear ODE, the characteristic equation (or auxiliary equation) is defined as

$$ \displaystyle \lambda^2 + a\lambda + b = 0. $$  [K2011 p54]

The problem gives us two distinct real-roots:

$$ \displaystyle \lambda_{1} = 2.6 $$ and $$ \displaystyle \lambda_{2} = -4.3. $$

Solving the quadratic equation above gives us $$ \displaystyle \lambda_{1} = \frac {-a + \sqrt{a^2 - 4b}} 2 = \frac {1}{2} (-a + \sqrt{a^2 - 4b}) = 2.6, $$ and $$ \displaystyle \lambda_{2} = \frac {-a - \sqrt{a^2 - 4b}} 2 = \frac {1}{2} (-a - \sqrt{a^2 - 4b}) = -4.3 $$

or $$ \displaystyle \lambda_{1} = (-a + \sqrt{a^2 - 4b}) = 5.2, $$ and $$ \displaystyle \lambda_{2} = (-a - \sqrt{a^2 - 4b}) = -8.6. $$

Summing $$ \displaystyle \lambda_{1} $$ and $$ \displaystyle \lambda_{2} $$ we can eliminate the discriminant $$ -a + \sqrt{a^2 - 4b} $$ and solve for $$ \displaystyle a $$ $$ \displaystyle -2a = -3.4 $$ $$ \displaystyle a = 1.7 $$

Plugging $$ \displaystyle a $$ back into either $$ \displaystyle \lambda_1 $$ or $$ \displaystyle \lambda_2, $$ we can solve for $$ \displaystyle b $$

$$ \displaystyle -1.7 + \sqrt{1.7^2-4b} = 5.2$$ $$ \displaystyle     \sqrt{2.89-4b} = 6.9$$ $$ \displaystyle 2.89 - 4b = 44.61$$ $$ \displaystyle -4b = 44.72$$ $$ \displaystyle b = -11.18$$

Now plugging $$ \displaystyle a $$ and $$ \displaystyle b $$ back into the general ODE, we have $$ \displaystyle y'' + 1.7y' - 11.18y = 0 $$

 17.

Given the basis: $$ \displaystyle e^{- \sqrt{5}x}, xe^{- \sqrt{5}x}, $$ we know we get only one root $$ \displaystyle \lambda = \lambda_1 = \lambda_2 = - \frac {a}{2}. $$ [K2011 p.55]

Thus, the discriminant $$ \displaystyle a^2 - 4b = 0. $$    (2)

Knowing $$ \displaystyle \lambda = - \frac {a}{2} $$ for the given basis, we can say $$ \displaystyle e^{- \frac {a}{2} x} = e^{- \sqrt{5}x} $$

to solve for unknown $$ \displaystyle a. $$

$$ \displaystyle - \frac {a}{2} = - \sqrt {5} $$ $$ \displaystyle a = 2 \sqrt {5} $$

Plugging $$ \displaystyle a $$ into  (2)  gives us $$ \displaystyle ({2 \sqrt {5}})^2 - 4b = 0 $$ $$ \displaystyle b = \frac {20}{4} = 5. $$

Now substituting $$ \displaystyle a $$ and $$ \displaystyle b $$ into  (1)  gives us our ODE: $$ \displaystyle y'' + 2 \sqrt{5} y' + 5y = 0 $$

title: User:Egm4313.s12.team8.colocar/R2.5


text:
File:Report2Plot.JPEG

Licensing
for school project

title: File:Report2Plot.JPEG


text:
Commutator

Let A, B, ... be operators. Then the commutator of A and B is defined as [A,B]=AB-BA.

Let a, b, ... be constants, then identities include [f(x),x]	=	0

[A,A]	=	0

[A,B]	=	-[B,A]

[A,BC]	=	[A,B]C+B[A,C]

[AB,C]	=	[A,C]B+A[B,C]

[a+A,b+B]	=	[A,B]

[A+B,C+D]	=	[A,C]+[A,D]+[B,C]+[B,D].

Let A and B be tensors. Then [A,B]=delAB-delBA.

There is a related notion of commutator in the theory of groups. The commutator of two group elements A and B is ABA-1B-1, and two elements A and B are said to commute when their commutator is the identity element. When the group is a Lie group, the Lie bracket in its Lie algebra is an infinitesimal version of the group commutator. For instance, let A and B be square matrices, and let α(s) and β(t) be paths in the Lie group of nonsingular matrices which satisfy α(0)=β(0)	=	I

 * $$\left.\frac{\partial\alpha}{\partial s}\right|_{s=0}=A$$

 * $$\left.\frac{\partial\beta}{\partial t}\right|_{t=0}=B$$

then
 * $$\left.\frac{\partial}{\partial s}\frac{\partial}{\partial s}\alpha(s)\beta(t)\alpha^{-1}(s)\beta^{-1}(t)\right|_{s=0,t=0}=[A,B].$$

title: Commutator


text:
User:Egm4313.s12.team8.tclamb/R2

=R2.8 - Two ODEs with Complex Roots=

Problem Statement
Find a general solution. Check your answer by substitution.

Part A:
 * $$\displaystyle y'' + y' + 3.25y = 0$$

Part B:
 * $$y'' + 0.54 y' + \left(0.0729 + \pi\right) y = 0$$

Finding the General Solution
Assume solutions are proportional to $$e^{rt}$$:
 * $$e^{rt}\left(r^2 + r + 3.25\right) = 0$$

Observe $$e^{rt} \neq 0$$ for any $$t$$:
 * $$\displaystyle r^2 + r + 3.25 = 0$$

Solve for $$r$$:
 * $$r = -\frac{1}{2} \pm \sqrt{3} i$$

Substitute in $$r$$ for initial assumption:
 * $$y(t) = c_1 e^{\left(-\frac{1}{2} + \sqrt{3} i\right) t} + c_2 e^{\left(-\frac{1}{2} - \sqrt{3} i\right) t}$$

Expand the solution using Euler's identity ($$e^{a + bi}=e^{a}\left(\cos{b} + i\sin{b}\right)$$):
 * $$\begin{align}

y(t) &= c_1 e^{-\frac{1}{2} t} \left(\cos{\sqrt{3} t} + i\sin{\sqrt{3} t}\right) \\ &+ c_2 e^{-\frac{1}{2} t} \left(\cos{\sqrt{3} t} - i\sin{\sqrt{3} t}\right) \end{align}$$ Collect like terms:
 * $$\begin{align}

y(t) &= \left(c_1 + c_2\right) e^{-\frac{1}{2} t} \cos{\sqrt{3} t} \\ &+ \left(c_1 - c_2\right) i e^{-\frac{1}{2} t} \sin{\sqrt{3} t} \end{align}$$ Define new constants $$c'_1 = c_1 + c_2$$ and $$c'_2 = \left(c_1 - c_2\right) i$$: $$y(t) = c'_1 e^{-\frac{1}{2} t} \cos{\sqrt{3} t} + c'_2 e^{-\frac{1}{2} t} \sin{\sqrt{3} t}$$

Checking by Substitution
Determine $$3.25 y$$, $$y'$$, and $$y''$$ from $$y(t)$$:
 * $$\begin{array}{rllrlrllcrrlcll}

3.25 & y  & = e^{-\frac{1}{2} t} (   &  \frac{13}{4} & c'_1 &               &        & \cos{\sqrt{3} t} & + &   &           &      &   \frac{13}{4} & c'_2   & \sin{\sqrt{3} t} ) \\ & y' & = e^{-\frac{1}{2} t} ( ( & -\frac{1}{2}  & c'_1 & + \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} & + & ( & -\sqrt{3} & c'_1 & - \frac{1}{2}  & c'_2 ) & \sin{\sqrt{3} t} ) \\ & y'' & = e^{-\frac{1}{2} t} ( ( & -\frac{11}{4} & c'_1 & - \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} & + & ( & \sqrt{3} & c'_1 & - \frac{11}{4} & c'_2 ) & \sin{\sqrt{3} t} ) \end{array}$$ Substitute into original ODE:
 * $$\begin{array}{rrrcrllrlrlll}

( e^{-\frac{1}{2} t} & ( ( & -\frac{11}{4} & c'_1 & - \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} + ( & \sqrt{3} & c'_1 & - \frac{11}{4} & c'_2 ) & \sin{\sqrt{3} t} ) ) & \\ + \; ( e^{-\frac{1}{2} t} & ( ( & -\frac{1}{2} & c'_1 & + \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} + ( & -\sqrt{3} & c'_1 & - \frac{1}{2}  & c'_2 ) & \sin{\sqrt{3} t} ) ) & \\ + \; ( e^{-\frac{1}{2} t} & (  &  \frac{13}{4} & c'_1 &               &        & \cos{\sqrt{3} t} +   &           &      &   \frac{13}{4} & c'_2   & \sin{\sqrt{3} t} ) ) & = 0 \end{array}$$ Reduce: $$\displaystyle 0 = 0$$

Finding the General Solution
Assume solutions are proportional to $$e^{rt}$$:
 * $$e^{rt}\left(r^2 + 0.54 r + \left(0.0729 + \pi\right)\right) = 0$$

Observe $$e^{rt} \neq 0$$ for any $$t$$:
 * $$r^2 + 0.54 r + \left(0.0729 + \pi\right) = 0$$

Rearrange by completing the square:
 * $$\left(r + 0.27\right)^2 + \pi = 0$$

Solve for $$r$$:
 * $$r = -0.27 \pm \sqrt{\pi} i$$

Substitute in $$r$$ for initial assumption:
 * $$y(t) = c_1 e^{\left(-0.27 + \sqrt{\pi} i\right) t} + c_2 e^{\left(-0.27 - \sqrt{\pi} i\right) t}$$

Expand the solution using Euler's identity ($$e^{a + bi}=e^{a}\left(\cos{b} + i\sin{b}\right)$$):
 * $$\begin{align}

y(t) &= c_1 e^{-0.27 t} \left(\cos{\sqrt{\pi} t} + i\sin{\sqrt{\pi} t}\right) \\ &+ c_2 e^{-0.27 t} \left(\cos{\sqrt{\pi} t} - i\sin{\sqrt{\pi} t}\right) \end{align}$$ Collect like terms:
 * $$\begin{align}

y(t) &= \left(c_1 + c_2\right) e^{-0.27 t} \cos{\sqrt{\pi} t} \\ &+ \left(c_1 - c_2\right) i e^{-0.27 t} \sin{\sqrt{\pi} t} \end{align}$$ Define new constants $$c'_1 = c_1 + c_2$$ and $$c'_2 = \left(c_1 - c_2\right) i$$: $$\displaystyle y(t) = c'_1 e^{-0.27 t} \cos{\sqrt{\pi} t} + c'_2 e^{-0.27 t} \sin{\sqrt{\pi} t}$$

Checking by Substitution
Determine $$( 0.0729 + \pi ) y$$, $$0.54 y'$$, and $$y''$$ from $$y(t)$$:
 * $$\begin{array}{rllrlrllcrrlcll}

( 0.0729 + \pi ) & y  & = e^{-0.27 t} (   &  ( 0.0729 + \pi )  & c'_1 &                        &        & \cos{\sqrt{\pi} t} & + &   &                     &      & ( 0.0729 + \pi )  & c'_2   & \sin{\sqrt{\pi} t} ) \\ & y' & = e^{-0.27 t} ( ( &            -0.27   & c'_1 & + \;               \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( &                -\sqrt{\pi} & c'_1 &       - \; 0.27   & c'_2 ) & \sin{\sqrt{\pi} t} ) \\ 0.54 & y' & = e^{-0.27 t} ( ( & -2 \times  0.0729  & c'_1 & + \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( & - 2 \times 0.27 \sqrt{\pi} & c'_1 & - \; 2 \times 0.0729 & c'_2 ) & \sin{\sqrt{\pi} t} ) \\ & y'' & = e^{-0.27 t} ( ( & ( 0.0729 - \pi ) & c'_1 & - \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( &  2 \times 0.27 \sqrt{\pi} & c'_1 & + \; (0.0729 - \pi) & c'_2 ) & \sin{\sqrt{\pi} t} ) \end{array}$$ Substitute into original ODE:
 * $$\begin{array}{rrrlrlllrrllll}

( e^{-0.27 t} & ( ( & ( -\pi + 0.0729 )  & c'_1 & - \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + \; ( &   2 \times 0.27 \sqrt{\pi} & c'_1 & + \; ( 0.0729 - \pi) & c'_2 ) & \sin{\sqrt{\pi} t} ) ) & \\ + \; ( e^{-0.27 t} & ( ( & -2 \times 0.0729   & c'_1 & + \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + \; ( & - 2 \times 0.27 \sqrt{\pi} & c'_1 & - \; 2 \times 0.0729 & c'_2 ) & \sin{\sqrt{\pi} t} ) ) & \\ + \; ( e^{-0.27 t} & (  &  ( 0.0729 + \pi  )  & c'_1 &                               &        & \cos{\sqrt{\pi} t} & +   &                            &      & ( 0.0729 + \pi )  & c'_2   & \sin{\sqrt{\pi} t} ) ) & = 0 \end{array}$$ Reduce: $$\displaystyle 0 = 0$$

title: User:Egm4313.s12.team8.tclamb/R2


text:
File:Capture111.png

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title: File:Capture111.png