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User talk:Bhaskaran

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XML Schema

XML Schemas express shared vocabularies and allow machines to carry out rules made by people. They provide a means for defining the structure, content and semantics of XML documents. in more detail. XML Schema was approved as a W3C Recommendation on 2 May 2001 and a second edition incorporating many errata was published on 28 October 2004; see reference list for pointers.

The XML Schema Working Group, part of the XML Activity, is responsible for the W3C's work on this topic -- for details see the Working Group's charter. W3C Members can also participate in the W3C XML Schema Interest Group, a forum for the discussion of technical issues relating to the development of XML Schema. It also has a charter.

The Bugzilla system should also be used to report errors with XML Schema 1.0.

title: XML Schema


text:
User:Egm6321.f10.team3.Sudheesh/Homeworks/Homework5

= Problem 3 - Solution of Euler L2-ODE - with Varying Coefficients and Constant Coefficients =

Given

 * Euler $$\displaystyle L2-ODE-VC $$,

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

 * $$\displaystyle {{a}_{2}}{{x}^{2}}{{y}^{''}}+{{a}_{1}}x{{y}^{'}}+{{a}_{0}}y=0$$
 * $$\displaystyle (Eq. 3.1)$$
 * }
 * }
 * }

 * Consider characteristic equation,

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

 * $${{\left( r-\lambda \right)}^{2}}=0$$
 * $$\displaystyle (Eq. 3.2)$$
 * }
 * }
 * }

 * Where $$\displaystyle \lambda =5$$

Find

 * (1) Find $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$ such $$\displaystyle (Eq. 3.2)$$ is a characteristic equation of $$\displaystyle (Eq. 3.1)$$

 * (1.2-1.4) Solve $$\displaystyle (Eq. 3.1)$$

 * (2) Solve $$\displaystyle {{b}_{2}}{{y}^{''}}+{{b}_{1}}{{y}^{'}}+{{b}_{0}}y=0$$

(1.1) Find $$\displaystyle {{a}_{2}}$$, $$\displaystyle {{a}_{1}}$$ and $$\displaystyle {{a}_{0}}$$

 * Assume, $$\displaystyle y={{x}^{r}}$$
 * Taking derivative,

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

 * $$\displaystyle {{y}^{'}}=r{{x}^{r-1}}$$
 * }
 * }
 * }

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

 * $${{y}^{''}}=r\left( r-1 \right){{x}^{r-2}}$$
 * }
 * }
 * }

 * Now plugg in $$\displaystyle y$$ and whose derivatives in $$\displaystyle (Eq. 3.1)$$,to obtain characteristic equation:

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

 * $${{a}_{2}}{{x}^{2}}r\left( r-1 \right){{x}^{r-2}}+{{a}_{1}}xr{{x}^{r-1}}+{{a}_{0}}{{x}^{r}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{a}_{2}}r\left( r-1 \right)+{{a}_{1}}r+{{a}_{0}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{a}_{2}}{{r}^{2}}+\left( {{a}_{1}}-{{a}_{2}} \right)r+{{a}_{0}}=0$$
 * $$\displaystyle (Eq. 3.3)$$
 * }
 * }
 * }

 * Expanding the given characteristic equation $$\displaystyle (Eq. 3.2)$$:

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

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

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

 * $$\Rightarrow {{r}^{2}}-10r+25=0$$
 * $$\displaystyle (Eq. 3.4)$$
 * }
 * }
 * }

 * Equating $$\displaystyle (Eq. 3.3)$$and $$\displaystyle (Eq. 3.4)$$ give coefficients as:

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

\begin{array}{*{20}l} {{a}_{2}}=1 & {{a}_{1}}=-9 & {{a}_{0}}=25\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Therefore $$\displaystyle Eq. 3.1$$ becomes:

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

 * $$\displaystyle {{x}^{2}}{{y}^{''}}-9x{{y}^{'}}+25y=0$$
 * $$\displaystyle (Eq. 3.5)$$
 * }
 * }
 * }

(1.2) First Homogenous solution

 * First homogenous solution is

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

$$\displaystyle {{y}_{1}}={{x}^{\lambda }}={{x}^{5}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 3.6)$$
 * }
 * }

(1.3) Second Homogenous solution

 * Second homogenous solution is

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

 * $${{y}_{2}}=U\left( x \right){{y}_{1}}=U\left( x \right){{x}^{5}}$$
 * $$\displaystyle (Eq. 3.7)$$
 * }
 * }
 * }

 * Taking derivatives:

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

 * $${{y}_{2}}^{'}={{x}^{5}}{{U}^{'}}\left( x \right)+5{{x}^{4}}U\left( x \right)$$
 * }
 * }
 * }

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

 * $${{y}_{2}}^{}={{U}^{}}\left( x \right){{x}^{5}}+10{{U}^{'}}\left( x \right){{x}^{4}}+20U\left( x \right){{x}^{3}}$$
 * }
 * }
 * }

 * Substitute above three equations in $$\displaystyle Eq. 3.6$$:

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

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

 * From lecture and definition, all $$\displaystyle U\left( x \right){{x}^{5}}$$ terms drop out because combined equation equal to zero

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

 * $$\Rightarrow {{U}^{''}}{{x}^{7}}+{{U}^{'}}{{x}^{6}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{U}^{''}}+\frac{x}=0$$
 * $$\displaystyle (Eq. 3.8)$$
 * }
 * }
 * }

 * Let $$\displaystyle {{U}^{'}}=S$$ and then $$\displaystyle (Eq. 3.8)$$ becomes an equation with order reduced by 1:

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

 * $${{S}^{'}}+\frac{S}{x}=0$$
 * $$\displaystyle (Eq. 3.9)$$
 * }
 * }
 * }

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

 * $$\Rightarrow \frac{\partial S}{\partial x}=-\frac{S}{x}$$
 * }
 * }
 * }

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

 * $$\Rightarrow \frac{\partial S}{S}=-\frac{\partial x}{x}$$
 * }
 * }
 * }

 * Integrating the above equation implies:

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

 * $$\displaystyle \ln S=-\ln x+c$$
 * }
 * }
 * }

 * Taking exponetial on both sides results in:

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

 * $$S={{C}_{1}}\frac{1}{x}$$
 * $$\displaystyle (Eq. 3.10)$$
 * }
 * }
 * }

 * But,

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

 * $$\displaystyle {{U}^{'}}=S$$
 * }
 * }
 * }

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

 * $$\Rightarrow U=\int{Sdx}+{{C}_{2}}$$
 * }
 * }
 * }

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

 * $$\Rightarrow U=\int{\left( {{C}_{1}}\frac{1}{x} \right)dx}+{{C}_{2}}$$
 * }
 * }
 * }

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

 * $$\Rightarrow U={{C}_{1}}\ln x+{{C}_{2}}$$
 * }
 * }
 * }

 * Substituting the above equation in $$\displaystyle (Eq. 3.7)$$ yields,

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

$${{y}_{2}}=\left( {{C}_{1}}\ln x+{{C}_{2}} \right){{x}^{5}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 3.11)$$
 * }
 * }

(1.4) General Homogenous solution

 * General homogenous solution is given by

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

 * $$\displaystyle y={{k}_{1}}{{y}_{1}}+{{k}_{2}}{{y}_{2}}$$
 * }
 * }
 * }

Substituting the $$\displaystyle Eq. 3.6$$ and $$\displaystyle Eq. 3.11$$ yields:

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

 * $$y={{k}_{1}}{{x}^{5}}+{{k}_{2}}\left( {{C}_{1}}\ln x+{{C}_{2}} \right){{x}^{5}}$$
 * }
 * }
 * }

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

 * $$y=\left( {{k}_{1}}+{{C}_{2}} \right){{x}^{5}}+{{k}_{2}}\left( {{C}_{1}}\ln x \right){{x}^{5}}$$
 * }
 * }
 * }

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

$$y={{K}_{1}}{{x}^{5}}+{{K}_{2}}\ln x\cdot {{x}^{5}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 3.12)$$
 * }
 * }

(2) Solve Euler L2-ODE-CC

 * Euler L2-ODE-CC:

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

 * $$\displaystyle {{b}_{2}}{{y}^{''}}+{{b}_{1}}{{y}^{'}}+{{b}_{0}}y=0$$
 * $$\displaystyle (Eq. 3.13)$$
 * }
 * }
 * }

 * Assume,

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

 * $$\displaystyle y={{e}^{rx}}$$
 * }
 * }
 * }

 * Taking derivative:

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

 * $$\displaystyle {{y}^{'}}=r{{e}^{rx}}$$
 * }
 * }
 * }

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

 * $$\displaystyle {{y}^{''}}={{r}^{2}}{{e}^{rx}}$$
 * }
 * }
 * }

Substituting the above three equations in $$\displaystyle Eq. 3.13$$:

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

 * $$\displaystyle {{b}_{2}}{{r}^{2}}+{{b}_{1}}r+{{b}_{0}}=0$$
 * $$\displaystyle (Eq. 3.14)$$
 * }
 * }
 * }

 * Comparing $$\displaystyle Eq. 3.14$$ with $$\displaystyle Eq. 3.4$$ give coefficients:

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

\begin{array}{*{20}l} {{b}_{2}}=1 & {{b}_{1}}=-10 & {{b}_{0}}=25\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Now $$\displaystyle Eq. 3.13$$ becomes:

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

 * $$\displaystyle {{y}^{''}}-10{{y}^{'}}+25y=0$$
 * $$\displaystyle (Eq. 3.15)$$
 * }
 * }
 * }

 * First homogenous solution is:

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

 * $$\displaystyle {{y}_{1}}={{e}^{5x}}$$
 * $$\displaystyle (Eq. 3.16)$$
 * }
 * }
 * }

 * Second homogenous solution is:

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

 * $$\displaystyle {{y}_{2}}=U\left( x \right){{e}^{5x}}$$
 *
$$\displaystyle (Eq. 3.17)$$
 * }
 * }
 * }

 * taking derivative of $$\displaystyle Eq. 3.17$$:

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

 * $$\displaystyle {{y}_{2}}^{'}={{U}^{'}}\left( x \right){{e}^{5x}}+5U\left( x \right){{e}^{5x}}$$
 * }
 * }
 * }

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

 * $$\displaystyle {{y}_{2}}^{}={{U}^{}}\left( x \right){{e}^{5x}}+10{{U}^{'}}\left( x \right){{e}^{5x}}+25U\left( x \right){{e}^{5x}}$$
 * }
 * }
 * }

Plugging the above three equations into $$\displaystyle Eq. 3.15$$ and dropping the terms with $$\displaystyle U\left( x \right)$$ as in the first case, results in:

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

 * $$\displaystyle {{U}^{''}}\left( x \right){{e}^{5x}}=0$$
 * }
 * }
 * }

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

 * $$\displaystyle \Rightarrow {{U}^{''}}\left( x \right)=0$$
 * }
 * }
 * }

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

 * $$\displaystyle \Rightarrow {{U}^{'}}\left( x \right)={{C}_{1}}$$
 * }
 * }
 * }

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

 * $$\displaystyle \Rightarrow U\left( x \right)={{C}_{1}}x+{{C}_{2}}$$
 * }
 * }
 * }

 * Therefore second homogenous solution is;

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

 * $$\displaystyle {{y}_{2}}=\left( {{C}_{1}}x+{{C}_{2}} \right){{e}^{5x}}$$
 *
$$\displaystyle (Eq. 3.18)$$
 * }
 * }
 * }

 * General Homogenous solution

 * General homogenous solution is given by

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

 * $$\displaystyle y={{k}_{1}}{{y}_{1}}+{{k}_{2}}{{y}_{2}}$$
 * }
 * }
 * }

 * Substituting the $$\displaystyle Eq. 3.16$$ and $$\displaystyle Eq. 3.18$$ yields:

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

 * $$y={{k}_{1}}{{e}^{5x}}+{{k}_{2}}\left( {{C}_{1}}x+{{C}_{2}} \right){{e}^{5x}}$$
 * }
 * }
 * }

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

 * $$y=\left( {{k}_{1}}+{{C}_{2}} \right){{e}^{5x}}+{{k}_{2}}{{C}_{1}}x\cdot {{e}^{5x}}$$
 * }
 * }
 * }

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

$$\displaystyle y={{K}_{1}}{{e}^{5x}}+{{K}_{2}}x{{e}^{5x}}$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 3.19)$$
 * }
 * }

= Problem 5 - Solution of Single Degree of Freedom Dynamic System =

Statement

 * Consider the Forced Single Degree of Freedom System shown below:

 * [[Image:Sudheesh hw5.jpg|center|650px]]

 * Equation of Motion of the system is given by

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

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

 * Where, $$\displaystyle{{a}_{2}}=m$$, $$\displaystyle{{a}_{1}}=c$$ and $$\displaystyle {{a}_{0}}=k$$

 * $$\displaystyle (Eq. 5.1)$$ is a $$\displaystyle Nonhomogenous-L2-ODE-CC$$

(1.1)Find PDEs for Integrating Factor $$\displaystyle h\left( t,y \right)$$

 * It could be proved that the $$\displaystyle Eq.1$$ in the given form is not exact. But it can be made exact using Integrating Factor. For that reason multiply $$\displaystyle Eq.5.1$$ by integrating factor $$\displaystyle h\left( t,y \right)$$.

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

 * $$h\left( t,y \right)\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=h\left( t,y \right)f\left( t \right)$$
 *
$$\displaystyle (Eq. 5.2.1)$$
 * }
 * }
 * }

 * Consider the LHS of the above equation and which is in the form as shown below:

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

 * $$h\left( t,y \right){{a}_{2}}{{y}^{}}+h\left( t,y \right)\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=f\left( t,y,p \right){{y}^{}}+g\left( t,y,p \right)$$
 * }
 * }
 * }

 * Where,

 * $$\displaystyle f\left( t,y,p \right)=h\left( t,y \right){{a}_{2}}$$
 * $$\displaystyle g\left( t,y,p \right)=h\left( t,y \right)\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)$$
 * $$\displaystyle p={{y}^{'}}$$

 * Now taking partial derivatives,

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

\begin{array}{*{20}l} {{f}_{tt}}={{h}_{tt}}{{a}_{2}} & {{f}_{ty}}={{h}_{ty}}{{a}_{2}}& {{f}_{y}}={{h}_{y}}{{a}_{2}}\\ {{f}_{yy}}={{h}_{yy}}{{a}_{2}} & {{f}_{tp}}=0 & {{f}_{yp}}=0 \\ {{g}_{tp}}={{h}_{t}}{{a}_{1}}& {{g}_{y}}={{h}_{y}}\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)+h{{a}_{0}} \\ {{g}_{yp}}={{h}_{y}}{{a}_{1}} & {{f}_{tp}}=0 & {{g}_{pp}}=0 \\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * Now consider the two relations of second conditions of exactness [[media:2010_09_23_14_52_54.djvu|Eq.1&2,page3,Mtg 15 (b)]]

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

 * $$\displaystyle {{f}_{tt}}+2p{{f}_{ty}}+{{p}^{2}}{{f}_{yy}}={{g}_{tp}}+p{{g}_{yp}}-{{g}_{y}}$$
 *
$$\displaystyle (Eq. 5.2.2)$$
 * }
 * }
 * }

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

 * $$\displaystyle {{f}_{tp}}+p{{f}_{yp}}+2{{f}_{y}}={{g}_{pp}}$$
 *
$$\displaystyle (Eq. 5.2.3)$$
 * }
 * }
 * }

 * First substitute partial derivatives in $$\displaystyle Eq. 5.2.3$$

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

 * $$\displaystyle 0+0+2{{h}_{y}}{{a}_{2}}=0$$
 * }
 * }
 * }

 * The above condiiton is valid only if $$\displaystyle {{h}_{y}}=0$$ implies $$\displaystyle h$$ is independent of $$\displaystyle y$$. that is

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

 * $$h\left( t,y \right)=h\left( t \right)$$
 * }
 * }
 * }

 * Now substitute partial derivatives in $$\displaystyle Eq. 5.2.2$$

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

 * $${{h}_{tt}}{{a}_{2}}+\underbrace{2p{{h}_{ty}}{{a}_{2}}}_{0}+\underbrace_{0}={{h}_{t}}{{a}_{1}}+\underbrace{p{{h}_{y}}{{a}_{1}}}_{0}-\underbrace{{{h}_{y}}\left( {{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)-}_{0}h{{a}_{0}}$$
 * }
 * }
 * }

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

$$\Rightarrow {{a}_{2}}{{h}_{tt}}-{{a}_{1}}{{h}_{t}}+{{a}_{0}}h=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.2.4)
 * }
 * }

 * The above equation is the PDE for $$\displaystyle h$$

Given

 * Consider trial solution for Integrating Factor,

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

$$
 * $$\displaystyle h\left( t \right)={{e}^{\alpha t}}$$
 *
$$\displaystyle (Eq.5.3.1)
 *
$$\displaystyle (Eq.5.3.1)
 * }
 * where, $$\displaystyle \alpha $$, is unknown, which has to be determined.
 * where, $$\displaystyle \alpha $$, is unknown, which has to be determined.

 * Multiplying $$\displaystyle Eq.5.1 $$ by $$\displaystyle Eq.5.3.1 $$ implies:

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

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

 * Integrating both sides gives:

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

.
 * $$\int{{{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)dt}=\int{{{e}^{\alpha t}}f\left( t \right)dt}$$
 *
$$\displaystyle (Eq. 5.3.2)$$
 * }
 * }
 * }

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

\parallel \\ 0 \end{smallmatrix}}{{y}^{''}}+{{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right)$$ $$
 * $$\int{{{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)dt}={{e}^{\alpha t}}\left( \underbrace_{\begin{smallmatrix}
 * $$\int{{{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)dt}={{e}^{\alpha t}}\left( \underbrace_{\begin{smallmatrix}
 *
$$\displaystyle (Eq. 5.3.3)
 * }
 * }

====(2.1)Find $$\displaystyle{{\overset{-}{\mathop{a}}\,}_{1}}$$, $$\displaystyle{{\overset{-}{\mathop{a}}\,}_{0}}$$ in terms of $$\displaystyle{{a}_{2}}$$, $$\displaystyle{{a}_{1}}$$, $$\displaystyle{{a}_{0}}$$====

 * Differentiating both sides of $$\displaystyle Eq.5.3.3 $$

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

 * $$\displaystyle {{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=\frac{d}{dt}\left[ {{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right) \right]$$
 * }
 * }
 * }

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

 * $${{e}^{\alpha t}}\left( {{a}_{2}}{{y}^{}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y \right)=\alpha {{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right)+{{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{}}+{{\overset{-}{\mathop{a}}\,}_{0}}{{y}^{'}} \right)$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{a}_{2}}{{y}^{}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y={{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{}}+\left( \alpha {{\overset{-}{\mathop{a}}\,}_{1}}+{{\overset{-}{\mathop{a}}\,}_{0}} \right){{y}^{'}}+\alpha {{\overset{-}{\mathop{a}}\,}_{0}}y$$
 * }
 * }
 * }

 * Now comparing the coefficents on both sides,

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

$${{\overset{-}{\mathop{a}}\,}_{1}}={{a}_{2}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.4)
 * }
 * }

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

 * $$\alpha {{\overset{-}{\mathop{a}}\,}_{1}}+{{\overset{-}{\mathop{a}}\,}_{0}}={{a}_{1}}$$
 * }
 * }
 * }

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

$$\Rightarrow {{\overset{-}{\mathop{a}}\,}_{0}}={{a}_{1}}-\alpha {{\overset{-}{\mathop{a}}\,}_{1}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.5)
 * }
 * }

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

 * $$\alpha {{\overset{-}{\mathop{a}}\,}_{0}}={{a}_{0}}$$
 * }
 * }
 * }

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

$$\Rightarrow {{\overset{-}{\mathop{a}}\,}_{0}}=\frac{\alpha }$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.6)
 * }
 * }

(2.2)Find Quadratic equation for $$\displaystyle\alpha $$

 * $$\displaystyle Eq.5.3.5$$ and $$\displaystyle Eq.5.3.6$$ implies:

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

 * $${{a}_{1}}-\alpha {{\overset{-}{\mathop{a}}\,}_{1}}=\frac{\alpha }$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{a}_{1}}\alpha -{{\overset{-}{\mathop{a}}\,}_{1}}{{\alpha }^{2}}={{a}_{0}}$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{\overset{-}{\mathop{a}}\,}_{1}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0$$
 * }
 * }
 * }

.
 * Now substituting $$\displaystyle Eq.5.3.4$$ into the above equation implies:

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

$$\displaystyle {{a}_{2}}{{\alpha }^{2}}-{{a}_{1}}\alpha +{{a}_{0}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.7)
 * }
 * }

.

(2.3)Find Reduced order equation

 * $$\displaystyle Eq.5.3.3$$ and $$\displaystyle Eq.5.3.4$$ implies:

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

 * $${{e}^{\alpha t}}\left( {{\overset{-}{\mathop{a}}\,}_{1}}{{y}^{'}}+{{\overset{-}{\mathop{a}}\,}_{0}}y \right)=\int{{{e}^{\alpha t}}f\left( t \right)dt}$$
 * }
 * }
 * }

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

$$
 * $$\Rightarrow {{y}^{'}}+\fracy=\underbrace{\frac\int{{{e}^{\alpha t}}f\left( t \right)dt}}_{F\left( t \right)}$$
 *
$$\displaystyle (Eq. 5.3.8)
 *
$$\displaystyle (Eq. 5.3.8)
 * }
 * }

(2.4)Use IFM to find $$\displaystyle y$$

 * Since $$\displaystyle Eq.5.3.8 $$ is $$\displaystyle L1-ODE-CC$$,the Integrating Factor is given by,

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

$$
 * $$h\left( t \right)={{e}^{\int{\beta dt}}}={{e}^{\beta t}}$$
 *
$$\displaystyle (Eq. 5.3.9)
 *
$$\displaystyle (Eq. 5.3.9)
 * }
 * }

 * where, $$\displaystyle \beta =\frac$$

 * Now $$\displaystyle y\left( t \right)$$ is,

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

 * $$y\left( t \right)=\frac{1}{h\left( t \right)}\int{h\left( t \right)F\left( t \right)dt}$$
 * }
 * }
 * }

 * Substituting $$\displaystyle h\left( t \right)$$ and $$\displaystyle F\left( t \right)$$ in the above equation implies:

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

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

Combine constants to form $$\displaystyle {{C}_{1}}$$ and $$\displaystyle {{C}_{2}}$$,

 * {| 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:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.10) $$
 * }
 * }

(2.5)Show $$\displaystyle\alpha \beta =\frac$$ and $$\displaystyle\alpha +\beta =\frac$$

 * We know $$\displaystyle\beta =\frac$$ and $$\displaystyle\alpha =\frac$$

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

$$\alpha \beta =\frac\cdot \frac=\frac=\frac$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.11)
 * }
 * }

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

 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\alpha +\beta =\frac+\frac=\frac{{{a}_{0}}{{\overset{-}{\mathop{a}}\,}_{1}}+{{\overset{-}{\mathop{a}}\,}_{0}}^{2}}=\frac{{{a}_{0}}{{a}_{2}}+{{\overset{-}{\mathop{a}}\,}_{0}}\left( \frac{\alpha } \right)}{\left( \frac{\alpha } \right){{a}_{2}}}=\frac{\alpha {{a}_{2}}+{{\overset{-}{\mathop{a}}\,}_{0}}}=\frac$$ $$
 *
$$\displaystyle (Eq. 5.3.12)
 * }
 * }

(2.6)Deduce expression for particular solution $$\displaystyle{{y}_{p}}$$ for general excitation $$\displaystyle f\left( t \right)$$

 * We can see from the solution for $$\displaystyle y,(Eq.5.3.10)$$ that it is a linear combination of homogenous solutions and particular solution

 * Where, homogenous solutions are:

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

 * $${{y}_{H}}^{1}={{e}^{-\alpha t}}$$
 * }
 * }
 * }

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

 * $${{y}_{H}}^{2}={{e}^{-\beta t}}$$
 * }
 * }
 * }

 * Particular solution is,

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

$${{y}_{P}}=\frac\int{{{e}^{(\beta -\alpha )t}}\left( \int{{{e}^{\alpha t}}f(t)dt} \right)dt}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.13)
 * }
 * }

(2.7)Verify particular solution with table [[media:egm6321.f09.mtg20.djvu|Mtg 20, Fall 2009]]for $$\displaystyle f\left( t \right)={{t}^{2}}$$

 * Substitute $$\displaystyle f\left( t \right)={{t}^{2}}$$ in $$\displaystyle (Eq.5.3.13)$$:

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

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

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

 * $${{y}_{P}}=\frac{{{\alpha }^{2}}{{\beta }^{2}}{{t}^{2}}-2\left( \alpha {{\beta }^{2}}+{{\alpha }^{2}}\beta \right)t+2\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta  \right)}$$
 * }
 * }
 * }

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

 * $$=\frac-\frac{2\left( \alpha {{\beta }^{2}}+{{\alpha }^{2}}\beta \right)t}+\frac{2\left( {{\left( \alpha +\beta  \right)}^{2}}-\alpha \beta  \right)}$$
 * }
 * }
 * }

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

.
 * $$=\frac{{{a}_{2}}\alpha \beta }-\frac{2\left( \alpha +\beta \right)t}+\frac{2\left( {{\left( \alpha +\beta  \right)}^{2}}-\alpha \beta  \right)}$$
 * }
 * }
 * }

 * Now subtituting $$\displaystyle(Eq.5.3.11)$$ and $$\displaystyle(Eq.5.3.12))$$ for$$\displaystyle\left( \alpha +\beta \right)$$ and $$\displaystyle\left( \alpha \beta  \right)$$ respectively in the above equation, implies:

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

$$
 * $${{y}_{P}}=\frac-\frac{2{{a}_{1}}t}{{{a}_{0}}^{2}}+2\left( \frac{{{a}_{1}}^{2}}{{{a}_{0}}^{3}}-\frac{{{a}_{0}}^{2}} \right)$$
 *
$$\displaystyle (Eq. 5.3.14)
 *
$$\displaystyle (Eq. 5.3.14)
 * }
 * }

Using the table for particular solution in Mtg 20, fall 2009, consider a trial solution,

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

$$
 * $$\displaystyle {{y}_{P}}=A{{t}^{2}}+Bt+C$$
 *
$$\displaystyle (Eq. 5.3.15)
 *
$$\displaystyle (Eq. 5.3.15)
 * }
 * }

 * where A,B,and C are unknown cofficients to be determined.

 * Taking derivative,

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

 * $$\displaystyle {{y}_{P}}^{'}=2At+B$$
 * }
 * }
 * }

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

 * $$\displaystyle {{y}_{P}}^{''}=2A$$
 * }
 * }
 * }

 * Substitute above equations in $$\displaystyle(Eq.5.1)$$

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

 * $$\displaystyle 2A{{a}_{2}}+\left( 2At+B \right){{a}_{1}}+\left( A{{t}^{2}}+Bt+C \right){{a}_{0}}={{t}^{2}}$$
 * }
 * }
 * }

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

 * $$\displaystyle {{a}_{0}}A{{t}^{2}}+\left( 2{{a}_{1}}A+{{a}_{0}}B \right)t+\left( 2{{a}_{2}}A+{{a}_{1}}B+{{a}_{0}}C \right)={{t}^{2}}$$
 * }
 * }
 * }

 * Equating coefficients on both side of equations gives:

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

 * $$\displaystyle A=\frac{1}$$
 * }
 * }
 * }

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

 * $$\displaystyle B=\frac{-2{{a}_{1}}}{{{a}_{0}}^{2}}$$
 * }
 * }
 * }

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

 * $$C=2\left( \frac{{{a}_{1}}^{2}}{{{a}_{0}}^{3}}-\frac{{{a}_{0}}^{2}} \right)$$
 * }
 * }
 * }

Plugg in A,B and C in $$\displaystyle(Eq.5.3.15)$$ gives

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

$$
 * $${{y}_{P}}=\frac-\frac{2{{a}_{1}}t}{{{a}_{0}}^{2}}+2\left( \frac{{{a}_{1}}^{2}}{{{a}_{0}}^{3}}-\frac{{{a}_{0}}^{2}} \right)$$
 *
$$\displaystyle (Eq. 5.3.16)
 *
$$\displaystyle (Eq. 5.3.16)
 * }
 * }

 * Evident from $$\displaystyle(Eq.5.3.14)$$ and $$\displaystyle(Eq.5.3.16)$$ that particular solution obtained using the deduced expression and table are same.

(2.8)Solve L2-ODE-CC with the forcing function $$\displaystyle f\left( t \right)={{e}^{-{{t}^{2}}}}$$ (Gaussian distribution)

 * Homogenous solutions will be same as determined earlier.
 * To determine particular solution, substitute $$\displaystyle f\left( t \right)={{e}^{-{{t}^{2}}}}$$ in $$\displaystyle (Eq.5.3.13)$$:

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

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

 * Simplifying,

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

$$
 * $${{y}_{P}}=\frac{\sqrt{\pi }\cdot {{e}^{\frac{4}}}}{2\left( \beta -\alpha \right){{a}_{2}}}\left( {{e}^{-\alpha t}}erf\left( t-\frac{\alpha }{2} \right)-erf\left( t-\frac{\beta }{2} \right) \right)$$
 *
$$\displaystyle (Eq. 5.3.17)
 *
$$\displaystyle (Eq. 5.3.17)
 * }
 * }

 * Now $$\displaystyle y$$ is obtained by combining homogenous solutions and particular solution ($$\displaystyle Eq.5.3.17$$)

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

$$y={{C}_{1}}{{e}^{-\alpha t}}+{{C}_{2}}{{e}^{-\beta t}}+\frac{\sqrt{\pi }\cdot {{e}^{\frac{4}}}}{2\left( \beta -\alpha \right){{a}_{2}}}\left( {{e}^{-\alpha t}}erf\left( t-\frac{\alpha }{2} \right)-erf\left( t-\frac{\beta }{2} \right) \right)$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\displaystyle (Eq. 5.3.18)
 * }
 * }

Find coefficients $$\displaystyle {{a}_{0}}$$, $$\displaystyle {{a}_{1}}$$, $$\displaystyle {{a}_{2}}$$

 * $$\displaystyle(a) Eq.5.1$$ accept a characteristic equation $$\displaystyle\left( r+1 \right)\left( r-2 \right)=0$$

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

 * $$\left( r+1 \right)\left( r-2 \right)=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{r}^{2}}-r-2=0$$
 * }
 * }
 * }

.
 * So the $$\displaystyle L2-ODE $$ is given by

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

 * $$\displaystyle {{y}^{''}}-y-2y=0$$
 * }
 * }
 * }

.
 * The coefficients are:

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

\begin{array}{*{20}l} {{a}_{2}}=1 & {{a}_{1}}=-1 & {{a}_{0}}=-2\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

.
 * $$\displaystyle(b) Eq.5.1$$ accept a characteristic equation $$\displaystyle{{\left( r-4 \right)}^{2}}=0$$

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

 * $${{\left( r-4 \right)}^{2}}=0$$
 * }
 * }
 * }

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

 * $$\Rightarrow {{r}^{2}}-8r+16=0$$
 * }
 * }
 * }

.
 * Now the $$\displaystyle L2-ODE $$ is given by

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

 * $$\displaystyle {{y}^{''}}-8y+16y=0$$
 * }
 * }
 * }

.
 * The coefficients are:

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

\begin{array}{*{20}l} {{a}_{2}}=1 & {{a}_{1}}=-8 & {{a}_{0}}=16\\ \end{array} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }


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


text:
User:Achayan

Rahul Johnson Palakkappilly (Payapilly)

title: User:Achayan


text:
User:Egm6321.f10.team4/HW5

= Problem 1 -  Solving multiple y derivative wrt x =

From Meeting 26, p 26-2.

Given
The multiple y derivatives wrt x that is given from F&9
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y_{x}=e^{-t}y_{t} $$ $$
 * $$\displaystyle (Eq. 1-3)
 * }
 * }

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

$$\displaystyle y_{xx}=e^{-2t}(y_{tt}-y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-4)
 * }
 * }

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

$$\displaystyle y_{xx}=e^{-3t}(y_{ttt}-3y_{tt}+2y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-5)
 * }
 * }

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

$$\displaystyle y_{xxxx}=e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}) $$ $$
 * $$\displaystyle (Eq. 1-6)
 * }
 * }

Two approaches to obtain y derivatives
method 1
 * $$\displaystyle \underbrace{\frac{d}{dt} y(x(t))}_{y_t=} = \underbrace{\frac{dx}{dt}}_{e^t} \cdot \underbrace{\frac{dy}{dx}}_{y_x} $$

rewriting the equation wrt y derivate with respect to x,
 * $$\displaystyle y_x = y_{t} \cdot e^{-t}$$

method 2
 * $$\displaystyle y_{x}= \underbrace{\frac{dy}{dx}}_{y_x=} = \underbrace{\frac{dy}{dt}}_{y_t} \cdot \underbrace{\frac{dt}{dx}}_{\frac{dx}{dt}=e^{-t}}$$

 * $$\displaystyle y_x = y_{t} \cdot e^{-t}$$

Find
Part a find $$\displaystyle y_{xxxxx}$$ in terms of derivative of y with respect to t

Part b Plug (eq 1.3) and (eq 1.4) into (eq 1.1) to obtain $$\displaystyle y_{tt} - 3y_{t} + 2y = 0$$

Solution for the part a
Solving for the 5th Derivative

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

$$\displaystyle y_{xxxxx} = \frac{d}{dx} (y_{xxxx}) $$ $$
 * $$\displaystyle (Eq. 1-7)
 * }
 * }

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

$$\displaystyle = \frac{dt}{dx} \cdot \frac{d}{dt} (y_{xxxx}) $$ $$
 * $$\displaystyle (Eq. 1-8)
 * }
 * }

Substituting (Eq 1-6) into the previous (Eq 1-8) yields,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y_{xxxxx} = \underbrace {\frac{dt}{dx}}_{e^{-t}} \cdot \frac{d}{dt} \left[e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) \right] $$ $$
 * $$\displaystyle (Eq. 1-9)
 * }
 * }

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

$$\displaystyle = e^{-t} \left[-4e^{-4t}(y_{tttt} - 6y_{ttt} + 11y_{tt} - 6y_t) + e^{-4t}(y_{ttttt} - 6y_{tttt} + 11y_{ttt} - 6y_{tt}) \right] $$ $$
 * $$\displaystyle (Eq. 1-10)
 * }
 * }

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

$$\displaystyle = e^{-5t} \left[-4y_{tttt} + 24y_{ttt} -44y_{tt} + 24y_{t} + y_{ttttt} - 6y_{tttt} + 11y_{ttt} - 6y_{tt}\right] $$ $$
 * $$\displaystyle (Eq. 1-11)
 * }
 * }

Collecting terms with reduction order of y derivate wrt t yields,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle y_{xxxxx} = e^{-5t}(y_{ttttt} - 10y_{tttt} + 35y_{ttt} - 50y_{tt} + 24y_{t}) $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * $$\displaystyle (Eq. 1-12)
 * }
 * }

Solution for the part b
Plugging the obtained y derivatives into the given equation to obtain the following:

First, taking a look at what we have to use,
 * {| style="width:100%" border="0" align="left"

$$\displaystyle x=e^t $$ $$
 * $$\displaystyle
 * }
 * }

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

$$\displaystyle y_{x}=e^{-t}y_{t} $$ $$
 * $$\displaystyle (Eq. 1-14)
 * }
 * }

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

$$\displaystyle y_{xx}=e^{-2t}(y_{tt}-y_{t}) $$ $$
 * $$\displaystyle
 * }
 * }

Substituting what we have into the coresponding variables yields,

reorganizing the equation by reduction order, we can solve the desired equation

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

 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle y_{tt} -2y_{t} + 2y = 0 $$ $$
 * $$\displaystyle (Eq. 1-17)
 * }
 * }

Egm6321.f10.team4.Yoon 00:00, 30 October 2010 (UTC) - Primary Author 68.17.105.125 01:01, 3 November 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 2: Method of Trial Solutions = From Meeting 26, p 26-3.

Given
Solve the given L2_ODE_VC((4)P25.3 using Method 2 i.e.) Trial Solution $$\displaystyle y=x^r$$ with the boundary condition as followed

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

x^2y'' - 2xy' + 2y = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

The given boundary conditions
 * {| style="width:100%" border="0" align="left"

y(_{x=}1) = -2, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-2)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y(_{x=}2) = 5. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-3)
 * }
 * }

required

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

y(x) = \sum^{n}_{i=1} C_ix^{i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Find
Part a Solve (Eq. 2-1) using method2(trial solution) to make it have the shown format below.

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

y(x) = C_1x^{r_1} + C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-4)
 * }
 * }

Part b Plot the homogeneous equation

Solution for Part a
First, taking the given trial solution y's 1st and 2nd deriviates,
 * {| style="width:100%" border="0" align="left"

y=x^r $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-5)
 * }
 * }

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

y'=rx^{r-1} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-6)
 * }
 * }

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

y''=r(r-1)x^{r-2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-7)
 * }
 * }

Substiuting (Eq 2-5), (Eq 2-6), and (Eq 2-7) into the given L2_ODE(Eq 2-1) yields
 * {| style="width:100%" border="0" align="left"

\cancelto{_1}{x^2} r \cdot x^{r- \cancelto{1}{2}}(r-1) - 2 \cancelto{_1}{x} \cdot x^{r-\cancelto{1}{1}}r + 2x^r = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-8)
 * }
 * }

Simplifying and eliminating non-zero constant ahead of polynomial,
 * {| style="width:100%" border="0" align="left"

x^r \left( r^2 -r - 2r + 2 \right) = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-9)
 * }
 * }

The charactoristic Equation of the ODE is
 * {| style="width:100%" border="0" align="left"

r^2- 3r + 2 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-10)
 * }
 * }

Solving for the unknown 'r's yields
 * {| style="width:100%" border="0" align="left"

r_1 = 1, r_2 = 2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-11)
 * }
 * }

The solution takes the form of a homogeneous solution, so substituting the rs into the equation,
 * {| style="width:100%" border="0" align="left"

y(x) = C_1x^{r_1}+C_2x^{r_2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-12)
 * }
 * }

Substituting the roots of the polynomial into the homogeneous format,
 * {| style="width:100%" border="0" align="left"

y(x) = C_1x^{1}+C_2x^{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-13)
 * }
 * }

plugging the given boundary conditions yields
 * {| style="width:100%" border="0" align="left"

y(_{x1=}1)= C_1 \cdot (_{x1=} 1)^{1}+C_2 \cdot (_{x1=}1)^{2} = -2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

y(_{x2=}2)= C_1 \cdot (_{x2=}2)^{1}+C_2 \cdot (_{x2=}2)^{2} = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-15)
 * }
 * }

Reorganizing the equation wrt $$\displaystyle C_1$$ and $$\displaystyle C_2$$,
 * {| style="width:100%" border="0" align="left"

C_1 + C_2 = -2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-16)
 * }
 * }

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

2C_1 + 4C_2 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-17)
 * }
 * }

Solving the above equations, we obtain $$\displaystyle C_1$$ and $$\displaystyle C_2$$,
 * {| style="width:100%" border="0" align="left"

C_1 = - \frac{13}{2}; C_2= \frac{9}{2} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2-18)
 * }
 * }

Therefore substiting the results from (Eq. 2-18) into the homogeneous solution format(Eq. 2-13)
 * {| style="width:100%" border="0" align="left"

 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |

$$ \displaystyle y(x) = - \frac{13}{2}x + \frac{9}{2}x^2 $$ $$
 *
$$\displaystyle (Eq. 2-19)
 * }
 * }

Solution for Part b
Plot the obtained equation,
 * {| style="width:100%" border="0" align="left"

y(x) = - \frac{13}{2}x + \frac{9}{2}x^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Egm6321.f10.team4.Yoon 00:00, 30 October 2010 (UTC) - Primary Author 68.17.105.125 01:00, 3 November 2010 (UTC)Egm6321.f10.team4.osentowski - Editor

= Problem 3 - Characteristic Equations for Euler L2-ODEs = From Meeting 27, p 27-1.

Part 1
Consider the Euler L2-ODE-VC:

Part 2
Consider the Euler L2-ODE-CC:

Part 1

 * 1) For Eq. 3-1, find $$\displaystyle a_2 $$, $$\displaystyle a_1 $$, and $$\displaystyle a_0 $$ such that $$\displaystyle \left( r-\lambda \right)^2=0 $$.
 * 2) Find the first homogeneous solution: $$\displaystyle y_1(x)=x^{\lambda} $$
 * 3) Find the second homogeneous solution: $$\displaystyle y_2(x)=u(x)x^{\lambda}=u(x)y_1(x) $$, where $$\displaystyle u(x) $$ is an unknown function of $$\displaystyle x $$. Also find the differential equation governing $$\displaystyle u(x) $$ and solve for $$\displaystyle u(x) $$, then $$\displaystyle y_2(x) $$.
 * 4) Find the general homogeneous solution.

Part 2

 * 1) For Eq. 3-2, repeat steps 1-4 in Part 1.

Part 1.1
Beginning with Method 2 (trial solution) from [http://upload.wikimedia.org/wikiversity/en/9/9d/2010_10_14_14_56_10.djvu Mtg. 25] p. 25-3, we will assume:

Thus the first and second derivatives of $$\displaystyle y $$ w.r.t. $$\displaystyle x $$ are:

Substituting into Eq. 3-1, we get:

Divide by $$\displaystyle x^r $$ to get:

We are trying to find the coefficients $$\displaystyle a_2 $$, $$\displaystyle a_1 $$, and $$\displaystyle a_0 $$ such that $$\displaystyle \left( r-\lambda \right)^2=0 $$, so we will set Eq. 3-6 equal to $$\displaystyle \left( r-\lambda \right)^2=0 $$:

We can see by inspection that this condition (i.e. Eq. 3-7) is satisfied if:

Part 1.2
The first homogeneous solution is $$\displaystyle y_1(x)=x^{\lambda} $$. We already know that $$\displaystyle (r-\lambda)^2=0 $$ (by definition), so we can find the roots immediately:

Thus we can write the first homogeneous solution as:

Part 1.3
The second homogeneous solution and its first two derivatives are:

Substituting Eqs. 3-8 into Eq. 3-1, we get:

Now, substitute Eqs. 3-11 into Eq. 3-12:

Divide through by $$\displaystyle x^{\lambda}$$: Distribute the terms outside the brackets: Now, rearrange and factor like terms: This reduces to the simple expression: Next, divide through by $$\displaystyle x $$: We can further simplify Eq. 3-18 by reducing its order if we define: Thus the differential equation governing $$\displaystyle u(x) $$ in the second homogeneous equation is: where $$\displaystyle v := u'(x) $$. In order to solve for $$\displaystyle v(x) $$, we will once again assume the form of: Substituting these into Eq. 3-20, we get: If we divide through by $$\displaystyle x^s $$, we obtain: Now, let's substitute this back into the expression for $$\displaystyle v(x)$$ in Eq. 3-21: We can now find $$\displaystyle u(x) $$ from:

Finally, substituting Eq. 3-25 into Eq. 3-11, we obtain:

Part 1.4
We can now write the general form of $$\displaystyle y(x) $$ as: Using Eqs. 3-10 and 3-26, and replacing the coefficient in Eq. 3-10, Eq. 3-27 becomes: The general homogeneous solution is:

 * Note: It is interesting that the form of the general homogeneous solution (Eq. 3-29) is identical to that of the second homogeneous solution (Eq. 3-26), despite having added the first homogeneous solution (Eq. 3-10). The difference is only in the magnitude of the constant.

Part 2
For the reader's convenience, Eq. 3-2 is reproduced below:

Part 2.1
Let us assume that $$\displaystyle y $$ has the form:

Its first and second derivatives are thus:

When we substitute these into Eq. 3-2, we get:

Divide through by $$\displaystyle e^{rx}b_2 $$:

Recall that we are trying to find the coefficients $$\displaystyle b_2 $$, $$\displaystyle b_1 $$, and $$\displaystyle b_0 $$ such that $$\displaystyle \left( r-\lambda \right)^2=0 $$, so we will set Eq. 3-32 equal to $$\displaystyle \left( r-\lambda \right)^2=0 $$:

By inspection, we see that Eq 3-33 is satisfied if the coefficients are:

Part 2.2
The first homogeneous solution is $$\displaystyle y_1(x)=e^{\lambda x} $$. We also know that, by definition, $$\displaystyle (r-\lambda)^2=0 $$, so we can find the roots immediately:

Thus we can write the first homogeneous solution as:

Part 2.3
In order to find the second homogeneous solution, we start with:

Its first two derivatives are :

Divide Eq. 3-2 by $$\displaystyle b_2 $$ to make it easier to work with:

Now we can substitute $$\displaystyle y_2(x) $$ (Eq. 3-37) and its derivatives, along with the b-coefficients (Eqs. 3-34) into Eq. 3-38:

Divide through by $$\displaystyle e^{\lambda x} $$ :

Consolidating the terms in Eq. 3-40, we obtain the differential equation governing $$\displaystyle u(x) $$:

Integrating Eq. 3-41 twice yields the expression for u(x):

Now we will substitute this back into Eq. 3-37 to obtain the second homogeneous equation:

Part 2.4
We can write the general form of $$\displaystyle y(x) $$ as:

Next, we substitute Eqs. 3-36 and 3-43, and replace the coefficient in 3-36 to get:

The general homogeneous solution is thus:

 * Note: Just as in Part 1, we see that the general solution (Eq. 3-45) has the same form as the second homogeneous equation (Eq. 3-43).

Egm6321.f10.team4.ejm 18:27, 1 November 2010 (UTC) Author

= Problem 4 - Particular solution using variation of parameters= From Meeting 27, p 27-2.

Find
The particular solution $$\displaystyle y_P $$ of $$\displaystyle y $$

Solution
In HW2 problem 8 we determined that the homogeneous solution to equation 4-1 was

Taking the derivative of equation 4-2,

where

substituting equation 4-5 into 4-4 yields

Now if we substitute equations 4-2 and 4-6 into equation 4-1

Which can be reduced to

Solving for $$\displaystyle {A(x)}' $$

Integrating both sides of equation 4-7

Substituting equation 4-8 back into equation 4-2

Since we know

$$\displaystyle y_P(x) $$ must be

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

$$\displaystyle y_P(x)= {C_1}{exp{\left [ - \int^{x} a_0(s) ds \right ]}} \int^{x}\left [b(s) exp \left [ \int^{s} a_0(r) dr \right ] \right ] $$ $$
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style="width:45%; padding:10px; border:2px solid #8888aa" |
 * style= |
 *
$$\displaystyle (Eq. 4-12)
 * }
 * }

Egm6321.f10.team4.Auerbach 04:16, 3 November 2010 (UTC) - Primary Author

= Problem 5 - Equation of Motion: Mass Spring Damper System= From Meeting 28, p 28-2 to Meeting 29, p 29-1.

Given
A spring-mass-damper system with an applied force:

Consider the nonhomogeneous L2_ODE_CC:

$${{a}_{2}}{{y}^{''}}+{{a}_{1}}{{y}^{'}}+{{a}_{0}}y=f\left( t \right)$$
$$\displaystyle (Eq. 5-1) $$

where $${a_2}$$ is the mass constant m, $${a_1}$$ is the damping constant c, $${a_0}$$ is the spring constant k, $$f\left ( t \right )$$ is the forcing function

Find
5.1 PDEs for the integrating factor $$\displaystyle h(t,y)$$

5.2 Trial solution for the integrating factor $$\displaystyle h(t,y)$$

 * 5.2.1 $$\displaystyle \bar{a}_1$$ and $$\displaystyle \bar{a}_2$$ in terms of $$\displaystyle a_2,a_1,a_0$$

 * 5.2.2 Quadratic equation for $$\displaystyle \alpha $$

 * 5.2.3 Reduced order equation using integrating factor $$e^{\alpha t}[\bar{a}_{1}y^{'}+\bar{a}_{0}y^{'}] = \int e^{\alpha t} f\left ( t \right )dt$$ $$\bar{a}_{1}y^{'}+\bar{a}_{0}y^{'} = e^{-\alpha t}\int e^{\alpha t} f\left ( t \right )dt$$

 * 5.2.4 By implementing IFM (Integrating Factor Method), find $$y\left ( t \right )$$ for general excitation $$f\left ( t \right )$$

 * 5.2.5 Show $$\displaystyle \alpha \beta =\frac{\alpha _{0}}{\alpha _{2}} $$ and $$\displaystyle \alpha + \beta =\frac{\alpha _{1}}{\alpha _{2}} $$

 * 5.2.6 Deduce an expression for the particular solution $$\displaystyle y_P $$ for a general excitation $$\displaystyle f(t) $$

 * 5.2.7 Verify with the table of particular solutions. Take into consideration that $$f\left ( t \right ) = t^{2}$$

 * 5.2.8 Solve the linear 2nd order ordinary differential equation with constant coefficients (L2_ODE_CC) with $$f(t)=exp(-t^2) $$. Find the coefficients $$\displaystyle a_2,a_1,a_0$$ such that the L2_ODE_CC accepts the following as characteristic equations:

 * 5.2.8.1 $$\displaystyle (r+1)(r-2)=0 $$

 * 5.2.8.2 $$\displaystyle (r-4)^2=0 $$

5.1
The equation of motion for a spring-mass-damper system can be written as:

Using an integrating factor $$\displaystyle h(t,y)$$, equation 5-1 becomes

If we then set $$\displaystyle p=y'$$, equation 5-2 can be written in a form that satisfies the first condition of exactness:

The second condition of exactness requires the following 2 conditions must be met:

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

$$\displaystyle {g_{pp}}=f_{tp}+pf_{yp}+2f_{y} $$ $$
 *
$$\displaystyle (Eq. 5-3)
 * }
 * }

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

$$\displaystyle {f_{tt}}+2pf_{ty}+p^2f_{yy}=g_{tp}+pg_{yp}-g_{y} $$ $$
 *
$$\displaystyle (Eq. 5-4)
 * }
 * }

Taking the partial derivatives in equation 5-3,

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

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

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

$$\displaystyle f_{tp}=0 $$
 * }
 * }

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

$$\displaystyle f f_{yp}=0 $$
 * }
 * }

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

$$\displaystyle f_{y}=h_ya_0 $$
 * }
 * }

Equation 5-3 reduces to

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

$$\displaystyle 0=0+p0+2h_ya_0 $$
 * }
 * }

so,

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

$$\displaystyle h_y=0 $$
 * }
 * }

Taking the partial derivatives in equation 5-4,

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

$$\displaystyle f_{tt}=h_{tt}a_2 $$
 * }
 * }

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

$$\displaystyle f_{ty}=h_{ty}a_2=0 $$
 * }
 * }

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

$$\displaystyle f_{yy}=h_{yy}a_2=0 $$
 * }
 * }

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

$$\displaystyle g_{tp}=h_{t}a_1 $$
 * }
 * }

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

$$\displaystyle g_{yp}=h_{y}a_1=0 $$
 * }
 * }

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

$$\displaystyle g_{y}=h_{y}a_1p+h_{y}a_0y+ha_0-h_yf(t)=ha_0 $$
 * }
 * }

Equation 5-4 reduces to

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

$$\displaystyle h_{tt}a_2+2p0+p^20=h_{t}a_1+p0-ha_0 $$
 * }
 * }

or

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

$$\displaystyle h_{tt}a_2-h_{t}a_1+ha_0=0 $$
 * }
 * }

5.2
Starting off with the trial solution for the intergration factor $$\displaystyle h(t)=exp(\alpha t)$$, equation 5-1 becomes

If we integrate both sides with respect to $$\displaystyle t$$

Assuming the result will be a reduction of order

5.2.1
Differentiating the LHS of equation 5-7 and the LHS of equation 5-6 and comparing:

Examining like $$\displaystyle y''$$ terms:

Thus

Examining like $$\displaystyle y'$$ terms:

Thus

Examining like $$\displaystyle y$$ terms:

Thus

5.2.2
Using equations 5-10 and 5-11

or

5.2.3
Starting with equation 5-7

Dividing both sides by $$\displaystyle \bar{a}_1$$

5.2.4
Using what was previously discussed in [http://upload.wikimedia.org/wikiversity/en/0/00/2010_09_14_15_00_52.djvu Mtg. 10],

where

Thus, using the equation presented in 10-3

5.2.5
From the above definitions of $$ \displaystyle \alpha $$ and $$ \displaystyle \beta $$, and since $$ \displaystyle \bar{a}_{1}=a_2 $$:

5.2.6
The solution to $$\displaystyle y $$ is made up of the particular solution, $$\displaystyle y_P $$, and the homogeneous solutions,$$\displaystyle y_{H1,2} $$. The solution of $$\displaystyle y $$ presented in equation 5-12 can be expanded to:

so

5.2.7
Let's start off by defining $$\displaystyle f\left ( t \right ) = t^{2}$$

substituting earlier definitions of $$\displaystyle \alpha$$ and $$\displaystyle \beta$$, we can reduce to

Using the table for particular solutions:

Substituting back in equation 5-1

Thus,

Which yields the same particular solution as before,

5.2.8
Let's start off by defining $$\displaystyle f\left ( t \right ) = exp(-t^{2})$$

5.2.8.1
Given the characteristic equation

or

so,

5.2.8.2
Given the characteristic equation

or

so,

Egm6321.f10.team4.Corella.RL 20:57, 5 November 2010 (UTC) - Co-author Egm6321.f10.team4.Auerbach 04:16, 3 November 2010 (UTC) - Co-author Egm6321.f10.team4.ejm 20:40, 5 November 2010 (UTC) Editor

= Problem 6 - Expressions for the Particular Solution= From Meeting 30, p 30-1.

Given
From our lecture notes:

$$\displaystyle y_p (x) =  u_1 (x) \int \frac {1}{h (x)} [ \int h (x) f (x) dx ]  dx $$
$$\displaystyle (Eq. 6-1) $$

From the text by King (pg.8, Eqn. 1.6)

$$\displaystyle y_p (x) =  \int^{x} f(s) \frac {u_1 (s) u_2 (x) - u_1 (x) u_2 (s)}{W(s)} ds $$
$$\displaystyle (Eq. 6-2) $$

where $$\displaystyle W(s) $$ is the Wronskian, defined as

$$\displaystyle W(s) =  u_1 u_2'  -  u_2 u_1' $$
$$\displaystyle (Eq. 6-3) $$

Show
that the given expressions (6-1) and (6-2) are equivalent.

Solution
We will begin by expressing $$\displaystyle h(x) $$ in terms of $$\displaystyle u_1 $$ and $$\displaystyle u_2 $$.

$$\displaystyle \frac {1}{h(x)} =  (\frac {u_2}{u_1})' $$
$$\displaystyle (Eq. 6-4) $$

By the quotient rule, this can be rewritten

$$\displaystyle \frac {1}{h(x)} =  \frac {u_1 u_2'  -  u_2 u_1'}{u_1^2} $$
$$\displaystyle (Eq. 6-5) $$

and it follows that

$$\displaystyle h(x) =  \frac {u_1^2}{u_1 u_2'  -  u_2 u_1'} $$
$$\displaystyle (Eq. 6-6) $$

By substituting the expressions from (6-5) and (6-6) into our given expression (6-1)

$$\displaystyle y_p(x) =  u_1 (x) \int \frac {u_1 u_2' - u_2 u_1'}{u_1^2} [\int^{x} f(s) \frac {u_1^2}{u_1 u_2' - u_2 u_1'} ds]  dx $$

where $$\displaystyle (s) $$ is a dummy variable.

Substituting the Wronskian into the second integral gives

$$\displaystyle y_p(x) =  u_1 (x) \int \frac {u_1 u_2' - u_2 u_1'}{u_1^2} [\int^{x} f(s) \frac {u_1^2}{W(s)} ds]  dx $$

Combining the two integrals into one gives

$$\displaystyle y_p(x) =  u_1(x) \int^{x} f(s)  \frac {u_1(s)^2}{W(s)}  \frac {u_1(s) u_2(x)  -  u_2(s) u_(x)}{u_1(s)^2}  ds $$

The primed terms are now expressed in terms of x, and the unprimed terms are expressed in terms of the dummy variable s.

The $$\displaystyle u_1 (s)^2 $$ terms will cancel, leaving the expression

$$\displaystyle y_p(x) = u_1(x) \int^{x} f(s) \frac {u_1(s) u_2(x)  -  u_2(s) u_1(x)}{W(s)}  ds $$

which is our equation (6-2), multiplied by the term $$\displaystyle u_1(x) $$. Had we started with the general solution similar to that as expressed in Meeting 10

$$\displaystyle y(x) =  \frac {1}{h (x)}  \int h (x) f (x) dx  $$

then there would be no additional $$\displaystyle u_1(x) $$ term. In order to get perfect reconciliation between the lecture notes and the text, we would have to modify our problem statement to not include the additional $$\displaystyle u_1(x) $$ term.

128.227.69.48 20:18, 2 November 2010 (UTC) Egm6321.f10.team4.osentowski - Author

= Problem 7 Find two homogeneous solutions= From Meeting 31, p 31-1.

Solution
First start by computing the first and second integrals of $$\displaystyle y $$

Substitute $$\displaystyle y,y', y''$$ into Eq. 8-1 to obtain the following

Divide Eq. 8-4 by $$\displaystyle e^{rx}$$ to obtain the characteristic equation.

We will obtain two roots by solving Eq. 8-5

First root

Second root

The two roots are real and distinct and will generate the following two homogeneous solutions

Egm6321.f10.team4.petralanda.n 18:39, 1 November 2010 (UTC)Naiara Petralanda

= Problem 8 Find a L2_ODE_VC through reverse engineering = From Meeting 31, p 31-3.

Given
Given the following characteristic equation

and the trial solution method form

Find
Using reverse engineering find a L2_ODE_VC that will satisfy Eq. 8-1 using the trial solution method on Eq. 8-2.

Solution
Take the first derivative of Eq. 8-2

Take the second derivative of Eq. 8-2

Generate the equation using preliminary functions $$\displaystyle a_0, a_1, a_2 $$

Divide both sides by $$\displaystyle y $$ and equate it with the given characteristic equation Eq. 8-1

We will obtain three equations based on the power of $$\displaystyle r $$

The first equation will solve for $$\displaystyle a_2 $$

The second equation will yield the value of $$\displaystyle a_1 $$

If we substitute the solution on Eq. 8-8

The third and last equation will allow us to solve for the third variable coefficient

If we insert the solutions shown in Eq. 8-8 and Eq. 8-10 we will get the form of $$\displaystyle a_o $$

If we combine Eq. 8-8, Eq. 8-10 and Eq. 8-12 we can form the L2_ODE_VC

Egm6321.f10.team4.petralanda.n 18:40, 1 November 2010 (UTC)Naiara Petralanda

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title: User:Godian


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User:Egm6321.f10.team2.zou/HW5

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

From the lecture slide Mtg 27-2

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

From the lecture slide Mtg 28 and Mtg 29-1

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

From the lecture slide Mtg 31-4

Solution
= References =

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


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User talk:Godian

 Hello Godian, 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:Godian


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User talk:Bhainbore

 Hello Bhainbore, 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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Nate!
Hay Nate as Bhainbore, I`m Aaqib. Mabuabsdd--Mabuabsdd 12:53, 9 February 2011 (UTC)

title: User talk:Bhainbore


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User:2blackpanthers

My two cats look like two black panthers.

title: User:2blackpanthers


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User talk:2blackpanthers

 Hello 2blackpanthers, 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:2blackpanthers


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Talk:Ideas in Geometry/Instructive examples/Section 3.2

I think you did a very good job explaining how to get the continued fraction expansion of 5/3. The steps made sense and it was easy to follow your solution. I think it would be even easier to see your solution if you put each step on separate lines instead of putting everything on the same line. Good job!

Michelle Hur


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Talk:Conspiracy theory criticism

An Introduction
Welcome!

I am Niko21, your internet guide to the strange and startling world of conspiracy theories. We will delve into an abyss of murder, treachery, and sometimes just plain nonsense. Everything from Paul McCartney to JFK, Pearl Harbor and 9/11. Even Reptilian Aliens posing as the Bush dynasty and the British Royals is not taboo. In fact, that's what we thrive on! So get ready for our first lesson:A Day in Dallas...

See also A Day in Dallas.... --mikeu talk 15:24, 25 December 2015 (UTC)

title: Talk:Conspiracy theory criticism


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User talk:Shnortz

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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, 31 October 2010 (UTC)

title: User talk:Shnortz


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User talk:Vipul Deorani

 Hello Vipul Deorani, 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 07:18, 31 October 2010 (UTC)

title: User talk:Vipul Deorani


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User:EGM6321.f10.team6.yoshioka/HW5

= Problem 3 Solution of the Euler L2_ODE_VC and L2_ODE_CC =

Given
The Euler L2_ODE_VC:

The Euler L2_ODE_CC:

And the characteristic equation

Find
1.1) $${{a}_{2}}$$, $${{a}_{1}}$$ and $${{a}_{0}}$$ such that equation 3.3 is the characteristic equation of 3.1.

1.2) The 1st homogeneous solution

1.3) The second homogeneous solution

1.4) The general homogeneous solution

2.1) $${{b}_{2}}$$, $${{b}_{1}}$$ and $${{b}_{0}}$$ such that equation 3.3 is the characteristic equation of 3.2.

2.2) The 1st homogeneous solution

2.3) The 2nd homogeneous solution

2.4) The general homogeneous solution

1.1) The coefficients of 3.1 can be found by using the trial solution method. The trial solution for the Euler equation with variable coefficient is:
The derivatives of the trial solution is defined as:

Substituting the equations 3.1.1, 3.1.2 and 3.1.3 into 3.1 we get:

Therefore we need to solve the following equation to find the coefficients:

Comparing the equation 3.1.7 to the equation 3.3 we have:

Therefore, comparing the coefficients of both side of the equation 3.1.8 we get:

1.3) Since $$\lambda $$ has two identical roots the second homogeneous solution can be calculate using the form:
Calculating the derivatives of 3.1.17:

Substituting 3.1.17, 3.1.18 and 3.1.19 into 3.1 we have:

Rearranging the terms of 3.1.20:

The equation 3.1.21 is simplified as:

Substituting the coefficients $${{a}_{2}}$$ (equation 3.1.10), $${{a}_{1}}$$ (equation 3.1.13), $${{a}_{0}}$$ (equation 3.1.14) and the 1st homogeneous solution with the respective derivatives presented in $${{y}_{1}}$$ (equation 3.1.1), $${{y}_{1}}'$$ (equation 3.1.2) and $${{y}_{1}}''$$ (equation 3.1.13), where r is replaced by $$\lambda $$ into 3.1.22:

Therefore we will solve for:

Using the reduction of order:

Substituting 3.1.28 into 3.1.29:

Integrating both sides of 3.1.32:

Using the exponential function in 3.1.34

Where $${{k}_{2}}$$ can be written as:

Therefore, z(x) can be written as:

Since the order of U(x) was reduced, we can get U(x) by integrating 3.1.38

Finally, substituting 3.1.42 and 3.1.16 into 3.1.17 the 2nd homogeneous solution is defined as:

1.4) The generalized homogeneous solution can be calculated by using the following equation:
Substituting equations 3.1.16 and 3.1.43 into 3.1.44 we have:

Rearranging the constants from equation 3.1.47, the generalized homogeneous solution can be expressed as:

2.1) The coefficients of 3.2 can be found by using the trial solution method. The trial solution for the Euler equation with constant coefficient is:
The derivatives of the trial solution is defined as:

Substituting the equations 3.2.1, 3.2.2 and 3.2.3 into 3.2 we get:

Therefore we need to solve the following equation to find the coefficients:

Comparing the equation 3.2.6 to the equation 3.3 we have:

Therefore, comparing the coefficients of both side of the equation 3.2.7 we get:

2.3) Since $$\lambda $$ has two identical roots the 2nd homogeneous solution can be calculate using the form:
Calculating the derivatives of 3.2.15:

Substituting 3.1.15, 3.1.16 and 3.1.17 into 3.2 we have:

Rearranging the terms of 3.2.18:

The equation 3.2.19 is simplified as:

Substituting the coefficients $${{b}_{2}}$$ (equation 3.2.9), $${{b}_{1}}$$ (equation 3.2.11), $${{b}_{0}}$$ (equation 3.2.12) and the 1st homogeneous solution with the respective derivatives presented in $${{y}_{1}}$$ (equation 3.2.1), $${{y}_{1}}'$$ (equation 3.2.2) and $${{y}_{1}}''$$ (equation 3.2.3), where r is replaced by $$\lambda $$, into 3.2.20:

Therefore we need to solve for:

Integrating both sides of 3.2.24 we have:

Integrating 3.2.25 we have U(x) as:

Finally, substituting 3.2.26 and 3.2.14 into 3.2.15 the 2nd homogeneous solution is defined as:

2.4) The generalized homogeneous solution can be calculated by using the following equation:
Substituting equations 3.2.14 and 3.2.27 into 3.2.28 we have:

Rearranging the constants from equation 3.2.31, the generalized homogeneous solution can be expressed as:

= Problem 7 =

= Find =

The 2 homogeneous solution $${{u}_{1}}$$ and $${{u}_{2}}$$ for (7.1) using the trial solution:

= Solution =

First we need to calculate the derivatives of the trial solution(7.1):

Substituting the equations 7.2, 7.3 and 7.4 into 7.1 we get:

Rearranging equation 7.7 we get:

Solving the equation 7.8 for r we get:

From equation 7.13 the root $${{r}_{1}}$$ can be calculated by:

From equation 7.13 the root $${{r}_{2}}$$ can be calculated by:

With the root $${{r}_{1}}$$ from equation 7.18 we can calculate the first homogeneous solution $${{u}_{1}}$$ by:

With the root $${{r}_{2}}$$ from the equation 7.23 we can calculate the second homogeneous solution $${{u}_{2}}$$ by:

title: User:EGM6321.f10.team6.yoshioka/HW5


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User:Peltirasia/tht

Tiedonhakutehtävä
Maijun muistiinpanoja:

1.STM:n sivuilta

STM ja Suomen Kuntaliitto tekivät ikäihmisten hoitoa ja palveluja koskevan laatusuosituksen 2001, joka on päivitetty 2008

''” Suomen perustuslain mukaan julkisen vallan on turvattava perus- ja ihmisoikeuksien toteutuminen, kuten oikeus yhdenvertaisuuteen ja välttämättömään huolenpitoon. Laatusuosituksessa määritellään ikääntyneiden palvelujen järjestämistä ohjaavat arvot ja eettiset periaatteet. Suositus sisältää strategisia linjauksia laadun ja vaikuttavuuden parantamiseksi kolmella osa-alueella: (1) hyvinvoinnin ja terveyden edistäminen ja palvelurakenteen kehittäminen, (2) henkilöstön määrä, osaaminen ja johtaminen, (3) asumis- ja hoitoympäristöt.”'' http://www.stm.fi/julkaisut/nayta/_julkaisu/1063089

Vanhustenhoitolakia valmistellaan, luonnos tarkoitus saada valmiiksi kevääseen 2011 mennessä

"Valtakunnallisina tavoitteina vuoteen 2012 mennessä on, että 75 vuotta täyttäneistä henkilöistä
 * 91–92 prosenttia asuu kotona itsenäisesti tai kattavan palvelutarpeen arvioinnin perusteella myönnettyjen tarkoituksenmukaisten sosiaali- ja terveyspalvelujen turvin
 * 13–14 prosenttia saa säännöllistä kotihoitoa
 * 5–6 prosenttia saa omaishoidon tukea
 * 5–6 prosenttia on tehostetun palveluasumisen piirissä
 * 3 prosenttia on hoidossa vanhainkodeissa tai pitkäaikaisessa hoidossa terveyskeskusten vuodeosastoilla

Kunnat asettavat omissa ikääntymispoliittisissa ohjelmissaan ikääntyneiden palveluille omat tavoitteensa valtakunnallisten tavoitteiden ja paikallisten tarpeiden pohjalta." http://www.stm.fi/sosiaali_ja_terveyspalvelut/ikaihmiset/palvelujen_laatu/laatusuositus

-- Hyödyllisiä asiakirjoja STM:n sivuilta --

Toimeentuloturva kehdosta keinutuoliin http://www.stm.fi/julkaisut/esitteita-sarja/nayta/_julkaisu/1067169#fi

Hyvä saattohoito Suomessa http://www.stm.fi/julkaisut/julkaisuja-sarja/nayta/_julkaisu/1528097#fi

Ikääntyneiden palveluiden uudet konseptit http://www.stm.fi/julkaisut/selvityksia-sarja/nayta/_julkaisu/1374605#fi

2. Tutkimukset ja hankkeet ikääntyneisiin liittyen:

Kunta- ja palvelurakenneuudistus (Paras)
 * tavoitteena taata kaikille suomalaisille yhdenvertaiset sosiaali- ja terveyspalvelut
 * edellytyksinä edellämainitulle ovat toimiva ja eheä kuntarakenne, palvelurakenteiden ja palveluiden järjestäminen nykyistä laajemmalle väestöpohjalle, kuntien välinen yhteistyö sekä uudet toimintatavat (tarkoittaen mitä???)
 * uudistuksen perusteena helmikuussa 2007 voimaan tullut ns. Paras-puitelaki
 * vastuu uudistuksesta kunnilla; kunnilta edellytetty päätöksiä puitelain velvoitteiden täyttymisestä

Terveys 2015 -kansanterveysohjelma
 * Pitkän aikavälinen terveyspoliittinen ohjelma, jonka tavoitteena on terveiden ja toimintakykyisten elinvuosien lisääminen ja väestöryhmien välisten terveyserojen kaventaminen
 * Ohjelmaa toteuttavat useat eri tahot, muun muassa kunnat, elinkeinoelämä ja järjestöt. Ohjelman taustalla on Maailman terveysjärjestön WHO:n Terveyttä kaikille -ohjelma.

Toimiva terveyskeskus -toimenpideohjelma
 * tarkoituksena on vahvistaa perusterveydenhuoltoa kehittämällä terveyskeskusten käytäntöjä, hallintoa ja johtamista sekä terveydenhuollon koulutusta ja tutkimusta
 * paneudutaan erityisesti terveydenhuoltohenkilöstön saatavuuteen ja riittävyyteen, sillä ne ovat terveyskeskusten toimivuuden avaintekijöitä
 * uudistukset otetaan terveyskeskuksissa vaiheittain käyttöön vuosina 2009-2011
 * hoitoon pääsyä helpotetaan muun muassa uudistamalla terveyskeskusten työn- ja vastuunjakoa (kuten esim. lisää hoitajien ja fysioterapeuttien vastaanottoja)

3. Luimme Helsingin Sanomien sunnuntainumeron

Hesarissa oli kaksi ikääntyneisiin liittyvää uutista: Erään kunnan eläkeläisten mökkivuokrien alentamisesta ja kummasteluartikkeli siitä, miten Trendi-lehdessä kaksi 70-vuotiasta voi pitää parisuhdeneuvopalstaa. Ikääntyneille kohdistettua mainontaa edusti senioripalveluasuntomainos.

4. Liittoja:

Vanhustyön Keskusliitto
 * runsaan 340 vanhusten hyväksi työskentelevän yhteisön keskusjärjestö
 * jäsenyhteisöillä mm. palvelutaloja, vanhainkoteja ja dementiayksiköitä ympäri maata, koti- ja tukipalveluita, vapaa-ajan toimintaa sekä vapaaehtoistyön ohjaamista
 * liitto toimii asiantuntijaorganisaationa vanhustenhuoltoon liittyvissä lainsäädännöissä ja muissa valmistelutöissä
 * järjestää koulutus- ja neuvottelupäiviä

Muunmuassa seuraavat järjestöt toimivat Vanhustyön Keskusliiton alaisina:

Deutsche Ev.Luth. Gemeinde - Saksalaisen seurakunnan seniorikoti http://www.deutschegemeinde.fi/seniorenwohnheim

 * Saksankielinen vanhainkoti

Jyväskylän Sotainvalidien Asuntosäätiö http://www.telkanpesa.fi/index.html

 * Asumispalveluja, päiväkeskus, ja kuntoutuspalveluja

Pohjois-Karjalan Dementiayhdistys ry http://www.dementiayhdistys.fi/

 * Muistihäiriöistä tietoa jakava etujärjestö, jonka toimialana koko Pohjois-Karjala ja jolla useita päättyneitä RAY:n projekteja

Vanhusten Turva ry http://www.dlc.fi/~vturva/

 * Vanhainkoti Onnitalon taustayhteisö

Apuomena ry, Lohja http://www.apuomena.fi/

 * Apuomena ry on lohjalaisten yhdistysten v. 1998 perustama yhteistyöjärjestö, jonka tarkoituksena on ehkäistä syrjäytymistä ja yksinäisyyttä tarjoamalla palveluja ja virikkeitä ikäihmisille ja erityisryhmille.

title: User:Peltirasia/tht