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Dominant group/Evolution

The origin of the term dominant group is as yet unknown. With an apparent earliest occurrence of "group" at ~1739, the term may have sprung into use from then forward. Perhaps its initial usage is social that crept into evolution. Perhaps its usage is the opposite.

Exploring possible occurrences within evolution theory may produce a discovery of an origin in another language way before ~1739. Part of the challenge of this type of experimental inquiry is that authors of today or even in the 1800s are likely to use "dominant group" to describe the past when the term itself was actually not used by an earlier author.

University level access to much earlier original documents may be required to confirm the term earlier than 1826, its current date of specific use in English by Kirby.

As Google scholar, for example, displays older and older manuscripts or books, especially before 1826, the probability increases of finding the term or eliminating its use by specific authors.

Evolution
Evolution may be little more than a process by which different kinds of living organisms have developed and diversified from earlier forms during the history of the Earth.

Dominant group
Examples from primary sources are to be used to prove or disprove each hypothesis. These can be collected per subject or in general.

 * 1) Accident hypothesis: dominant group is an accident of whatever processes are operating.
 * 2) Artifact hypothesis: dominant group may be an artifact of human endeavor or may have preceded humanity.
 * 3) Association hypothesis: dominant group is associated in some way with the original research.
 * 4) Bad group hypothesis: dominant group is the group that engages in discrimination, abuse, punishment, and additional criminal activity against other groups. It often has an unfair advantage and uses it to express monopolistic practices.
 * 5) Control group hypothesis: there is a control group that can be used to study dominant group.
 * 6) Entity hypothesis: dominant group is an entity within each field where a primary author of original research uses the term.
 * 7) Evolution hypothesis: dominant group is a product of evolutionary processes, such groups are the evolutionary process, produce evolutionary processes, or are independent of evolutionary processes.
 * 8) Identifier hypothesis: dominant group is an identifier used by primary source authors of original research to identify an observation in the process of analysis.
 * 9) Importance hypothesis: dominant group signifies original research results that usually need to be explained by theory and interpretation of experiments.
 * 10) Indicator hypothesis: dominant group may be an indicator of something as yet not understood by the primary author of original research.
 * 11) Influence hypothesis: dominant group is included in a primary source article containing original research to indicate influence or an influential phenomenon.
 * 12) Interest hypothesis: dominant group is a theoretical entity used by scholarly authors of primary sources for phenomena of interest.
 * 13) Metadefinition hypothesis: all uses of dominant group by all primary source authors of original research are included in the metadefinition for dominant group.
 * 14) Null hypothesis: there is no significant or special meaning of dominant group in any sentence or figure caption in any refereed journal article.
 * 15) Object hypothesis: dominant group is an object within each field where a primary author of original research uses the term.
 * 16) Obvious hypothesis: the only meaning of dominant group is the one found in Mosby's Medical Dictionary.
 * 17) Original research hypothesis: dominant group is included in a primary source article by the author to indicate that the article contains original research.
 * 18) Primordial hypothesis: dominant group is a primordial concept inherent to humans such that every language or other form of communication no matter how old or whether extinct, on the verge of extinction, or not, has at least a synonym for dominant group.
 * 19) Purpose hypothesis: dominant group is written into articles by authors for a purpose.
 * 20) Regional hypothesis: dominant group, when it occurs, is only a manifestation of the limitations within a region. Variation of those limitations may result in the loss of a dominant group with the eventual appearance of a new one or none at all.
 * 21) Source hypothesis: dominant group is a source within each field where a primary author of original research uses the term.
 * 22) Term hypothesis: dominant group is a significant term that may require a 'rigorous definition' or application and verification of an empirical definition.

Theoretical evolution
Def. "gradual directional change especially one leading to a more advanced or complex form" is called evolution.

Def. "[t]he change in the genetic composition of a population over successive generations" is called evolution.

Def. any change across successive generations in the heritable characteristics of biological populations is called evolution.

Essentialism
Def. "[t]he view that objects have properties that are essential to them" is called essentialism.

With the beginnings of biological taxonomy in the late 17th century, Western biological thinking was influenced by essentialism, the belief that every species has essential characteristics that are unalterable, a concept from medieval Aristotelian metaphysics that fit well with natural theology.

Essentialism is the view that, for any specific kind of entity, there is a set of characteristics or properties all of which any entity of that kind must possess. All things can therefore be precisely defined or described. In this view, it follows that terms or words should have a single definition and meaning.

The essential properties of a tiger are those without which it is no longer a tiger. Other properties, such as stripes or number of legs, are considered inessential or 'accidental'.

"Biologist Ernst Mayr epitomizes the effect of such an essentialist character of Platonic forms in biology: "Flesh-and-blood rabbits may vary, but their variations are always to be seen as flawed deviation from the ideal essence of rabbit". For Mayr, the healthful antithesis of essentialism in biology is "population thinking".

Recent work by historians of systematics ... Mary P. Winsor, Ron Amundson and Staffan Müller-Wille [argues that] the usual suspects (such as Linnaeus and the Ideal Morphologists) were very far from being essentialists, and it appears that the so-called "essentialism story" (or "myth") in biology is a result of conflating the views expressed by philosophers from Aristotle onwards through to John Stuart Mill and William Whewell in the immediately pre-Darwinian period, using biological examples, with the use of terms in biology like species.

"An essentialist view of a given species is committed to there being some property which all and only the members of that species possess." "The essentialist requires that a species be defined in terms of the characteristics of the organisms which belong to it." "Diversity [may] be accounted for as the joint product of natural regularities and interfering forces."

Higher type has been associated with the Natural State Model of essentialism.

"Some dominant group members could invoke essentialism as a legitimizing ideology to justify existing power relations, but dominant group members with egalitarian worldviews might invoke essentialism to resist dominant ideologies."

"[E]ssentialism is the psychological lynchpin that anchors one's personal and social identity (Mahalingam & Rodriguez, 2006), and enables us to construct theories of social groups that often legitimize the dominant group's privileged status and (sometimes physical) control over the nondominant group; internalizing such essentialist beliefs may also improve the psychological well-being of the dominant group members in this system."

"Thus the cultural fundamentalism of the British or French right-wing in its talk about immigrants is clearly a dangerous sort of essentialism, partly because it is being wielded by a dominant group which has some power to impose its views on others."

Evolutionary processes
Evolutionary processes give rise to diversity at every level of biological organisation, including [biodiversity] species, individual organisms and [molecular evolution] molecules such as DNA and proteins.

With respect to dominant groups, evolutionary processes
 * 1) produce dominant groups,
 * 2) are the same as dominant groups (dominant groups are the evolutionary process),
 * 3) are produced by dominant groups, or
 * 4) are independent of dominant groups (i.e., dominant groups are an artifact).

"There is no evidence that the superiority of any existing dominant group is based on any thing but an accident, and any attempt to maintain that dominance by reason, is merely the rationalization of a "myth."

Evolutionary progressivism
In 1960, Huxley wrote "Improved organization gives biological advantage. Accordingly, the new type becomes a successful or dominant group. It spreads and multiplies and differentiates into a multiplicity of branches. This new biological success is usually achieved at the biological expense of the older dominant group from which it sprang or whose place it had usurped."

It has been asserted that the usage of dominant group in evolutionary progessivism fails "adequately to exclude any group of organisms."

"[B]iologists have demonstrated innumerable modes of adaptation and specialization. Such diversity indicates the great utility and economy of these concepts ... in the evolution of life there may be a number of related kinds of progress ... relating to complexity, adaptability, numerical increase, efficiency, etc."

Extinction events
Def. "[t]he action of making or becoming" "[n]o longer in existence; having died out" is called extinction.

Def. the apparent percentage (not the absolute number) of marine animal genera represented by those that are readily fossilized becoming extinct during any given time interval is called the genus extinction intensity.

"Many of the extinction events appear to be somewhat extended in time. In at least some cases this is the result of a paleontological artifact known as the Signor-Lipps effect (Signor & Lipps 1982). Briefly, this is the observation that inadequate sampling can cause a taxon to seem to disappear before its actual time of extinction. This has the effect of making an extinction event appear extended even if it occurred quite rapidly. Hence, when estimating the true magnitude of an extinction event it would be common to combine together the events occurring over several preceding bins as long as they also show excess extinctions. This explains why many estimates of the magnitude of an extinction event may be larger than the 20-30% shown as the largest single bin for most of the extinctions shown here."

On page 212 of a 1944 book by Simpson is "[i]n the history of life it is a striking fact that major changes in the taxonomic groups occupying various ecological positions do not, as a rule, result from direct competition of the groups concerned in each case and the survival of the fittest, as most students would assume a priori. On the contrary, the usual sequence is for one dominant group to die out, leaving the zone empty, before the other group becomes abundant." "Simpson noted that major extinctions provide opportunities (space, ecological niches, etc.) for later diversification by the survivors."

Modern evolutionary synthesis
"Other previously dominant groups of organisms that also became extinct at the end of the Cretaceous are many marine taxa, such as most nautiloids and the ammonities, both of whom had been previously highly successful organisms." Page 320 of the 2001 book: "Ernst Mayr is the biologist largely responsible for shaping the modern synthesis of genetics and evolutionary theory."

"The central problem with the synthesis is its failure to show (or to provide distinct signs) that natural selection of random mutations could account for observed levels of adaptation."

Natural selection
The earliest use of "dominant groups" by Charles Darwin is on page 343, “The dominant species of the larger dominant groups tend to leave many modified descendants, and thus new sub-groups and groups are formed.”

Later in 1860, “Under the many conditions of life which this world affords, any group which is numerous in individuals and species and is widely distributed, may properly be called dominant", a dominant group. [Letter 110. To W.H. Harvey, August, 1860]

Evolution by means of natural selection is the process by which genetic mutations that enhance reproduction become and remain, more common in successive generations of a population. It has often been called a "self-evident" mechanism because it necessarily follows from three simple facts:
 * Heritable variation exists within populations of organisms.
 * Organisms produce more offspring than can survive.
 * These offspring vary in their ability to survive and reproduce.

These conditions produce competition between organisms for survival and reproduction. Consequently, organisms with traits that give them an advantage over their competitors pass these advantageous traits on, while traits that do not confer an advantage are not passed on to the next generation.

The central concept of natural selection is the evolutionary fitness of an organism. Fitness is measured by an organism's ability to survive and reproduce, which determines the size of its genetic contribution to the next generation. However, fitness is not the same as the total number of offspring: instead fitness is indicated by the proportion of subsequent generations that carry an organism's genes. For example, if an organism could survive well and reproduce rapidly, but its offspring were all too small and weak to survive, this organism would make little genetic contribution to future generations and would thus have low fitness.

Natural selection produces dominant groups of organisms. But, do dominant groups of organisms then produce new species?

Over time, this process can result in populations that specialize for particular ecological niches and may eventually result in the emergence of new species. "Over time, this process can result in adaptations that specialize organisms for particular ecological niches and may eventually result in the emergence of new species, an event known as speciation."

Neo-Lamarckism
Def. a theory of inheritance based on a modification and extension of Lamarckism, essentially maintaining the principle that genetic changes can be influenced and directed by environmental factors is called Neo-Lamarckism.

"Perhaps more revealing would be to ask this group of molecular geneticists to describe Lamarckism versus Mendelism. ... These examples characterize a lack of integrative capability that exists in the contemporary and dominant group of gene specialists."

Orthogenesis
Def. "[a] series of similar mutations in successive generations, producing evolutionary change" is called orthogenesis.

Orthogenesis is the hypothesis that life has an innate tendency to evolve in a unilinear fashion due to some internal or external "driving force". The hypothesis is based on essentialism and cosmic teleology and proposes an intrinsic drive which slowly transforms species. Classic proponents of orthogenesis have rejected the theory of natural selection as the organising mechanism in evolution, and theories of speciation for a rectilinear model of guided evolution acting on discrete species with "essences". The term orthogenesis was popularized by Theodor Eimer, though many of the ideas are much older (Bateson 1909).

"Many other principles, such as those concerned with reversion and orthogenesis, are gradually being formulated and it is only a matter of time and more extended observation before the science of phylogeny will be placed on a much more uniform and exact footing. ... The ancestors of our modern hexapods achieved their first success through some advance in specialization over this more primitive type, but the improvements which gave them ascendancy and which enabled them to found a distinct and dominant group were certain unknown changes, doubtless in plastic and functionally important characters which were of great value for survival at the time, but which, having isolated the family and put it on its feet, so to speak, continued to change and may be possessed by few or no living descendants." Bold added.

Reversion
Def. "[t]he return of a genetic characteristic after a period of suppression" is called a reversion.

In botany, a sport or bud sport is a part of a plant (normally a woody plant, but sometimes in herbs as well) that shows morphological differences from the rest of the plant. Sports may differ by foliage shape or color, flowers, or branch structure. Sports with desirable characteristics are often propagated vegetatively to form new cultivars that retain the characteristics of the new morphology. Such selections are often prone to "reversion", meaning that part or all of the plant reverts to its original form.

A back mutation or reversion is a point mutation that restores the original sequence and hence the original phenotype.

Saltationism
Def. "belief that evolution operates by the sudden development of new species or biological features from one generation to the next" is called saltationism.

Def. "[a] sudden change from one generation to the next" is called saltation.

Polyploidy (most common in plants but not unknown in animals) is considered a type of saltation But, Dufresne never used the term "saltation" in his article.

"Work of systematists [2] and geneticists [3 and 4], on the stages of speciation, confirmed Darwin's view that speciation rarely involves discontinuous 'saltations' (polyploidy being the principal exception)."

Gradualism
Def. "belief that evolution proceeds at a steady pace, without the sudden development of new species or biological features from one generation to the next" is called gradualism.

Punctuationism
Def. "belief that evolution does not proceed at a steady pace, but instead is characterized by periods of stasis, punctuated by brief (within several hundred-thousand years) periods of rapid change" is called punctuationism.

Teleology
Def. "[t]he study of the purpose or design of natural occurrences" is called teleology.

A teleology is any philosophical account which holds that final causes exist in nature, meaning that design and purpose analogous to that found in human actions are inherent also in the rest of nature. In modern science, an explanation that relies on teleology is avoided, because whether they are true or false is argued to be beyond the ability of human perception and understanding to judge. "The received intellectual tradition has it that, in the sixteenth and seventeenth centuries, revolutionary philosophers began to curtail and reject the teleology of the medieval and scholastic Aristotelians, abandoning final causes in favor of a purely mechanistic model of the Universe." But using teleology as an explanation style, in particular within evolutionary biology, is still controversial. Apparent teleology is a recurring issue in evolutionary biology, much to the consternation of some writers. Statements which imply that nature has goals, for example where a species is said to do something "in order to" achieve survival, appear teleological, and therefore invalid.

Transmutation of species
Def. the process of changing, transforming or converting one species into another, or from one species into another, is called transmutation of species.

In the early 19th century, Jean-Baptiste Lamarck proposed his theory of the transmutation of species, the first fully formed theory of evolution.

"Lamarck became convinced of the transmutation of species during the period 1797-1800 ... He used Trigonia as a prime example of a once-dominant group [in the Mesozoic] which had been drastically restricted in diversity, abundance, and geographic range" down to one genus Neotrigonia and five living species. All are found off the coast of Australia.

Vitalism
Def. "the doctrine that life involves some immaterial "vital force", and cannot be explained scientifically" is called vitalism.

Vitalism, as defined by the Merriam-Webster dictionary, is

 * 1) a doctrine that the functions of a living organism are due to a vital principle distinct from biochemical reactions
 * 2) a doctrine that the processes of life are not explicable by the laws of physics and chemistry alone and that life is in some part self-determining.

Other vitalists included Johannes Reinke and Oscar Hertwig. Reinke used the word neovitalism to describe his work, he claimed that it would be eventually verified through experimentation and wanted an improvement over the other vitalistic theories. The work of Reinke was an influence for Carl Jung.

Hypotheses

 * 1) A dominant group is as likely to go extinct as any group.

title: Dominant group/Evolution


text:
Controversies in Science/Are humans causing global warming/A Critique of Biological consequences of global warming: is the signal already apparent

(Review Paper) Cited in

Global Warming is Already Happening
Data that was researched has indicated that global warming was already happening in the past and humans are not solely the cause. Irregular climate from the past has already affected the physiology and distribution of some species. .

Methods
Scientists have conducted studies where intensive monitoring has occurred over an extended period. The patterns from many plants, marine animals, and aphids were taken into affect. .

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


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University of Florida/Egm4313/s12.team5.R3

Question
Consider the following L2-ODE-CC:

 * $$\displaystyle y^{''}-10y^{'}+25y=7e^{5x}-2x^{2} $$

With initial conditions y(0)=4, y'(0)=-5

Find the solution. Plot this solution and the solution in the example on p.7-3

Solution
General solution of our ODE:

 * $$\displaystyle y^{''}-10y^{'}+25y=0 $$

 * $$\displaystyle a^{2}-4b=10^{2}-4(25)=0 \Rightarrow $$ double real roots

 * $$\displaystyle \lambda =\lambda _{1}=\lambda_{2}=-\frac{a}{2}=-\frac{-10}{2}=5 $$

Giving us a general solution of:

 * $$\displaystyle y_{g}=(c_{1}+c_{2}x)e^{5x} $$

Particular solution of our ODE by method of undetermined coefficients:

Since our excitation r(x) is of the form $$\displaystyle r(x)=ke^{\gamma x}-kx^{n} $$

Our two particular solutions will be of the form:

 * $$\displaystyle y_{p1}=Cx^{2}e^{\gamma x}, y_{p2}=K_{2}x^{2}+K_{1}x+K_{0} $$

 * Have to multiply $$\displaystyle y_{p1} $$ by $$\displaystyle x^{2} $$ because the $$\displaystyle e^{5x} $$ term already appears in the general solution.

Taking derivatives of $$\displaystyle y_{p1} $$:

 * $$\displaystyle y_{p1} = Cx^{2}e^{5x} $$
 * $$\displaystyle y^{'}_{p1} = 2Cxe^{5x}+5Cx^{2}e^{5x} $$
 * $$\displaystyle y^{''}_{p1} = 2Ce^{5x}+10Cxe^{5x}+10Cxe^{5x}+25Cx^{2}e^{5x} $$

Substituting derivatives into the original ODE and collecting like terms:

 * $$\displaystyle C[2+10x+10x+25x^{2}-20x-50x^{2}+25x^{2}]e^{5x}=7e^{5x} $$

 * $$\displaystyle 2C=7 \Rightarrow C=\frac{7}{2} $$

Giving us our first particular solution:

 * $$\displaystyle y_{p1}=\frac{7}{2}x^{2}e^{5x} $$

Taking derivatives of $$\displaystyle y_{p2} $$:

 * $$\displaystyle y_{p2}=K_{2}x^{2}+K_{1}x+K_{0} $$
 * $$\displaystyle y^{'}_{p2}=2K_{2}x+K_{1} $$
 * $$\displaystyle y^{''}_{p2}=2K_{2} $$

Substituting derivatives into original ODE and collecting like terms:

 * $$\displaystyle 25K_{2}=-2\Rightarrow K_{2}=-\frac{2}{25} $$
 * $$\displaystyle -20K_{2}+25K_{1}=0\Rightarrow K_{1}=-\frac{8}{125} $$
 * $$\displaystyle 2K_{2}-10K_{1}+25K_{0}=0 \Rightarrow K_{0}=-\frac{12}{625} $$

Giving us our second particular solution:

 * $$\displaystyle y_{p2}=-\frac{2}{25}x^{2}-\frac{8}{125}x-\frac{12}{625} $$

With a final solution being the sum of the general and particular solutions:

 * $$\displaystyle y = y_{g}+y_{p1}+y_{p2} $$
 * $$\displaystyle y = (c1+c2x)e^{5x}+\frac{7}{2}x^{2}e^{5x}-\frac{2}{25}x^{2}-\frac{8}{125}x-\frac{12}{625} $$

Using initial conditions to solve for c1 and c2:

 * $$\displaystyle y(0)=4 \Rightarrow 4=c_{1}-\frac{12}{625} \Rightarrow c_{1}=\frac{2512}{625} $$

 * $$\displaystyle y^{'}(0)=-5 $$ where $$\displaystyle y^{'}=e^{5x}[5c_{1}+c_{2}+5c_{2}x+7x+\frac{35}{2}x^{2}]-\frac{4}{25}x-\frac{8}{125} $$

 * $$\displaystyle -5=5c_{1}+c_{2}-\frac{8}{125} \Rightarrow c_{2}=-\frac{3129}{125} $$

Final solution:

$$\displaystyle y=(\frac{2512}{625}-\frac{3129}{125}x)e^{5x}+\frac{7}{2}x^{2}e^{5x}-\frac{2}{25}x^{2}-\frac{8}{125}x-\frac{12}{625} $$

Matlab Code:

 * x=0:0.001:10;
 * y=((2512/625)-(3129/125).*x).*exp(5*x)+(7/2).*x.^2.*exp(5*x)-(2/25).*x.^2-(8/125).*x-(12/625);
 * plot(x,y),xlabel('x'),ylabel('y(x)')

Solution in example on P.7-3

$$\displaystyle y=4e^{5x}-25xe^{5x}+\frac{7}{2}x^{2}e^{5x} $$

Matlab code:

 * x=0:0.001:10;
 * y=4.*exp(5*x)-25.*x.*exp(5*x)+(7/2).*x.^2.*exp(5*x);
 * plot(x,y),xlabel('x'),ylabel('y(x)')

Author
This problem was solved and uploaded by: Joshua House

This problem was proofread by: David Herrick

Part 1
Find the homogeneous L2-ODE-CC having the following roots: $$\displaystyle \lambda _{1}=\lambda ,\lambda _{2}=\lambda +\varepsilon $$

Part 2
Show that the following is a homogeneous solution: $$\displaystyle \frac{e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon } $$

Part 3
Find the limit of the homogeneous solution as $$\displaystyle \varepsilon \to 0 $$

Part 4
Take the derivative of $$\displaystyle e^{\lambda x} $$ with respect to $$\displaystyle \lambda $$

Part 5
Compare the results in parts (3) and (4), and relate to the result by using variation of parameters

Part 6
Compute (2) using $$\displaystyle \lambda =5 $$ and $$\displaystyle \varepsilon =0.001 $$ and compare to the value obtained from the exact 2nd homogeneous solution

Part 1

 * Letting $$\displaystyle \lambda _{1}=x,\lambda _{2}=x+\epsilon $$

 * $$\displaystyle (\lambda -x)(\lambda -[x+\varepsilon ])=0 $$

 * $$\displaystyle \lambda ^{2}-\lambda x-\lambda \varepsilon -\lambda x+x^{2}+x\varepsilon =0 $$

 * $$\displaystyle \lambda ^{2}-\lambda [2x+\varepsilon ]+x[x+\varepsilon ]=0 $$

 * Replacing the x's with lambda's:

$$\displaystyle y^{''}-y^{'}[2\lambda +\varepsilon ]+y\lambda [\lambda +\varepsilon ]=0 $$

Part 2

 * $$\displaystyle y=\frac{e^{(\lambda+\varepsilon )x }-e^{\lambda x}}{\varepsilon }, y'=\frac{(\lambda +\varepsilon )e^{(\lambda +\varepsilon )x}-\lambda e^{\lambda x}}{\varepsilon },y''=\frac{(\lambda +\varepsilon )^{2}e^{(\lambda +\varepsilon )x}-\lambda ^{2}e^{\lambda x}}{\varepsilon } $$

Plugging this into the original ODE $$\displaystyle y^{''}-y^{'}[2\lambda +\varepsilon ]+y\lambda [\lambda +\varepsilon ]=0 $$ and collecting like terms:

 * $$\displaystyle e^{(\lambda +\varepsilon )x}[\lambda ^{2}+2\lambda \varepsilon +\varepsilon ^{2}-2\lambda ^{2}-2\lambda \varepsilon -\lambda \varepsilon -\varepsilon ^{2}+\lambda ^{2}+\varepsilon \lambda ]+e^{\lambda x}[-\lambda ^{2}+2\lambda ^{2}+\lambda \varepsilon -\lambda ^{2}-\lambda \varepsilon ]=0 $$

 * Much of the trivial algebra was left out in order to greatly reduce coding time. These intermediate steps, which were done on paper, can be presented if necessary.

From inspection, all of the bracketed terms add up to zero, thus verifying we were given a correct homogeneous solution.

Part 3
$$\displaystyle \lim_{\varepsilon \rightarrow 0}\frac{e^{(\lambda +\varepsilon)x }-e^{\lambda x}}{\varepsilon } $$

Since the limit of the numerator and the limit of the denominator are both 0 as epsilon approaches 0, we can use L'Hopitals rule to find the limit of the entire function

L'Hopital's Rule:

If: $$\displaystyle   \lim_{\varepsilon \rightarrow 0}\frac{f'(\varepsilon )}{g'(\varepsilon )} $$ exists

And: $$\displaystyle \lim_{\varepsilon \rightarrow 0}f(\varepsilon)=\lim_{\varepsilon \rightarrow 0}g(\varepsilon ) =0 $$ Then: $$\displaystyle \lim_{\varepsilon \rightarrow 0}\frac{f(\varepsilon )}{g(\varepsilon )}=\lim_{\varepsilon \rightarrow 0}\frac{f'(\varepsilon )}{g'(\varepsilon )} $$

Taking derivatives of the numerator and denominating with respect to epsilon, and finding the limit as epsilon approaches 0:

$$\displaystyle \lim_{\varepsilon \rightarrow 0}\frac{xe^{(\lambda +\varepsilon )x}}{1}=xe^{\lambda x} $$

Part 4
$$\displaystyle \frac{d[e^{\lambda x}]}{d\lambda }=xe^{\lambda x} $$

Part 5
The results from (3) and (4) show that $$\displaystyle \frac{d[e^{\lambda x}]}{d\lambda }=\lim_{e\rightarrow 0}\frac{e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon }=xe^{\lambda x} $$

Part 6
From $$\displaystyle \frac{e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon } $$ and letting $$\displaystyle \lambda =5, \varepsilon =0.001 $$

$$\displaystyle \frac{e^{5.001x}-e^{5x}}{0.001} $$

Comparing this to the value of the exact 2nd homogeneous solution (values were plugged into the expression in part 2)

$$\displaystyle e^{5.001x}[5^{2}+2(5)(0.001)+0.001^{2}-2(5^{2})-2(5)(0.001)-5(0.001)-(0.001^{2})+5^{2}+0.001(5)] +e^{5x}[-5^{2}+2(5^{2})+5(0.001)-5^{2}-5(0.001)] = 0 $$

Author
This problem was solved and uploaded by: Joshua House

This problem was proofread by: Michael Wallace

Question
Find the complete solution for equation 5 from pg. 7.7 in the notes:

$$ \displaystyle \ y'' - 2y' + 3y = 4x^2\ $$

with initial conditions given as:

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

Plot the solution.

Solution
First, find the homogenous solution to this differential equation, i.e. where r(x) = 0.

$$ \displaystyle \ y''_h - 3y'_h + 2y_h = 0\ $$

$$ \displaystyle \ (\lambda - 1) (\lambda-2) = 0\ $$

$$ \displaystyle \ \lambda = 1, 2\ $$

Thus, the homogenous solution is of the form $$ \displaystyle \ y_h = C_1e^{\lambda_1x} + C_2e^{\lambda_2x}\ $$

Plugging in the answers for lambda yields:

$$ \displaystyle \ y_h = C_1e^{x} + C_2e^{2x}\ $$

Now to solve for the particular solution, we must look at the excitation: $$ \displaystyle \ r(x) = 4x^2\ $$

A particular solution to an excitation of the form $$ \displaystyle \ Cx^n\ $$ is defined as $$ \displaystyle \ \sum_{j=0}^{n}c_jx^j\ $$

Thus, $$ \displaystyle \ y_p = c_0 + c_1x + c_2x^2\ $$

To solve for the 3 unknown constants, plug the particular solution into the differential equation:

$$ \displaystyle \ y_p'' - 3y_p' + 2y_p = 4x^2\ $$

$$ \displaystyle \ (c_0 + c_1x + c_2x^2)'' -3(c_0 + c_1x + c_2x^2)' + 2(c_0 + c_1x + c_2x^2) = 4x^2\ $$

$$ \displaystyle \ y_p' = 2c_2x + c_1\ $$

$$ \displaystyle \ y_p'' = 2c_2\ $$

Substituting these relations into the equation we get:

$$ \displaystyle \ 2c_2 - 3(2c_2x + c_1) + 2(c_2x^2 + c_1x + c_0) = 4x^2\ $$

Grouping like terms we get:

$$ \displaystyle \ 2c_2x^2 + (2c_1 - 6c_2)x + (2c_2 - 3c_1 + 2c_0) = 4x^2\ $$

Matching up the coefficients for the $$ \displaystyle \ x^2\ $$ term on the left and right sides we get:

$$ \displaystyle \ 2c_2 = 4\ $$

Therefore $$ \displaystyle \ c_2 = 2\ $$

Matching up the coefficients for the $$ \displaystyle \ x\ $$ term on the left and right sides we get:

$$ \displaystyle \ 2c_1 - 6c_2 = 0\ $$

Therefore $$ \displaystyle \ c_1 = 6c_2/2 = 6(2)/2 = 6\ $$

Finally, matching up the coefficients for the constant terms on the left and right sides we get:

$$ \displaystyle \ 2c_2 - 3c_1 + 2c_0 = 0\ $$

Therefore $$ \displaystyle \ c_0 = (3c_1 - 2c_2)/2 = (3(6) - 2(2) )/2 = (18-4)/2 = 7\ $$

The particular solution is then: $$ \displaystyle \ y_p = 2x^2 + 6x + 7\ $$

The general solution is the sum of the homogenous solution and the particular solution, therefore:

$$ \displaystyle \ y(x) = y_h + y_p = C_1e^{x} + C_2e^{2x} + 2x^2 + 6x + 7\ $$

To solve for the 2 remaining unknown constants, we use our initial conditions from the problem statement.

$$ \displaystyle \ y(0) = 1 = C_1 + C_2 + 7\ $$

To use the second initial condition, we must take a derivative of our general solution.

$$ \displaystyle \ y'(x) = C_1e^{x} + 2C_2e^{2x} + 4x + 6\ $$

$$ \displaystyle \ y'(0) = 0 = C_1 + 2C_2 + 6\ $$

Rewriting the equations we get:

$$ \displaystyle \ C_1 + 2C_2 + 6 = 0\ $$

$$ \displaystyle \ C_1 + C_2 + 6 = 0\ $$

Subtracting these equations we get:

$$ \displaystyle \ C_2 = 0\ $$

Therefore: $$ \displaystyle \ C_1 = -6 - C_2 = -6\ $$

Thus the complete solution is:

$$ \displaystyle \ y(x) = 2x^2 + 6x + 7 -6e^{x}\ $$

MATLAB code;

x = 0:0.001:10;

y = 2*x.^2 + 6*x + 7 - 6*exp(x);

plot(x,y)

xlabel( 'x')

ylabel( 'y(x)')

Author
This problem was solved and uploaded by: David Herrick

This problem was proofread by: Michael Wallace

Question
Use the Basic Rule (1) and the Sum Rule (3) on p.7-2 of the notes to show that the appropriate particular solution for:
 * $$ \ y'' - 3y' +2y = 4x^2 - 6x^5\ $$

is of the form:
 * $$ \ y_p(x) = \sum_{j=0}^5 c_j x^j\ $$

with n = 5, i.e. (1) on p.7-12.

Solution
From the Basic Rule:
 * $$ \ r_1 (x) = 4x^2\ $$
 * $$ \ r_2 (x) = 6x^5\ $$

Therefore, from Table 2.1:
 * $$ y_{p_1} (x) = K_2 x^2 + K_1 x + K_0 $$ since k = 4, x = x, and n = 2.
 * $$ y_{p_2} (x) = K_5 x^5 + K_4 x^4 + K_3 x^3 + K_2 x^2 + K_1 x + K_0 $$ since k = 6, x = x, and n = 5.

For $$ y_{p_1} (x) $$:
 * $$ \ y_{p_1}'(x) =2K_2x+K_1 \ $$
 * $$ \ y_{p_1}''(x) = 2K_2 \ $$

Substituting $$ y_{p_1} (x) $$ into the original equation gives:
 * $$ \ (2K_2) - 3(2K_2x+K_1) + 2(K_2 x^2 + K_1 x + K_0) = 4x^2 -6x^5 \ $$
 * $$ \ = x^2 (2K_2) + x (2K_1 + 3K_2) + (2K_2 + 3K_1 + 2K_0) = 4x^2 - 6x^5 \ $$

Comparing the $$ x^2 $$, $$ x $$, and $$ x^0 $$ coefficients gives: For $$ x^2 $$: $$ \ 2K_2 = 4 \ $$ Therefore $$ K_2 = 2 $$. For $$ x $$: $$ \ 2K_1 + 3K_2 = 0 \ $$ Therefore $$ K_1 = -3 $$. For $$ x^0 $$: $$ \ 2K_2 + 3K_1 + 2K_0 = 0 \ $$ Therefore $$ K_0 = 2.5 $$. Therefore, $$ y_{p_1} (x) = 2x^2 - 3x + 2.5 $$. For $$ y_{p_2} (x) $$:
 * $$ \ y_{p_2}' (x) = 5K_5x^4 + 4K_4x^3 + 3K_3x^2 + 2K_2x + K_1 \ $$
 * $$ \ y_{p_2}'' (x) = 20K_5x^3 + 12K_4x^2 + 6K_3x+2K_2 \ $$

Substituting $$ y_{p_1} (x) $$ into the original equation gives:
 * $$ \ (20K_5x^3 + 12K_4x^2 + 6K_3x+2K_2) - 3(5K_5x^4 + 4K_4x^3 + 3K_3x^2 + 2K_2x + K_1) + 2(K_5 x^5 + K_4 x^4 + K_3 x^3 + K_2 x^2 + K_1 x + K_0) = 4x^2 -6x^5 \ $$

Simplifying yields:
 * $$ \ x^5 (K_5) + x^4 (K_4 - 15K_5) + x^3 (K_3 - 12K_4 + 40K_5) + x^2 (K_2 - 9K_3 + 24K_4) + x (K_1 - 6K_2 + 12K_3) + (K_0 - 3K_1 + 4K_2) = 4x^2 -6x^5 \ $$

Comparing the $$ x^5 $$, $$ x^4 $$, $$ x^3 $$, $$ x^2 $$, $$ x $$, and $$ x^0 $$ coefficients gives: For $$ x^5 $$: $$ \ K_5 = -6 \ $$ Therefore $$ K_5 = -6 $$. For $$ x^4 $$: $$ \ K_4 - 15K_5 = 0 \ $$ Therefore $$ K_4 = -90 $$. For $$ x^3 $$: $$ \ K_3 - 12K_4 + 40K_5 = 0 \ $$ Therefore $$ K_3 = 840 $$. For $$ x^2 $$: $$ \ K_2 - 9K_3 + 24K_4 = 0 \ $$ Therefore $$ K_2 = -9724 $$. For $$ x $$: $$ \ K_1 - 6K_2 + 12K_3 = 0 \ $$ Therefore $$ K_1 = 59425 $$. For $$ x^0 $$: $$ \ K_0 - 3K_1 + 4K_2 = 0 \ $$ Therefore $$ K_0 = -217171 $$. Therefore, $$ y_{p_2} (x) = -6x^5 - 90x^4 + 840x^3 - 9724x^2 + 59425x - 217171 $$ By the Sum Rule:
 * $$ \ y_p (x) = y_{p_1} (x) + y_{p_2} (x) \ $$
 * $$ \ y_p (x) = (2x^2 - 3x + 2.5) + (-6x^5 - 90x^4 + 840x^3 - 9724x^2 + 59425x - 217171) \ $$

Simplifying gives:
 * $$ \ y_p (x) = -6x^5 - 90x^4 + 840x^3 - 9722x^2 + 59422x - 217168.5 \ $$

Choose:
 * $$ \ K_5 = -6 \ $$
 * $$ \ K_4 = -90 \ $$
 * $$ \ K_3 = 840 \ $$
 * $$ \ K_2 = -9722 \ $$
 * $$ \ K_1 = 59422 \ $$
 * $$ \ K_0 = -217168.5 \ $$

Then the equation becomes of the form: $$ \ y_p (x) = \sum_{j=0}^n (c_j x^j) \ $$ where n = 5.

Author
This problem was solved and uploaded by John North.

This problem was proofread by Michael Wallace

Question
Complete the solution for $$ \ y'' - 3y' + 2y = 4x^2 - 6x^5\ $$ ( (2) p. 7-11) )  as follows:

Part 1
1) Obtain equations (2) - (4) and (6) p. 7-14

(2) Coefficients of $$ \ x\ $$:

(3) Coefficients of $$ \ x^2\ $$:

(4) Coefficients of $$ \ x^3\ $$:

(6) Coefficients of $$ \ x^5\ $$:

Part 2
Verify all equations by long hand expansion of the series in (4) p. 7-12, instead of using the series in (2) p. 7-13.

$$ \ \sum_{j = 2}^{5} c_j \cdot j \cdot (j-1) \cdot x^{j-2} - 3\sum_{j = 1}^{5} c_j \cdot j \cdot x^{j-1} + 2 \sum_{j = 0}^{5} c_j x^j = 4x^2 - 6x^5\ $$

Part 3
Put the system of equations for $$ \ \left \{ c_0,...,c_5 \right \}\ $$ in matrix form.

Part 4
Solve for the coefficients $$ \ \left \{ c_0,...,c_5 \right \}\ $$ by back substitution.

Part 5
Consider the initial conditions:

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

Find the solution y(x) and plot it.

Part 1
For all the coefficient equations, we will use the series from (2) p. 7-13

$$ \ \sum_{j = 0}^{3} \left [ c_{j+2} (j+2) (j+1) - 3c_{j+1} (j+1) + 2c_j\right] x^j - 3c_5(5)x^4 + 2 \left [c_4x^4 + c_5x^5 \right ] = 4x^2 - 6x^5\ $$

Solving for the coefficients of x, we use j = 1.

$$ \ (c_3 (3)(2) - 3c_2 (2) + 2c_1)x\ $$.

These are the coefficients for x on the left hand side of the equation, and the coefficients for x on the right hand side = 0.

Coefficients of x: $$ \ 6c_3 - 6c_2 + 2c_1 = 0\ $$

Solving for the coefficients of $$ \ x^2\ $$ we use j = 2.

$$ \ (c_4 (4) (3) - 3c_3 (3) + 2c_2)x^2\ $$

These are the coefficients for $$ \ x^2\ $$ on the left hand side of the equation, and the coefficients for $$ \ x^2\ $$ on the right hand side = 4

Coefficients of $$ \ x^2\ $$: $$ \ 12c_4 - 9c_3 + 2c_2 = 4\ $$

Solving for the coefficients of $$ \ x^3\ $$ we use j = 3.

$$ \ (c_5 (5) (4) - 3 c_4 (4) + 2c_3)x^3\ $$

These are the coefficients for $$ \ x^3\ $$ on the left hand side of the equation, and the coefficients for $$ \ x^3\ $$ on the right hand side = 0

Coefficients of $$ \ x^3\ $$: $$ \ 20c_5 - 12c_4 + 2c_3 = 0\ $$

Solving for the coefficients of $$ \ x^5\ $$, all we have is $$ \ 2c_5x^5\ $$

This is the only coefficient of $$ \ x^5\ $$ on the left hand side of the equation, and the coefficients for $$ \ x^5\ $$ on the right hand side = -6

Coefficients of $$ \ x^5\ $$: $$ \ 2c_5 = -6\ $$

Part 2
Expanding the 3 series using j = 0 to 5, we get:

$$ \ 2c_0 + 2c_1 x - 3c_1 + 2c_2 x^2 - 3c_2 (2) x + 2c_2 + 2c_3x^3 - 3(3)c_3x^2 + 3(2)c_3x + 2c_4x^4 - 3(4) c_4 x^3 + 4(3) c_4 x^2 + 2c_5x^5 -3(5)c_5x^4 + 5(4)c_5x^3 = 4x^2 - 6x^5\ $$

Simplifying and grouping like terms we get:

$$ \ (2c_0 - 3c_1 + 2c_2) + (2c_1 - 6c_2 + 6c_3)x + (2c_2 - 9c_3 + 12c_4)x^2 + (2c_3 -12c_4 + 20c_5)x^3 + (2c_4 - 15c_5)x^4 + 2c_5x^5 = 4x^2 - 6x^5\ $$

Compared to the answers of part 1, it can be seen that all the coefficients are equivalent, verifying that the two series are equivalent.

Part 3
Matrix form of the solution:

$$ \ \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0\\ 0 & 0 & 0 & 2 & -12 & 20\\ 0 & 0 & 0 & 0 & 2 & -15\\ 0 & 0 & 0 & 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1\\ c_2\\ c_3\\ c_4\\ c_5 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 4\\ 0\\ 0\\ -6 \end{bmatrix}\ $$

Part 4
Solving by back substitution, the first constant we solve for is c5

$$ \ 2c_5 = -6\ $$

$$ \ c_5 = -3\ $$

$$ \ -15c_5 + 2c_4 = 0\ $$

$$ \ -15(-3) + 2c_4 = 0\ $$

$$ \ c_4 = -22.5\ $$

$$ \ 20c_5 - 12c_4 + 2c_3 = 0\ $$

$$ \ 20(-3) - 12(-22.5) + 2c_3 = 0\ $$

$$ \ c_3 = -105\ $$

$$ \ 12c_4 - 9c_3 + 2c_2 = 4\ $$

$$ \ 12(-22.5) - 9(-105) + 2c_2 = 4\ $$

$$ \ c_2 = -335.5\ $$

$$ \ 6c_3 - 6c_2 + 2c_1 = 0\ $$

$$ \ 6(-105) - 6(-335.5) + 2c_1 = 0\ $$

$$ \ c_1 = -691.5\ $$

$$ \ 2c_2 - 3c_1 + 2c_0 = 0\ $$

$$ \ 2(-335.5) - 3(-691.5) + 2c_0 = 0\ $$

$$ c_0 = -701.75\ $$

Part 5
The particular solution is of the form $$ \ y_p = c_0 + c_1x + c_2x^2 + c_3x^3 + c_4x^4 + c_5x^5\ $$

Therefore, the particular solution to the differential equation = $$ \ y_p = -701.75 - 691.5x - 335.5x^2 - 105x^3 - 22.5x^4 -3x^5\ $$

To find the general solution, we need to add the particular solution to the homogenous solution.

The homogenous solution is given by:

$$ \ (\lambda_1 - 1) (\lambda_2 - 2) = 0\ $$

$$ \ y_h = C_1e^{\lambda_1x} + C_2e^{\lambda_2}x = C_1e^x + C_2e^{2x}\ $$

Therefore the general solution is:

$$ \ y_h + y_p = C_1e^x + C_2e^{2x} -701.75 -691.5x - 335.5x^2 - 105x^3 - 22.5x^4 - 3x^5\ $$

Now using the initial conditions to solve for the remaining constants we need the general solution and the derivative of the general solution.

$$ \ y'(x) = C_1e^x + 2C_2e^{2x} - 691.5 - 671x - 315x^2 - 90x^3 - 15x^4\ $$

$$ \ y(0) = 1 = C_1 + C_2 - 701.75\ $$

$$ \ y'(0) = 0 = C_1 + 2C_2 - 691.5\ $$

We can move the 1 over, and then subtract the equations to solve for $$ \ C_2\ $$

$$ \ -C_2 -11.25 = 0 \Rightarrow C_2 = -11.25\ $$

$$ \ C_1 + C_2 - 701.75 = 1\ $$

$$ \ C_1 = 1+701.75 - -11.25 = 714\ $$

Therefore the final general solution y(x) is:

$$ \ y(x) = 714e^x -11.25e^{2x} - 701.75 -691.5x - 335.5x^2 - 105x^3 - 22.5x^4 - 3x^5\ $$

Matlab code: x = 0:0.001:100;

y = 714*exp(x) - 11.25*exp(2*x) - 701.75 - 691.5*x - 335.5*x.^2 - 105*x.^3 - 22.5*x.^4 - 3*x.^5;

plot(x,y)

xlabel('x')

ylabel('y(x)')

Author
This problem was solved and uploaded by: David Herrick

This problem was proofread by Michael Wallace

Question

 * Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODE-CC (see p. 7-2b)

 * $$ y_{p,1}''-3y_{p,1}'+2y_{p,1}=r_1(x)=4x^2 \!$$

 * $$ y_{p,2}''-3y_{p,2}'+2y_{p,2}=r_2(x)=-6x^5 \!$$

 * The particular solution to $$ y_{p,1} \!$$ had been found in R3.3 p.7-11.
 * Find the particular solution $$ y_{p,2} \!$$, and then obtain the solution $$ y \!$$ for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.

Given

 * First particular solution:

 * $$ y_{p,1}=2x^2+6x+7 \!$$

 * Initial Conditions:

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

Particular Solution

 * Because of the specific excitation $$ r_2(x)=-6x^5 \!$$, using table 2.1 from K 2011 p.82, the correct form for the particular solution is $$ y_p(x)=\sum_{j=0}^{n}K_jx^j \!$$

 * Then the following is presented:

 * $$ y_{p,2}=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 \!$$
 * $$ y_{p,2}'=c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4 \!$$
 * $$ y_{p,2}''=2c_2+6c_3x+12c_4x^2+20c_5x^3 \!$$

 * Plug these equations back into the original $$ y_{p,2}''-3y_{p,2}'+2y_{p,2}=-6x^5 \!$$

 * $$ (2c_2+6c_3x+12c_4x^2+20c_5x^3)-3(c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4)+2(c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5)=-6x^5 \!$$

 * Use coefficient matching, the following equations are a result:

 * $$ c:  2c_2-3c_1+2c_0=0 \!$$
 * $$ x:  6c_3-6c_2+2c_1=0 \!$$
 * $$ x^2:  12c_4-9c_3+2c_2=0 \!$$
 * $$ x^3:  20c_5-12c_4+2c_3=0 \!$$
 * $$ x^4:  -15c_5+2c_4=0 \!$$
 * $$ x^5:  2c_5=-6 \!$$

 * Use back substitution method to solve for every coefficient, starting with $$ c_5 \!$$

 * $$ 2c_5=-6 \rightarrow c_5=-3 \!$$
 * $$ 2c_4-15(-3)=0 \rightarrow c_4=-\frac{45}{2} \!$$
 * $$ 2c_3-12(-\frac{45}{2})+20(-3)=0 \rightarrow c_3=-105 \!$$
 * $$ 2c_2-9(-105)+12(-\frac{45}{2})=0 \rightarrow c_2=-\frac{675}{2} \!$$
 * $$ 2c_1-6(-\frac{675}{2})+6(-105)=0 \rightarrow c_1=-\frac{1395}{2} \!$$
 * $$ 2c_0-3(-\frac{1395}{2})+2(-\frac{675}{2})=0 \rightarrow c_0=-\frac{2835}{4} \!$$

 * Plug these values back into $$ y_{p,2}=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 \!$$ :

$$ y_{p,2}=-\frac{2835}{4}-\frac{1395}{2}x-\frac{675}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$

 * Superposition principle applies ( L2-ODE-CC ), $$ y_p=y_{p,1}+y_{p,2} \!$$ gives the general particular solution:

$$ y_p(x)=-\frac{2807}{4}-\frac{1383}{2}x-\frac{671}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$

Homogeneous Solution

 * $$ y_{h}''-3y_{h}'+2y_{h}=0 \!$$

 * Because of the linearity, a combination of two linear independent solutions is also a solution to this homogeneous equation:

 * $$ y_{h,1}''-3y_{h,1}'+2y_{h,1}=0 \!$$
 * $$ y_{h,2}''-3y_{h,2}'+2y_{h,2}=0 \!$$

 * With the solutions as:

 * $$ y_{h,1}=e^{\lambda_1x} \!$$
 * $$ y_{h,2}=e^{\lambda_2x} \!$$

 * To determine the value of $$ \lambda_{1,2} \!$$, the characteristic equation must be determined from the homogeneous equation

 * $$ \lambda^2-3\lambda+2=0 \!$$

 * $$ \lambda_{1,2}=\frac{-a\pm \sqrt{a^2-4b}}{2} \!$$
 * $$ \lambda_{1,2}=\frac{-(-3)\pm \sqrt{(-3)^2-4(2)}}{2} \!$$
 * $$ \lambda_1=2 \; \; \; \; \; \; \; \lambda_2=1\!$$

 * The solutions for each distinct linearly independent homogeneous equation become:

 * $$ y_{h,1}=e^{2x} \!$$
 * $$ y_{h,2}=e^{x} \!$$

 * The combination of the previous two equations, multiplied by two constants that satisfy two initial conditions, is also a solution:

$$ y_h(x)=C_1e^{2x}+C_2e^x \!$$

General Solution

 * The general or overall solution for the L2-ODE-CC:

 * $$ y(x)=y_h+y_p \!$$
 * $$ y(x)=C_1e^{2x}+C_2e^x-\frac{2807}{4}-\frac{1383}{2}x-\frac{671}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$
 * $$ y'(x)=2C_1e^{2x}+C_2e^x-\frac{1383}{2}-671x-315x^2-90x^3-15x^4 \!$$

 * Use initial conditions to solve for $$ C_1 \!$$ and $$ C_2 \!$$:

 * $$ 1=C_1+C_2-\frac{2807}{4} \!$$
 * $$ 0=C_1+C_2-\frac{1383}{2} \!$$
 * $$ C_1=-\frac{45}{4} \; \; \; C_2=714 \!$$

 * The general solution:

$$ y(x)=-\frac{45}{4}e^{2x}+714e^x-\frac{2807}{4}-\frac{1383}{2}x-\frac{671}{2}x^2-105x^3-\frac{45}{2}x^4-3x^5 \!$$

Author
This problem was solved and uploaded by Derik Bell

This problem was proofread by Michael Wallace

Question
Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities.

Equation (1)

$$ \displaystyle \sum_{j=2}^{5}c_j j(j-1)x^{j-2} = \sum_{j=0}^{3}c_{j+2}(j+2)(j+1)x^j $$

Equation (2)

$$ \displaystyle \sum_{j=1}^{5}c_j jx^{j-1} = \sum_{j=0}^{4}c_{j+1}(j+1)x^j $$

Solution
Expanding both sides of Equation (1) yields the following expression:

$$ \displaystyle c_2(2)(1)x^0+c_3(3)(2)x+c_4(4)(3)x^2+c_5(5)(4)x^3 = 2c_2+6c_3x+12c_4x^2+20c_5x^3 $$

Therefore, they are equivalent.

Expanding both sides of Equation (2) yields the following expression: $$ \displaystyle c_1(1)x^0 + c_2(2)x^1 + c_3(3)x^2+c_4(4)x^3+c_5(5)x^4 = c_1 + 2c_2x + 3c_3x^2+4c_4x^3+5c_5x^4 $$

Therefore, they are equivalent.

Author
This problem was solved and uploaded by: William Knapper

This problem was proofread by: Joshua House

Question
K 2011 p84 pbs. 5,6 Find a (real) general solution. State which rule you are using. Show each step of your work.

Problem 5 Solution

 * $$ \ y'' + 4y' + 4y = e^{-x}\cos(x) \ $$

Step 1: General Solution of the Homogeneous ODE Characteristic Equation:
 * $$ \ \lambda^2 + 4\lambda + 4 = 0 \ $$

Solve for the determinate:
 * $$ \ a^2 - 4b = 16 - 16 = 0 \ $$

Indicates a double real root, therefore:
 * $$ \ y_h = e^{\frac{-a} {2} x} (C_1 + C_2x) \ $$
 * $$ = e^{\frac{-4} {2} x} (C_1 + C_2x) = e^{2x} (C_1 + C_2x) \ $$

Step 2: Particular solution $$ \ y_p \ $$ of the non-homogeneous ODE. Using the Basic Rule and Table 2.1:
 * $$ \ r(x) = e^{-x} cos(x) \ $$
 * $$ \ y_p (x) = e^{-x} (K\cos(x) + M\sin(x)) \ $$ since k = 1, alpha = -1, and omega = 1.

 * $$ \ y_p' (x) = e^{-x} (-K\sin(x) + M\cos(x)) - e^{-x} (K\cos(x) + M\sin(x)) \ $$
 * $$ \ = e^{-x} (\cos(x) (M-K) - \sin(x) (K+M)) \ $$
 * $$ \ y_p'' (x) = e^{-x} (-\sin(x) (M-K) - \cos(x) (K+M)) - e^{-x} (\cos(x) (M-K) - \sin(x) (K+M)) \ $$
 * $$ \ = e^{-x} (\sin(x) (2K) + \cos(x) (2M)) \ $$

Plugging particular solution to original equation:
 * $$ \ e^{-x} (\sin(x) (2K) + \cos(x) (2M)) + 4(e^{-x} (\cos(x) (M-K) - \sin(x) (K+M))) + 4(e^{-x} (K\cos(x) + M\sin(x))) = e^{-x}\cos(x) \ $$

Simplifying and canceling gives:
 * $$ \ \sin(x) (2K) + \cos(x) (2M) + \cos(x) (4M - 4K) - \sin(x) (4K + 4M) + \cos(x) (4K) + \sin(x) (4M) = \cos(x) \ $$

Further simplification gives:
 * $$ \ \sin(x) (-2K) + \cos(x) (6M) = \cos(x) \ $$

Comparing sin(x) and cos(x) terms gives:
 * $$ \ -2K = 0 \ $$

Therefore K = 0.
 * $$ \ 6M = 1 \ $$

Therefore M = $$ \ \frac{1}{6} \ $$. Therefore the General Solution is:
 * $$ \ y_g (x) = y_h (x) + y_p (x) \ $$

$$ \ y_g (x) = e^{-2x} (C_1 + C_2x) + e^{-x}(\frac{1}{6} \sin(x)) \ $$

Problem 6 Solution

 * $$ \ y'' + y' + (\pi^2 + \frac{1}{4})y = e^{-\frac{x}{2}}\sin(\pi x) \ $$

Step 1: Find the homogeneous solution to the ODE. Characteristic Equation:
 * $$ \ \lambda^2 + \lambda + \pi^2 + \frac{1}{4} = e^{-\frac{x}{2}}\sin(\pi x) \ $$

Find the Determinate:
 * $$ a^2 - 4 b = 1 - 4(\frac{1}{4} + \pi^2) = -4 \pi $$

Indicates unreal solutions. Therefore, real homogeneous solution is:
 * $$ \ y_h = e^{-2x} (A\cos(wx) + B\sin(wx)) \ $$

where:
 * $$ \ w = (b - \frac{1}{4}a^2)^\frac{1}{2} \ = \pi \ $$

 * $$ \ y_h = e^{-2x} (A\cos(\pi x) + B\sin(\pi x)) \ $$

Step 2: Solution $$ \ y_p \ $$ of the non-homogeneous ODE. Using the Basic Rule and Table 2.1:
 * $$ \ r(x) = e^{-\frac{x}{2}}\sin(\pi x) \ $$
 * $$ \ y_p (x) = e^{-\frac{x}{2}} (K\cos(\pi x) + M\sin(\pi x)) \ $$

since alpha = $$ \ \frac {-1}{2} \ $$, w = $$ \ \pi \ $$, and k = 1. Therefore:
 * $$ \ y_p' (x) = -\frac{1}{2} e^{-\frac{x}{2}} (K\cos(\pi x) + M\sin(\pi x)) + e^{-\frac{x}{2}} (-K\pi\sin(\pi x) + M\pi\cos(\pi x)) \ $$
 * $$ \ = e^{-\frac{x}{2}} (\cos(\pi x) (M\pi - \frac{1}{2}K) - \sin(\pi x) (K\pi + \frac{1}{2}M)) \ $$
 * $$ \ y_p'' (x) = -\frac{1}{2} e^{-\frac{x}{2}} (\cos(\pi x) (M\pi - \frac{1}{2}K) - \sin(\pi x) (K\pi + \frac{1}{2}M)) + e^{-\frac{x}{2}} (-\sin(\pi x) (M\pi^2 - \frac{1}{2}K\pi) - \cos(\pi x) (K\pi^2 + \frac{1}{2}M\pi)) \ $$
 * $$ \ = e^{-\frac{x}{2}} (\cos(\pi x) (-M\pi +\frac{1}{4}K + K\pi^2) + \sin(\pi x) (\frac{1}{4}M + K\pi - M\pi^2)) \ $$

Substituting in to the original equation:
 * $$ \ (e^{-\frac{x}{2}} (\cos(\pi x) (-M\pi +\frac{1}{4}K + K\pi^2) + \sin(\pi x) (\frac{1}{4}M + K\pi - M\pi^2))) + (e^{-\frac{x}{2}} (\cos(\pi x) (M\pi - \frac{1}{2}K) - \sin(\pi x) (K\pi + \frac{1}{2}M))) + (\pi^2 + \frac{1}{4}) (e^{-\frac{x}{2}} (K\cos(\pi x) + M\sin(\pi x))) = e^{-\frac{x}{2}}\sin(\pi x) \ $$

Simplifying and Canceling gives:
 * $$ \ \cos(\pi x) (K\pi^2 - \frac{1}{4}K + M\pi^2 + \frac{1}{4}M) + \sin(\pi x) (-\frac{1}{4}M - M\pi^2 + K\pi^2 + \frac{1}{2}K) = \sin(\pi x) \ $$

Comparing $$ \ \sin(\pi x) \ $$ and $$ \ \cos(\pi x) \ $$ terms gives:
 * $$ \ K\pi^2 + -\frac{1}{4}K + M\pi^2 + \frac{1}{4}M = 0 \ $$
 * $$ \ -\frac{1}{4}M - M\pi^2 + K\pi^2 + \frac{1}{2}K = 1 \ $$

Solve for K and M:
 * $$ \ M\pi^2 = \frac{1}{4}K - \frac{1}{4}M - K\pi^2 \ $$

Therefore:
 * $$ \ \frac{1}{4}M + K\pi^2 + \frac{1}{2}K - \frac{1}{4}K + \frac{1}{4}M + K\pi^2 - \frac{1}{2}M = 1 \ $$

Reducing gives:
 * $$ \ 2K\pi^2 + \frac{1}{4}K = 1 \ $$

Solving for K gives K = 0.05. Therefore:
 * $$ \ 0.05\pi^2 - \frac{1}{2}(0.05) + M\pi^2 + \frac{1}{4}M = 0 \ $$

Solving for M gives M = -0.046. Therefore:
 * $$ \ y_g (x) = y_h (x) + y_p (x) \ $$

$$ \ y_g (x) = e^{-2x} (A\cos(\pi x) + B\sin(\pi x)) + e^{-\frac{x}{2}} (0.05\cos(\pi x) - 0.046\sin(\pi x)) \ $$

Author
This problem was solved and uploaded by: John North

This problem was proofread by: Michael Wallace

Question
Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail. 13. $$ \displaystyle 8y''-6y'+y=6coshx, y(0)=0.2 , y'(0)=0.05 $$ 14. $$ \displaystyle y''+4y'+4y=e^{-2x}sin2x, y(0)=1 , y'(0)=-1.5 $$

Problem 13
First, we find the homogeneous ODE solution
 * $$ \displaystyle 8\lambda^2-6\lambda+1=0 $$
 * $$ \displaystyle (4\lambda-1)(2\lambda-1)=0 $$
 * $$ \displaystyle \lambda=\frac{1}{4}, \lambda=\frac{1}{2} $$

which gives the general solution:
 * $$ \displaystyle y_h=c_1e^{\frac{1}{4}x}+c_2e^{\frac{1}{2}x} $$

Second, we get the particular solution of the non homogeneous ODE. We can substitute $$ coshx=\frac{e^x+e^{-x}}{2} $$ Using the sum rule we know that $$ y_p=y_{p1}+y_{p2} $$. Then using Table 2.1 we can solve:
 * $$ \displaystyle y_{p1}=c_3e^x, y_{p2}=c_4e^{-x} $$

Then we get
 * $$ \displaystyle y_p=c_3e^x+c_4e^{-x} $$
 * $$ \displaystyle y_p'=c_3e^x-c_4e^{-x} $$
 * $$ \displaystyle y_p''=c_3e^x+c_4e^{-x} $$

The original ODE becomes:
 * $$ \displaystyle 8y''-6y'+y=3e^x+3e^{-x} $$

Substituting $$ y_p $$ into the original ODE and separating into the two components we get:
 * $$ \displaystyle 8c_3-6c_3+c_3=3 $$

and
 * $$ \displaystyle 8c_4+6c_4+c_4=3 $$

Solving these two equations we get:
 * $$ \displaystyle c_3=1 $$
 * $$ \displaystyle c_4=\frac{1}{5} $$

Thus,
 * $$ \displaystyle y_p=e^x+\frac{1}{5}e^{-x} $$

We know that $$ y=y_h+y_p $$. so,
 * $$ \displaystyle y=c_1e^{\frac{1}{4}x}+c_2e^{\frac{1}{2}x}+e^x+\frac{1}{5}e^{-x} $$

Finding the solution to the initial value problem:
 * $$ \displaystyle y(0)=0.2=c_1(1)+c_2(1)+1+\frac{1}{5} $$
 * $$ \displaystyle -1=c_1+c_2 $$
 * $$ \displaystyle c_2=-c_1-1 $$
 * $$ \displaystyle y'=\frac{1}{4}c_1e^{\frac{1}{4}x}+\frac{1}{2}c_2e^{\frac{1}{2}x}+e^x-\frac{1}{5}e^{-x} $$
 * $$ \displaystyle y'(0)=0.05=\frac{1}{4}c_1(1)+\frac{1}{2}c_2(1)+1-\frac{1}{5} $$
 * $$ \displaystyle -0.75=\frac{1}{4}c_1+\frac{1}{2}c_2 $$

Substituting the known value for $$ c_2 $$
 * $$ \displaystyle -0.75=\frac{1}{4}c_1+\frac{1}{2}(-c_1-1) $$
 * $$ \displaystyle -0.75=\frac{1}{4}c_1-\frac{1}{2}c_1-\frac{1}{2} $$
 * $$ \displaystyle -0.25=-\frac{1}{4}c_1 $$

Solving this we find that:
 * $$ \displaystyle c_1=1 $$
 * $$ \displaystyle c_2=-2 $$

The final solution is: $$ \displaystyle y=e^{\frac{1}{4}x}-2e^{\frac{1}{2}x}+e^x+\frac{1}{5}e^{-x} $$

Problem 14
First, we find the homogeneous ODE solution
 * $$ \displaystyle \lambda^2+4\lambda+4=0 $$
 * $$ \displaystyle (\lambda+2)(\lambda+2)=0 $$
 * $$ \displaystyle \lambda=-2, \lambda=-2 $$

which gives the general solution:
 * $$ \displaystyle y_h=c_1e^{-2x}+c_2xe^{-2x} $$

Second, we get the particular solution of the non homogeneous ODE. Using the sum rule and using Table 2.1 we can solve:
 * $$ \displaystyle y_p=e^{-2x}(Kcos2x+Msin2x) $$

$$ e^{-2x} $$ is a homogeneous solution so $$ y_p $$ becomes:
 * $$ \displaystyle y_p=xe^{-2x}(Kcos2x+Msin2x) $$

Then we find:
 * $$ \displaystyle y_p'=(e^{-2x}-2xe^{-2x})(Kcos2x+Msin2x) + xe^{-2x}(-2Ksin2x+2Mcos2x) $$
 * $$ \displaystyle y_p'=Ke^{-2x}cos2x-2Kxe^{-2x}cos2x+Me^{-2x}sin2x-2Mxe^{-2x}sin2x-2Kxe^{-2x}sin2x+2Mxe^{-2x}cox2x$$
 * $$ \displaystyle y_p'=Ke^{-2x}cos2x+Me^{-2x}sin2x+(-2K-2M)xe^{-2x}sin2x+(-2K+2M)xe^{-2x}cos2x $$

And,
 * $$ \displaystyle y_p''=K[-2e^{-2x}cos2x-2e^{-2x}sin2x]+M[-2e^{-2x}sin2x+2e^{-2x}cos2x]+(-2K+2M)[(e^{-2x}-2xe^{-2x})(cos2x)+(-2xe^{-2x}sin2x)]$$ $$ \displaystyle +(-2K-2M)[(e^{-2x}-2xe^{-2x})(sin2x)+(2xe^{-2x}cos2x)] $$
 * $$ \displaystyle y_p''=-2Ke^{-2x}cos2x-2Ke^{-2x}sin2x-M2e^{-2x}sin2x+2Me^{-2x}cos2x-2Ke^{-2x}cos2x+4Kxe^{-2x}sin2x+4Kxe^{-2x}cos2x $$ $$ \displaystyle +2Me^{-2x}cos2x-4Mxe{^-2x}cos2x-4Mxe^{-2x}sin2x-2Ke^{-2x}sin2x+4Kxe^{-2x}sin2x-4Kxe^{-2x}cos2x $$ $$ \displaystyle -2Me^{-2x}sin2x+4Mxe^{-2x}sin2x-4Mxe^{-2x}cos2x $$
 * $$ \displaystyle y_p''=(-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x+8Kxe^{-2x}sin2x-8Mxe^{-2x}cos2x $$

Substituting $$ y_p $$ into the original ODE:
 * $$ \displaystyle (-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x+8Kxe^{-2x}sin2x-8Mxe^{-2x}cos2x-4Ke^{-2x}cos2x+4Me^{-2x}sin2x $$ $$ \displaystyle +(-8K-8M)xe^{-2x}sin2x+(-8K+8M)xe^{-2x}cos2x+4Kxe^{-2x}cos2x+4Mxe^{-2x}sin2x=e^{-2x}sin2x $$
 * $$ \displaystyle (-8K+4M)e^{-2x}cos2x+-4Ke^{-2x}sin2x=e^{-2x}sin2x $$

Separating into components we get:
 * $$ \displaystyle -4K=1 $$
 * $$ \displaystyle -8K+4M=0 $$

Solving the two equations we get:
 * $$ \displaystyle K=-\frac{1}{4} $$
 * $$ \displaystyle M=-\frac{1}{2} $$

Thus,
 * $$ \displaystyle y_p=e^{-2x}(-\frac{1}{4}cos2x-\frac{1}{2}sin2x) $$

We know that $$ y=y_h+y_p $$. so,
 * $$ \displaystyle y=c_1e^{-2x}+c_2xe^{-2x}+e^{-2x}(-\frac{1}{4}cos2x-\frac{1}{2}sin2x)$$

Finding the solution to the initial value problem:
 * $$ \displaystyle y(0)=1=c_1(1)+(1)[-\frac{1}{4}(1)] $$

We solve,
 * $$ \displaystyle c_1=-\frac{5}{4} $$

Also,
 * $$ \displaystyle y'=-2c_1e^{-2x}-2c_2xe^{-2x}+c_2e^{-2x}-2e^{-2x}(-\frac{1}{4}cos2x-\frac{1}{2}sin2x)+e^{-2x}(\frac{1}{2}sin2x-cos2x) $$
 * $$ \displaystyle y'(0)=-1.5=-2(-\frac{5}{4})(1)+c_2(1)-2(1)[-\frac{1}{4}(1)]+(1)(-1) $$

We solve,
 * $$ \displaystyle c_2=-3.5=-\frac{7}{2} $$

The final solution is: $$ \displaystyle y=-\frac{5}{4}e^{-2x}-\frac{7}{2}xe^{-2x}+e^{-2x}(-\frac{1}{4}cos2x-\frac{1}{2}sin2x) $$

Author
This problem was solved and uploaded by: Radina Dikova

This problem was proofread by: David Herrick

Contribution Summary
Problems 3 and 5 were solved and Problems 1 and 9 were proofread by David Herrick

Problems 1 and 2 were solved and Problem 7 was proofread by Joshua House

Problem 9 was solved by Radina Dikova

Problems 4 and 8 were solved by John North

Problem 7 was solved by William Knapper

Problem 6 was solved by Derik Bell

Problems 2, 3, 4, 5, 6, and 8 were proofread by Michael Wallace


text:
File:R3.3.png

Licensing
for school project

title: File:R3.3.png


text:
Research Based Strategies for Dealing with Students with Behavior Issues: Token Reinforcement Systems

Behavior management is an extremely challenging task for teachers to accomplish. There are prevention strategies and intervention strategies. Prevention strategies are techniques that are used in the classroom in order to prevent students from acting out or behaving poorly. These techniques include creating a positive and structured classroom environment and to continuously check on student's behavior throughout the day ' '(http://www.aft.org/pdfs/tools4teachers/CT-BehaviorMgmt0310.pdf)' '. Intervention strategies are techniques that teachers must use once the poor behavior has already taken place. Some of these strategies include self-instruction and relaxation. While there are many strategies for dealing with students with behavior issues, we will be focusing on a token reinforcement system. Token reinforcement systems are generally used for all students within a classroom. Some articles warn, however, that before implementing a token reinforcement system, you should exhaust all your other options (Cash & Woollens 1). Before implementing a token economy, teachers should attempt to use other forms of feedback, such as praise. According to one source, "only use a token economy system when necessary, and not when desirable behavior can be maintained with the use of less systematic and controlling techniques" (Cash & Woollens 1).

Articles warn about using this type of reinforcement system because they don't want students to rely on material objects in order to reinforce their behavior. Although sources say to try other methods first, Behavioradvisor.com states that this form of behavior management is effective for students who resist other methods (McIntyre). According to Laura Peters, the best way a token economy works is when, "the behavior to be reinforced addresses the unique needs of the students and is clearly defined, procedures are devised by giving out a token each time the behavior occurs initially, a set of rules is explained to govern the exchange of tokens for reinforcing events or objects, and there is a plan for fading the use of tokens" (Peters 1). This is important because students need to understand when and why someone earns a token and how the process works.

What is a token reinforcement system?

A token reinforcement system is a way to praise students who are displaying a desired behavior. Immediately after a student has shown a specific behavior, teachers will give students a token of some kind, which is a temporary reinforcer. Once the student has accumulated a certain number of tokens, they can exchange their tokens for a back-up reinforcer, which is something the child wants to earn (http://education.calumet.purdue.edu/). An example of a token is a gold star and an example of a back-up reinforcer can be a pizza party or a pencil. A token economy is considered to be a secondary conditioned reinforcement because the actual token is of little significance (Wallin 2). The actual token has no meaning, but it is paired with the ability to trade it in and receive a prize, therefore making the token seem reinforcing. This type of motivation can motivate students to display the appropriate behavior because they want to receive the tokens. A token economy is often compared to money in our society because we can exchange currency for goods that we want or need.

What are the advantages of a token economy? What are the disadvantages of a token economy?
 * increased ability to delay gratification (builds a child's ability to wait for gratification)
 * increased sense of time (student's are more aware of time and how long tasks take)
 * lessened satiation (always changing so students won't become bored with this system)
 * increased teaching rate (quick way to praise students without interrupting the teaching flow)
 * more natural reinforcement (allows students to know they are doing something good without receiving a prize every time they do so)
 * increased selection of reinforcers (since prizes are not being handed out after each good behavior, the prizes will be more meaningful)
 * ways to establish new behaviors (by reinforcing students to do a specific behavior)
 * maintain behavior (students will maintain the positive behavior)
 * long-term outcomes (they will be conditioned to continue to repeat the behavior)
 * personally satisfying (students are competing with themselves so they feel accomplished once they earn tokens)
 * ease of administration
 * lack of competition
 * fidelity of implementation
 * dependency on the system or difficulty fading
 * perceptions of the intent of the motivator
 * issues associated with taking away tokens (damaging for the child)
 * could cause ethical problems
 * delays gratification
 * students may cheat the system
 * students are only productive when tokens are given out

Research
 * Homework Success is a program in which students with ADHD can get help with their homework. In this program, a clinical trial was done to show the effects of a token economy. The token economy, "helped to reduce the amount of behavioral problems while completing homework which lead to the decreased stress levels" (http://wik.ed.uiuc.edu/index.php/Token_economy).
 * A summer sports program for students with ADHD implemented a token system with three different colored chips. The results of the study showed that the tokens increased attentive behavior and sportsmanlike conduct, while decreasing unsportsmanlike conduct (http://wik.ed.uiuc.edu/index.php/Token_economy).
 * An individual study was done on a girl named Tasha, who had severe behavior disorders. The study determined her behavior on the time it took her to complete an assignment, the number of assignments completed, and the frequency of inappropriate behavior. The results of the study found that the tokens were effective in both Tasha's academic and social behavior (https://docs.google.com/viewer?a=v&q=cache:3RcpEvlxvQIJ:www.internationalsped.com/documents/9Klimas%2520McLaughlin%2520%2520token%2520economy.DOC+&hl=en&gl=us&pid=bl&srcid=ADGEESh4rpqDbHyowyQsQ7CpSLVh38LgxwyQr5RjvA4aiiFnbu2yT3I15KJ-t85wWdb6JltjZTIWvAYksovVW773DWj3y20uRfSa7ArOjLxpaPU12KA-FZ-eOhcuNYMJbiAbT-AjGoeY&sig=AHIEtbTVzyDuE-EtM6IQ2KStzR9lzBnvQw ).

Bibliography

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2648534/

http://education.calumet.purdue.edu/vockell/edPsybook/Edpsy10/edpsy10token.htm

http://www.polyxo.com/visualsupport/tokeneconomies.html

https://docs.google.com/viewer?a=v&q=cache:n0FBoQjk638J:www.udel.edu/education/masters/psychology/handbook/recommendations/TokenEconomy.doc+&hl=en&gl=us&pid=bl&srcid=ADGEESi-tuM8NkwjhyM5nhX0b8TVCO3eFsjoVdIbwy0N5iK-JvSelc3VQXFAwJ4yBllxkGgPwfivEcvvgsHN2wr0kwwn-bly6Zz92ivV5P1bmrLFnjfx224345STSUSDtNf0DX3tL5D3&sig=AHIEtbR_WecU2aO_em6RVdQ6Zx5tz45kLQ

http://www.pbis.vcu.edu/2009/11/using-a-t.html

http://wik.ed.uiuc.edu/index.php/Token_economy

http://www.behavioradvisor.com/Tokens.html https://docs.google.com/viewer?a=v&q=cache:3RcpEvlxvQIJ:www.internationalsped.com/documents/9Klimas%2520McLaughlin%2520%2520token%2520economy.DOC+&hl=en&gl=us&pid=bl&srcid=ADGEESh4rpqDbHyowyQsQ7CpSLVh38LgxwyQr5RjvA4aiiFnbu2yT3I15KJ-t85wWdb6JltjZTIWvAYksovVW773DWj3y20uRfSa7ArOjLxpaPU12KA-FZ-eOhcuNYMJbiAbT-AjGoeY&sig=AHIEtbTVzyDuE-EtM6IQ2KStzR9lzBnvQw

http://www.aft.org/pdfs/tools4teachers/CT-BehaviorMgmt0310.pdf


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Wikiversity:Vision/2012

Things which could be changed right now

 * more prominent linking to chat, e.g. from navigation a link to: Chat
 * , Erkan Yilmaz uses the Chat + Identi.ca 20:09, 15 February 2012 (UTC)

Needs more things to be accomplished

 * more interactivity (more mumblings here + here), Erkan Yilmaz uses the Chat + Identi.ca 20:14, 15 February 2012 (UTC)
 * a playground to try out new software (features), see e.g. Sandbox server + replies, Erkan Yilmaz uses the Chat + Identi.ca 20:18, 15 February 2012 (UTC)

title: Wikiversity:Vision/2012


title: Category:2012


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User talk:Carterartist

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User talk:Bounchanh

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User talk:Shravan~enwikiversity

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Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

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User talk:BellsBastianz

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title: File:CORDIC.Background.4.A.20120214.pdf


text:
Engineering Experience 4: Design a Small Solar Vehicle/Nl/2012: Team AM1

INTRODUCTIE
Voor het EE4-project creëren wij, leden van Team Helios, een Small Solar Vehicle. Dit project houdt in dat we in groepen van zeven personen een Small Solar Vehicle (SSV) ontwerpen dat volledig afhankelijk is van zonne-energie. Hiervoor kriigt elk team een zonnepaneel en een DC-motor, zodat iedereen van dezelfde basis kan beginnen. Na tien weken wordt er een race georganiseerd waaraan alle zonnewagens het tegen elkaar opnemen. De SSV moet niet alleen degelijk werken, ook is het belangrijk om een mooie en innovatieve wagen te ontwerpen.

Samenwerkingscontract en afspraken
Vergaderschema:

- Wanneer: Het team vergadert elke week. Dit telkens op een vaste datum. Elke dinsdag ochtend, zal het team samenzitten om het werk te overlopen.

- Waar: De teamleden komen samen in het cafetaria van GroepT.

- Voorzitter en notulist: De teamleiders zitten de vergadering voor en fungeren als notulisten.

- Tijdens de vergadering wordt de Gantt Chart gebruikt om de stand van zaken na te gaan, en wordt het geüpdated indien nodig.

Communicatieregels:

- Iedereen moet elke dag zijn mails lezen.

- Mededelingen zullen gebeuren via de facebookgroep, mail of sms.

- Alle bestanden zullen te vinden zijn op de gemeenschappelijke dropbox account.

Algemene afspraken:

-Iedereen moet alles nalezen voor het wordt doorgestuurd naar de coach.

-Bij afwezigheid of te laat komen wordt de teamleider verwittigd.

- Wanneer een teamlid zijn verwachte uren niet haalt, wordt er verwacht dat hij deze inhaalt.

- Minstens 1 maal per week wordt er vergaderd of samengekomen voor het werken aan het project.

- Iedereen is verantwoordelijk voor het op tijd en kwaliteitsvol afwerken van zijn opgelegde taken.

- 3 maal te laat is traktatie.

Week 1 [13/02 - 19/02]
In deze eerste week krijgen we een inleiding van het project en heeft iedereen voor het eerst kennis kunnen maken met zijn teamgenoten.

Na deze korte kennismaking zijn we er echter onmiddellijk ingevlogen aangezien er vrijdag al de eerste deadline is.

We hebben eerst afgesproken wie voor wat verantwoordelijk is en daarna is iedereen begonnen met zijn taak. De onderdelen die klaar moeten zijn voor de deadline zijn: WBS, gantt chart, plan van aanpak, samenwerkingscontract en teamlogo & -naam (zie hierboven).

Week 2 [20/02 - 26/02]
Na de inleiding van de zonnepaneel hebben we met de team effectief één gekregen en getest (ook een DC motor) en we hebben hiervan de stroom-spanning karakteristiek van bepaald met variërende belasting.

Nadat we de gewenste UI grafiek bekomen waren, hebben we aan de hand van gegeven formules een poging gedaan om een maximaal vermogen te berekenen. UPDATE : We hebben onze nieuwste teamlid verwelkomd : Maxim van Cappellen.

Week 3 [27/02 - 4/03]
De seminarie over optimale overbrengingen werd gegeven door een international coach. Een reeks aantal formules gevolgd door grafieken hebben ons ongeveer een beeld gegeven over wat ons te wachten staat. Tijdens de teamvergadering hebben we besproken welke LOGO we kiezen voor ons team. We hebben nu nog enkel een teamfoto nodig. We hebben specifiekere taken ingedeeld zodat we allemaal exacter konden werken. Zo gaat er tegen volgende weken al wat opzoekwerk gedaan worden naar versnellingsbakken.

Week 4 [5/03 - 11/03]
Vandaag werd extra aandacht vereist, aangezien we voor het eerst de programma Simulink te zien kregen. Er werd beslist dat we de versnellingsbak het best in LEGO kunnen maken, aangezien we het geheel niet te zwaar mag wegen. Op Solid Edge werd er een voorbeeld gemaakt van hoe het wagentje er uit zal zien. Verder zal er tegen volgende week de manual bekeken worden op toledo om een en ander te weten te komen over factoren die ons wagen kan beïnvloeden zoals bv : luchtweerstand, wrijving en dergelijke. Een definitieve keuze over de wielen is ook bijna beslist, en minidiscs krijgen voorlopig de voorkeur voor de achterwielen. Tenslotte werd ook een LOGO uitgekozen voor team Helios.

Week 5 [12/03 - 18/02]
Een interessante seminarie over FABLAB werd gegeven door personeel van de KUL en diezelfde dag zijn we ook eens een piekje gaan nemen aan het departement van werktuigkunde van de burgerlijke ingenieurs. We hadden een design van solid edge bij en hebben met de lasercutting device een template uitgesneden van ons ontwerp. In eerste instantie een leerrijke uitstap, want we hebben eens gezien wat voor materiaal er beschikbaar was - en hebben meteen ook de knoop doorgehakt voor onze definitieve keuze, namelijk Balsahout. Een veel lichter materiaal t.o.v van onze originele keuze plexiglas (factor 10). Tenslotte hebben we kort vergaderd over de details van onze 1e verslag.

Week 6 [19/03 - 25/03]
Geen seminarie deze week, maar geen reden om te pauzeren. Na een korte Coach evaluatie over de stand van zaken, konden we weer vertrekken. De focus van deze week was voornamelijk het in orde brengen van het verslag en wat experimenteren met de materiaal die we vorig week ook al hadden voor een deel zoals tandwielen, discwielen etc... Een frustrurende fout die niet heel zichtbaar was in Simulink nam wat tijd in beslag, maar uiteindelijk werd alles gecorrigeerd een aangepast. De motor werd vandaag ook getest in combinatie met ons zonnepaneel om een idee te krijgen over hoe we best onze paneel konden aansluiten aan het voertuig. Enkele afmetingen in Solid Edge werden veranderd voor het aerodynamisch gunstiger te maken voor het geheel. Naar volgende week toe willen we al proberen verschillende componenten te assembleren.

Week 7 [26/03 - 1/04]
De laatste week voor de paasvakantie, en een feedbacksessie van de coach over ons verslag. Los van slordige taalfouten, hadden we blijkbaar inhoudelijk gezien enkele problemen bij de Sankey Diagram.

De simulink gedeelte van het verslag viel aardig goed mee en dit was in feite ook een van de moeilijkere stukken in het EE4 project. Met een fotokopie met al deze opmerkingen, zijn we terug aan slag gegaan en hebben we een economische analyse en een businessplan opgesteld. Het verslag werd nog eens nagelezen en de fouten uit de opmerkingen werden aangepast. Na de solidedge files waren aangepast van de vorige week, zijn we weer naar fablab getrokken om met balsahout materiaal uit te snijden voor de verdere assemblage van onze zonnewagen. Een probleem op dit punt is nog altijd de afwezigheid van de definitieve tandwielen, die passen op de assen die wij nu in ons bezit hebben. De hobbywinkels in Leuven, hebben ons daar niet echt verder mee geholpen, dus we hopen dat we iets vinden op het internet.

Week 8 [16/04 - 22/04]
Met de paasvakantie achter ons, is team Helios klaar om de draad weer op te pikken. De dag begon met een test om ons kennis te peilen in verband met alles wat we tot nu al hadden gezien in de college's van EE4. De overbrengingen zijn gevonden en we hebben bijna alles om de assemblage te vervolledigen. Het enige wat ons nu nog rest is een manier bedenken om het zonnepaneel flexibel te doen bewegen op ons wagenvlak. Tegen volgende week zou dit ook moeten in orde gebracht worden, tevens is het volgende week ook een eerste testrit met onze zonnewagen, dus we kijken al zeker uit naar volgende week dinsdag.

Week 9 [23/04 - 29/04]
De laatste seminarie van het EE4 project over technisch tekenen werd vandaag toegelicht. Er wacht ons een hele opgave om een technische tekening te maken van ons wagen, en gedurende deze week wordt er ook een testbaan beschikbaar gesteld in de garage van groepT, zodat elke team zijn wagentje kan testen.

De business plan wordt vandaag al reeds nagelezen voor eventuele taalfouten. Het 2e deel van het SSV report moet in week 11 binnen zijn, dus de krachtberekeningen, waar men vorige week al mee bezig was, moeten we trachten tot een goede einde te brengen. De gebruikte discs als "banden" worden ook getest op de baan, en achteraf moeten we bepalen of er eventueel rekkers moeten gevestigd worden, voor meer grip op de baan. Een andere probleem bestaat erin de minimale afwijking te corrigeren die het wagen ondervindt op zijn weg naar beneden, zodat hij perfect op een rechte lijn kan blijven rijden. Met de mogelijkheid op de race binnen twee weken (afhankelijk van de weeromstandigheden) moeten we proberen zoveel mogelijk te testen.

Week 10 [30/04 - 06/05]
We naderen de deadlines voor het finaal rapport en de finale realisatie van het SSV wagen. Team Helios is zeer enthousiast naar de race die een van de volgende weken zal plaatsvinden, en we hopen op minstens één award. De berekeningen zijn quasi gedaan, de proces report is geschreven, met hier & daar aanpassingen aan de Gantt chart geeft ons voldoende tijd om nog nodige aanpassingen aan ons wagentje te doen. We geraken bijna tot helemaal boven, en proberen nog als team eventuele oplossingen te bedenken, zoals 'vermindering' van massa. Het is vrij moeilijk om in te schatten hoe dat het race zal verlopen omdat we enkel kunnen testen met 9 V batterijen, maar met een mooi zonnig weer om ons te leiden, kunnen we wellicht hogere voltages bereiken om zo de top met gemak te bereiken. De tijd zal uitwijzen, wanneer we kunnen rekenen op de zonnegod, waar onze teamnaam uiteindelijk naar verwijst. We wensen in de komende weken, alle teams succes!


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