text:
User:Egm6321.f10.team3.franklin/Hwk5

(2.6)
I expanded my solution for 2.4,

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

$$ \displaystyle \begin{align} y & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+{{k}_{1}}\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}d\tau } \right] \\ & =\frac\left[ \int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+\frac{\beta -\alpha }{{e}^{(\beta -\alpha )t}} \right] \\ & =\frac{(\beta -\alpha ){_{2}}}{{e}^{-\alpha t}}+\frac{{e}^{-\beta t}}+\frac{{{{{a}}}_{2}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \end{align}$$ $$
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 * style="width:95%" |
 * $$\displaystyle (Eqs. 5.6.1)
 * }

Combine constants to form k1 and k2,

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

$$ \displaystyle y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (Eq. 5.6.2)
 * }

From above we seen the homog. and particular solutions,

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

$$ \displaystyle \begin{align} & y^{1}={{e}^{-\alpha t}} \\ & y^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ \end{align}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (Eqs. 5.6.3)
 * }

title: User:Egm6321.f10.team3.franklin/Hwk5


text:
User talk:HAH

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

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title: User talk:HAH


text:
User talk:Latonic

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

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title: User talk:Latonic


text:
User talk:Micki

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

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Electrical diagram
Thanks for the heads up. --Draicone (talk) 14:26, 28 November 2010 (UTC)

title: User talk:Micki


text:
User:Egm6341.s11/Lecture plan

 Under construction throughout the semester. Egm6321.f10 21:49, 31 October 2010 (UTC)

= All versions =

Spring 2011, Spring 2010

EML 5526, Spring 2011, Lecture plan

= Recorded lectures, TA user page =

Recorded lectures in E-Learning at UF

= Lecture notes, report table =

Additional presentations (video, wiki, static html) made in class may not be recorded on these transparencies.

A mtg number n followed by a version letter in parentheses ([a-z]) indicates that this pdf file had been updated with a 2nd version "b", 3rd version "c", etc. The second letter "x" after a version letter designates an extra lecture; this letter became "m" after the extra lecture had become a make-up lecture. If you had downloaded this pdf file before, you want to clear the cache of your browser so to get the new version.

djvu: (install the viewers DjView4 or evince) [[media:nm1.s11.mtg1.djvu|Mtg 1 (b)]], [[media:nm1.s11.mtg2.djvu|Mtg 2]], [[media:nm1.s11.mtg3.djvu|Mtg 3 (e)]], [[media:nm1.s11.mtg4.djvu|Mtg 4 (b)]], [[media:nm1.s11.mtg5.djvu|Mtg 5 (d)]], [[media:nm1.s11.mtg6.djvu|Mtg 6 (b,m)]], [[media:nm1.s11.mtg7.djvu|Mtg 7 (d,m)]], [[media:nm1.s11.mtg8.djvu|Mtg 8 (d)]], [[media:nm1.s11.mtg9.djvu|Mtg 9 (e)]], [[media:nm1.s11.mtg10.djvu|Mtg 10 (b,m)]], [[media:nm1.s11.mtg11.djvu|Mtg 11 (c)]], [[media:nm1.s11.mtg12.djvu|Mtg 12]], [[media:nm1.s11.mtg13.djvu|Mtg 13 (b)]], [[media:nm1.s11.mtg14.djvu|Mtg 14 (b)]], [[media:nm1.s11.mtg15.djvu|Mtg 15 (b)]], [[media:nm1.s11.mtg16.djvu|Mtg 16 (b,m)]], [[media:nm1.s11.mtg17.djvu|Mtg 17 (c)]], [[media:nm1.s11.mtg18.djvu|Mtg 18 (c)]], [[media:nm1.s11.mtg19.djvu|Mtg 19 (c)]], [[media:nm1.s11.mtg20.djvu|Mtg 20 (b)]], [[media:nm1.s11.mtg21.djvu|Mtg 21 (c)]], Mtgs 22+23: Exam 1 (thus the Mtg no. from 22 to 26 should be incremented by 2, i.e., [[media:nm1.s11.mtg22.djvu|Mtg 22]] should be numbered Mtg 24, ..., [[media:nm1.s11.mtg26.djvu|Mtg 26]] should be numbered Mtg 28) [[media:nm1.s11.mtg22.djvu|Mtg 22 (b,m)]], [[media:nm1.s11.mtg23.djvu|Mtg 23 (a,m)]], [[media:nm1.s11.mtg24.djvu|Mtg 24 (d)]], [[media:nm1.s11.mtg25.djvu|Mtg 25 (c)]], [[media:nm1.s11.mtg26.djvu|Mtg 26 (b)]], (Mtg numbers are readjusted from Mtg 29 on) [[media:nm1.s11.mtg29.djvu|Mtg 29 (b,m)]] [[media:nm1.s11.mtg29.old.djvu|{old}]], [[media:nm1.s11.mtg30.djvu|Mtg 30 (3,m)]], [[media:nm1.s11.mtg31.djvu|Mtg 31 (b)]], [[media:nm1.s11.mtg32.djvu|Mtg 32 (b,x)]], [[media:nm1.s11.mtg33.djvu|Mtg 33 (b)]], [[media:nm1.s11.mtg34.djvu|Mtg 34 (b)]], [[media:nm1.s11.mtg35.djvu|Mtg 35]], [[media:nm1.s11.mtg36.djvu|Mtg 36 (b,x)]], [[media:nm1.s11.mtg37.djvu|Mtg 37 (b)]], [[media:nm1.s11.mtg38.djvu|Mtg 38]], [[media:nm1.s11.mtg39.djvu|Mtg 39]], [[media:nm1.s11.mtg40.djvu|Mtg 40 (c)]] ([[media:egm6341.s10.mtg38.djvu|S10 Mtg 38]], [[media:egm6341.s10.mtg39.djvu|S10 Mtg 39]], [[media:egm6341.s10.mtg40.djvu|S10 Mtg 40]], [[media:egm6341.s10.mtg41.djvu|S10 Mtg 41]]), [[media:nm1.s11.mtg41.djvu|Mtg 41 (c)]] ([[media:egm6341.s10.mtg41.djvu|S10 Mtg 41]]), [[media:nm1.s11.mtg42.djvu|Mtg 42 (b)]] ([[media:egm6341.s10.mtg38.djvu|S10 Mtg 38]], [[media:egm6341.s10.mtg39.djvu|S10 Mtg 39]], [[media:egm6341.s10.mtg40.djvu|S10 Mtg 40]], [[media:egm6341.s10.mtg41.djvu|S10 Mtg 41]], [[media:egm6341.s10.mtg42.djvu|S10 Mtg 42]]), Mtgs 43+44: Exam 2

Report table

Course billboard

Inspiring video on teamwork and learning: A hole in the wall: How children learn without a teacher

Mourning the Death of Handwriting, By Claire Suddath. Time Magazine, Monday, Aug. 03, 2009.

Op-Art: The Write Stuff, by Inga Dubay and Barbara Getty, NY Times, 8 Sep 2009.

= References =

$\displaystyle \color{red}{\spadesuit}$NEW:Symbolic computation
=== $\displaystyle \color{red}{\spadesuit}$NEW:Fast Clenshaw-Curtis quadrature, Matlab central ===

by Greg von Winckel, 12 Feb 2005 (Updated 16 Feb 2005). This extremely fast and efficient algorithm uses MATLAB's ifft routine to compute the Clenshaw-Curtis nodes and weights in linear time. The routine appears optimal for 2^N+1 points. Running on an average laptop, this routine computed N=2^20+1 (1048577 points) in about 4.5 seconds. Great for integrating highly oscillatory functions.

Books
$$\displaystyle \clubsuit$$ K. Atkinson, An Introduction to Numerical Analysis, Wiley, 1989. UF library QA297 .A84 1989 Google Amazon

$$\displaystyle \clubsuit$$ Suli & Mayers, An introduction to numerical analysis, Cambridge, 2003. UF library E-book Google Amazon

$$\displaystyle \color{red}{\clubsuit}$$  NEW:

NIST Digital Library of Mathematical Functions, companion of the NIST Handbook of Mathematical Functions, ed. by F.W.J. Olver, et al., Cambridge U. Press, 2010. "Together these works represent a successor to the highly successful Handbook of Mathematical Functions (M. Abramowitz and I. Stegun, Eds.; see below), which was published by the National Bureau of Standards in 1964." NA Digest, Vol.10, No.19, May 2010.

$$\displaystyle \clubsuit$$ M. Abramowitz & I. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, New York: Dover Publications, 1972. Read online, Download Wikipedia

$$\displaystyle \clubsuit$$ L.N. Trefethen, Spectral Methods in Matlab, SIAM, Philadelphia, 2000. google matlab codes, e.g., clencurt;

$$\displaystyle \color{red}{\clubsuit}$$  NEW:

see also Fast Clenshaw-Curtis quadrature by von Winckel, 2005.

$$\displaystyle \clubsuit$$ L.N. Trefethen, Approximation Theory and Approximation Practice: A 21st Century Treatment in the Form of 32 Executable Chebfun M-Files, 2010 draft of first 8 chapters, pdf.

$$\displaystyle \clubsuit$$ J.D. Murray, Mathematical biology I: An introduction, 3rd edition, Springer, 2002. UF library QH323.5.M88 2002 google amazon

$$\displaystyle \clubsuit$$ R.H. Battin, An Introduction to the Mathematics and Methods of Astrodynamics, AIAA, 1999. google amazon. Battin was a speaker in the symposium Apollo: Reflections and Lessons (40th anniversary of the first moon landing), MIT, 11 Jun 2009.

$$\displaystyle \clubsuit$$ D. Zwillinger, Handbook of Differential Equations, Third Edition, Academic Press, 1998. ISBN-10: 0127843965. ISBN-13: 978-0127843964. UF library QA371.Z88 1989, 2 copies, one for in-library use. Google books Amazon.com

Web references
$$\displaystyle \spadesuit$$ Algorithm (wikipedia)

$$\displaystyle \spadesuit$$ Related MIT OpenCourseWare courses

$$\displaystyle \spadesuit$$ Wolfram|Alpha is ... "learning resource available to your students at no cost that works as a computational knowledge engine. Wolfram|Alpha is not a search engine like Google or Yahoo!, because unlike a traditional search engine, Wolfram|Alpha has the capability to instantly compute the answer to previously unasked questions instead of scouring the web and returning links to pages that already exist. The results are displayed in an easy-to-read, understandable format that can be used as a primary source for educational and academic purposes."

$$\displaystyle \color{red}{\spadesuit}$$  NEW:

Stephen Wolfram: Computing the theory of everything, video on TED.com

$$\displaystyle \spadesuit$$ EqWorld, The World of Mathematical Equations. It is a good idea to verify the sources, as the site is not responsible for accuracy and correctness; see Rights and obligations of contributors and website administration.
 * 2nd-order linear ODEs
 * 2nd-order non-linear ODEs
 * ODE education

$$\displaystyle \spadesuit$$ ODE (Wikipedia): Be careful; always verify the sources.

$$\displaystyle \color{red}{\spadesuit}$$  NEW:

Sitmo: Resources for financial engineers, latex equation editor, Clenshaw-Curtis quadrature, etc.

Papers
$$\displaystyle \clubsuit$$ L.N. Trefethen and Z. Battles, The Chebfun project, Oxford.
 * "Vision statement": Trefethen, Computing numerically with functions instead of numbers (Math. in Comp. Sci., 2007)
 * One-page summaries: Trefethen, Chebfun and Chebop (IMAJNA Newsletter, 2008), Approximation theory and practice  (Oxford Maths. Inst. Newsletter, 2009)
 * Survey of chebfun and chebop: Platte and Trefethen, Chebfun: a new kind of numerical computing (ECMI Proceedings, 2009)
 * Original chebfun paper: Battles and Trefethen, An extension of Matlab to continuous functions and operators (SIAM J. Sci. Comp., 2004)
 * Introduction of Chebops: Driscoll, Bornemann and Trefethen, The chebop system for automatic solution of differential equations (BIT Numer. Math., 2008)

$$\displaystyle \clubsuit$$ P. Kessler, Trapezoidal rule error, Math 128a, Berkeley, 2006. These notes were referred to in a paper by S.G. Johnson, 2008.

$$\displaystyle \clubsuit$$ S. G. Johnson, Notes on the convergence of trapezoidal-rule quadrature, online MIT course notes (2008). This document mentioned Kessler's notes in a footnote.

$$\displaystyle \clubsuit$$ Lloyd N. Trefethen, Is Gauss quadrature better than Clenshaw-Curtis?, SIAM Review 50 (1), 67-87 (2008).

$$\displaystyle \clubsuit$$ B.A. Cipra, The Best of the 20th Century: Editors Name Top 10 Algorithms, SIAM News, May 16, 2000. A look at ten algorithms named some of the "best" of the computer age. The Fast Fourier Transform (FFT) is one of them.

$$\displaystyle \clubsuit$$ D.N. Rockmore, The FFT — an algorithm the whole family can use, Computing in Science & Engineering, January/February 2000, Volume 2, Number 1, pp. 60--6. Contained a historical account of the development of the FFT.

$$\displaystyle \clubsuit$$ Sri Welaratna, 30 years of FFT Analyzers, Sound and Vibration (January 1997, 30th anniversary issue). A historical review of hardware FFT devices. See Fast Fourier transform (wikipedia).

$$\displaystyle \clubsuit$$ B.A. Conway, The choices available to space trajectory optimizers, seminar given at UF on Tue, 23 Feb 2010. By listening to this talk, I noticed an application of the Simpson's rule; I thought about teaching this application and searched the literature as shown below. Prof. Conway later graciously shared his talk slides with me (I did the literature search below, and found the paper by Hargraves & Paris 1987 and other papers before I received his talk slides, which I also shared with the students in the class). Another seminar on the use of the Hermite-Simpson algorithm was given at UF on 30 Mar 2010.

$$\displaystyle \clubsuit$$ Gauss, Legendre, ... pseudospectral methods
 * Web of Science literature search: Topic=(gauss legendre pseudospectral method*) Timespan=All Years. Databases=SCI-EXPANDED, SSCI, A&HCI.
 * Hermite-Legendre-Gauss-Lobatto Direct Transcription in Trajectory Optimization. Author(s): Williams P. Source: JOURNAL OF GUIDANCE CONTROL AND DYNAMICS  Volume: 32   Issue: 4   Pages: 1392-1395   Published: JUL-AUG 2009
 * Direct trajectory optimization using nonlinear programming and collocation, HARGRAVES, C. R., ; PARIS, S. W., Boeing Aerospace Co., Seattle, WA, Journal of Guidance, Control, and Dynamics 1987 0731-5090 vol.10 no.4 (338-342)
 * Direct optimization using collocation based on high-order Gauss-Lobatto quadrature rules, Herman, A.L., Conway, B.A., Journal of Guidance, Control, and Dynamics 1996 0731-5090 vol.19 no.3 (592-599).
 * Computational optimal control of the terminal bunt manoeuvre - Part 1: minimum altitude case, S. Subchan, R. Zbikowski, Volume 28 Issue 5, Pages 311 - 395 (September/October 2007). Bunt (maneuver): "2. (Engineering / Aeronautics) to cause (an aircraft) to fly in part of an inverted loop or (of an aircraft) to fly in such a loop".
 * Bryson, Denham, A steepest-ascent method for solving optimum programming problems, Journal of Applied Mechanics, Jun 1962, p.247-257.
 * Direct Trajectory Optimization by a Chebyshev Pseudospectral Method Fariba Fahroo, ; I. Michael Ross, Journal of Guidance, Control, and Dynamics 2002 0731-5090 vol.25 no.1 (160-166)
 * Rational Legendre pseudospectral approach for solving nonlinear differential equations of Lane-Emden type Author(s): Parand K, Shahini M, Dehghan M Source: JOURNAL OF COMPUTATIONAL PHYSICS  Volume: 228   Issue: 23   Pages: 8830-8840   Published: DEC 10 2009
 * Leaky mode analysis of optical waveguides by legendre and laguerre pseudospectral method Author(s): Huang CC Source: MICROWAVE AND OPTICAL TECHNOLOGY LETTERS  Volume: 50   Issue: 10   Pages: 2507-2509   Published: OCT 2008
 * Pseudospectral methods for infinite-horizon optimal control problems Author(s): Fahroo F, Ross IM Source: JOURNAL OF GUIDANCE CONTROL AND DYNAMICS  Volume: 31   Issue: 4   Pages: 927-936   Published: JUL-AUG 2008

= News =

$\displaystyle \color{red}{\clubsuit}$ NEW: Is openness in the digital space killing creativity?, Digital Planet, BBC, 15 Mar 2011.

$$\displaystyle \color{red}{\spadesuit}$$  NEW:

Bing, Now With More Wolfram, NY Times, November 11, 2009. ... Microsoft today announced that it will begin integrating statistics and other data from computational engine WolframAlpha into its search engine Bing.

$$\displaystyle \color{red}{\spadesuit}$$  NEW:

Ask.com to Return to Old Service, NY Times, November 9, 2010. ... Other search engines have taken on Google with little success. Cuil, one of the upstarts, closed in September after two years. But new challengers include Blekko and WolframAlpha.

$$\displaystyle \color{red}{\spadesuit}$$  NEW:

Fun With WolframAlpha, NY Times, June 5, 2009.

= Motivation: Simulations =

$$\displaystyle \color{red}{\spadesuit}$$  NEW:

Humans to Asteroids: Watch Out!, NY Times, By RUSSELL SCHWEICKART Published: October 25, 2010. Deflect That Asteroid, NY Times, November 2, 2010

An asteroid breakup 160 Myr ago as the probable source of the K/T impactor, William F. Bottke, David Vokrouhlický & David Nesvorný, Nature 449, 48-53 (6 September 2007).

Asteroids: Spun in the sun, William F. Bottke, Nature 446, 382-383 (22 March 2007).

Asteroids: How to make a flying saucer, William F. Bottke, Nature 454, 173-174 (10 July 2008).

Asteroid, NASA. Pictures of asteroids Ida, Eros, and Chicxulub impact (Yucatan peninsula), extinction of dinausaurs.

Large asteroid impacting the Earth, simulation

Hubble Space Telescope Captures Rare Jupiter Collision, 07.24.09

Near Earth Objects Program, JPL.

Solar system collision: Set Target Earth (land only), Projectile Rock, Projectile diameter 10 km, Projectile velocity 60 km / sec; click Kaboom; see photo of Chicxulub impact.

= Numerical integration of functions: Preliminaries =

Atkinson, 1989, p.249

Norms
Atkinson, 1989, p.10

Integral mean value theorem
Atkinson, 1989, p.4

Mean value theorem (wikipedia)

Simpson's rule
On the history of the Newton-Raphson method:

Computation, comparison

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

$$  \displaystyle I = \int\limits_0^x \frac{e^t - 1}{t} d t = Ei(x) - ln(x) - \gamma
 * style="width:95%" |
 * style="width:95%" |

$$  (1)
 *

 * }

The exact result of the above integral is given in Abramowitz & Stegun, p.230, with the Euler's constant $$\gamma\,\;$$ given in Abramowitz & Stegun, p.255.

A more accurate Euler's constant can be found at Wolfram|Alpha.

See also Euler–Mascheroni constant (wikipedia) and Euler–Mascheroni constant (wolfram).

Since the exponential integral is defined as

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

$$  \displaystyle Ei(x) := ln(x) + \gamma + \sum_{n=1}^{\infty} \frac{x^n}{n n!} $$  (2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

the integral in Eq.(1) is nothing but the series in the 3rd term in Eq.(2), which can also be obtained by integrating the Taylor series.

Newton-Cotes formula
On Thu, 11 Feb 2010, I listened to the following interesting NPR program at the University of Illinois at Urbana-Champaign; of course Newton plays an important role in the current course (and in many other areas):

Composite rule
= Lagrange interpolation error: Theorem =

Atkinson, 1989, p.134

Derivative mean-value theorem
Atkinson, 1989, p.4

Mean value theorem (wikipedia)

Rolle's theorem
Rolle's theorem (wikipedia)

Application: Cauchy distribution
Lorentzian distribution of spectral line width (Lecture 08), in Astronomy course on Nebulae, K. Wood, Univ. of St Andrews, Scotland, UK.

Portrait of Gauss on the Deutsch mark: Notice the normal distribution curve next to Gauss' portrait. Gauss "developed the Gaussian Bell Curve, to account for the statistical variation in individual observations of stellar locations." So even though Gauss was not the first to have come across the normal distribution, he may have invented it independently of de Moivre, and applied it to astronomy.

See also the quotation on Pythagoras' theorem (and the corresponding NPR program in which there was a mention of Stigler's law of misnomy).

Composite rules
= Higher-order error analysis of trapezoidal rule =

Euler-MacLaurin series
Euler–Maclaurin formula

googled for "matlab bernoulli number"

 * Matlab function for Bernoulli numbers (matlab central)

 * MATLAB routines for computation of Special Functions CETA (Center for Electromagnetic Theory and Applications), MIT.

Corrected trapezoidal rules $$CT_k (n)$$
Another notation $$CT_k (n) \equiv CT^{(2i+1)}_n (f)$$ with (2i+1)th derivative and (n+1) integration points.

Elliptic integral of second kind

Ellipse (wikipedia)

Orbital mechanics
The 18th-century battle over lunar motion, Siegfried Bodenmann, Physics Today, Jan 2010. In a dispute with more than just scientific import, Alexis Clairaut, Leonhard Euler, and Jean le Rond d’Alembert each employed their own strategies to establish that they were the first to understand a puzzling feature of the Moon’s orbit.

The Sky Is Falling; The Threat of Near Earth Objects (audio) (From the US National Academies) The US spends approximately $4 million each year searching for near-Earth objects to detect those that may collide with Earth. What is the true threat that we are facing and what can we do about it?

The Phuto Files (video): An excellent, funny, lively, and highly informative NOVA program on the history of the discovery and the current status of the (dwarf) planet Pluto.

The case for Pluto (talk video), Alan Boyle, MSNBC, Distinctive voices @ the Beckman Center, National Academies, 2010. While watching this presentation, it may be useful to take a look at the excellent article Pluto (wikipedia) now and then to identify the names of different people mentioned in the talk; the main story of the talk is also recorded in this Wikipedia article.

Clenshaw-Curtis quadrature
The Clenshaw-Curtis quadrature is used in spectral methods for efficient integration, and the spectral methods are among the important numerical methods of the 20th century:

L.N. Trefethen, Approximation Theory and Approximation Practice: A 21st Century Treatment in the Form of 32 Executable Chebfun M-Files, 2010 draft of first 8 chapters, pdf.

Discrete cosine transform
Fourier transform
 * Sine and cosine transforms
 * Discrete cosine transform
 * Computation, Fast Fourier transform

S.A. Khayam, The Discrete Cosine Transform (DCT): Theory and Application (pdf), 2003: Application of DCT to image/video compression.

Chebyshev polynomial expansion
= Application: Optimal control of trajectory =

Supersonic interceptor at minimum time
Bryson, Denham, A steepest-ascent method for solving optimum programming problems, Journal of Applied Mechanics, Jun 1962, p.247-257.

Minimum altitude aircraft bunting maneuver
Computational optimal control of the terminal bunt manoeuvre - Part 1: minimum altitude case, S. Subchan, R. Zbikowski, Volume 28 Issue 5, Pages 311 - 395 (September/October 2007). Bunt (maneuver): "2. (Engineering / Aeronautics) to cause (an aircraft) to fly in part of an inverted loop or (of an aircraft) to fly in such a loop".

Application: Population dynamics
J.D. Murray, Mathematical biology I: An introduction, 3rd edition, Springer, 2002. UF library QH323.5.M88 2002 google amazon

Butterfly effects
= Orthogonal polynomials =

Orthogonality
 Lecture plan of other courses

title: User:Egm6341.s11/Lecture plan


text: User:Ha98574/Railroad/Korea/Honam Express Line


text: User:Ha98574/Railroad/Korea/Honam Express Line/Template


text:
User:Egm6321.f10.team2.dube/HW5

=Problem 5.1: Finding fifth derivative of y=

From the lecture slide Mtg 26-2

5.1.1 Given
Solving Euler L2_ODE_VC has to methods of solution. For the first method we have 2 stages. In mtheod one the first stage is transformation of variables in which we change dependent variable. The second stage we try to find solution for dependent variable. In second method we directly try to find solution for dependent variable. But know we discuss first stage for the first method.

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

$$  \displaystyle \sum_{i=0}^{n}a_{i}x^{i}y^{(i)}=0 $$     (Eq 1.1)
 *
 * }

This equation involves coefficients need to determined.After changing independent variable

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

$$  \displaystyle y=y(x(t)) $$     (Eq 1.2)
 *
 * }

Let's take
 * {| style="width:100%" border="0"

$$  \displaystyle x=e^{t} $$     (Eq 1.3)
 *
 * }

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

$$  \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} x} $$ (Eq 1.4)
 *
 * }

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

$$  \displaystyle y_{x}=y_{t}e^{-t} $$     (Eq 1.5)
 *
 * }

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

$$  \displaystyle y_{xxxx}=e^{-4t}\left [ y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t} \right ] $$     (Eq 1.6)
 *
 * }

5.1.2 Solution

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

$$  \displaystyle y_{xxxxx}=\frac{\mathrm{d} }{\mathrm{d} x}y_{xxxx}=\frac{\mathrm{d} t}{\mathrm{d} x}\frac{\mathrm{d} }{\mathrm{d} t}\left ( e^{-4t}y_{tttt}-6e^{-4t}y_{ttt}+11e^{-4t}y_{tt}-6e^{-4t}y_{t} \right ) $$     (Eq 1.7)
 *
 * }

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

$$  \displaystyle y_{xxxxx}=e^{-t}\left [ e^{-4t}y_{ttttt}-10e^{-4t}y_{tttt}+35e^{-4t}y_{ttt}-50e^{-4t}y_{tt}+24e^{-4t}y_{t} \right ] $$     (Eq 1.8)
 *
 * }

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

$$  \displaystyle y_{xxxxx}=e^{-5t}\left [ y_{ttttt}-10y_{tttt}+35y_{ttt}-50y_{tt}+24y_{t} \right ] $$     (Eq 1.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Also note that the coefficients before those derivatives are typical in polynominals like $$\displaystyle a(a-1)(a-2)......$$, then we can have a general form of derivatives of $$\displaystyle y$$ as,

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

$$ \displaystyle {{y}^{(n)}}={{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y$$ (Eq 1.10)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *
 * }

where $$\displaystyle D$$ denotes the operator "$$\displaystyle \frac{d}{dt}$$" and $$\displaystyle n\in \mathbb{N}$$.

We can prove Eq 1.10 by mathematical induction.

When $$\displaystyle n=1$$, Eq 1.5 has prove this case.

Now let's assume Eq 1.10 is valid at some $$\displaystyle n\in \mathbb{N}$$, meaning,

 * $$\displaystyle {{y}^{(n)}}={{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y$$

then,

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

$$ \displaystyle

\begin{align} {{y}^{(n+1)}} & =\frac{d}{dx}{{y}^{(n)}} \\ & =\frac{1}\frac{d}{dt}{{y}^{(n)}} \\ & ={{e}^{-t}}\frac{d}{dt}\left[ {{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y \right] \\ & ={{e}^{-t}}\left[ (-n){{e}^{-nt}}D(D-1)(D-2)......(D-n+1)y \right. \\ & \left. {{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }{{^ – }^ – }^ – +{{e}^{-nt}}{{D}^{2}}(D-1)(D-2)......(D-n+1)y \right] \\ & ={{e}^{-t}}{{e}^{-nt}}({{D}^{2}}-nD)(D-1)(D-2)......(D-n+1)y \\ & ={{e}^{-(n+1)t}}D(D-1)(D-2)......(D-(n+1)+1)y \\ \end{align}$$

(Eq 1.11)
 *
 * }

So we have proved Eq 1.10 is valid throughout any $$\displaystyle n\in \mathbb{N}$$.

=Problem 5.2:Using Trial Solution to Solve Euler's Equation=

From the lecture slide Mtg 26-3

5.2.1 Given
Given the ODE:
 * {| style="width:100%" border="0"

$$  \displaystyle x^{2}{y}''+2x{y}'+2y=0 $$     (Eq 2.1) and boundary conditions:
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y(1)=-2 $$     (Eq 2.3)
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y(2)=5 $$     (Eq 2.4) solve using
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y=x^{r} $$     (Eq 2.5) and the Trial Solution method.
 *
 * }

5.2.2 Solution
In the trial solution use the given equation and take the first two derivatives:
 * {| style="width:100%" border="0"

$$  \displaystyle y=x^{r} $$     (Eq 2.6)
 *
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle y^{'}=rx^{r-1} $$     (Eq 2.7)
 *

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

$$  \displaystyle y^{''}=r(r-1)x^{r-2} $$     (Eq 2.8) Substitute these values into Eq. 2.1 above:
 *

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

$$  \displaystyle x^{2}r(r-1)x^{r-2}-2xrx^{r-1}+2y=0 $$     (Eq 2.9) From this equation it is easy to eliminate constants and we end up with:
 *

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

$$  \displaystyle x^{r}(r^{2}-3r+2)=0 $$     (Eq 2.10) Meaning that our equation of interest is:
 *

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

$$  \displaystyle (r^{2}-3r+2)=0 $$     (Eq 2.11) Solving this the answers are:
 *

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

$$  \displaystyle r_{1}=1 $$     (Eq 2.12)
 *

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

$$  \displaystyle r_{2}=2 $$     (Eq 2.13) Using the equation:
 *

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

$$  \displaystyle y(x)=C_{1}x^{r_{1}}+C_{2}x^{r_{2}} $$     (Eq 2.14) we plug in r_{1} and r_{2} to get:
 *

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

$$  \displaystyle y(x)=C_{1}x+C_{2}x^{2} $$     (Eq 2.15) Using the above boundary conditions we get two equations:
 *

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

$$  \displaystyle -2=C_{1}+C_{2} $$     (Eq 2.16) and
 *

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

$$  \displaystyle 5=2C_{1}+4C_{2} $$     (Eq 2.17) These are then solved:
 *

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

$$  \displaystyle \frac{9}{2}=C_{2} $$     (Eq 2.18)
 *

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

$$  \displaystyle -\frac{13}{2}=C_{1} $$     (Eq 2.19) Plugging back in we get:
 *

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

$$  \displaystyle y(x)=-\frac{13}{2}x+\frac{9}{2}x^{2} $$     (Eq 2.20)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=Problem 5.3: Euler L2_ODE_VC=

From the lecture slide Mtg 27

5.3.1.Given
Characteristic equation is given in meeting 27-1 as:

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

$$  \displaystyle \left ( r-\lambda \right )^2=0 $$     (Eq 3.1)
 *

 * }

Where $$\displaystyle \lambda=5$$

The form of equation with unknown coeffficients to be determined

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

$$  \displaystyle a_{2}x^{2}{y}''+a_{1}x{y}'+a_{0}y=0 $$     (Eq 3.2)
 *

 * }

And the given first homogeneous solution :

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

$$  \displaystyle y_{1}=x^{\lambda} $$     (Eq 3.3)
 *

 * }

5.3.2.1. Homogeneous equation with varying coefficients

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

$$  \displaystyle \left ( r-5 \right )^{2}=0 $$     (Eq 3.4)
 *

 * }

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

$$  \displaystyle r^{2}-10r+25=0 $$     (Eq 3.5)
 *

 * }

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

$$  \displaystyle y=x^{r} $$     (Eq 3.6)
 *

 * }

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

$$  \displaystyle {y}'=rx^{(r-1)} $$     (Eq 3.7)
 *

 * }

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

$$  \displaystyle {y}''=r(r-1)x^{(r-2)} $$     (Eq 3.8) Subtituting the trial solution and its derivatives into Eq 3.2 yields :
 *

 * }

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

$$  \displaystyle a_{2}x^{2}r(r-1)x^{(r-2)}+a_{1}x^{1}rx^{(r-1)}+a_{0}x^{r}=0 $$     (Eq 3.9)
 *

 * }

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

$$  \displaystyle x^{r}\left \lfloor a_{2}r(r-1)+a_{1}r+a_{0} \right \rfloor=0 $$     (Eq 3.10)
 *

 * }

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

$$  \displaystyle a_{2}r^{2}+(a_{1}-a_{2})r+a_{0}=0 $$     (Eq 3.11)
 *

 * }

Comparing with Eq 3.5 we can see that :

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

$$  \displaystyle a_{2}=1 ,a_{1}=-9 ,a_{0}=25 $$     (Eq 3.12)
 *

 * }

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

$$  \displaystyle x^{2}{y}''-9x{y}'+25y=0 $$     (Eq 3.13)
 *

 * }

Let the general solution of homogeneous solution to be,
 * {| style="width:100%" border="0"

$$  \displaystyle y=U(x)x^{5} $$     (Eq 3.14)
 *

 * }

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

$$  \displaystyle {y}'={U(x)}'x^{5}+5x^{4}U(x) $$     (Eq 3.15)
 *

 * }

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

$$  \displaystyle {y}={U(x)}x^{5}+10x^{4}{U(x)}'+20x^{3}U(x) $$     (Eq 3.16)
 *

 * }

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

$$  \displaystyle x^{2}\left [ {U(x)}''x^{5}+10x^{4}{U(x)}'+20x^{3}U(x) \right ]-9x\left [ x^{5}{U(x)}'+5x^{4}U(x) \right ]+25x^{5}U(x)=0 $$     (Eq 3.17)
 *

 * }

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

$$  \displaystyle {U(x)}''x^{5}+{U(x)}'\left [ 10x^{4}-9x^{4} \right ]+U(x)\left [ 20x^{3}-45x^{3}+25x^{3} \right ]=0 $$     (Eq 3.18)
 *

 * }

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

$$  \displaystyle {U(x)}''+\frac{1}{x}{U(x)}'=0 $$     (Eq 3.19)
 *

 * }

Now we can use the variable separation to reduce the order.See:Separation Of Variables To decrease our equation order.

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

$$  \displaystyle {Z}'+\frac{1}{x}Z=0 $$     (Eq 3.20)
 *

 * }

Where $$ \displaystyle Z={U}'$$

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

$$  \displaystyle \int\frac{dZ}{Z}=-\int \frac{dx}{x} $$     (Eq 3.21)
 *

 * }

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

$$  \displaystyle logz=logk_{1}-logx $$     (Eq 3.22)
 *

 * }

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

$$  \displaystyle z=\frac{k_{1}}{x} $$     (Eq 3.23)
 *

 * }

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

$$  \displaystyle U=\int Z dx+k_{2} $$     (Eq 3.24)
 *

 * }

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

$$  \displaystyle U=k_{1}logx+k_{2} $$     (Eq 3.25)
 *

 * }

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

$$  \displaystyle y=U(x)y_{1}(x)=\left ( k_{1}logx+k_{2} \right )x^{5} $$     (Eq 3.26)
 *

 * }

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

$$  \displaystyle y_{2}(x)=k_{1}x^{5}logx $$     (Eq 3.27)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

5.3.2.2Homogeneous function with constant coefficients
Euler L2_ODE_CC is given as follows ;
 * {| style="width:100%" border="0"

$$  \displaystyle b_{2}{y}''+b_{1}{y}'+b_{0}y=0 $$     (Eq 3.28) Let the trial solution to be
 *

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

$$  \displaystyle y=e^{rx} $$     (Eq 3.29)
 *

 * }

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

$$  \displaystyle {y}'=re^{rx} $$     (Eq 3.30)
 *

 * }

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

$$  \displaystyle {y}''=r^{2}e^{rx} $$     (Eq 3.31)
 *

 * }

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

$$  \displaystyle e^{rx}\left [ b_{2}r^{2}+b_{1}r+b_{0} \right ]=0 $$     (Eq 3.32)
 *

 * }

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

$$  \displaystyle b_{2}r^{2}+b_{1}r+b_{0}=r^{2}-10r+25 $$     (Eq 3.33) It can obciously seen that the unknowns are
 *

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

$$  \displaystyle b_{2}=1 ,b_{1}=-10 ,b_{0}=25 $$     (Eq 3.34)
 *

 * }

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

$$  \displaystyle y=U(x)e^{5x} $$     (Eq 3.35)
 *

 * }

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

$$  \displaystyle {y}'={U(x)}'e^{5x}+5U(x)e^{5x} $$     (Eq 3.36)
 *

 * }

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

$$  \displaystyle {y}=e^{5x}{U(x)}+10e^{5x}{U(x)}'+25e^{5x}U(x) $$     (Eq 3.37)
 *

 * }

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

$$  \displaystyle e^{5x}{U(x)}''+10e^{5x}{U(x)}'+25e^{5x}U(x)-10\left [ e^{5x}{U(x)}'+5e^{5x}U(x) \right ]+25e^{5x}U(x)=0 $$     (Eq 3.38)
 *

 * }

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

$$  \displaystyle {U(x)}''+{U(x)}'\left [ 10-10 \right ]+U(x)\left [ 25+25-50 \right ]=0 $$     (Eq 3.39)
 *

 * }

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

$$  \displaystyle {U(x)}''=0 $$     (Eq 3.40)
 *

 * }

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

$$  \displaystyle U=k_{1}x+k_{2} $$     (Eq 3.41)
 *

 * }

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

$$  \displaystyle y(x)=U(x)y_{1}(x)=k_{1}xe^{5x}+k_{2}e^{5x} $$     (Eq 3.42)
 *

 * }

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

$$  \displaystyle y_{2}=k_{1}xe^{5x} $$     (Eq 3.43)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5.4 Particular Solution by using Variation of Parameters Method =

From the lecture slide Mtg 27-2

Given
Linear, nonhomogeneous, first order differential equation

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

$$ \displaystyle \frac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$$ (Eq 4.1)
 *

 * }

Find
Particular solution using $$y\displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)$$

To find homogeneous solution:
Homogeneous differential equation

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

$$ \displaystyle \frac{d{{y}_{h}}}{dx}+P\left( x \right){{y}_{h}}=0$$ (Eq 4.2)
 *

 * }

and the solution of homogeneous differential equation is

 * $${{y}_{h}}=A{{e}^{-\int\limits_ – ^{x}{P\left( t \right)dt}}}=A{{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$

where A is a constant of integration (King, page 512).

To find particular solution:
Assuming

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

$$ \displaystyle \left( x \right)=A\left( x \right){{y}_{h}}(x)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.3)
 *

 * }

is solution of nonhomogeneous ODE (Eq4.1).

We determine A(x) by substituting Eq4.3 and its first derivative into Eq4.1. Differentiating Eq4.3, we obtain,

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

$$ \displaystyle \frac{d}{dx}=={A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.4)
 *

 * }

Now $$\displaystyle $$ must satisfy (Eq4.1). We need to substitute $$\displaystyle $$ and $$\displaystyle $$ according to Eq4.3 and Eq4.4 into Eq4.1.

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

$$ \displaystyle \frac{d}{dx}+P\left( x \right){{y}_{p}}=Q\left( x \right)$$ (Eq 4.5)
 *

 * }

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

$$ \displaystyle \left[ {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}-A(x)P(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right]+P(x)\left( A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}} \right)=Q(x)$$ (Eq 4.6)
 *

 * }

Second term and third term in the left side of Eq4.7 eliminate each other and we obtain

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

$$ \displaystyle {A}'(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}=Q(x)$$ (Eq 4.7)
 *

 * }

We rearrange Eq4.7

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

$$ \displaystyle {A}'(x)=Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}$$ (Eq 4.8)
 *

 * }

If we integrate both sides Eq4.8, A(x) will be

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

$$ \displaystyle A(x)=\int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx}$$ (Eq 4.9)
 *

 * }

If we put Eq4.9 into Eq4.3 in order to get the particular solution, we obtain

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

$$ \displaystyle \left( x \right)=A(x){{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\left( \int{Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx+C} \right)$$ (Eq 4.10)
 *

 * }

Eliminate the constant $$\displaystyle C$$ we can derive particular solution,

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

$$ \displaystyle {{y}_{P}}={{e}^{-\int\limits_ – ^ – {P\left( x \right)dx}}}\int Q(x){{e}^{\int\limits_ – ^ – {P\left( x \right)dx}}}dx$$ (Eq 4.11)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we can solve homework equation in Mtg10-4 as an example,

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

$$ \displaystyle \frac{dy}{dx}+xy=2x+3$$ (Eq 4.12)
 *

 * }

Thus we have,

 * $$\displaystyle \left\{ \begin{align}

& P\left( x \right)=x \\ & Q\left( x \right)=2x+3 \\ \end{align} \right.$$

If we solve the homogenous equation for $$\displaystyle P\left( x \right)=x$$,

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

$$ \displaystyle {{y}_{h}}=A{{e}^{-\frac{2}}}$$
 *

 * }

then we have(see Kreyszig's book ),

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

$$ \displaystyle ={{e}^{-\int\limits_ – ^{x}{tdt}}}\int\limits_ – ^{x}{(2s+3){{e}^{\int\limits_ – ^{s}{tdt}}}ds}={{e}^{-\frac{2}}}\left[ \int\limits_ – ^{x}{2s{{e}^{^{\frac{2}}}}ds}+\int\limits_ – ^{x}{3{{e}^{^{\frac{2}}}}ds} \right]$$ (Eq 4.13)
 *

 * }

Integrating inside of bracket by using Wolfram Alpha, we obtain $$\displaystyle $$ shown below

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

$$ \displaystyle =2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)+{{C}_{1}}{{e}^{-\frac{2}}}$$ (Eq 4.14)
 *

 * }

where erfi(x) is the imaginary error function.

Then we extract particular solution from above,

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

$$ \displaystyle {{y}_{p}}=2+3\sqrt{\frac{\pi }{2}}.{{e}^{-\frac{2}}}.erfi\left( \frac{x}{\sqrt{2}} \right)$$ (Eq 4.15)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5.5 Solving Non-homogeneous Linear Second Order ODE with Constant Coefficients=

From the lecture slide Mtg 28 and Mtg 29-1

Given
The spring-dashpot system has the Equation of Motion like follows:

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

$$ \displaystyle {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y=f(t)$$ (Eq 5.1)
 *

 * }

where $$\displaystyle y$$ is a function of $$\displaystyle t$$, and $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ are all constants making Eq 5.1 a linear second-order ODE with constant coeffiients.

Find
(1)Find PDEs for integrating factor $$\displaystyle H(t,y)$$ using condition 2nd of exactness stated in Mtg15

(2)Solve Eq 5.1 using trial solution for integrating factor such that $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha $$ is some constant. .

(1) Finding PDEs for integrating factor $$\displaystyle H(t,y)$$
Multiply $$\displaystyle H(t,y)$$ to both sides of the Eq 5.1,

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

$$ \displaystyle H(t,y){{a}_{2}}{y}''+H(t,y){{a}_{1}}{y}'+H(t,y){{a}_{0}}y=H(t,y)f(t)$$ (Eq 5.2)
 *

 * }

If we assume firstly $$\displaystyle H(t,y)$$ to be a function of only $$\displaystyle t$$(see Mtg 10), then we can just find some $$\displaystyle H(t)$$ to make the L.H.S of Eq 5.2 exact in order to change its form as,

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

$$ \displaystyle \frac{d\phi }{dt}=H(t)f(t)$$ (Eq 5.3)
 *

 * }

then we can integrate both sides to reduce the order of Eq 5.1. However, if the assumption "$$\displaystyle H(t,y)$$ is a function of only $$\displaystyle t$$" is not satisfid, we can't integrate the R.H.S of Eq 5.3. In this case, we should move $$\displaystyle f(t)$$ from R.H.S to L.H.S and try to find any $$\displaystyle H(t,y)$$ to make the whole thing exact.

Thus let

 * $$\displaystyle F=H(t){{a}_{2}}{y}+H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y=f(t,y,p){y}+g(t,y,p)$$

in which,

 * $$\displaystyle \begin{align}

& f=H(t){{a}_{2}} \\ & g=H(t){{a}_{1}}{y}'+H(t){{a}_{0}}y \\ & p={y}' \\ \end{align}$$

Then we have,

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

$$ \displaystyle \begin{align} & {{f}_{tt}}={{a}_{2}}{{H}_{tt}} \\ & {{f}_{ty}}={{a}_{2}}{{H}_{ty}} \\ & {{f}_{tp}}={{f}_{yp}}=0 \\ & {{f}_{y}}={{a}_{2}}{{H}_{y}} \\ & {{f}_{yy}}={{a}_{2}}{{H}_{yy}} \\ & {{g}_{tp}}={{a}_{1}}{{H}_{t}} \\ & {{g}_{yp}}={{a}_{1}}{{H}_{y}} \\ & {{g}_{pp}}=0 \\ & {{g}_{y}}={{a}_{1}}p{{H}_{y}}+{{a}_{0}}H+{{a}_{0}}y{{H}_{y}} \\ \end{align}$$ (Eqs 5.4)
 *

 * }

Substituting Eqs 5.4 into (1)&(2) in Mtg15, we have,

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

$$ \displaystyle {{a}_{2}}{{H}_{tt}}+2p{{a}_{2}}{{H}_{ty}}+{{p}^{2}}{{a}_{2}}{{H}_{yy}}={{a}_{1}}{{H}_{t}}-{{a}_{0}}H-{{a}_{0}}y{{H}_{y}}$$ (Eq 5.5)
 *

 * }

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

$$ \displaystyle 2{{a}_{2}}{{H}_{y}}=0$$ (Eq 5.6)
 *

 * }

From Eq 5.6 we can see that the basic assumption is satisfied and if we substitute Eq 5.6 in to Eq 5.5, we have,

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

$$ \displaystyle {{a}_{2}}{{H}_{tt}}-{{a}_{1}}{{H}_{t}}+{{a}_{0}}H=0$$ (Eq 5.7)
 *

 * }

or

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

$$ \displaystyle {{a}_{2}}{H}''-{{a}_{1}}{H}'+{{a}_{0}}H=0$$ (Eq 5.8)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Since the integrating factor $$\displaystyle H(t,y)$$ turns out to be $$\displaystyle H(t)$$, the PDE governing $$\displaystyle H(t,y)$$ become an ODE given by Eq 5.8.

(2) Solve Eq 5.1 using trial solution for integrating factor $$\displaystyle h(t)={{e}^{\alpha t}}$$
Let's try an integrating factor like $$\displaystyle h(t)={{e}^{\alpha t}}$$ where $$\displaystyle \alpha$$ is a constant.

Multiply Eq 5.1 by $$\displaystyle h(t)={{e}^{\alpha t}}$$, we have,

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

$$ \displaystyle {{e}^{\alpha t}}\left( {{a}_{2}}{y}''+{{a}_{1}}{y}'+{{a}_{0}}y \right)={{e}^{\alpha t}}f(t)$$ (Eq 5.9)
 *

 * }

In order to reduce the order, we integrating Eq 5.9 according to $$\displaystyle t$$ and assume that the result can be written in such form:

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

$$ \displaystyle {{e}^{\alpha t}}\left( {{{\bar{a}}}_{1}}{y}'+{{{\bar{a}}}_{2}}y \right)=\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.10)
 *

 * }

where $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{2}}$$ are assumed to be constants.

(2.1) Determine $$\displaystyle {{{\bar{a}}}_{1}}$$ and $$\displaystyle {{{\bar{a}}}_{1}}$$ according to $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$
Differentiate Eq 5.10 we have,

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

$$ \displaystyle {{e}^{\alpha t}}\left[ {{{\bar{a}}}_{1}}{y}''+(\alpha {{{\bar{a}}}_{1}}+{{{\bar{a}}}_{0}}){y}'+\alpha {{{\bar{a}}}_{0}}y \right]={{e}^{\alpha t}}f(t)$$ (Eq 5.11)
 *

 * }

Compare 5.11 and 5.9 we have,

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$$ \displaystyle \begin{align} & {{{\bar{a}}}_{1}}={{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ \end{align}$$ (Eqs 5.12)
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 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 *

 * }

from which we can see that the validity of Eq 5.10 relies on the value of $$\displaystyle \alpha $$.

(2.2) Find quadratic equation for $$\displaystyle \alpha $$
From Eqs 5.12 we have,

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$$ \displaystyle \begin{align} & {{a}_{1}}-\alpha {{a}_{2}}={{a}_{0}}/\alpha \\ \Rightarrow & {{a}_{2}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0 \\ \end{align}$$ (Eq 5.13)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

(2.3) Derive a reduced - order equation from Eq 5.10
From Eq 5.10 we have,

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

$$ \displaystyle {y}'+\beta y=\frac{1}\int{{{e}^{\alpha t}}f(t)dt}$$ (Eq 5.14)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$\displaystyle \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}}$$

(2.4) Solve Eq 5.14 using Integrating Factor Method
From Mtg10 we know the integrating factor(for Eq 5.14) is,

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

$$ \displaystyle i(t)=\exp \int{\beta dt}={{e}^{\beta t}}$$ (Eq 5.15)
 *

 * }

and the solution is:

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

$$ \displaystyle \begin{align} y & =\frac{1}{i(t)}\int_ – ^{t}{\frac{i(\tau )}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \end{align}$$ (Eq 5.16)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

(2.5) Show the relation between $$\displaystyle \alpha $$ and $$\displaystyle \beta $$
Since,

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

$$ \displaystyle \begin{align} & \beta ={{{\bar{a}}}_{0}}/{{{\bar{a}}}_{1}} \\ & {{{\bar{a}}}_{0}}={{a}_{1}}-\alpha {{a}_{2}} \\ & {{{\bar{a}}}_{0}}={{a}_{0}}/\alpha \\ & {{{\bar{a}}}_{1}}={{a}_{2}} \\ \end{align}$$ (Eqs 5.17)
 *

 * }

So we have,

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

$$ \displaystyle \begin{align} & \alpha \beta =\frac \\ & \alpha +\beta =\frac \\ \end{align}$$ (Eq 5.18)
 *

 * }

which means $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are the two roots of,

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

$$ \displaystyle {{\lambda }^{2}}-(\alpha +\beta )\lambda +\alpha \beta =0$$
 *

 * }

or

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

$$ \displaystyle {{\lambda }^{2}}-\frac\lambda +\frac=0$$ (Eq 5.19)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

which is identical to Eq 5.13. And from Eq 5.17~Eq 5.19 we can conclude that $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are inseparable pairs so that we can't distiguish them from each other through the their governing equation and the sequense of two integrating factor $$\displaystyle h(t)$$ and $$\displaystyle i(t)$$ is arbitrary. We can also conclude that the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, meaning exchanging the position of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ won't change the result.

When $$\displaystyle \alpha \ne \beta $$
Eq 5.16 can be expanded as,

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

$$ \displaystyle \begin{align} y & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{e}^{(\beta -\alpha )\tau }}{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+\frac{\beta -\alpha }{{e}^{(\beta -\alpha )t}} \right] \\ & =\frac{(\beta -\alpha ){_{1}}}{{e}^{-\alpha t}}+\frac{{e}^{-\beta t}}+\frac{{{{\bar{a}}}_{1}}}\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \end{align}$$ (Eq 5.20)
 *

 * }

Note that Eq 5.20 is under the assumption that $$\alpha \ne \beta $$.

After conbination of constants, we have,

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

$$ \displaystyle y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ (Eq 5.21)
 *

 * }

In which we can derive,

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$$ \displaystyle \begin{align} & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ \end{align}$$
 *

 * }

However, $$\displaystyle {{y}_{P}}$$ can be furthuer deduced as,

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

$$ \displaystyle \begin{align} {{y}_{P}} & =\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)d\left( {{e}^{(\beta -\alpha )t}} \right)} \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left[ {{e}^{(\beta -\alpha )t}}\int{{{e}^{\alpha t}}f(t)dt}-\int{{{e}^{(\beta -\alpha )t}}{{e}^{\alpha t}}f(t)dt} \right] \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \end{align}$$
 *

 * }

This proved my statement "the final structure of solution must be symetric with respect to $$\displaystyle \alpha $$ and $$\displaystyle \beta $$".

In sum we have (when $$\displaystyle \alpha \ne \beta $$),

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

$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}={{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt} \\ \end{align}$$ (Eq 5.22)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

When $$\displaystyle \alpha =\beta $$
Substitute the relation $$\displaystyle \alpha =\beta $$ into Eq 5.16,

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

$$ \displaystyle \begin{align} y= & \frac\int_ – ^{t}{{{e}^{(\alpha -\alpha )\tau }}(\int_ – ^{\tau }{{{e}^{\alpha s}}f(s)ds}})d\tau \\ & =\frac\int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau }+{{k}_{1}} \right)d\tau } \\ & =\frac\left[ \int_ – ^{t}{\left( \int{{{e}^{\alpha \tau }}f(\tau )d\tau } \right)d\tau }+\int_ – ^{t}{{{k}_{1}}d\tau } \right] \\ & =\frac\left[ \int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}+{{k}_{2}}+{{k}_{1}}t \right] \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\int{\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt} \\ & =\fract{{e}^{-\alpha t}}+\frac{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \end{align}$$ (Eq 5.20a)
 *

 * }

After conbination of constans we have

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$$ \displaystyle y={{C}_{1}}t{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\alpha t}}+\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right)$$ (Eq 5.21a)
 *

 * }

In which we can derive,

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$$ \displaystyle \begin{align} & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$
 *

 * }

In sum we have (when $$\displaystyle \alpha =\beta $$),

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

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 * style="width:90%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle \begin{align} & y={{C}_{1}}y_{H}^{1}+{{C}_{2}}y_{H}^{2}+{{y}_{P}} \\ & y_{H}^{1}=t{{e}^{-\alpha t}} \\ & y_{H}^{2}={{e}^{-\alpha t}} \\ & {{y}_{P}}=\frac\left( t\int{{{e}^{\alpha t}}f(t)dt}-\int{t{{e}^{\alpha t}}f(t)dt} \right) \\ \end{align}$$ (Eq 5.22a)
 *

 * }

(2.7) Verify particular solution according to table in Mtg20, Fall2009
If $$\displaystyle \alpha \ne \beta$$, then from Eq 5.22 we know,

 * $$\displaystyle {{y}_{P}}=\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\alpha t}}f(t)dt}-\frac{(\beta -\alpha ){{a}_{2}}}\int{{{e}^{\beta t}}f(t)dt}$$

If we substitute $$\displaystyle f(t)={{t}^{2}}$$ into the above equation, we have,

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

$$ \displaystyle \begin{align} {{y}_{P}} & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}f(t)dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}f(t)dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\alpha t}}{{t}^{2}}dt-\frac{(\beta -\alpha ){{a}_{2}}}\int {{e}^{\beta t}}{{t}^{2}}dt \\ & =\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2} \right)-\frac{(\beta -\alpha ){{a}_{2}}}\left( \frac{{{\beta }^{2}}{{t}^{2}}-2\beta t+2} \right) \\ & =\frac{{{\alpha }^{2}}{{\beta }^{2}}{{t}^{2}}-(2{{\alpha }^{2}}\beta +2\alpha {{\beta }^{2}})t+2{{\alpha }^{2}}+2{{\beta }^{2}}+2\alpha \beta } \\ & =\frac{{{(\alpha \beta )}^{2}}{{t}^{2}}-2(\alpha +\beta )t+2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{\frac{{{a}_{0}}^{2}}{{{a}_{2}}^{2}}{{t}^{2}}-2\frac{{{a}_{2}}^{2}}t+2\frac{{{a}_{1}}^{2}}{{{a}_{2}}^{2}}-2\frac{{{a}_{2}}^{2}}}{\frac{{{a}_{0}}^{3}}{{{a}_{2}}^{3}}{{a}_{2}}} \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23)
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 * style="width:90%; padding:10px; border:2px solid #8888aa" |
 *

 * }

in which we utilize,

 * $$\displaystyle \begin{align}

\int{{{e}^{\alpha t}}{{t}^{2}}dt} & =\frac{1}{\alpha }\int{{{t}^{2}}d{{e}^{\alpha t}}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{1}{\alpha }\int{2t{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}\int{td}{{e}^{\alpha t}} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}\int{{{e}^{\alpha t}}dt} \\ & =\frac{\alpha }{{e}^{\alpha t}}-\frac{2}t{{e}^{\alpha t}}+\frac{2}{{e}^{\alpha t}} \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-2\alpha t+2}{{e}^{\alpha t}} \\ \end{align}$$

Compare Eq 5.23 with Mtg20 slide of Fall 2009, we can see that Eq 5.23 has the same form with that in the slide which is a polynominal with at least order 2 according to $$\displaystyle t$$.

If $$\displaystyle \alpha=\beta$$, then from Eq 5.22a we know,

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

$$ \displaystyle \begin{align} & {{y}_{P}}=\frac\left( t\int {{e}^{\alpha t}}{{t}^{2}}dt-\int {{e}^{\alpha t}}{{t}^{3}}dt \right) \\ & =\frac\left( \frac{{{e}^{\alpha t}}({{\alpha }^{2}}{{t}^{3}}-2\alpha {{t}^{2}}+2t)}-\frac{{{e}^{\alpha t}}({{\alpha }^{3}}{{t}^{3}}-3{{\alpha }^{2}}{{t}^{2}}+6\alpha t-6)} \right) \\ & =\frac{{{\alpha }^{2}}{{t}^{2}}-4\alpha t+6} \\ & =\frac{\alpha \beta }{{t}^{2}}-\frac{2(\alpha +\beta )}t+\frac{2{{(\alpha +\beta )}^{2}}-2\alpha \beta } \\ & =\frac{1}{{t}^{2}}-\frac{2{{a}_{1}}}{{{a}_{0}}^{2}}t+\frac{2{{a}_{1}}^{2}-2{{a}_{0}}{{a}_{2}}}{{{a}_{0}}^{3}} \\ \end{align}$$ (Eq 5.23a)
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 *

 * }

We can see that Eq 5.23a still has the same form with Eq 5.23 and that in the slide Mtg20. Thus the particular solution is verified.

(2.8) Solve L2-ODE-CC with $$\displaystyle f(t)=\exp (-{{t}^{2}})$$
If $$\displaystyle f(t)=\exp (-{{t}^{2}})$$, then Eqs 5.22 and Eqs 5.22a we derive

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$$ \displaystyle \begin{align} & y_{H}^{1}=\left\{ \begin{align} & {{e}^{-\alpha t}}....................................................................................(\alpha \ne \beta ) \\ & t{{e}^{-\alpha t}}...................................................................................(\alpha =\beta ) \\ \end{align} \right. \\ & y_{H}^{2}={{e}^{-\beta t}} \\ & {{y}_{P}}=\left\{ \begin{align} & \frac{\sqrt{\pi }\left[ {{e}^{\frac{4}-\alpha t}}erf(\frac{\alpha }{2}-t)-{{e}^{\frac{4}-\beta t}}erf(\frac{\beta }{2}-t) \right]}{2{{a}_{2}}(\alpha -\beta )}........................(\alpha \ne \beta ) \\ & -\frac{1}{2{{a}_{2}}}\sqrt{\pi }{{e}^{\frac{4}-\alpha t}}\left[ (t-\frac{\alpha }{2})erf(\frac{\alpha }{2}-t)-\frac{1}{\sqrt{\pi }}{{e}^{-\frac{4}}} \right].........(\alpha =\beta ) \\ \end{align} \right. \\ \end{align}$$ (Eq 5.24)
 *

 * }

It has been mentioned that governing equation of $$\displaystyle \alpha $$ is indentical to Eq 5.19, and it has also been mentioned that the sequence of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ is arbitrary.

Case 1: $$\displaystyle (\gamma +1)(\gamma -2)=0$$
From the characteristic equation we can get information of $$\displaystyle \alpha $$ and $$\displaystyle \beta $$, which are the ratio of $$\displaystyle {{a}_{0}}$$ and $$\displaystyle {{a}_{1}}$$ with respect to $$\displaystyle {{a}_{2}}$$. However we can't derive information of $$\displaystyle {{a}_{2}}$$. In fact $$\displaystyle {{a}_{2}}$$ has nothing to do with characteristic equation since in the homogeneous case $$\displaystyle {{a}_{2}}$$ can always be the divisor of both sides(the R.H.S is zero so $$\displaystyle {{a}_{2}}$$ disappeared only left some ratios) and has no effect on the results if $$\displaystyle \alpha $$ and $$\displaystyle \beta $$ are fixed. But in the non-homogeneous case $$\displaystyle {{a}_{2}}$$ can be seen as part of the R.H.S(non-zero) and do have influence on the final result. As we can't get enough information to determine the value of $$\displaystyle {{a}_{2}}$$, we take it as 1:

 * $$\displaystyle {{a}_{2}}=1$$

From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:

 * $$\displaystyle {y}''-{y}'-2y={{e}^{-{{t}^{2}}}}$$

then

 * $$\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-1 \\ & \alpha \beta =\frac=-2 \\ \end{align} \right. \\ & \Rightarrow \left\{ \begin{align} & \alpha =1 \\ & \beta =-2 \\ \end{align} \right. \\ \end{align}$$

Substitute these into Eq 5.24 we have the solution as,

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

$$ \displaystyle y={{C}_{1}}{{e}^{-t}}+{{C}_{2}}{{e}^{2t}}+\frac{\sqrt{\pi }}{6}\left[ {{e}^{0.25-t}}erf(0.5-t)-{{e}^{1+2t}}erf(-1-t) \right]$$ (Eq 5.25)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

Case 2: $$\displaystyle {{(\gamma -4)}^{2}}=0$$
From the characteristic equation we have the following L2-ODE-CC by refering Eq 5.13 and Eq 5.1:

 * $$\displaystyle {y}''-8{y}'+16y={{e}^{-{{t}^{2}}}}$$

then,

 * $$\displaystyle \begin{align}

& \left\{ \begin{align} & \alpha +\beta =\frac=-8 \\ & \alpha \beta =\frac=16 \\ \end{align} \right. \\ & \Rightarrow \alpha =\beta =-4 \\ \end{align}$$

Substitute this into Eq 5.24 we have,

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

$$ \displaystyle y={{C}_{1}}t{{e}^{4t}}+{{C}_{2}}{{e}^{4t}}-\frac{\sqrt{\pi }}{2}{{e}^{4+4t}}\left[ (t+2)erf(2+t)-\frac{1}{\sqrt{\pi }}{{e}^{-{{(2+t)}^{2}}}} \right]$$ (Eq 5.26)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

where $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$ are some constants.

= Problem 5.6: Show Agreement of Particular Integrals =

From the lecture slide Mtg 30-1

Given
Show that the following equation:

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$$ \displaystyle

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}

$$ (Eq 6.1)
 *

 * }

is equivalent to the following particular integral found in the Differential Equations text book by A.C.King:
 * {| style="width:100%" border="0"

$$ \displaystyle

{{y}_{p}}\left( x \right)=\oint^x{f\left( s \right)\left\{ \frac{{{u}_{1}}(s) {{u}_{2}}(x)-{{u}_{1}}(x) {{u}_{2}}(s)}{W(s)} \right\}}\,ds

$$ (Eq 6.2)
 *

 * }

given that the following equations holds true:
 * {| style="width:100%" border="0"

$$ \displaystyle

W={{u}_{1}} {{u}_{2}}'-{{u}_{1}}' {{u}_{2}}=\left| \begin{matrix} {{u}_{1}} & {{u}_{2}} \\ {{u}_{1}}' & {{u}_{2}}' \\ \end{matrix} \right|

$$ (Eq 6.3)
 *

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

$$ \displaystyle

(\frac{u_{2}}{u_{1}})'=\frac{1}{h}

$$ (Eq 6.4)
 *

 * }

And considering the following conversion between the notation used in the lectures and A.C. King's notation:
 * {| style="width:100%" border="0"

$$ \displaystyle

\oint^x{f\left( s \right)}\,ds=\int^x{f\left(s\right)}\,ds =\int{f\left(x\right)}\,dx

$$  w/o constant (Eq 6.5)
 *

 * }

Solution
The first step to tackling this problem requires the application of the Integration by Parts formula to equation 6.1

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$$\displaystyle \int w\,dv=wv-\int v\,dw $$

(Eq 6.6)
 *

 * }

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

$$\displaystyle

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{w}\underbrace{\frac{1}{h\left( x \right)}\,dx}_{dv}}$$

(Eq 6.7) The remaining unknowns in the right hand side of equation 6.6 are shown below:
 *

 * }

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

$$\displaystyle dw=h\left( x \right) f\left( x \right)dx

$$

(Eq 6.8)
 *

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

$$\displaystyle v= \int {\frac{1}{h\left( x \right)}}\,dx

$$

(Eq 6.9) Equations 6.7, 6.8 and 6.9 can be combined and rewritten into the form shown in equation 6.5 to yield the following:
 *

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

$$\displaystyle \begin{align} {{y}_{p}}\left( x \right)= & {{u}_{1}}\left( x \right)\left[ \left( \mathop{\oint }^{x}h\left( s \right)f\left( s \right)\,ds \right)\int \frac{1}{h\left( x \right)}\,dx \right. \\ & \left. -\mathop{\oint }^{x}\left( \mathop{\oint }^{x}\frac{1}{h\left( s \right)}\,ds \right)h\left( s \right)f\left( s \right)\,ds \right] \\ \end{align}$$

(Eq 6.10)
 *

 * }

The relation shown in equation 6.4 can be used to simplify equation 6.10 as follows:
 * {| style="width:100%" border="0"

$$\displaystyle \begin{align} {{y}_{p}}\left( x \right)= & {{u}_{1}}\left( x \right)\left[ \left( \mathop{\oint }^{x}\frac{u_{1}^{2}\left( s \right)}{{{u}_{{{2}'}}}\left( s \right){{u}_{1}}\left( s \right)-{{u}_{2}}\left( s \right){{u}_}\left( s \right)}f\left( s \right)\,ds \right)\frac{{{u}_{2}}\left( x \right)}{{{u}_{1}}\left( x \right)} \right. \\ & \left. -\left[ \mathop{\oint }^{x}\frac{{{u}_{2}}\left( s \right)}{{{u}_{1}}\left( s \right)}\frac{{{u}_{1}}^{2}\left( s \right)}{{{u}_{{{2}'}}}\left( s \right){{u}_{1}}\left( s \right)-{{u}_{2}}\left( s \right){{u}_}\left( s \right)}f\left( s \right)\,ds \right] \right] \\ \end{align}$$ (Eq 6.11) Now, multiplying through by u1(x) and using the relation shown in equation 6.3 yields the following:
 *

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

$$ \displaystyle \begin{align} {{y}_{p}}\left( x \right) & =\left( \mathop{\oint }^{x}\frac{{{u}_{1}}\left( s \right){{u}_{2}}\left( x \right)}{W\left( s \right)}f\left( s \right)\,ds \right)-\left( \mathop{\oint }^{x}\frac{{{u}_{2}}\left( s \right){{u}_{1}}\left( x \right)}{W\left( s \right)}f\left( s \right)\,ds \right) \\ & =\mathop{\oint }^{x}f\left( s \right)\left\{ \frac{{{u}_{1}}(s){{u}_{2}}(x)-{{u}_{1}}(x){{u}_{2}}(s)}{W(s)} \right\}\,ds \\ \end{align}$$ (Eq 6.12)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5.7: Find Two Homogeneous Solutions using a Trial Solution =

From the lecture slide Mtg 31-1

Given
Find u1 and u2 of Equation 7.1 using the trial solution shown in equation 7.2:

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

$$ \displaystyle

(x +1) y'' - (2x+3)y' + 2y = 0

$$ (Eq 7.1)
 *

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

$$ \displaystyle

y(x)=e^{rx}

$$ (Eq 7.2)
 *

 * }

Solution
The first step is to differentiate the trial solution as follows:

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

$$ \displaystyle y(x) = e^{rx} $$ (Eq 7.3)
 *

 * }

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

$$ \displaystyle y'(x) = r e^{rx} $$ (Eq 7.4)
 *

 * }

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

$$ \displaystyle y''(x) = r^2 e^{rx} $$ (Eq 7.5)
 *

 * }

Next, plug the values into equation 7.1 to yield:

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

$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$ (Eq 7.6) Since erx will not equal zero for any value of x, divide equation 7.6 by this value and factor out the x to yield:
 *

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

$$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0 $$ (Eq 7.7) In order to satisfy equation 7.7 for any x, the following equation is solved:
 *

 * }

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

$$ \displaystyle r(r-2)=(r-2)(r-1)=0 $$ (Eq 7.8)
 *

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

$$ \displaystyle r=2 $$ (Eq 7.9) By plugging equation 7.9 into the trial solution in equation 7.3 we have obtained the first homogeneous solution:
 *

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

$$ \displaystyle u_1(x)=e^{2x} $$ (Eq 7.10)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

The second homogeneous solution is obtained by the following:

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

$$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$$
 *

(Eq 7.11)
 * }

Equation 7.1 is rearranged into the following form:

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

$$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0 $$ (Eq 7.12)
 *

 * }

The value for a1 shown in equation 7.12 is then plugged into equation 7.11 along with the first homogeneous solution to solve for the second solution as shown:

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

$$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$$ (Eq 7.13)
 *

 * }

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

$$ \displaystyle \Rightarrow u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$$ (Eq 7.14)
 *

 * }

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

$$ \displaystyle \Rightarrow u_2(x)=\frac{-1}{4}(2x+3)$$ (Eq 7.15)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

= Problem 5.8 L2_ODE_VC (Homogeneous) by using Reverse Engineering =

From the lecture slide Mtg 31-4

Given
Trial Solution: $$\displaystyle y=\frac$$ and Characteristic Equation: $$\displaystyle {{r}^{2}}+3=0$$

Find
Find L2_ODE_VC according to a trial solution and the characteristic equation given above

Solution
Wolfram Alpha used to find the derivatives of the trial solution $$\displaystyle y$$

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

$$ \displaystyle \left[ y=\frac \right].{{a}_{0}}(x)$$ (Eq 8.1)
 *

 * }

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

$$ \displaystyle \left[ \frac{d}{dx}\left( \frac \right)=\left( r-\frac{2}{x} \right)\left( \frac \right) \right].{{a}_{1}}(x)$$ (Eq 8.2)
 *

 * }

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

$$ \displaystyle \left[ \frac{d{{x}^{2}}}\left( \frac \right)=\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)\left( \frac \right) \right].{{a}_{2}}(x)$$ (Eq 8.3)
 *

 * }

Adding Eq8.1, Eq8.2 and Eq8.3 each other with taking common parenthesis of $$\displaystyle y$$ will be equal to $$\displaystyle y$$ multiply by characteristic equation.

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

$$ \displaystyle \frac\left[ {{a}_{2}}\left( {{r}^{2}}-\frac{4r}{x}+\frac{6} \right)+{{a}_{1}}\left( r-\frac{2}{x} \right)+{{a}_{0}} \right]=\frac\left( {{r}^{2}}+3 \right)=0$$ (Eq 8.4)
 *

 * }

First expanding the inside of brackets and then rearranging and taking common parentheses of $$\displaystyle {{r}^{2}}$$, $$\displaystyle r$$ and 1, it will be

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

$$ \displaystyle {{r}^{2}}({{a}_{2}})+r\left( -\frac{4}{x}{{a}_{2}}+{{a}_{1}} \right)+1\left( \frac{6}{{a}_{2}}-\frac{2}{x}{{a}_{1}}+{{a}_{0}} \right)={{r}^{2}}+0.r+3$$ (Eq 8.5)
 *

 * }

From the equality of both sides, we find the values of coefficients $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$.

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

$$ \displaystyle \begin{align} & {{a}_{2}}=1 \\ & {{a}_{1}}=\frac{4}{x} \\ & {{a}_{0}}=3+\frac{2} \\ \end{align}$$
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$ \displaystyle {{r}^{2}}+3=0\Rightarrow {{r}_{1,2}}=\mp i\sqrt{3}$$ (Eq 8.6)
 *

 * }

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

$$ \displaystyle {{a}_{2}}(x){y}''+{{a}_{1}}(x){y}'+{{a}_{0}}(x)y=0$$ (Eq 8.7)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

is the general form of a second-order ordinary differential equation with coefficients.

If we substitute $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{0}}$$ into Eq8.7, we obtain

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

$$ \displaystyle {y}''+\frac{4}{x}{y}'+(3+\frac{2})y=0$$ (Eq 8.8)
 *

 * }

Multiply all equation with $$\displaystyle {{x}^{2}}$$ to eliminate denominators

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

$$ \displaystyle {{x}^{2}}{y}''+4x{y}'+\left( 3{{x}^{2}}+2 \right)y=0$$ (Eq 8.9)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 *

 * }

=Contributions Of Team Members=

 * Problem 1:    Solved by: Oztekin,Zou

Author: Oztekin

Proofread by: Oztekin, Dube, Zou

 * Problem 2:    Solved by: Dube

Author: Dube

Proofread by: Oztekin,Ismail

 * Problem 3:    Solved by: Oztekin

Author: Oztekin

Proofread by: Dube

 * Problem 4:    Solved by: Ismail,Yaakov

Author: Zou, Ismail

Proofread by: Zou,Oztekin

 * Problem 5:    Solved by: Zou, Oztekin

Author: Zou

Proofread by: Oztekin,Yaakow

 * Problem 6:    Solved by: Patterson

Author: Patterson

Proofread by: Yaakov, Oztekin

 * Problem 7:    Solved by: Patterson

Author: Patterson

Proofread by: Oztekin,Yaakov,Zou

 * Problem 8:    Solved by: Ismail,Yaakov

Author: Zou, Ismail

Proofread by: Zou,Oztekin,Dube

= References =

title: User:Egm6321.f10.team2.dube/HW5


text:
Motivation/Self-concept

Possible topics:
 * How does self-concept relate to motivation?
 * What self-conceptions about our motivational interests and behaviours do we have?
 * Does self-concept cause motivation or vice-versa or are they reciprocal
 * Vocational motivation and self-concept

title: Motivation/Self-concept


text: File:Cognitive dissonance.JPG

title: File:Cognitive dissonance.JPG


text:
Talk:Motivation and emotion/Textbook/Motivation/Self-concept

Feedback
-- Jtneill - Talk - c 04:14, 1 November 2010 (UTC)
 * 1) Overall this is a promising draft so far, with lots of detailed and interesting information. The main areas to improve are to reduce the overall length, separate the examples from the main body, shorten overly long paragraphs. The introduction and conclusion need to work on establishing the key foci for the chapter. Make it clear in each section how it addresses the chapter's stated aims.
 * 2) I started separating the Liz and Ben example from the body text - it is a really good ideas to have a rich example like this, but I think it will work better if you run the example in separate side-boxes. I suggest do something similar throughout - makes it easier for a reader to digest.
 * 3) Avoid overly long paragraphs - generally, 3 to 5 sentences is a good paragraph length.
 * 4) Self-concept motivates emotions - is this heading accurate? (I think it could better reflect the content) (you only need to relate self-concept to motivation - not emotions. So, maybe simplify.)
 * 5) is self-concept seen here as motivating emotions - or just causing them? Do the cognitions causes emotions then cause motivation? Try to explain a bit more how this section connects with the overall topic (are there focus questions to be answered?). Or perhaps this is a bit of a side detour here?
 * 6) I think this might work better if you have the pop songs and related commentary in a separate side box and then have a short section which focuses on relevant theory and research (with citations).
 * 7) Self-Discrepancy Theory: In Focus - I'd suggest running this in a sidebox too, running down the page with the theory/research content next to it (as for the previous sections).
 * 8) Self-Discrepancy + Possible selves = almost 2000 words! (Self-esteem is another 700 words). I didn't read all of this - I get the sense that it could be abbreviated. What is the focus question you are trying to answer? Explain this question first and why it's important - then provide an answer drawing on and explaining relevant theory and research, with an example in a sidebox.
 * 9) Self Determination Theory could be relevant in many places here - e.g., the need for competence relates to self-discrepancy and possible selves
 * 10) Self-Evaluation Maintenance Model - there was no previous announcement of this model or particularly flow/connection. As you redraft make sure to set up in your introduction what the key questions and topics are going to be and why the chapter focuses on these ones in particular.
 * 11) References
 * 12) May be of interest: Showers, C. J., Abramson, L. Y., & Hogan, M. E. (1998). The dynamic self: How the content and structure of the self-concept change with mood. Journal of Personality and Social Psychology, 75, 478-493.

Feedback 2
Well done on advancing this draft. It is already a significant and substantial work. I've looked quickly through it again - my rough sense is this is DI-quality with potential for higher. Feedback below: -- Jtneill - Talk - c 02:51, 5 November 2010 (UTC)
 * 1) Introduction
 * 2) Shaping up nicely, I gave it a tweak.
 * 3) Explain or link to info about those two types of well-being.
 * 4) Check the link at the end.
 * 5) Add an overview of the focus questions the chapter seeks to address
 * 6) Boxes - I like these; simple, clear interesting. I'd probably increase their widths a bit (it will look different on different screens - here they are a bit narrowly squished).
 * 7) Definitions - Rich and useful; well done
 * 8) How Does the Self-Concept Develop? onwards - the rest of this content is substantial, interesting, and well referenced with clear focus on theory. The main areas to work on are
 * 9) flow and integration between the sections (e.g., by explaining the focus questions in the introduction and the intro/conclusion to each section.
 * 10) My other main suggestion is to as much as possible keep drawing the connection and relevant to motivation (in all sections).
 * 11) Where possible, keep the overall length shorter - not a big deal and I know this is hard. But for one other chapter I've suggested putting some related but supplementary material into subpages (but their draft was much longer).
 * 12) Since the chapter is already quite substantial I'd be inclined to have relatively short sections for "Do Self-Concepts Vary Across Cultures?" and "How Stable is the Self-Concept and Why is Stability Important?" - these are important topics for self-concept, but probably not as important as How does SC develop and how can it be changed for this chapter. If some whole sections were to be left out (an option) then it would be these two sections - or they could be treat them as supplementary topics.
 * 13) The pop-songs are useful - leave them in - but I would just separate them out of the main text body into a separate box (as you've done)
 * 14) I think there is some literature about different types of self-discrepancy, although a lot focuses on actual-ideal. I'm not how different actual-ought discrepancies would be from actual-ideal discrepancies - they sound quite similar to me.
 * 15) Keep the self-maintenance model but yes, make it clear in the intro/conclusion how it connects in with the focus questions and rest of the content - and draw connection with motivation.

-- Jtneill - Talk - c 05:12, 23 November 2010 (UTC)

-- Jtneill - Talk - c 23:24, 13 December 2010 (UTC)


text:
User:Egm6321.f10.team2.patterson/Homework 5

= Problem 6: Show Agreement of Particular Integrals =

Given
Show that the following equation:

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

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

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}

$$ (eq 6.1)
 *
 * }

is equivalent to the following particular integral found in the Differential Equations text book by A.C.King:
 * {| style="width:100%" border="0"

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

{{y}_{p}}\left( x \right)=\oint^x{f\left( s \right)\left\{ \frac{{{u}_{1}}(s) {{u}_{2}}(x)-{{u}_{1}}(x) {{u}_{2}}(s)}{W(s)} \right\}}\,ds

$$ (eq 6.2)
 *
 * }

given that the following equations holds true:
 * {| style="width:100%" border="0"

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

W={{u}_{1}} {{u}_{2}}'-{{u}_{1}}' {{u}_{2}}=\left| \begin{matrix} {{u}_{1}} & {{u}_{2}} \\ {{u}_{1}}' & {{u}_{2}}' \\ \end{matrix} \right|

$$ (eq 6.3)
 *
 * }
 * {| style="width:100%" border="0"

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

(\frac{u_{2}}{u_{1}})'=\frac{1}{h}

$$ (eq 6.4)
 *
 * }

And considering the following conversion between the notation used in the lectures and A.C. King's notation:
 * {| style="width:100%" border="0"

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

\oint^x{f\left( s \right)}\,ds=\int^x{f\left(s\right)}\,ds =\int{f\left(x\right)}\,dx

$$  w/o constant (eq 6.5)
 *
 * }

Solution
The first step to tackling this problem requires the application of the Integration by Parts formula to equation 6.1

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

$$\displaystyle \int w\,dv=wv-\int v\,dw $$
 * style="width:95%" |
 * style="width:95%" |

(eq 6.6)
 *
 * }

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

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

{{y}_{p}}\left( x \right)={{u}_{1}}\left( x \right)\int{\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{w}\underbrace{\frac{1}{h\left( x \right)}\,dx}_{dv}}$$

(eq 6.7) The remaining unknowns in the right hand side of equation 6.6 are shown below:
 *
 * }

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

$$\displaystyle dw=h\left( x \right) f\left( x \right)dx
 * style="width:95%" |
 * style="width:95%" |

$$

(eq 6.8)
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle v= \int {\frac{1}{h\left( x \right)}}\,dx
 * style="width:95%" |
 * style="width:95%" |

$$

(eq 6.9) Equations 6.7, 6.8 and 6.9 can be combined and rewritten into the form shown in equation 6.5 to yield the following:
 *
 * }
 * {| style="width:100%" border="0"

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

{{y}_{p}}\left( x \right)=u_{1}\left( x \right)\left[\left( \oint^x{h\left(s \right)f\left(s \right)}\,ds \right)\int{\frac{1}{h\left(x \right)}}\,dx- \left[\oint^x{\left(\oint^x{\frac{1}{h\left(s\right)}}\,ds\right) h\left(s\right)f\left(s\right)}\,ds\right] \right]

$$

(eq 6.10)
 *
 * }

The relation shown in equation 6.4 can be used to simplify equation 6.10 as follows:
 * {| style="width:100%" border="0"

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

{{y}_{p}}\left( x \right)=u_{1}\left( x \right)\left[\left( \oint^x{\frac{u_{1}^{2}\left(s\right)}{u_{2}'\left(s\right)u_{1}\left(s\right)-u_{2}\left(s\right)u_{1}'\left(s\right)}f\left(s \right)}\,ds \right)\frac{u_{2}\left(x\right)}{u_{1}\left(x\right)}- \left[\oint^x{\frac{u_{2}\left(s\right)}{\cancel{u_{1}\left(s\right)}} \frac{u_{1}^\cancel\left(s\right)}{u_{2}'\left(s\right)u_{1}\left(s\right)-u_{2}\left(s\right)u_{1}'\left(s\right)}f\left(s\right)}\,ds\right] \right]

$$

(eq 6.11) Now, multiplying through by u1(x) and using the relation shown in equation 6.3 yields the following:
 *
 * }
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

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

{{y}_{p}}\left( x \right)=\left( \oint^x{\frac{u_{1}\left(s\right)u_{2}\left(x\right)}{W\left(s\right)}f\left(s \right)}\,ds \right)- \left(\oint^x{\frac{u_{2}\left(s\right)u_{1}\left(x\right)}{W\left(s\right)}f\left(s\right)}\,ds\right) = \oint^x{f\left( s \right)\left\{ \frac{{{u}_{1}}(s) {{u}_{2}}(x)-{{u}_{1}}(x) {{u}_{2}}(s)}{W(s)} \right\}}\,ds

$$

(eq 6.12)
 *
 * }

= Problem 7: Find Two Homogeneous Solutions using a Trial Solution =

Given
Find u1 and u2 of Equation 7.1 using the trial solution shown in equation 7.2:

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

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

(x +1) y'' - (2x+3)y' + 2y = 0

$$ (eq 7.1)
 *
 * }
 * {| style="width:100%" border="0"

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

y(x)=e^{rx}

$$ (eq 7.2)
 *
 * }

Solution
The first step is to differentiate the trial solution as follows:
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y(x) = e^{rx} $$ (eq 7.3)
 *
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle y'(x) = r e^{rx} $$ (eq 7.4)
 *
 * }

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

$$ \displaystyle y''(x) = r^2 e^{rx} $$ (eq 7.5) Next, plug the values into equation 7.1 to yield:
 *

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

$$ \displaystyle e^{rx}\left[(x+1)r^2-(2x+3)r+2\right]=0 $$ (eq 7.6) Since erx will not equal zero for any value of x, divide equation 7.6 by this value and factor out the x to yield:
 *

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

$$ \displaystyle x(r^2-2r)+(r-2)(r-1)=0 $$ (eq 7.7) In order to satisfy equation 7.7 for any x, the following equation is solved:
 *

 * }

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

$$ \displaystyle r(r-2)=(r-2)(r-1)=0 $$ (eq 7.8)
 *

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

$$ \displaystyle r=2 $$ (eq 7.9) By plugging equation 7.9 into the trial solution in equation 7.3 we have obtained the first homogeneous solution:
 *

 * }
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle u_1(x)=e^{2x} $$ (eq 7.10) The second homogeneous solution is obtained by the following:
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$ \displaystyle u_2(x)=u_1(x)\int{u_1^{-2}(x)e^{-\int{a_1(x)}\,dx}}\,dx$$ (eq 7.11) Equation 7.l is rearranged into the following form:
 *

 * }

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

$$ \displaystyle y''-\underbrace{\frac{2x+3}{x+1}}_{a_1}y'+\underbrace{\frac{2}{x+1}}_{a_0}y=0 $$ (eq 7.12) The value for a1 shown in equation 7.12 is then plugged into equation 7.11 along with the first homogeneous solution to solve for the second solution as shown:
 *

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

$$ \displaystyle u_2(x)=e^{2x}\int{e^{-4x}e^{\int{\frac{2x+3}{x+1}}\,dx}}\,dx$$ (eq 7.13)
 *

 * }

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

$$ \displaystyle u_2(x)=e^{2x}\int{(x+1)e^{-2x}}\,dx$$ (eq 7.14)
 *

 * }
 * {| style="width:92%; padding:10px; border:2px solid #8888aa"

$$ \displaystyle u_2(x)=\frac{-1}{4}(2x+3)$$ (eq 7.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }


text:
Talk:Motivation and emotion/Book/2010/Indigenous Australians and motivation

Feedback
Here's some feedback on this chapter so far: -- Jtneill - Talk - c 06:02, 1 November 2010 (UTC)
 * 1) Well done on getting this underway. I've Moodle messaged and emailed you some more information. Some other comments follow.
 * 2) I've added some related focus questions which you might consider. Make sure (for the final draft) that the focus questions match the content.
 * 3) Add at least some very brief history, to set the context - some links to relevant Wikipedia articles could be appropriate here.
 * 4) Table 1. Some major comparisons between cultures - between which cultures? Is this specific to indigenous Australians?
 * 5) Image - expand the caption.
 * 6) Check grammar: "Therefore, motivations will reflect a collective nature within Indigenous Australians populations."
 * 7) Some more about Schwartz's (1991) results could be useful/interesting.
 * 8) Key references could include major government reports and reviews e.g., related to employment.
 * 9) Make sure you do a good search of the Australian Journal of Psychology for relevant references.

I've added an intro paragraph to help shape the context of this chapter. Feel free of course to modify, but I hope it might help. -- Jtneill - Talk - c 22:58, 4 November 2010 (UTC)

Feedback 2
-- Jtneill - Talk - c 08:43, 5 November 2010 (UTC)
 * 1) Introduction
 * 2) I've redrafted the intro paragraph - and also shifted the focus questions to immediately follow the introductory paragraph. This opening could be further improved by leading into and explaining/justifying the focus questions.
 * 3) The body text is taking shape now. Probably the area that deserves the most work I think will be the intro and conclusions to each section to help put each section in context and in relation to the focus questions, and to help flow between the sections. But get your body content down first, then you can work on smoothing and integrating it.

-- Jtneill - Talk - c 08:27, 11 December 2010 (UTC)

-- Jtneill - Talk - c 03:40, 14 December 2010 (UTC)


text:
Talk:Motivation and emotion/Textbook/Emotion/Music

Feedback
-- Jtneill - Talk - c 07:08, 1 November 2010 (UTC)
 * 1) Well done on getting this underway. I've had a quick look over the current content, tweaked a few parts, and made some suggestions below.
 * 2) Overall, the current draft does rely too heavily on too few references. Here's some suggested references and links you may be interested to follow up - http://www.delicious.com/jtneill/emotion+music
 * 3) Good image, but the license for File:Youth-runs-wild-440.jpg hasn't been specified - this image will most likely get deleted because it probably doesn't have a free to use license (otherwise, I imagine that the Youth Runs Wild wikipedia article would use this image.
 * 4) Different Styles of Music is promising but probably lacks sufficient connection to emotion or referencing. Therefore, perhaps summarise more briefly - and/or add an overall summary - what are the main implications of different types of music and their impacts on emotion?
 * 5) The relationship between emotion and music - Bridging the gap between Emotion and Music - This is probably the most important section to be expanded - this will be mainly where you are likely to address the two key marking criteria (theory and research) - the other criteria is quality of expression.
 * 6) Make sure to do a careful copy-edit - I've fixed some typos that I found.
 * 7) I found this in some of my notes: "Labbe, Schmidt and Babin (2007) found that listening to self-selected music or classical music following exposure to a stressor significantly reduced the negative emotional states and physiological arousal compared to sitting in silence or listening to music chosen by others

Feedback 2
This is shaping up - I've fiddled a bit with a simpler structure - see what you think - change back anything you don't like etc. Hope this helps. -- Jtneill - Talk - c 08:18, 5 November 2010 (UTC)
 * 1) Introduction
 * 2) Explain the aims, focus questions, and what will be covered in the chapter.
 * 3) The other content could be shifted possibly into other sections to allow the introduction to be more of an overview.
 * 4) APA style - check citation style e.g., & should only be used inside brackets and for subsequent citations within a paragraph don't include the year
 * 5) I put the info about musical styles into a subpage and added a link to this from the What is music section with the suggestion that you offer a brief summary about music styles. I wonder if somewhere later on it might be possible to create a table which shows which styles correspond mostly typically which type of emotions (I'm not sure what's been done in this area).
 * 6) Check out if you haven't found it already:
 * 7) http://www.users.muohio.edu/shermalw/edp101as07/Group1a_Gi-project09.doc - looks like some good ideas and references here.
 * 8) http://gupea.ub.gu.se/bitstream/2077/9604/2/gupea_2077_9604_2.pdf
 * 9) Remember the marking criteria is Theory 33% and Research 33% so make sure you cover these in as substantial and explicit way as possible - tell the reader in the intro what will be covered and why, then remind in the summary/conclusion what the answers to the focus questions are. The other 33% is written expression, APA style, layout etc.

-- Jtneill - Talk - c 03:29, 14 December 2010 (UTC)


text:
User talk:Osilion

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

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Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:17, 1 November 2010 (UTC)

title: User talk:Osilion


text:
User talk:Fasil.pc

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

 * Take a guided tour and learn to edit.
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 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 07:17, 1 November 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 07:17, 1 November 2010 (UTC)

title: User talk:Fasil.pc


text:
User:Leutha/links

These are my useful Links:
 * DARnet wiki

title: User:Leutha/links


text:
User:Leutha/glossary

This is my glossary:
 * Bloom's Taxonomy
 * Gibbs Reflective Cycle (DARnet)

title: User:Leutha/glossary