text:
User:Algamicagrat

This is my user page. Thank you.

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

title: User:Algamicagrat


text:
Fish

Fishes are animals that live in the water. All fishes are vertebrates and have a spinal column made of either cartilage or bone. Fishes breathe in the water through structures called gills, which are located on either side of the head.

Gills act as the "lungs" of fishes. Gills are fine structures with lots of tiny blood vessels, called capillaries, close to their surface. From these capillaries fish take in oxygen, and release carbon dioxide and other waste products when water passes over them.

Most fishes are covered in protective scales. These are different than the scales of reptiles and birds. Fish scales actually grow inside the skin of the fish, while reptile and bird scales grow on the surface of the skin.

Most fishes are ectothermic, meaning they rely on the environment around them for their body temperature. Some fishes like tuna are endothermic, and produce their own internal body heat.

Fishes have a simple two-chambered heart with one atrium and one ventricle. Blood is pumped from the heart to the gills.

In addition to gills, some fishes can take in oxygen from the air. These include lungfishes, gouramies and bettas, and some catfishes. They may swallow air and take in oxygen through their intestines (catfishes), or have specialized organs that are similar but no the same thing as our lungs.

Species of Fish
There are over 30,000 known living species of fish.

Some commonly known species of fish include: and many others.
 * Anchovy
 * Clownfish
 * Salmon
 * Sharks
 * goldfish
 * guppies

Fish-Human Relationships
Fish and humans have been interacting for many thousands of years.

The scientific study of fishes is called ichthyology, and a scientist that studies fishes is called an ichthyologist.

Fish are a very important food source for humans. In many times and locations, fish have provided the majority of the animal protein that people have access to.

Fish have been kept in captivity by humans for several thousand years. They have mostly been kept as a ready supply of food.

In the last few hundred years, fish have also been kept as ornamental pets. Goldfish, the first fish known to have been kept for ornamental purposes, have been kept and bred for their beauty in China for about 1000 years.

Angling, or fishing, has been practiced as a recreational sport for about 500 years, though no one is certain how it developed from a technique solely for collecting food into the funs sport it has become in many parts of the world today.

With human population growth, in some places more fish have been captured for food than wild populations can replace. Some species of food fish are threatened or in danger of extinction.

Fish vs. Fishes
In English, the word "fish" is properly used to refer to one individual fish, or many of the same kind of fish. For example, a salmon is a fish, and salmon are fish.

The word "fishes" refers to a group of fish of more than one species. For example, salmon and anchovies are fishes, and a group of salmon and anchovies are fishes.

External Resources

 * [FishBase] with information on almost every known species of fish.
 * [How to become an ichthyologist] with information on how to have a career working with fishes.
 * [Fish] on Wikipedia.

title: Fish


text:
Google

We use Google when we search the Internet. Google was originally called BackRub and was created by Larry Page and Sergey Brin, two Stanford University graduates. Google's basic technology is called PageRank.

Google is one of the helpful program in a computer and even in mobile phones. Google gives everything and anything you asked for. It is very helpful especially for school, office or business purposes. You can search everything in google even the answer to your assignment. Google contains everything like Facebook, Twitter, Tumblr, Instagram and YouTube, these are just an example of those famous sites in Google. You can also make your own site in google if you want to do a business, advertisement and even a personal site for your blogs or articles that you have written or if you want to share something to everyone. Google also provides images to visualize something you wanna know or to view something. Even daily News updates are in google, if you want to know everything from where you at or from where your love one's at. In google almost everything is possible.

Google is very useful when it comes for literature especially for those students who are doing their thesis or research. There's a lot of studies and research papers that was published in google that can help other to have their background study. It is also one way to lessen your stress in thinking of anything because google can give you definitions of words that sometimes confuse you. Google is a very useful source, it helps everyone with everything. Just keep in mind that Google was made to help everyone that is in need, just use google in a manner way not to humiliate or don't use Google in your evil plans.

Contents
List of all.
 * /App engine/
 * /Matrix/
 * /Search/
 * /Search and Wikiversity/
 * /SketchUp 6, review/
 * /Summer of Code 2010 Proposal: GIFT Conversion/
 * /Wave for education/
 * /Web designer/
 * /Wikiversity/

Synopsis
Many people think that if they create a website with a catchy name, millions of people are going to flood to their site from search engines like Google. Unfortunately, this is just a dream. People quickly realize their site does not even show up when they search Google.

At the heart of Google software is a system called PageRank. This gives every site on the Internet a rank from 0–10. So, how is this calculated? Well, the PageRank of your site is determined by the links to your web site. Each time somebody adds a link to your web site, Google interprets this as a vote for your site. The more links you have on your site, the more votes you get. But Google also looks a little deeper than just volumes of links. It analyzes the importance of the web site that has cast a vote for your site. Sites that Google assigns a high PageRank are considered important. When a highly ranked website links to your site it also increases the PageRank of your website.

To make your website visible on search engines, highly ranked websites must link to you.

title: Google


text:
User talk:Hydrogen 007

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title: User talk:Hydrogen 007


text:
Learning the abc's/A

A is the first letter in the alphabet.
 * Many words start with A such as:

Apple

Alphabet

Amethyst

Amtrack

Airplane

 * Words also have A in the middle such as:

Has

Man

Can

Fan

Jam

 * Words with A in the end:

Banana

Camera

title: Learning the abc's/A


text:
Learning the abc's/B

B is the second letter in the alphabet. Baby
 * Many words start with B such as:

Bottle

Bird

Box

Ball

Boat
 * Words also have B in the middle such as:

Cabin

Lobby

Mobile

Robin

Sobbing
 * Words with B in the end:

Verb

Curb

Sob

Job

Mob

title: Learning the abc's/B


text:
User:Egm4313.s12.team1.stewart

Chris Stewart's Wiki Page

Team Page

Course Wiki

Reports
Report 1

Report 2

6.6
From [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec10-1.djvu Lecture 10 Pg. 9]

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$$  \displaystyle y''_{p}+4y'_{p}+13y_{p}=2e^{-2x}cos(3x) $$     (6.1) To find the particular solution of an ODE we can use the method of undetermined coefficients. In report 3 we used the table 2.1 below to find the standard form of the particular solution given the form of the excitation $$r(x)\!$$. Table 2.1
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So the general form of the particular solution is:
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$$  \displaystyle y_{p}(x)=xe^{-2x}[Mcos3x+Nsin3x] $$     (6.2)
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Task 1 asks to simplify they first term of the given ODE which is $$y_{p}''(x)\!$$. This is done by taking the second derivative of $$y_{p}(x)\!$$ and plugging it into equation 6.1
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$$  \displaystyle y''_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.3)
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$$y_{p}''(x)\!$$ simplifies to:
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$$  \displaystyle y''_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.4)
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Now we must do the same for the second term $$4y_{p}'(x)\!$$.
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$$  \displaystyle 4y'_{p}(x)=\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.5)
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$$  \displaystyle 4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)] $$     (6.6)
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Add the second term to the first to continue building the original ODE (equation 6.1).
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$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=\frac{\partial }{\partial x}\frac{\partial (xe^{-2x}[Mcos3x+Nsin3x])}{\partial x}+\frac{\partial (4xe^{-2x}[Mcos3x+Nsin3x])}{\partial x} $$ (6.7)
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$$  \displaystyle y''_{p}(x)+4y'_{p}(x)=4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)]) $$     (6.8)
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Finally add the third term $$13y_{p}(x)\!$$ to the other two to have the lhs of the ODE 6.1
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$$  \displaystyle 13y_{p}(x)=13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.9)
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$$  \displaystyle y''_{p}(x)+4y'_{p}(x)+13y_{p}(x)=e^{-2x}(sin(3x)[6M(2x-1)-N(5x+4)]-cos(3x)[M(5x+4)+6N(2x-1)])+4e^{-2x}[sin(3x)(-3Mx-2Nx+N)+cos(3x)(-2Mx+M+3Nx)]+13xe^{-2x}[Mcos3x+Nsin3x] $$     (6.10)
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This simplifies to the particular solution below:
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$$  \displaystyle y_{p}(x)=6e^{-2x}[Ncos(3x)-Msin(3x)] $$     (6.11)
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R5
'''[https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec8b.djvu See R5.7 Lect. 8b pg. 11]:'''

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$$\mathbf{v}=4\mathbf e_1+2\mathbf e_2=c_1\mathbf b_1+c_2\mathbf b_2$$
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The oblique basis vectors b1,b2 are:
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$$\mathbf b_1 =2\mathbf e_1 + 7\mathbf e_2$$
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$$\mathbf b_2 =1.5\mathbf e_1 + 3\mathbf e_2$$ 1. Find the components c1, c2 using the Gramian matrix. 2. Verify the result found above.
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R4.3

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$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 & -12 & 20 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & -15 & 30 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 2 & -18 & 42 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 &-21 & 56 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -24 & 72 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -27 & 90\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -30\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{bmatrix} \cdot \begin{pmatrix} {A}_{0}\\ {A}_{1}\\ {A}_{2}\\ {A}_{3}\\ {A}_{4}\\ {A}_{5}\\ {A}_{6}\\ {A}_{7}\\ {A}_{8}\\ {A}_{9}\\ {A}_{10}\\ \end{pmatrix} = \begin{Bmatrix} 0\\ 1\\ -1/2\\ 1/6\\ -1/24\\ 1/120\\ -1/720 \\ 1/5040\\ 1/40320\\ 1/362880\\ 1/3628800\\ \end{Bmatrix} $$     (3.30)
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Statement
4.1 from lecture notes [https://elearning2.courses.ufl.edu/access/content/group/UFL-EGM4313-5641-12012/Lecture%20Notes/iea.s12.sec7c-1.djvu R4.1 Lect. 7c pgs. 19-22]

Given the general form of polynomial excitation.

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$$  \displaystyle y''+ay'+by=\sum_{j=0}^{n}d_{j}x^{j} $$     (1.1)
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 *
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The particular solution that satisfies:

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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x^{j} $$     (1.2)
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 *
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The first and second derivative of the particular solution that solves the original polynomial excitation equation.

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$$  \displaystyle y_{p}'(x)=\sum_{j=0}^{n-1}c_{j+1}(j+1) x^{j} $$     (1.3)
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 *
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$$  \displaystyle y_{p}''(x)=\sum_{j=0}^{n-2}c_{j+2}(j+2)(j+1) x^{j} $$     (1.4)
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The particular solutions are put into the polynomial excitation equation to give the general summation form:

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$$  \displaystyle \sum_{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}(j+1)+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\sum_{j=0}^{n}d_{j}x^{j} $$     (1.5)
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Obtain the equations associated with $$d_{1}\!$$, coefficients of $$x\!$$; $$d_{2}\!$$, coefficients of $$x^{2}\!$$; $$d_{n-2}\!$$, coefficients of $$x^{n-2}\!$$; $$d_{n-1}\!$$, coefficients of $$x^{n-1}\!$$; $$d_{n}\!$$, coefficients of $$x^{n}\!$$. Five total equations for coefficients.

Also set up the matrix $$\mathbf{A}\!$$ that satisfies $$\mathbf{A}\mathbf{c}=\mathbf{d}\!$$.

Solution
The given equation associated with $$d_{0}\!$$ taking $$j=0\!$$

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$$  \displaystyle 2c_{2}+ac_{1}+bc_{0}=d_{0} $$     (1.6)
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Taking $$j=1\!$$
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$$  \displaystyle [c_{1+2}(1+2)(1+1)+ac_{1+1}(1+1)+bc_{1}]x^{1}=d_{1}x^{^{1}} $$     (1.7)
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$$  \displaystyle [6c_{3}+2ac_{2}+bc_{1}]x=d_{1}x $$     (1.8)
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The equation associated with $$d_{1}\!$$, coefficients of $$x\!$$
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$$  \displaystyle 6c_{3}+2ac_{2}+bc_{1}=d_{1} $$     (1.9)
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Taking $$j=2\!$$
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$$  \displaystyle [c_{2+2}(2+2)(2+1)+ac_{2+1}(2+1)+bc_{2}]x^{2}=d_{2}x^{^{2}} $$     (1.10)
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$$  \displaystyle [12c_{4}+3ac_{3}+bc_{2}]x^{2}=d_{2}x^{2} $$     (1.11)
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The equation associated with $$d_{2}\!$$, coefficients of $$x^{2}\!$$
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$$  \displaystyle 12c_{4}+3ac_{3}+bc_{2}=d_{2} $$     (1.12)
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Taking $$j=n-2\!$$
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$$  \displaystyle [c_{(n-2)+2}(n-2+2)(n-2+1)+ac_{(n-2+1)}(n-2+1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2} $$     (1.13)
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$$  \displaystyle [c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}]x^{n-2}=d_{n-2}x^{n-2} $$     (1.14)
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The equation associated with $$d_{n-2}\!$$, coefficients of $$x^{n-2}\!$$
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$$  \displaystyle c_{n}n(n-1)+ac_{(n-1)}(n-1)+bc_{n-2}=d_{n-2} $$     (1.15)
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Taking $$j=n-1\!$$
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$$  \displaystyle [c_{(n-1)+2}(n-1+2)(n-1+1)+ac_{(n-1)+1}(n-1+1)+bc_{(n-1)}]x^{n-1}+ac_{n}nx^{n-1}+bc_{n-1}x^{n-1}=d_{n-1}x^{n-1} $$     (1.16)
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$$  \displaystyle [c_{n+1}(n+1)(n)+2ac_{n}n+2bc_{(n-1)}]x^{n-1}=d_{n-1}x^{n-1} $$     (1.17)
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The equation associated with $$d_{n-1}\!$$, coefficients of $$x^{n-1}\!$$
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$$  \displaystyle c_{n+1}(n+1)(n)+2ac_{n}n+2bc_{(n-1)}=d_{n-1} $$     (1.18)
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Taking $$j=n\!$$.
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$$  \displaystyle [c_{n+2}(n+2)(n+1)+ac_{n+1}(n+1)+bc_{n}]x^{n}+bc_{n}x^{n}=d_{n}x^{^{n}} $$     (1.19)
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The equation associated with $$d_{n}\!$$, coefficients of $$x^{n}\!$$
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$$  \displaystyle c_{n+2}(n+2)(n+1)+ac_{n+1}(n+1)+2bc_{n}=d_{n} $$     (1.20)
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Now set up the equation:
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$$  \displaystyle \mathbf{A}\mathbf{c}=\mathbf{d} $$     (1.21)
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$$  \displaystyle \begin{bmatrix} Equation for d_{0}\\ Equation for d_{1}\\ Equation for d_{2}\\ ...\\ Equation for d_{n-2}\\ Equation for d_{n-1}\\ Equation for d_{n} \end{bmatrix}\begin{bmatrix} c_{0}\\ c_{1}\\ c_{2}\\ ...\\ c_{n-2}\\ c_{n-1}\\ c_{n} \end{bmatrix}=\begin{bmatrix} d_{0}\\ d_{1}\\ d_{2}\\ ...\\ d_{n-2}\\ d_{n-1}\\ d_{n} \end{bmatrix} $$     (1.22)
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Therefore the matrix $$\mathbf{A}\!$$ that satisfies the matrix equation is:
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$$  \displaystyle \mathbf{A}=\begin{bmatrix} b& a& 2&  &  &  & \\ & b&  2a&  6&  &  & \\ & &  b&  3a&  12&  & \\ ...& ...&  ...&  ...&  ...&  ...& ...\\  &  &  &  &  b&  a(n-1)& n(n-1)\\ & &  &  &  &  2b& 2an\\ & &  &  &  &  & 2b \end{bmatrix} $$     (1.23)
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Author
Solved and Typed By - Chris Stewart User:Egm4313.s12.team1.stewart -- 21:12, 11 March 2012 (UTC)

Reviewed By -

R3
Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for
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$$  \displaystyle y''-3y'+2y=4x^{2}-6x^{5} $$     (4.1) is of the form
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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_{j}x_{j} $$     (4.2) i.e.,
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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{5}c_{j}x_{j} $$     (4.3)
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Basic Rule: Select $$y_{p}(x)\!$$ from the table and determine the coefficients by substituting $$y_{p}(x)\!$$ in

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$$  \displaystyle y''+ay'+by=r(x) $$     (4.4) Sum Rule: If $$r(x)\!$$ is the sum of the terms in the 1st column of table 2.1 then $$y_{p}(x)\!$$ is the sum of the corresponding terms in the 2nd column of this table. Table 2.1
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Solution
According to the table the Homogeneous equation has two $$r(x)\!$$ values of the form $$kx^{n}(n=0,1,2...)$$. Using the Basic Rule this means that there is a particular solution of the form $$K_{n}x^{n}+K_{n-1}x^{n-1}+...+K_{1}x+K_{0}\!$$ for each $$r(x)\!$$.

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$$  \displaystyle r_{1}(x)=4x^{2} $$     (4.5)
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So the particular solution for this $$r_{1}(x)\!$$ should be:

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$$  \displaystyle y_{p1}= 4x^{2}+4x^{1}+4 $$     (4.6)
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Which simplifies to:

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$$  \displaystyle y_{p1}= \sum_{j=0}^{2}4x^{j} $$     (4.7)
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For the second $$r_{2}(x)\!$$:

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$$  \displaystyle r_{2}(x)=-6x^{5} $$     (4.8)
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So the particular solution for this $$r_{2}(x)\!$$ should be:

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$$  \displaystyle y_{p2}= -6x^{5}-6x^{4}-6x^{3}-6x^{2}-6x-6 $$     (4.9)
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Which simplifies to:

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$$  \displaystyle y_{p1}= \sum_{j=0}^{5}-6x^{j} $$     (4.10)
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Now using the Sum Rule which just states if there are two $$r(x)\!$$ values in any form on the left side of the table, then the particular solution is the sum of the solutions for $$r(x)\!$$ on the right side of the table.

So the particular solution for $$y''-3y'+2y=4x^{2}-6x^{5}\!$$ is:

 * {| style="width:100%" border="1"

$$  \displaystyle y_{p}=\sum_{j=0}^{2}4x^{j}+ \sum_{j=0}^{5}-6x^{j} $$     (4.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Statement
From K 2011 pg. 59 #3,4

Find the General Solution to the two ODEs below. (See Sec.5 Pg.5-6 notes)

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

$$  \displaystyle y''+6y'+8.96y=0 $$
 * style="width:95%" |
 * style="width:95%" |

 * }

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

$$  \displaystyle y''+4y'+(\pi ^2 +4)y=0 $$
 * style="width:95%" |
 * style="width:95%" |

 * }

Solution
Given:

The Homogeneous equation is given:

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

$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & -6 & 6 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 & -12 & 20 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 2 & -15 & 30 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 2 & -18 & 42 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 &-21 & 56 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -24 & 72 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -27 & 90\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & -30\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 \\ \end{bmatrix} \cdot \begin{pmatrix} {A}_{0}\\ {A}_{1}\\ {A}_{2}\\ {A}_{3}\\ {A}_{4}\\ {A}_{5}\\ {A}_{6}\\ {A}_{7}\\ {A}_{8}\\ {A}_{9}\\ {A}_{10}\\ \end{pmatrix} = \begin{Bmatrix} 0\\ 1\\ -1/2\\ 1/6\\ -1/24\\ 1/120\\ -1/720 \\ 1/5040\\ 1/40320\\ 1/362880\\ 1/3628800\\ \end{Bmatrix} $$     (3.0)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The first step is to change the equation into the characteristic equation.

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

$$  \displaystyle \lambda ^2+6\lambda +8.96=0 $$     (3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Next use either completing the square or the quadratic formula to solve for $$\lambda\!$$ (the roots of the characteristic equation). In this case completing the square is used.

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

$$  \displaystyle (\lambda ^2+6\lambda +9)+8.96=0+9 $$     (3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (\lambda +3)^2=.04 $$     (3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The roots of the characteristic equation:

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

$$  \displaystyle \lambda =-3\pm .2 $$     (3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now that the roots are known we can use the Homogeneous solutions in this equation:

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

$$  \displaystyle y=C_{1}e^{\lambda x}+C_{2}e^{\lambda x} $$ (3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Finally plug in the known roots and this is the final answer and general solution:

 * {| style="width:100%" border="1"

$$  \displaystyle y=C_{1}e^{-2.8x}+C_{2}e^{-3.2x} $$     (3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The Homogeneous equation is given:

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

$$  \displaystyle y''+4y'+(\pi ^2 +4)y=0 $$     (3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Again, the first step is to convert to the characteristic equation.

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

$$  \displaystyle \lambda ^2+4\lambda +(\pi ^2+4)=0 $$     (3.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

In order to solve for the roots ($$\lambda\!$$) either use the quadratic formula or completing the square. Once again in this case completing the was used to solve for $$\lambda\!$$.

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

$$  \displaystyle (\lambda ^2+4\lambda +4)+(\pi ^2+4)=0+4 $$     (3.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (\lambda +2)^2=4-(\pi ^2+4) $$     (3.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (\lambda +2)^2=-\pi ^2 $$     (3.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The root takes the square root of a negative one, therefore the complex number $$i\!$$ is in the solution. The roots of the characteristic equation is shown:

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

$$  \displaystyle \lambda =-2\pm \pi i $$ (3.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For complex roots the Homogeneous solution is different and the root must broken into components and plugged into the complex Homogeneous solution equation below.

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

$$  \displaystyle \lambda =a\pm bi $$ (3.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y=C_{1}e^{ax}cos(bx)+C_{2}e^{ax}sin(bx) $$     (3.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When plugging in $$\lambda\!$$ the final answer and general solution is:

$$  \displaystyle y=C_{1}e^{-2x}cos(\pi x)+C_{2}e^{-2x}sin(\pi x) $$ (3.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author
Solved and typed by - Egm4313.s12.team1.stewart 02:03, 4 February 2012 (UTC) Reviewed By -

Edited By -

title: User:Egm4313.s12.team1.stewart


text:
Big bang/A Crituqe of Assessing Big-Bang Nucleosynthesis

(Review Paper) Cited in

The 4He Measurements Predict Big Bang
They have found that the 4He measurements, which the theory underestimated is now vindicated by a more accurate theory.

Methods
They used Bayesian Methods to predict what the Baryon percentages are.

Results
They found the new theoretical amounts are 95% accurate with the measured amounts


text:
User:Vickyschneider gricar

Hi, My name is Vicky Schneider-Gricar and I work at The Genome Analysis Centre (TGAC) in Norwich, UK as Head of Training and Outreach. Prior to this I was the User Training Coordinator at EMBL-EBI. Before joining the EMBL-EBI in 2007 I held an Assistant Professor position at the University of Bern and the Institute for Aquatic Sciences (EAWAG) and postdocs at the University of Zurich and University of Rome (Torvergata). I studied biology at the University of Rome and obtained her PhD on the evolutionary biology at the University of Leiden (NL) and Lyon (France).

title: User:Vickyschneider gricar


text:
User:Egm4313.s12.team10.sa

http://en.wikiversity.org/wiki/User:Egm4313.s12.team10

title: User:Egm4313.s12.team10.sa


text:
User:Egm4313.s12.team5.wknapper

Test

title: User:Egm4313.s12.team5.wknapper


text:
User:Egm4313.s12.team5.bell

This is a test. 1 2 3

Figure 1

Figure 2

title: User:Egm4313.s12.team5.bell


text:
User:Egm4313.s12.team15.r1

<Team 15

REPORT 1

= R1.1 =

Given
Derive the equation of motion of a spring-dashpot system in parallel (see Figure 1 below), with a mass and applied force $$f(t).$$ Figure 1

Find
The equation of motion of the system.

Solution
Kinematics, Kinetics, and Relations equations as interpreted from picture

Note: Notation of variables based upon p.1-4 notes.

Kinematics: $$y=y_{k}= y_{c}\!$$

Kinetics: $$f(t)=my'' + f_{I}\!$$ $$f_{I}\!$$ is the total damping force $$f_{I}=f_{k}+f_{c}\!$$

Relations:

-Force due to spring: $$f_{k}=ky_{k}\!$$ -Force due to dash-pot: $$f_{c}=cy'_{c}\!$$

Putting kinematics, kinetics, and relations together:

$$y=y_{k}=y''_{c}\!$$

$$f_{I}=f_{k}+f_{c}\!$$ Using kinetic relationship, replace variables to include relations equations: $$f(t)=my''_{c}+f_{k}+f_{c}\!$$

$$f_{k}=ky_{k}=ky_{c}\!$$ $$f_{c}=cy'_{c}\!$$

$$f(t)=my''_{c}+ky_{c}+cy'_{c}\!$$

'''Final Equation $$f(t)=my_{c}+ky_{c}+cy'_{c}\!$$'

Author
Solved and typed by Tim Pham 21:35, 25 January 2012 (UTC)

Reviewed By -Jenny Schulze

= R1.2 =

Given
Fig.53 K 2011 p.85

Find
Derive the equation of motion of the spring-mass-dashpot in Fig.52, in K 2011 p.85, with an applied force r(t) on the ball.

Solution
for ball:

$$F=ma=m\frac{\partial^2 y}{\partial t^2}=my''$$

for spring: (from hooke's law)

$$F=ky\!$$

for dashpot: (good approx for small velocities)

$$F=cy'\!$$

In this case, since there is an external force given it must be in equilibrum with the internal forces, it is called the driving force:

$$\sum F=r(t)$$

Thus:

$$my''+cy'+ky=r(t)\!$$

Author
Solved and typed by James Moncrief 01:28, 1 February 2012 (UTC)

Reviewed By - Jenny Schulze

= R1.3 =

Given
Spring-Dashpot-Mass System from Sec 1 lecture notes.

Find
Draw the free body diagrams for each component and derive the equation of motion for the mass. (Assume no additional force due to gravity.)

Solution
Free Body Diagrams

Equation of Motion for Mass

From Newton's Second Law:

$$\sum F = m\,\frac{\mathrm{d}v}{\mathrm{d}t} = ma$$

By definition:

$$a={y}''\!$$

So:

$$\sum F = m{y}''$$

From Mass FBD:

$$\sum F = f(t)- f_{I}$$

Combining the two equations:

$$m{y}'' = f(t)- f_{I}\!$$

'''Final equation after rearranging $$m{y} + f_{I} = f(t)\!$$'

Author
Solved and typed by Neil Tidwell 00:14, 29 January 2012 (UTC)

Reviewed By - Jenny Schulze

= R1.4 =

Given
$$ V=LC\frac{d^{2}v_{c}}{dt^{2}}+RC\frac{dv_{c}}{dt}+v_{c} \! $$ (1)

Find
Refer to Lecture 2 notes.

Derive (3) and (4) from (2).

Solution
Using $$I=C\frac{dv_{c}}{dt} \! $$ (2) & $$I'=C\frac{d^{2}v_{c}}{dt^{2}} \! $$(3)

After substituting equations (2 and (3) equation (1) becomes: $$V=LI'+RI+v_{c} \! $$ (4)

Deriving equation (4) yields: $$V'=LI''+RI'+\frac{1}{C}I\! $$

Note:

$$ \frac{dv_{c}}{dt}=\frac{1}{C}I\! $$

Using $$I=\frac{dQ}{dt} \! $$(5)

Deriving equation (5) becomes: $$I'=\frac{dQ^{2}}{dt^{2}} \! $$(6)

Substituting equations (5) and (6) into equation (4) yields: $$V=L\frac{dQ^{2}}{dt^{2}}+R\frac{dQ}{dt}+\frac{1}{C}Q\! $$

Note:

$$v_{c}=\frac{1}{C}Q=\frac{1}{C}\int I dt\! $$

Author
Solved and typed by - Jerry Shugart Reviewed By - Jenny Schulze

= R1.5 =

Given
Refer to the textbook Advanced Engineering Mathematics, 10th edition by Kreyszig, page 59.

$$ y''+ay'+by=0 \! $$

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

Roots:

$$ \lambda _1=\frac{1}{2}\sqrt{-a+(a^2-4b)} \!$$

$$ \lambda _2=\frac{1}{2}\sqrt{-a-(a^2-4b)} \!$$

Imaginary:

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

Double:

$$ y=(c_1+c_2 x)e^{-ax/2} \! $$

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

5) $$ y''+2\pi y'+\pi ^{2}y=0 \!$$

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

$$y{}''+ay{}'+by=0 \!$$

$$a=4 \!$$

$$b=(\pi ^{2}+4) \!$$

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

$$\lambda^2+4\lambda+(\pi ^{2}+4)=0 \!$$

$$\lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b}) \!$$

$$\lambda_1=\frac{1}{2}(-4+\sqrt{4^2-4(\pi ^{2}+4)})=\frac{1}{2}(-4+\sqrt{-39.4784})=-2+\frac{1}{2}i(-39.4784) \!$$

$$\omega=39.4784 \!$$

$$\lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b}) \!$$

$$\lambda_2=\frac{1}{2}(-4-\sqrt{4^2-4(\pi ^{2}+4)})=\frac{1}{2}(-4-(-39.4784)^.5)=-2-\frac{1}{2}i(-39.4784) \!$$

$$y=e^{-ax/2}(Acos(\omega x)+Bsin(\omega x))=e^{-2x}(Acos(39.4784x)+Bsin(39.4784x)) \!$$

5) $$y{}''+2\pi y{}'+\pi ^{2}y=0 \!$$

$$y''+ay'+by=0 \!$$

$$a=2\pi \!$$

$$ b=\pi ^{2} \!$$

$$\lambda^{2}+a\lambda+b=0 \!$$

$$\lambda^{2}+(2\pi)\lambda+(\pi ^{2})=0 \!$$

$$\lambda_1=\frac{1}{2}(-a+\sqrt{a^2-4b}) \!$$

$$\lambda_1=\frac{1}{2}(-2\pi +\sqrt{(2\pi) ^{2}-4(\pi ^{2})})= -\pi\!$$

$$\omega=39.4784 \!$$

$$\lambda_2=\frac{1}{2}(-a-\sqrt{a^2-4b}) \!$$

$$\lambda_2=\frac{1}{2}(-2\pi -\sqrt{(2\pi) ^2-4(\pi^{2})})= -\pi\!$$

$$ y=(c_1+c_2 x)e^{-ax/2} \! $$

$$ y=(c_1+c_2 x)e^{-(2\pi)x/2}= (c_1+c_2 x)e^{-(\pi)x}\! $$

Author
Solved and typed by - Kristin Howe

Reviewed By - Jenny Schulze

= R1.6 =

Given
R1.6.1 $$ y''= g = const \!$$ R1.6.2 $$ mv'= mg -bv^2 \!$$ R1.6.3 $$ h'=-k\sqrt{h}\!$$ R1.6.4 $$ my''+ky = 0 \!$$ R1.6.5 $$ y'' +w_o^2y=coswt    \  \ \  \ \  \ \  \ \  \ \  \ \  \  w_o = w \!$$ R1.6.6 $$ LI'' + RI' + \frac{1}{c}I =E' \!$$ R1.6.7 $$ EI_y^{iv} = f(x) \!$$ R1.6.8 $$ L{\theta }'' + gsin\theta = 0 \!$$

Find
For each O.D.E. in Fig. 2 ( except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show whether the principle of superposition can be applied.

Solution
Refer to the lecture notes for additional information on the Superposition Principle: Section 2 Lecture Notes

 * Note : Lecture notes were referred to obtain the solutions below.

R1.6.1

Order: 2

Linearity: Linear

Superposition : Yes

$$ y''= g = const  \!$$   (Eq.1) $$ y''_h=0 \!$$       Homogeneous solution (Eq.2) $$ y''_p=g \!$$      Particular solution (Eq.3) $$ \bar{y}=y_h + y_p = (y_h + y_p) = 0 +g = g \!$$  (Eq.4) $$ \bar{y''}=g \!$$

Since (Eq.4) = (Eq.1) Therefore, superposition can be applied.

R1.6.2

Order: 1

Linearity: Non-linear

Superposition: No $$ mv'= mg - bv^2  \!$$  (Eq.0) $$ mv' + bv^2 = mg  \!$$  (Eq.0 Rearranged ) $$ mv_h'+b(v_h)^2=0 \!$$ Homogeneous solution (Eq. 1 ) $$ mv_p'+b(v_p)^2=mg \!$$ Particular solution (Eq. 2 )

By solving we obtain the following:

$$\bar{v}=v_h+v_p\!$$

Summing Eq.1 & Eq. 2 :

$$ m(v_h'+v_p')+ b[(v_h)^2 + (v_p)^2]=mg \!$$

$$m(v_h +v_p)' +b[(v_h)^2 +(v_p)^2]=mg\!$$

$$\bar{v'} + b(v_h+v_p)^2=mg\!$$

$$(v_h+v_p)^2\neq\bar{v}^2 \!$$

Therefore, superposition cannot be applied.

R1.6.3

Order: 1

Linearity:Non- linear

Superposition: No

$$ h'=-k\sqrt{h}\!$$  (Eq.0) $$ h'+k\sqrt{h}=0\!$$  (Eq.0 Rearranged) $${h_h}'+k\sqrt{h_h}= 0 \!$$ Homogeneous solution (Eq. 1 ) $${h_p}'+k\sqrt{h_p}= 0\!$$ Particular solution (Eq. 2)

By solving the following is obtained:

$$\bar{h}=h_h+h_p\!$$

Eq.1 & Eq. 2 are added:

$${h_h}'+k\sqrt{h_h}+{h_p}'+k\sqrt{h_p}=0\!$$

$$({h_h}'+{h_p}')+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$({h_h}+{h_p}) '+k(\sqrt{h_h}+\sqrt{h_p})=0\!$$

$$(\sqrt{h_h}+\sqrt{h_p})\neq \sqrt\bar{h}\!$$

Thus, superposition cannot be applied.

R1.6.4

Order:2

Linearity: Linear

Superposition: Yes

$$ my''+ky = 0 \!$$  (Eq.0) $$m{y}_h''+ky_h= 0\!$$ Homogeneous solution (Eq. 1 ) $$m{y}_p''+ky_p= 0\!$$ Particular solution (Eq. 2)

By solving the following is obtained:

$$\bar{y}=y_h+y_p\!$$

$$\bar{y}=y_h+y_p''\!$$

Summing Eq.1 & Eq. 2 :

$$m({y}_h+{y}_p)''+k(y_h+y_p) = 0\!$$

$$m\bar{y}''+k\bar{y}= 0\!$$

$$m\bar{y}+k\bar{y}$$ = $$ my+ky \!$$

Therefore, superposition can be applied.

R1.6.5

Order: 2

Linearity: Linear

Superposition: Yes

$$ y'' +w_o^2y=cos wt   \  \ \  \ \  \ \  \ \  \ \  \ \  \  w_o = w \!$$

$$ y_h'' +w_o^2y_h=0\!$$
Homogeneous solution (Eq. 1 )

$$ y_p'' +w_o^2y_p=cos wt \!$$
Particular solution (Eq. 2)

By solving we obtain the following:

$$\bar{y}=y_h+y_p\!$$

$$\bar{y}=y_h+y_p''\!$$

Adding Eq.1 & Eq.2 :

$$y_h+w_o^2y_h+y_p+ w_o^2y_p = cos wt\!$$

$$( y_h+y_p)''+w_o^2(y_h+y_p)= cos wt\!$$

$$\bar{y}''+w_o^2\bar{y}= cos wt\!$$

Therefore, superposition can be applied.

R1.6.6

Order: 2

Linearity : Linear

Superposition: Yes

$$ LI'' + RI' + \frac{1}{c}I =E' \!$$

$$ LI_h'' + RI_h' + \frac{1}{c}I_h =0 \!$$
Homogeneous solution (Eq. 1 )

$$ LI_p'' + RI_p' + \frac{1}{c}I_p =E' \!$$
Particular solution (Eq. 2 )

By solving we obtain the following:

$$\bar{I}= (I_h+I_p)\!$$

$$\bar{I}'= (I_h+I_p) '\!$$

$$\bar{I}= (I_h+I_p) \!$$

Adding Eq.1 & Eq. 2:

$$L(I_h+I_p) ''+R(I_h+I_p) ' + \frac{1}{c}(I_h+I_p) = E'\!$$

$$L\bar{I}''+R\bar{I}'+\frac{1}{c}\bar{I}= E'\!$$

Therefore, superposition can be applied.

R1.6.7

Order: 4

Linearity :Linear

Superposition : Yes

$$ EIy^{iv} = f(x) \!$$

$$ y_h= EIy_h^{iv}=0\!$$
 Homogeneous solution (Eq. 1 )

$$ y_p= EIy_p^{iv}=f(x)\!$$
 Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{y}=y_h+y_p\!$$

Summing Eq.1 & Eq. 2 :

$$ EIy_h^{iv} + EIy_p^{iv} = f(x)\!$$

$$ EI(y_h + y_p)^{iv}= f(x)\!$$

$$ EI\bar{y}^{iv} = f(x)\!$$

Therefore, superposition can be applied.

R1.6.8

Order: 2

Linearity: Non-linear

Superposition: No

$$ L{\theta }'' + gsin\theta = 0 \!$$ $$ L{\theta_h }'' + gsin\theta_h = 0 \!$$
 Homogeneous solution (Eq. 1 ) $$ L{\theta_p }'' + gsin\theta_p = f(x) \!$$
 Particular solution (Eq. 2 )

By solving the following is obtained:

$$\bar{\theta}= {\theta}_h + {\theta}_p \!$$

Summing Eq. 1 & Eq. 2 :

$$L{\theta_h + \theta_p}''+gsin{\theta_h+\theta_p}=0\!$$

$$\bar{\theta}\neq\ {\theta}_h + {\theta}_p \!$$

Therefore, superposition cannot be applied.

Author
Solved and typed by - Cynthia Hernandez and Jenny Schulze

Reviewed By - Jenny Schulze

= Contributing Members =

Table code referenced from Team 1

title: User:Egm4313.s12.team15.r1


text:
User:Egm4313.s12.team1.silvestri

Emotion Silvestri Team page

Course wiki

Model reports

Lecture notes/video

equation editor

$$ \sum_{j=2}^{5} C_{j} j(j-1) x^{^{j-2}}=\sum_{j=0}^{3} C_{j+2} (j+2)(j+1) x^{^{j}} \!$$ Egm4313.s12.team1.silvestri 19:31, 8 February 2012 (UTC)

title: User:Egm4313.s12.team1.silvestri


text:
User:Mattski au/Creativity

This page looks at "creativity" in relation to Creativity and Mindfulness research proposal.

What is it?
Creativity is a subjective experience, which is both process and product orientated. It is usually seen as a person’s capacity to produce an idea, product, insight or artistic substance which are evaluated by experts as being of high scientific, aesthetic, social or technical value. However, there is no universal agreement on creativity and the term involves many factors and exisits across many disciplines, domains or fields.

According to Csikszentmihalyi creativity is a social construct and is an interaction among a domain, a person and a field.

title: User:Mattski au/Creativity


text:
User:Mattski au/Mindfulness

This page looks at mindfulness in relation to Creativity and Mindfulness research proposal.

What is it?
Mindfulness is a non-judgmental psychological state, where a person is aware and accepting of each thought and sensory perception that arises on a moment-to-moment basis. This state allows a person to be consciously aware of their thoughts and feelings, and open to new experiences.

title: User:Mattski au/Mindfulness


text:
User:Egm4313.s12.team10.R1

To return to the Team 10 site click here = Problem 1: Equation of motion for a spring-dashpot system in parallel =

Given
Refer to Lecture slide [1-5] for problem statement. A spring-dashpot system in parallel, with a mass and an applied force f(t)

Find
The equation of motion of the system.

Solution
Free Body Diagram:

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

 * style="width:100%" |
 * $$\displaystyle {y}={y_k}+{y_c} $$
 * $$\displaystyle {y}={y_k}+{y_c} $$

 * style= |
 * }

Kinetics:

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{f_k}=k\overline{y} $$
 * $$\displaystyle \overline{f_k}=k\overline{y} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{f_c}=c\overline{{y}'} $$
 * $$\displaystyle \overline{f_c}=c\overline{{y}'} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle m\overline{a}=m\overline{{y}''} $$
 * $$\displaystyle m\overline{a}=m\overline{{y}''} $$

 * style= |
 * }

Final Solution:

By summing forces in the Y direction we obtain:
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \overline{f(t)}=m\overline{{y}''} +c\overline{{y}'}+k\overline{y} $$
 * $$\displaystyle \overline{f(t)}=m\overline{{y}''} +c\overline{{y}'}+k\overline{y} $$

 * style= |
 * }

= Problem 2: Equation of motion for a spring-dashpot system in series=

Given
Refer to Lecture slide [1-5] for problem statement.

Find
The equation of motion of the spring mass-dashpot.

Solution
Free Body Diagram

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

 * style="width:100%" |
 * $$\displaystyle {y}={y_k}+{y_c} $$
 * $$\displaystyle {y}={y_k}+{y_c} $$

 * style= |
 * }

Kinetics:

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle m\overline{a}=m\ddot{y}=m\overline{y}''$$
 * $$\displaystyle m\overline{a}=m\ddot{y}=m\overline{y}''$$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=k\overline{y}$$
 * $$\displaystyle \overline{F_k}=k\overline{y}$$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=C\overline{y'}$$
 * $$\displaystyle \overline{F_c}=C\overline{y'}$$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \sum \overline{F_y}=m\overline{a} $$
 * $$\displaystyle \sum \overline{F_y}=m\overline{a} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \sum\overline{F_y}=-\overline{F_c}-\overline{F_k}+\overline{F}_{r}=m\overline{a} $$
 * $$\displaystyle \sum\overline{F_y}=-\overline{F_c}-\overline{F_k}+\overline{F}_{r}=m\overline{a} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle m\overline{y''}+\overline{y'}c+\overline{y}k=\overline{r(t)} $$
 * $$\displaystyle m\overline{y''}+\overline{y'}c+\overline{y}k=\overline{r(t)} $$

 * style= |
 * }

= Problem 3: Equation of motion for a spring-dashpot system in series =

Given
Refer to Lecture slide [1-6] for problem statement.

Find
The equation of motion of the spring mass-dashpot.

Solution
Free Body Diagram

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

 * style="width:100%" |
 * $$\displaystyle {y}={y_k}+{y_c} $$
 * $$\displaystyle {y}={y_k}+{y_c} $$

 * style= |
 * }

Kinetics:

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$
 * $$\displaystyle \overline{F_c}=\text{Force Due to Dashpot} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$
 * $$\displaystyle \overline{F_k}=\text{Force Due to Spring} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$
 * $$\displaystyle \overline{F_r}=\text{Force Applied}\,\overline{r(t)} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_k}=k\overline{y} $$
 * $$\displaystyle \overline{F_k}=k\overline{y} $$

$$
 * $$\displaystyle
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{F_c}=c\overline{{y}'} $$
 * $$\displaystyle \overline{F_c}=c\overline{{y}'} $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{y}=\overline{y_k}+\overline{y_c} $$
 * $$\displaystyle \overline{y}=\overline{y_k}+\overline{y_c} $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{{y}}=\overline{{y_k}}+\overline{{y_c}''} $$
 * $$\displaystyle \overline{{y}}=\overline{{y_k}}+\overline{{y_c}''} $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline=\overline{F_k}=\overline{F_c} $$
 * $$\displaystyle \overline=\overline{F_k}=\overline{F_c} $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle m\overline{{y}''}+\overline{F_I}=\overline{F}(t) $$
 * $$\displaystyle m\overline{{y}''}+\overline{F_I}=\overline{F}(t) $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle k\overline{y_k}=c\overline{{y_c}'} $$
 * $$\displaystyle k\overline{y_k}=c\overline{{y_c}'} $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle \overline{{y_c}'}=\frac{k}{c}\overline $$
 * $$\displaystyle \overline{{y_c}'}=\frac{k}{c}\overline $$

|| $$\displaystyle $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$
 * $$\displaystyle m(\overline{{y}_k''}+\frac{k}{c}\overline{y}_k)+k\overline{y}_k=\overline{f}(t) $$

|| $$\displaystyle $$
 * }

= Problem 4: Equation Derivation =

Given
Refer to Lecture slide [2-2] for problem statement. The components of the circuit can be related by the following formulas:

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

 * style="width:100%" |
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$

|| $$\displaystyle (Eq. 4.1) $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$

|| $$\displaystyle (Eq. 4.2) $$
 * }

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

 * style="width:100%" |
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$

|| $$\displaystyle (Eq. 4.3) $$
 * }

Find
Derive equations (3) and (4) from equation (2).

Equation (3)

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

 * style="width:100%" |
 * $$\displaystyle I=C\frac{dV_c}{dt}$$
 * $$\displaystyle I=C\frac{dV_c}{dt}$$

$$
 * $$\displaystyle (Eq. 4.1a)
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{I}{C}=\frac{dV_c}{dt} $$
 * $$\displaystyle \frac{I}{C}=\frac{dV_c}{dt} $$

$$
 * $$\displaystyle (Eq. 4.2a)
 * }

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

 * style="width:100%" |
 * $$\displaystyle {V}'=\frac{dV}{dt} $$
 * $$\displaystyle {V}'=\frac{dV}{dt} $$

$$
 * $$\displaystyle (Eq. 4.3a)
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}(I)=\frac{d}{dt}(C\frac{dV_c}{dt})=C\frac{d^2V_c}{dt}={I}' $$
 * $$\displaystyle \frac{d}{dt}(I)=\frac{d}{dt}(C\frac{dV_c}{dt})=C\frac{d^2V_c}{dt}={I}' $$

$$
 * $$\displaystyle (Eq. 4.4a)
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}({I}')=\frac{d}{dt}(C\frac{d^2V_c}{dt^2})=C\frac{d^3V_c}{dt^3}={I}'' $$
 * $$\displaystyle \frac{d}{dt}({I}')=\frac{d}{dt}(C\frac{d^2V_c}{dt^2})=C\frac{d^3V_c}{dt^3}={I}'' $$

$$
 *
$$\displaystyle (Eq. 4.5a)
 * }

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

 * style="width:100%" |
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$

$$ Taking the derivative of equation (2) yields
 *
$$\displaystyle (Eq. (2) )
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}(V)=\frac{d}{dt}(LC\frac{d^2V_c}{dt^2})+\frac{d}{dt}(RC\frac{dV_c}{dt})+\frac{d}{dt}(V_c) $$
 * $$\displaystyle \frac{d}{dt}(V)=\frac{d}{dt}(LC\frac{d^2V_c}{dt^2})+\frac{d}{dt}(RC\frac{dV_c}{dt})+\frac{d}{dt}(V_c) $$

$$
 *
$$\displaystyle (Eq. 4.6a)
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{dV}{dt}=LC\frac{d^3V_c}{dt^3}+RC\frac{d^2V_c}{dt^2}+\frac{dV_c}{dt} $$
 * $$\displaystyle \frac{dV}{dt}=LC\frac{d^3V_c}{dt^3}+RC\frac{d^2V_c}{dt^2}+\frac{dV_c}{dt} $$

$$
 *
$$\displaystyle (Eq. 4.7a)
 * }

Substituting in from Eq. 4.3a,4.4a, and 4.5a yields:

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

 * style="width:100%" |
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$
 * $$\displaystyle {V}'=L{I}''+R{I}'+\frac{I}{C} $$

$$
 *
$$\displaystyle (Eq. (3))
 * }

Equation (4)

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

 * style="width:100%" |
 * $$\displaystyle Q=CV_c $$
 * $$\displaystyle Q=CV_c $$

$$
 *
$$\displaystyle (Eq. 4.1b)
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}(Q)=\frac{d}{dt}(CV_c)=\frac{dV_c}{dt}C={Q}' $$
 * $$\displaystyle \frac{d}{dt}(Q)=\frac{d}{dt}(CV_c)=\frac{dV_c}{dt}C={Q}' $$

$$
 *
$$\displaystyle (Eq. 4.2b)
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d}{dt}({Q}')=\frac{d}{dt}(C\frac{dV_C}{dt})=C\frac{d^2V_c}{dt^2}={Q}'' $$
 * $$\displaystyle \frac{d}{dt}({Q}')=\frac{d}{dt}(C\frac{dV_C}{dt})=C\frac{d^2V_c}{dt^2}={Q}'' $$

$$
 *
$$\displaystyle (Eq. 4.3b)
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{Q}{C}=V_c\ $$
 * $$\displaystyle \frac{Q}{C}=V_c\ $$

$$
 *
$$\displaystyle (Eq. 4.4b)
 * }

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

 * style="width:100%" |
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$
 * $$\displaystyle V=LC\frac{d^2V_c}{dt^2}+RC\frac{dV_c}{dt}+V_c $$

$$
 *
$$\displaystyle (Eq. (2))
 * }

Substituting in from equations 4.2b,4.3b, and 4.4b yields.

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

 * style="width:100%" |
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$
 * $$\displaystyle V=L{Q''}+R{Q}'+\frac{Q}{C} $$

$$
 *
$$\displaystyle (Eq. (4) )
 * }

= Problem 5: Find the general solution. Check solution by substitution. =

Given
Refer to Lecture slide [2-5] for problem statement.

Solve problems 4 and 5 from page 59 of Advanced Engineering Mathematics

Problem (4)

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

 * style="width:100%" |
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$

 * style= |
 * }

Problem (5)

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

 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$

 * style= |
 * }

Find
General solution

Problem (4)
Given Equation
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle {y}'' + 4y' + (\pi^2 + 4)y = 0 $$

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

 * style="width:100%" |
 * $$\displaystyle a = 4 $$
 * $$\displaystyle a = 4 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle b = (\pi^2 + 4) $$
 * $$\displaystyle b = (\pi^2 + 4) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$
 * $$\displaystyle \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda = \frac{dy}{dx} $$
 * $$\displaystyle \lambda = \frac{dy}{dx} $$

Substituting Lambda for y Values
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \lambda^2 + f\lambda + (\pi^2 + 4) = 0 $$
 * $$\displaystyle \lambda^2 + f\lambda + (\pi^2 + 4) = 0 $$

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

 * style="width:100%" |
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$
 * $$\displaystyle \omega^2 = b-\frac{1}{4}a^2 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \omega^2 = \pi^2 $$
 * $$\displaystyle \omega^2 = \pi^2 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$
 * $$\displaystyle \omega^2 = (\pi^2 + 4) - 4 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \omega = \pi $$
 * $$\displaystyle \omega = \pi $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega = -2\, \pm \pi i $$
 * $$\displaystyle \lambda = - \frac{1}{2}a\, \pm i\omega = -2\, \pm \pi i $$

Solving For y1 and y2
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$
 * $$\displaystyle y_{1} = e^{\frac{-ax}{2}}\cos\omega x $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$
 * $$\displaystyle y_{2} = e^{\frac{-ax}{2}}\sin\omega x $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$
 * $$\displaystyle y_{1} = e^{-2x}\cos\pi x $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$
 * $$\displaystyle y_{2} = e^{-2x}\sin\pi x $$

 * style= |
 * }

The general equation is:

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

 * style="width:100%" |
 * $$\displaystyle y = e^\frac{-ax}{2}(A\cos\omega x + B\sin\omega x)$$
 * $$\displaystyle y = e^\frac{-ax}{2}(A\cos\omega x + B\sin\omega x)$$

 * style= |
 * }

Solving for our equation, we get:

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

 * style="width:100%" |
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$
 * $$\displaystyle y' = {-2e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)+ e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 2e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) - 2e^{-2x}(-\pi{C_1}\sin\pi{x} + \pi{C_2}\cos\pi{x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$
 * $$\displaystyle y'' = {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) $$

Pluging y", y' and y into Initial Equation
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$
 * $$\displaystyle {4e^{-2x}}(C_1\cos\pi x + C_2\sin\pi x)- 4e^{-2x}(-\pi{C_1}\sin\pi {x} + \pi{C_2}\cos\pi {x}) + e^{-2x}(-\pi^2{C_1}\cos\pi{x} - \pi^2{C_2}\sin\pi{x}) - 8e^{-2x}({C_1}\cos\pi{x}+{C_2}\sin\pi{x}) + 4e^{-2x}(-\pi{C_1}\sin\pi{x}+\pi{C_2}\cos\pi{x}) + (\pi^2 +4)[e^{-2x}({C_1}\cos\pi{x} + {C_2}\sin\pi{x})] = 0 $$

 * style= |
 * }

Therefore:

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

 * style="width:100%" |
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$
 * $$\displaystyle y = e^{-2x}(C_1\cos\pi x + C_2\sin\pi x) $$

 * style= |
 * }

is the solution

Problem (5)

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

 * style="width:100%" |
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$
 * $$\displaystyle y''+2 \pi y'+ \pi^2y=0 $$

 * style= |
 * }

The ODE above can be simplified by using the variables below.

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

 * style="width:100%" |
 * $$\displaystyle a = 2\pi $$
 * $$\displaystyle a = 2\pi $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle b = \pi ^2 $$
 * $$\displaystyle b = \pi ^2 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' + a y' + by = 0 $$
 * $$\displaystyle y'' + a y' + by = 0 $$

 * style= |
 * }

The equation is now in the general format for linear ODE’s.

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

 * style="width:100%" |
 * $$\displaystyle y = e^{\lambda x} $$
 * $$\displaystyle y = e^{\lambda x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y' = \lambda e^{\lambda x} $$
 * $$\displaystyle y' = \lambda e^{\lambda x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$
 * $$\displaystyle y'' = \lambda ^2 e^{\lambda x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$
 * $$\displaystyle \lambda ^2 e^{\lambda x} + 2 \pi \lambda e ^{\lambda x} + \pi ^2 e ^{ \lambda x} = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$
 * $$\displaystyle e^{\lambda x} (\lambda ^2 + 2\pi\lambda + \pi ^2) = 0 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$
 * $$\displaystyle \lambda ^2 + 2\pi\lambda + \pi ^2 = 0 $$

Solving For Lambda Values
 * style= |
 * }
 * {| style="width:100%" border="0"

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a \pi \sqrt{a^2 - 4 b}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{(2\pi)^2 - 4 \pi ^2}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi \pm \sqrt{4\pi^2 - 4 \pi ^2}}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-2\pi}{2} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$
 * $$\displaystyle \lambda_1,\lambda_2 = \frac{-a}{2} $$

 * style= |
 * }

This means that the characteristic equation has a doubleroot.

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

 * style="width:100%" |
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$
 * $$\displaystyle y_1 = e ^{(ax/2)} = e^{\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_2 = u y_1 $$
 * $$\displaystyle y_2 = u y_1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle u = c_2 x + c_1, c_1 = 0, c_2 = 1 $$
 * $$\displaystyle u = c_2 x + c_1, c_1 = 0, c_2 = 1 $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$
 * $$\displaystyle y_2 = x y_1 = x e^{-\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y = e^{-\pi x}(c_1 + c_2 x) $$
 * $$\displaystyle y = e^{-\pi x}(c_1 + c_2 x) $$

 * style= |
 * }

This is the general solution to the ordinary differential equation.

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

 * style="width:100%" |
 * $$\displaystyle y' = -\pi c_1 e^{-\pi x}+c_2e^{-\pi x}-\pi c_2 x e^{-\pi x} $$
 * $$\displaystyle y' = -\pi c_1 e^{-\pi x}+c_2e^{-\pi x}-\pi c_2 x e^{-\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle y'' = \pi ^2 c_1 e^{-\pi x}-\pi c_2e^{-\pi x}+\pi ^2 c_2 x e^{-\pi x} - \pi c_2e^{-\pi x} $$
 * $$\displaystyle y'' = \pi ^2 c_1 e^{-\pi x}-\pi c_2e^{-\pi x}+\pi ^2 c_2 x e^{-\pi x} - \pi c_2e^{-\pi x} $$

 * style= |
 * }

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

 * style="width:100%" |
 * $$\displaystyle e^{-\pi x} [\pi ^2 c_1 -\pi c_2 + \pi ^2 c_2 x - \pi - 2\pi^2 c_1 + 2\pi c_2 - 2\pi ^2 c_2 x + c_1 \pi ^2 + \pi ^2 c_2x] = 0 $$
 * $$\displaystyle e^{-\pi x} [\pi ^2 c_1 -\pi c_2 + \pi ^2 c_2 x - \pi - 2\pi^2 c_1 + 2\pi c_2 - 2\pi ^2 c_2 x + c_1 \pi ^2 + \pi ^2 c_2x] = 0 $$

 * style= |
 * }

All the terms inside the bracket cancel out and equal zero, so the general solution holds.

= Problem 6: =

Given
Refer to Lecture slide [2-2] for problem statement.

Problem (A)

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

$$\displaystyle y'' = g = const. $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (B)

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

$$\displaystyle mV' = mg-bV^2$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (C)

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

$$\displaystyle h' = -k \sqrt h $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (D)

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

$$\displaystyle m{Y''} + K{Y} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (E)

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

$$\displaystyle {Y''} + \omega_o^2 Y = \cos \omega {t} $$, $$\displaystyle \omega_o \approx \omega $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (F)

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

$$\displaystyle LI''+RI'+\frac{I}{C}=E $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (G)

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

$$\displaystyle {E}{I}{y^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (H)

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

$$\displaystyle L\theta''+g\sin{\theta}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Find
Determine the order of the differential equations for the first 8 figures above. For each equation, determine if it's linear and whether the superposition principle can be applied.

Solution
This is the general format for a linear equation:

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

$$\displaystyle y''+p(x)y'+q(x)y=r(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Problem (A)

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

$$\displaystyle y'' = g = const. $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

The ODE consists of only additive combinations of the dependent variable y and its derivatives. Therefore, the ODE is linear.

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

$$\displaystyle y''$$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogenous Equation:

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

$$\displaystyle y_h = 0$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:

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

$$\displaystyle y_p = g$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:

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

$$\displaystyle \overline{y} = y_h+y_p$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \overline{y} = g$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle y_h+y_p = g = \overline{y}$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (B)

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

$$\displaystyle mV' = mg-bV^2$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle mV' = mg-bV^2 \Rightarrow mV'+bV^2 = mg$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle V^2 \neq V$$ The general format of a linear equation is not followed since the variable, so this is a nonlinear ODE
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle V'$$ is the highest order derivative. Therefore, this is a first-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogenous Equation:

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

$$\displaystyle V_h=m{V_h}'+b{V_h}^2=0$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:

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

$$\displaystyle V_p=m{V_p}'+b{V_p}^2=mg$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:

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

$$\displaystyle \overline{V}=V_h+V_p$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle \overline{V}=m(V_h+V_p)'+b({V_h}^2+2 V_pV_h+V_p^2)=mg$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\displaystyle V_h+V_p=m(V_h'+V_p')+b(V_h^2+V_p^2)=mg\neq\overline{V}$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle cannot be applied.

Problem (C)

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

$$\displaystyle h' = -k \sqrt h $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \sqrt h = h^{1/2} \neq h$$ Therefore, this is a nonlinear ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle h' $$ is the highest order derivative. Therefore, this is a first-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous Equation:

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

$$\displaystyle h_h=h_h'+k\sqrt{h_h}=0$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:

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

$$\displaystyle h_p=h_p'+k\sqrt{h_p}=r(t) $$, where r(t) is some function of time
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{h}=h_h+h_p=(\overline{h})'+k\sqrt{\overline{h}}=r(t)$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:

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

$$\displaystyle \overline{h}=h_h+h_p$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{h}'=(h_h'+h_p')=(h_h+h_p)'$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle h_h+h_p=(h_h'+h_p')+k\sqrt{h_h}+k\sqrt{h_p}=(\overline{h})'+k(\sqrt(h_h)+\sqrt{h_p})=r(t)$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle h_h+h_p\neq\overline{h}$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle cannot be applied.

Problem (D)

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

$$\displaystyle m{Y''} + K{Y} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

This is a linear ODE due to the fact that it consists of only additive combinations of one dependent variable and its derivatives.

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

$$\displaystyle y'' $$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous equation:

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

$$\displaystyle m{Y_h''} + K{Y_h} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:

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

$$\displaystyle m{Y_p''} + K{Y_h} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving, we get:

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

$$\displaystyle \overline{Y} = {Y_h} + {Y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle m({Y_h} + {Y_p}) + K({Y_h}+ {Y_p}) = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle m(\overline{Y''}) + K(\overline{Y})=f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{Y} = {Y_h} + {Y_p''} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{Y} = {Y_h} + {Y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (E)

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

$$\displaystyle {Y''} + \omega_o^2 Y = \cos \omega {t} $$, $$\displaystyle \omega_o \approx \omega $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

This is a linear ODE.

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

$$\displaystyle y'' $$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \omega_o \approx \omega $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous equation:

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

$$\displaystyle {Y_h''} + \omega_o^2 Y_h = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:

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

$$\displaystyle {Y_p''} + \omega_o^2 Y_p = \cos \omega {t} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving, we get:

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

$$\displaystyle \overline{Y} = {Y_h} + {Y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle ({Y_h} + {Y_p}) + \omega_o^2 ({Y_h} + {Y_p}) = \cos\omega{t} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle (\overline{Y''}) + \omega_o^2 (\overline{Y}) = \cos\omega{t} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (F)

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

$$\displaystyle LI''+RI'+\frac{I}{C}=E $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

This is a linear ODE.

$$\displaystyle I'' $$ is the highest order derivative. Therefore, this is a second-order ODE.

Homogeneous equation:

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

$$\displaystyle I_h=L{I_h}''+R{I_h}'+\frac{I_h}{C}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:

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

$$\displaystyle I_p=L{I_p}''+R{I_p}'+\frac{I_p}{C}=E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving, we get:

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

$$\displaystyle \overline{I}=L\overline{I}''+R\overline{I}'+\frac{\overline{I}}{C}=E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{I}=L({I_p+I_h})''+R({I_p+I_h})'+\frac{I_p+I_h}{C}=E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle I_p+I_h=L(I_h+I_p)+R(I_h'+I_p')+\frac{(I_p+I_h)}{C}=0+E' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle (I_h+I_p)=(I_h+I_p)'' $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle (I_h'+I_p')=(I_h+I_p)'$$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{I}=I_p+I_h $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (G)
This is a linear ODE.

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

$$\displaystyle {E}{I}{y^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle y^{iv} $$ is the highest order derivative. Therefore, this is a fourth-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous equation:

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

$$\displaystyle {y_h} = {E}{I}{y_h^{IV}} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular equation:

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

$$\displaystyle {y_p} = {E}{I}{y_p^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:

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

$$\displaystyle \overline{y} = {y_h} + {y_p} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{y} = {E}{I}({y_h}+{y_p})^{IV} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

If the superposition principle applies:

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

$$\displaystyle {E} {I}\overline{y}^{IV} = {E}{I}({y_h}+{y_p})^{IV} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Equation 1:

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

$$\displaystyle {E}{I}{y_h^{IV}} = 0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Equation 2:

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

$$\displaystyle {E}{I}{y_p^{IV}} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Equation 1 + Equation 2:

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

$$\displaystyle {E}{I}({y_h^4}+{y_p^{IV}}) = 0 + f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Linearity of Derivative:

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

$$\displaystyle -(y^{IV} + x^{IV}) = (y+x)^{IV} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle -{E}{I}({y_h}+{y_p})^{IV} = f(x) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle -{E}{I}(\overline{y})^{IV} = f(x) = EI({y_h} + {y_p})^{IV} $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle can be applied.

Problem (H)

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

$$\displaystyle L\theta''+g\sin{\theta}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \sin\theta \neq \theta$$ Therefore, this is a nonlinear ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \theta'' $$ is the highest order derivative. Therefore, this is a second-order ODE.
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle L\theta''+g\sin{\theta}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Homogeneous Equation:

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

$$\displaystyle \theta_h=L\theta_h''+g\sin{\theta_h}=0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Particular Equation:

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

$$\displaystyle \theta_p=L\theta_p''+g\sin{\theta_p}=r(t) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Solving we get:

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

$$\displaystyle \overline{\theta}=\theta_p+\theta_h=L\overline\theta''+g\sin{\overline{\theta}}=r(t) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{\theta}=L(\theta_h+\theta_p)''+g\sin({\theta_p+\theta_h}) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \theta_p+\theta_h=L(\theta_p+\theta_h)+g\sin{\theta_h}+g\sin{\theta_p}=r(t)+0 $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \theta_p+\theta_h=L(\theta_p+\theta_h)''+g\sin{\theta_h}+g\sin{\theta_p}=r(t) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \theta_p+\theta_h=L(\overline{\theta})''+g\sin{\theta_h}+g\sin{\theta_p}=r(t) $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

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

$$\displaystyle \overline{\theta}\neq\theta_p+\theta_h $$
 * style="width:100%" |
 * style="width:100%" |
 * style= |
 * }

Therefore, the superposition principle cannot be applied.

title: User:Egm4313.s12.team10.R1


text:
User:Egm4313.s12.team5.R1

test

title: User:Egm4313.s12.team5.R1