text:
User:Egm6321.f10.team03/Hwk5

=Problem 1: Higher Order Derivatives =

Given
Given a fourth order differential equation in the form
 * {| style="width:100%" border="0"

$$ {y}_{xxxx}= exp \left(-4t \right) \left( {y}_{tttt}-6{y}_{ttt}+11{y}_{tt}-6{y}_{t} \right) $$
 * $$\displaystyle (Eq. 1.1) $$
 * }
 * }

Find
Find the fifth order derivative, $${y}_{xxxxx}$$

Solution
Equation 1.1 was found by assuming $$x=exp\left(t\right) $$, therefore the derivative $$ \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$, where $$ \frac{dx}{dt}= exp\left(t\right) $$. Rearranging gives us
 * {| style="width:100%" border="0"

$$ \frac{dy}{dx}=\frac{dt}{dx}\frac{dy}{dt}$$ and
 * $$\displaystyle (Eq. 1.2) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{d}{dx} {y}_{xxxx}=\frac{dt}{dx}\frac{d}{dt} {y}_{xxxx} $$ Substituting $$ \frac{dx}{dt}= exp\left(t\right) $$ into equation 1.3 yields
 * $$\displaystyle (Eq. 1.3) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{d}{dx} {y}_{xxxx}=exp \left(-t \right)\frac{d}{dt} {y}_{xxxx} $$ Taking the derivative with respect to $$t $$ of equation 1.1 gives
 * $$\displaystyle (Eq. 1.4) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{d}{dt} {y}_{xxxx}=exp \left(-4t \right) \left({y}_{ttttt}-10{y}_{tttt}+35 {y}_{ttt}-50 {y}_{tt}+24 {y}_{t} \right) $$ Plugging equation 1.5 into equation 1.4 gives us the solution:
 * $$\displaystyle (Eq. 1.5) $$
 * }
 * }
 * {| style="width:100%" border="0"

$$  {y}_{xxxxx}=exp \left(-5t \right) \left({y}_{ttttt}-10{y}_{tttt}+35 {y}_{ttt}-50 {y}_{tt}+24 {y}_{t} \right) $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\,(Eq. 1.6) $$
 * style= |
 * }

Contributing Members

 * Solved and posted by James Roark 16:51, 27 October 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 15:34, 2 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.Sudheesh 16:14, 2 November 2010 (UTC)
 * Proofread by Egm6321.f10.team3.franklin 01:57, 3 November 2010 (UTC)

=Problem 2: Solving Euler Equation using method of trial solutions=

Given
Given an ODE of the form
 * {| style="width:100%" border="0"

$$ {x}^{2} \ddot{y}-2x \dot{y}+2y=0 $$ With boundary conditions $$y\left( 1\right)=-2 $$ and $$y \left(2 \right)=5 $$
 * $$\displaystyle (Eq. 2.1) $$
 * }
 * }

Find
Solve equation 2.1 using the method of trial solution.

Solution
Assume a solution of the form $$\,y= {x}^{r}$$, where $$r$$ is the roots of the characteristic equation. Therefore, $$\dot{y}=r{x}^{r-1}$$ and $$\ddot{y}=r(r-1) {x}^{r-2}$$. Plugging these equations into equation 2.1, we obtain the characteristic equation

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

$$\, {r}^{2}-3r+2=0 $$
 * $$\displaystyle (Eq. 2.2) $$
 * }
 * }

Which yields roots of unity and 2. The solution can then be written as a linear combination of each
 * {| style="width:100%" border="0"

$$ y \left(x \right)= {c}_{1}x+ {c}_{2}{x}^{2} $$ Substituting the boundary conditions,
 * $$\displaystyle (Eq. 2.3) $$
 * }
 * }

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

$$\displaystyle {{c}_{1}}+{{c}_{2}}=-2$$
 * }
 * }

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

$$\displaystyle 2{{c}_{1}}+4{{c}_{2}}=5$$
 * }
 * }

Solving the above equations give our constants.

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

$$\displaystyle {{c}_{1}}=-\frac{13}{2}$$
 * }
 * }

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

$$\displaystyle {{c}_{2}}=\frac{9}{2}$$
 * }
 * }

Hence the solution becomes

Plot of solution

 * [[Image:Sudheesh hw5 PLOT1.jpg|center|650px]]

Contributing Members

 * Solved and posted by James Roark 16:51, 27 October 2010 (UTC)
 * Proofread byEgm6321.f10.team03.sigillo 16:01, 2 November 2010 (UTC)
 * Plotted the graph and Proofread solution byEgm6321.f10.team3.Sudheesh 16:37, 2 November 2010 (UTC)

= 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 its 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 $$\displaystyle Eq. 3.5$$:

 * {| 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" |
 *
$$\displaystyle (Eq. 3.19)$$
 * }
 * }

Contributing members

 * Solved by Egm6321.f10.team03.schulze (Kurt)
 * Posted and rechecked the solution by Egm6321.f10.team3.Sudheesh 04:25, 2 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 17:36, 2 November 2010 (UTC)

=Problem 4: Find a particular solution using the method "Variation of Constants"=

Given
The following equation is given from page 11-2 out of the class notes:
 * {| style="width:100%" border="0"

$$\,{y}'+xy=2x+3$$
 *
$$\,(Eq. 4.1) $$
 * style= |
 * }

Find
The particular solution $${{y}_{p}}$$ has to be found using the method of variation of constants after having found the homogeneous solution $${{y}_{h}}$$.

Solution
First we find the homogeneous solution to the homogeneous equation out of Eq. 4.1, that is:
 * {| style="width:100%" border="0"

$$\,{y}'+xy=0$$ With the separation of variables we get:
 *
$$\,(Eq. 4.1) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{1}{y}dy=-x\,dx$$ The result of the integration on both sides of the equation yields to:
 *
$$\,(Eq. 4.2) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{\frac{1}{y}dy}=\int{-x\,dx+C\Leftrightarrow \log \left( y \right)=\frac{-{{x}^{2}}}{2}+C}$$ hence the homogeneous solution $${{y}_{h}}$$ is: The result of the integration on both sides of the equation yields to:
 *
$$\,(Eq. 4.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{y}_{h}}={{e}^{\frac{-{{x}^{2}}}{2}}}\cdot A$$ with
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 *
$$\,(Eq. 4.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,A={{e}^{C}}$$ The next Step is the variation of the constant $$\, A$$ that now becomes a function of the variable $$\, x$$. The solution $$\, y$$ and its derviation can thus be written as:
 *
$$\,(Eq. 4.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$y=A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}$$
 *
$$\,(Eq. 4.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'={A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}-A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x$$ Inserting Eq. 4.5 and Eq. 4.6 into Eq. 4.1 the given equation can be written as:
 *
$$\,(Eq. 4.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}-A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x+A(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}\cdot x=2x+3$$ The second and third term of the left-hand-side cancel out and Eq. 4.7 can be written as:
 *
$$\,(Eq. 4.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${A}'(x)\cdot {{e}^{\frac{-{{x}^{2}}}{2}}}=2x+3$$ To find the particular solution for $$y$$ the second last step is to find an expression for $$A(x)$$. This can be obtained by integrating $${A}'(x)$$ of Eq. 4.9:
 *
$$\,(Eq. 4.9) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=\int{\left( 2x+3 \right)}\cdot {{e}^{\frac{2}}}\,dx+k$$
 *
$$\,(Eq. 4.10) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=3\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,ds}+2\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,\cdot s\,ds}+k$$
 *
$$\,(Eq. 4.11) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=3\cdot \int\limits_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}\,ds}+2\cdot {{e}^{\frac{2}}}+k$$
 *
$$\,(Eq. 4.12) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$A(x)=3\sqrt{\frac{\pi }{2}}\cdot erfi\left( \frac{x}{\sqrt{2}} \right)+2{{e}^{\frac{2}}}+k$$ Inserting Eq. 4.13 into Eq. 4.6 gives the particular soluton $${{y}_{p}}$$:
 *
$$\,(Eq. 4.13) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{y}_{p}}=2+3\sqrt{\frac{\pi }{2}}\cdot erfi\left( \frac{x}{\sqrt{2}} \right)+{{e}^{\frac{-{{x}^{2}}}{2}}}+k$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\,(Eq. 4.14) $$
 * style= |
 * }

Contributing Members
Solved and posted byEgm6321.f10.team03.sigillo 01:33, 31 October 2010 (UTC)

= 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 $$\displaystyle Eq.5.1$$ 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" align="left"

$$\displaystyle {{a}_{2}}{{h}_{tt}}-{{a}_{1}}{{h}_{t}}+{{a}_{0}}h=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 *
$$\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" align="left"

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

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

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

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

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

Contributing Members

 * Solved and posted byEgm6321.f10.team3.Sudheesh 00:47, 2 November 2010 (UTC)
 * Solved by Egm6321.f10.team3.franklin 01:55, 3 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 22:47, 2 November 2010 (UTC)

=Problem 6: Equivalence of particular solutions=

Given
The particular solution for a L2-ODE-VC developed in class during Mtg 30 is given as:
 * {| style="width:100%" border="0"

$${{y}_{p}}={{u}_{1}}\left( x \right)\int{\frac{1}{h\left( x \right)}\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)\,dx}$$ Where $$h(x)$$ is equal to:
 *
$$\,(Eq. 6.1) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$h(x)=\frac{{{u}_{1}}^{2}(x)}{{{u}_{2}}^{\prime }(x)\cdot {{u}_{1}}(x)-{{u}_{2}}(x)\cdot {{u}_{1}}^{\prime }(x)}$$
 *
$$\,(Eq. 6.2) $$
 * style= |
 * }

Find
Show that the developed particular solution is equal to the particular solution in King, Billingham and Otto page 8 Eq. 1.6 as:
 * {| style="width:100%" border="0"

$${{y}_{p}}=\oint{f\left( s \right)\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{W(s)} \right)}\,ds$$ Where $$W(s)$$ is the Wronskian and given as:
 *
$$\,(Eq. 6.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$W(s)={{u}_{1}}(s)\cdot {{u}_{2}}^{\prime }(s)-{{u}_{1}}^{\prime }(s)\cdot {{u}_{2}}(s)=\left| \begin{matrix} {{u}_{1}}(x) & {{u}_{2}}(x) \\ {{u}_{1}}^{\prime }(x) & {{u}_{2}}^{\prime }(x) \\ \end{matrix} \right|$$
 *
$$\,(Eq. 6.4) $$
 * style= |
 * }

Solution
The right hand side of Eq. 6.1 can be integrated using the partial integration method:
 * {| style="width:100%" border="0"

$$\int(t)\cdot r(t)\,dt=r(t)\cdot v(t)-\int{{r}'(t)\cdot v(t)}\,dt$$ Where in the specific case of Eq. 6.1, $$r(t)$$ and $$v(t)$$ can be found as:
 *
$$\,(Eq. 6.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{u}_{1}}\left( x \right)\int{\underbrace{\frac{1}{h\left( x \right)}}_\underbrace{\left( \int{h\left( x \right)f\left( x \right)\,dx} \right)}_{r}}\,dx$$ Applying the partial integration method to Eq. 6.1 the right hand side can be rewritten as:
 *
$$\,(Eq. 6.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{u}_{1}}(x)\cdot \left[ \underbrace{\int{h(x)f(x)\,dx}}_{r}\cdot \underbrace{\frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}}_{v}-\int{\underbrace{h(x)\cdot f(x)}_\cdot }\underbrace{\frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}}_{v}\,dx \right]$$ Applying Eq. 6.2, $$h(x)$$ can be rewritten as:
 *
$$\,(Eq. 6.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{u}_{1}}(x)\cdot \left[ \int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds\cdot \frac{{{u}_{2}}(x)}{{{u}_{1}}(x)}-\int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{2}}(s)}{{{u}_{1}}(s)}\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds \right]$$ Multiplying $${{u}_{1}}$$ into the squared brackets we get:
 *
$$\,(Eq. 6.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\left[ \int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{1}}^{2}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds\cdot \frac{{{u}_{2}}(x)}{1}-\int\limits_ – ^{x}{f(s)\cdot \frac{{{u}_{2}}(s)}{1}\cdot \frac{{{u}_{1}}(x)\cdot {{u}_{1}}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)}}\,ds \right]$$ Which is equal to the contour integral:
 *
$$\,(Eq. 6.9) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\oint\limits_{x}{f(s)\cdot }\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{{{u}_{2}}^{\prime }(s)\cdot {{u}_{1}}(s)-{{u}_{2}}(s)\cdot {{u}_{1}}^{\prime }(s)} \right)\,ds=\oint\limits_{x}{f\left( s \right)\left( \frac{{{u}_{1}}(s)\cdot {{u}_{2}}(x)-{{u}_{1}}(x)\cdot {{u}_{2}}(s)}{W(s)} \right)\,ds}$$ '''Eq. 6.10, thus agrees with Eq. 1.6 in King, Billingham and Otto page 8.'''
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 *
$$\,(Eq. 6.10) $$
 * style= |
 * }

Contributing Members

 * Solved and posted byEgm6321.f10.team03.sigillo 19:42, 31 October 2010 (UTC)
 * Proofread by James Roark 18:40, 3 November 2010 (UTC)

=Problem 7: Find $$ {u}_{1},  {u}_{2} $$ using trial solution and variation of parameters=

Given
Give the reverse engineered equation on [|Mtg 31-1] of the lecture notes

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

$$ \left(x+1 \right) {y}^{''} - \left(2x+3 \right) {y}^{'} + 2y=0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle (Eq. 7.1)
 * }

Find
Find $$ {u}_{1},  {u}_{2} $$ 2 Homogeneous solutions of $$\displaystyle (Eq 7.1) $$ using the trial solution:

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

y= {e}^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.2)
 * style= |
 * }

Solution
To solve we first find the roots to the characteristic equation by plugging $$\displaystyle (Eq 7.2) $$ into $$\displaystyle (Eq 7.1) $$ to obtain

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

\left(x+1 \right) {r}^{2} - \left(2x+3 \right) r + 2=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.3)
 * style= |
 * }

Find the roots by using the quadratic formula
 * {| style="width:100%" border="0"

r= \frac{\left(2x+3 \right) \pm \sqrt{{\left(2x+3 \right)}^{2}-8\left(x+1 \right)}} {2 \left( x+1\right)}= \frac{\left(2x+3 \right) \pm \left(2x+1 \right) } {2 \left( x+1\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.4)
 * style= |
 * }

Solving we get the two roots to the characteristic equation:

 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{r}_{1} = 2 ; {r}_{2}=  \frac{1}{ \left( x+1\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.5)
 * style= |
 * }

Now we use the trial solution $$\displaystyle y= {e}^{rx} $$ and the constant root to obtain:

 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{u}_{1} = exp(2x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.6)
 * style= |
 * }

To find $$\displaystyle {u}_{2}(x) $$ we will use the variation of parameters method.

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

y(x)=U(x) {u}_{1}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.7)
 * style= |
 * }

Plugging $$\displaystyle (Eq. 7.7)$$ into $$\displaystyle (Eq. 7.1)$$ gives us

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

{U}^{'}(x) \left[ -\left(2x+3 \right) {u}_{1}(x)+ \left( x+1\right) {u}^{'}_{1}(x) \right] + \left( x+1\right) {u}_{1}(x) {U}^{''}(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.8)
 * style= |
 * }

Now use the reduction of order method 0 to find the expression for $$\displaystyle {u}_{2}$$. Using equation 1 from [|Mtg 31-1] we can now solve for $$\displaystyle {u}_{2}(x)$$

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

{u}_{2}(x)= {u}_{1}(x) \int \frac{1}{ {u}^{2}_{1}}exp \left[ - \int {a}_{1}dx\right]dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.9)
 * style= |
 * }

where $$\displaystyle {a}_{1}:= \frac{2x+3}{x+1} $$. Solving the integral for $$\displaystyle {a}_{1}$$ first we have
 * {| style="width:100%" border="0"

\int {a}_{1}dx= \int \frac{2x+3}{x+1} dx = \left[2x+ log(x+1)\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.10)
 * style= |
 * }

Now plug $$\displaystyle (Eq. 7.10)$$ into $$\displaystyle (Eq. 7.9) $$ to obtain

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

{u}_{2}(x)= {u}_{1}(x) \int \frac{x+1}{ {u}^{2}_{1}}exp \left[ -2x\right]dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.11)
 * style= |
 * }

Substituting in our expression for $$\displaystyle {u}_{1}(x)=exp\left(2x\right)$$ we have

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

{u}_{2}(x)= exp \left(2x \right) \int \left(x+1\right) {exp \left(-2x \right)}^{2}exp \left[ -2x\right]dx= exp \left(2x \right) \int \left(x+1\right) exp \left[ -6x\right]dx $$ $$ Solving the above equation by parts yields the solution for $$\displaystyle {u}_{2}(x)$$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.11)
 * style= |
 * }

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

{u}_{2}(x)= exp \left(2x \right) \left[ - \frac{1}{6}x exp \left(-6x \right)- \frac{1}{36}exp \left(-6x \right)-\frac{1}{6}exp \left(-6x \right)\right]= - \frac{1}{36} \left( 6x+7\right)exp \left( -4x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7.12)
 * style= |
 * }

The two homogeneous solutions are:

 * {| style="width:25%; padding:1px; border:2px solid #0000FF"

{u}_{1} = exp(2x); {u}_{2}(x)= - \frac{1}{36} \left( 6x+7\right)exp \left( -4x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * style= |
 * }

Contributing Members
Solved and posted by--Egm6321.f10.team3.cook 02:12, 2 November 2010 (UTC)

Proof read by Egm6321.f10.team03 21:39, 2 November 2010 (UTC) Schulze

=Problem 8 - Create L2-ODE-VC (Reverse Engineering) =

Given
Trial solution

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

$$
 * $$ y(x) = \frac {e^{rx}}{x^2} $$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

And char.eq.

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

r^{2} + 3 = 0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle
 * }
 * }

Find
Find L2-ODE-VC

Solution

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

$$
 * $$ \frac{\partial }{\partial x} \left( \frac {e^{rx} }{x^2}\right) = \left( \frac {r e^{rx} }{x^2} - \frac {2 e^{rx} }{x^3} \right) = \frac {e^{rx} }{x^2} \left( r - \frac {2 }{x}\right)$$
 *
$$\displaystyle
 *
$$\displaystyle
 * }
 * }

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

(r-\frac{2}{x})\right)$$ $$
 * $$ \frac{\partial^2 }{\partial^2 x} \left( \frac {e^{rx} }{x^2}\right) = \left( \frac {2e^{rx} }{x^4} - \frac {2e^{rx}(r-\frac {2} {x}) }{x^3} + \frac {re^{rx}(r-\frac {2}{x}) }{x^2} \right) = \frac {e^{rx} }{x^2} \left(  \frac{2}{x^2} -\frac{2}{x}(r-\frac{2}{x}) +r
 * $$ \frac{\partial^2 }{\partial^2 x} \left( \frac {e^{rx} }{x^2}\right) = \left( \frac {2e^{rx} }{x^4} - \frac {2e^{rx}(r-\frac {2} {x}) }{x^3} + \frac {re^{rx}(r-\frac {2}{x}) }{x^2} \right) = \frac {e^{rx} }{x^2} \left(  \frac{2}{x^2} -\frac{2}{x}(r-\frac{2}{x}) +r
 *
$$\displaystyle
 * }
 * }

To find L2-ODE-VC, multiply (Eqt 5.8.1), (Eqt 5.8.2) and (Eqt 5.8.3) by the fuctions of x
 * {| style="width:100%" border="0"

$$ \left[ \frac{e^{rx}}{x^2} \right ] * a_{0}(x) $$ ||
$$\displaystyle (Eq. 5.8.1) $$
 * }
 * }

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

$$ \left[ \frac{\partial }{\partial x} \left( \frac {e^{rx} }{x^2}\right) =  \frac {e^{rx} }{x^2} \left( r - \frac {2 }{x} \right) \right] * a_{1}(x)  $$ ||
$$\displaystyle (Eq. 5.8.2) $$
 * }
 * }

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

$$ \left[ \frac{\partial^2 }{\partial^2 x} \left( \frac {e^{rx} }{x^2}\right)  =  \frac {e^{rx} }{x^2} \left(  \frac{2}{x^2} -\frac{2}{x}(r-\frac{2}{x}) +r(r-\frac{2}{x})\right)\right]  * a_{2}(x) $$ ||
$$\displaystyle (Eq. 5.8.3) $$
 * }
 * }

Do the sum of (Eqt 5.8.1), (Eqt 5.8.2) and (Eqt 5.8.3)

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

$$ \frac {e^{rx}}{x^2} \left( a_{2}(x)\left [ \frac {2}{x^2} - \frac {2}{x}(r-\frac {2}{x}) + r(r-\frac {2}{x})\right ] + a_{1}(x)\left[ r-\frac{2}{x} \right] + a_{0}(x) \right ) = \frac {e^{rx}}{x} (r^2 + 3) = 0$$ ||
$$\displaystyle $$
 * }
 * }

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

$$ \frac {e^{rx}}{x^2} \left( a_{2}(x) r^2 + \left [ a_{1}(x) - \frac{4}{x}a_{2}(x)\right ]r + \left[\frac{6}{x^2}a_{2}(x) - \frac{2}{x}a_{1}(x) \right ] + a_{0}(x) \right ) = \frac {e^{rx}}{x} (r^2 + 3) = 0$$ ||
$$\displaystyle (Eq. 5.8.4) $$
 * }
 * }

from (Eqt. 5.8.4),

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

$$ \therefore a_{2}(x) = 1$$ ||
$$\displaystyle $$
 * }
 * }

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

$$ \therefore a_{1}(x) = \frac{4}{x}$$ ||
$$\displaystyle $$
 * }
 * }

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

$$ \therefore a_{0}(x) = 3 + \frac{2}{x^2}$$ ||
$$\displaystyle $$
 * }
 * }

Therefore,

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

$$ {{y}'}' + \frac {4}{x}{y}' + \left (3+\frac{2}{x^2}\right )y = 0$$ ||
$$\displaystyle (Eq. 1.8.) $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style= |
 * }

Contributing Members
Solved and posted byEgm6321.f10.team03.Hong SJ 16:25, 01 November 2010 (UTC)

Proof read by Egm6321.f10.team03 21:41, 2 November 2010 (UTC)Schulze

Proofread by Egm6321.f10.team3.Sudheesh 23:46, 2 November 2010 (UTC)

=Contributing members=
 * Egm6321.f10.team3.Sudheesh 04:31, 2 November 2010 (UTC) Solved prob #5, posted prob #3 and #5, rechecked solution of prob #3, Proof read prob #1,2,8
 * Egm6321.f10.team03.sigillo 17:56, 2 November 2010 (UTC) Solved problem 4 and 6. Proofread problem 1,2,3,5
 * Egm6321.f10.team3.Schulze 04:31, 2 November 2010 (UTC) Solved prob 3, proof read 7 and 8.
 * Egm6321.f10.team3.franklin 01:59, 3 November 2010 (UTC) Solved prob 5

=Notes and references=

title: User:Egm6321.f10.team03/Hwk5


text:
Open education and research at the University of Canberra

Presentation by Leigh Blackall and James Neill, University of Canberra, at the Annual Conference of Knowledge Commercialisation Australasia, November 11, 2010.

Key points

 * 1) Universities have vast amounts of unrealised intellectual capital - most of it has very little commercial viability - but much of it has potentially significant scientific, educational, and social value which, if openly shared, could bring significant local, national and international benefits.
 * 2) Does a protectionist stance, for commercial purposes alone, return enough to compensate for these lost opportunities? At any rate, how does a university become aware of projects with commercial potential?
 * 3) We think the proposed IP policy for the University of Canberra gives us a more reliable trigger, while ensuring maximum opportunity gains for the majority of IP at the university.

Open academia

James Neill presented about open academia
 * 1) Open academia = Principles of openness (philosophy) plus acts of sharingness (practice)
 * 2) Collaborative process of developing and sharing in a knowledge commons
 * 3) A university's job is to openly develop and share knowledge to fuel an adaptive sustainable society
 * 4) Five pillars: Open access, open license, open formats, open software, open governance
 * 5) Invert current situation where the default is openness, and closed is the exception
 * 6) In practice: openness is vital to authentic scholarship
 * 7) Comment: Could we make more reference to using popular online media for network learning engagement and research dissemination?

Examples

 * 1) Open information initiatives: Federal Gov Australia, NZ Gov Framework, European Commission, US Federal Research Public Access Act
 * 2) Open journals and textbooks
 * 3) Otago Polytechnic Return on Investment - "It is estimated to cost $4000 to train one person how to use social media for open educational practices. At Otago Polytechnic, such a person goes on to return over $8000 worth of savings and gains to their organisation in the first year."

OpenUC proposal
Leigh Blackall presented about The OpenUC Proposal, that was under development 2010-2011.
 * An IP Policy that returns to individual ownership, defaults the organisation's copyright to Creative Commons Attribution, preserves options for individuals to opt out, recognises Indigenous autonomy, Coordinated units who offer IP support services - triggered by opt-out function.
 * An education and research policy that encourages open academic practices, makes greater use of the contemporary Internet, engages in socially networked communications and informal learning, and plays a role developing digital network literacy in the community
 * A performance review policy that recognises and rewards open academic practices, especially supporting early career researchers into open academic practices
 * Business Unit Service Level Agreements that support the development of open academic practices

On Knowledge Commercialisation Australasia's Annual Conference

 * The content of the conference focused heavily on very large profit making ventures, referring mainly to success stories in US Universities, deriving profitable income from patent royalties.
 * Presenters on these themes acknowledged that profit derived from patent royalties involves a long waiting period (between 4-12 years), before returns on the costs of the initial registration and subsequent protection and monitoring are experienced in less than 6% of cases. The element of luck plays a significant part in experiencing such returns.
 * Given such low odds for a returns on the patent royalties approach to commercialisation, why does KCA and related commercialisation offices focus so much on this approach, and have a seemingly "hard nosed" perspective on protecting intellectual property? Some presenters proposed unrealised values in open methods, and a shared commons perspective.
 * Through conversations with people from other universities and publicly funded research institutions, the dominant paradigm for IP management with KCA associates seems to be for their institutions to claim ownership over the IP generated by their staff and students. This is done on the belief that it simplifies IP management, and positions the institution more favourably to potential industry and commercial interests. Such an approach however, has a significant breakages and weaknesses. An institutional claim to ownership and control is only effective (in terms of simplification and attracting commercial interests) if all involved in a project are, and always have been employed by that institution. Employees and others, who join the project and institution at different points in a projects time, bring with them various encumbrances on their contributions, returning the situation to the complexity that the institutional claim to ownership was designed to fix. Claiming ownership over even that outside work would surely create bad will and delays. Further, the quality of innovation in that institution's work is vulnerable to the quality of the working conditions and workplace morale. Should they be low, then naturally staff and students will want to do their best work outside the institutions, where ownership and control cannot be exerted, and they retain freedom and flexibility to leave and take their work with them. It seems to us therefore, that these breakages and weakness far outweigh the potential benefits, and that another way ought to be considered. One that respects individual ownership, but affords maximum sharing and stimulation of innovation and rapid development.

Problems/Questions/Issues
Refine to the best 3:
 * How can we approach a paradigm shift proposal: What Is This Thing Called Science? Chapter 8 - Theories as Structures 1
 * How can we find time for staff to properly appreciate or develop open academic practices
 * Current UC Policy, infrastructure, culture and practice is deeply embedded with closed academic practice
 * Student expectations and prior learning experience is based on closed academic practice
 * It is financially difficult to quantify open acadamemic practices, because all costs, including social and ecological, are not yet accounted for
 * The Open Society and Its Enemies
 * Are we having discussions about the real purpose of universities?
 * Mission/Values/Strategies etc. - to what extent is/isn't and should/shouldn't openness be reflected within our understandings of unversities?
 * If openness was adopted more widely within universities (which is likely), what implications could this have?
 * To what extent should university entrepreneurship be directed towards commercialisation and to what extent should should university entrepreneurship be directed towards openness?
 * IF entering into a restrictive condition, should the party calling for the restriction compensate for the "lost opportunity"?

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


text:
File talk:Wikiversity Problem 2.2.3.pdf

Katelin, I think you have done a wonderful job. Your uploaded file is both effective and creative. Your explanation is broke down enough for anyone to understand without prior knowledge of if then statements. It is hard to find something for you to edit or change. Great job.

Jamie Balayti

I think your charts and explanation were very clear and explained the problem very well. You went into great detail, which would be very beneficial to people who have never been exposed to truth-tables before. If I wasn't in this class and had never seen a truth-table before, I think I would have understood this problem from reading your work on how you got to your conclusion. Good job! -Katy Anderson

I agree with the other two critiques that this diagram and explanation are very clear-cut and and excellent way for someone who has difficulties with if-then charts to have the basics laid out explicitly. The only specific line that could use a little more explanation is the if F then T statement where it was mentioned that If you do not stop then the light is red. That one could be confusing to a reader. But really, that this was excellent and you should have written this section for the course reader.

Rehus1 01:53, 3 November 2010 (UTC)

title: File talk:Wikiversity Problem 2.2.3.pdf


text: Latin/Conjugations

title: Latin/Conjugations


text:
Talk:Ideas in Geometry/Instructive examples/Section 2.1 Basic Set Theory number 10

This problem looks really good! I just had a couple suggestions. There is a spelling error the first time you use the word “denote” (just that it has an “s” on the end). For step 1 (finding X U(Y∩Z)) I might suggest making the visual a little easier to follow just by putting a fully shaded X + what you found for (Y∩Z) then showing how they come together. I feel this might help people follow along more easily especially for part 2 of the problem, which is a little more difficult. Also at the end, possibly showing the two sides of the equation side by side again for visual affect and clarity. The problem looks great, the written explanation describes your process very well, the organization of steps is clear, and your images looks really good! Nice work!

I liked how you had pictures for explaining the problem. Your wording on a lot of things was really confusing though. Some pictures are not clearly explained, make sure you explain each photo, not one explanation for two photos. Like explain the picture XU(Y∩Z), it seemed like you only really explained the picture Y∩Z. show how you got from that picture to the new one.

-Allison Chapello


text:
Conjugating a Latin Verb in Different Tenses

Please conjugate the verb Make sure you use lower case letters!

{ Conjugate the verb poto, potare, potavi, potus-drink, in Latin in the IMPERFECT tense. Then translate it into English. 1st S ? { potabam }
 * type="{}"}

{ 2nd s { potabas }
 * type="{}"}

{ 3rd s { potabat }
 * type="{}"}

{ 1st pl { potabamus }
 * type="{}"}

{ 2nd pl { potabatis }
 * type="{}"}

{ 3rd pl { potabant }
 * type="{}"}

{ Now in ENglish. 1st S ? { I was drinking }
 * type="{}"}

{ 2nd s { you were drinking }
 * type="{}"}

{ 3rd s { he was drinking|she was drinking|it was drinking }
 * type="{}"}

{ 1st pl { we were drinking }
 * type="{}"}

{ 2nd pl { you were drinking|yall were drinking|y'all were drinking|ya'll were drinking }
 * type="{}"}

{ 3rd pl { they were drinking }
 * type="{}"}

{ Conjugate the verb poto, potare, potavi, potus-drink, in Latin in the PERFECT tense. Then translate it into English. 1st S ? { potavi }
 * type="{}"}

{ 2nd s { potavisti }
 * type="{}"}

{ 3rd s { potavit }
 * type="{}"}

{ 1st pl { potavimus }
 * type="{}"}

{ 2nd pl { potavistis }
 * type="{}"}

{ 3rd pl { potaverunt }
 * type="{}"}

{ Now in ENglish. 1st S ? { I have drunk }
 * type="{}"}

{ 2nd s { you have drunk }
 * type="{}"}

{ 3rd s { he had drunk|she had drunk|it had drunk }
 * type="{}"}

{ 1st pl { we had drunk }
 * type="{}"}

{ 2nd pl { you had drunk|yall had drunk|y'all had drunk|ya'll had drunk }
 * type="{}"}

{ 3rd pl { they had drunk }
 * type="{}"}

Go to the Latin stream/Conjugations page for more information on different conjugations. Go to the /Practice/ page to practice your endings!


text:
Talk:Ideas in Geometry/Instructive examples/Section 3.2 Problem 42 - Lattice Area, Pick's Theorem, and Probability

Awesome work, Joe. The organization and clarity is fantastic. I like how you used the wikipedia format as opposed to just scanning in an image (like I did), it makes it much easier to read. The background information was certainly sufficient and if I did not know this material beforehand I would be able to look at your page and understand. Nice job.


text:
User talk:Mawais

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 17:11, 27 October 2010 (UTC)

title: User talk:Mawais


text:
User talk:Krutbulle~enwikiversity

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

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 17:11, 27 October 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Krutbulle. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Krutbulle~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:41, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:13, 19 April 2015 (UTC)

title: User talk:Krutbulle~enwikiversity


text:
User talk:Waseem khan

 Hello Waseem khan, 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:Waseem khan


text:
Boundary Value Problems/Lesson 10

http://en.wikipedia.org/wiki/Spherical_harmonics

title: Boundary Value Problems/Lesson 10


text:
User:Kychot/CV

 Curriculum vitae of the User:Kychot

Born

 * Date: November 26, 1953
 * Place: Prague, Czech Republic (at this time: Czechoslovakia), Europe

Family

 * Parents:
 * Mother: Věra Heřmanová (Dohnalová) (*1925)
 * Father: Stanislav Heřman (*1930)
 * Brothers and sisters: none
 * State: married
 * Wife: Alena Heřmanová (*1956)
 * Childrens:
 * Dalibor (*1977)
 * Vít (*1979)
 * Slávek (1984)
 * Matěj (*1992)

Education

 * 9 years Primary school: 1960–1969
 * 4 years High school (electro-technical): 1969–1973
 * 5 years Charles University (Math-Physical Faculty): 1973–1978
 * title: RNDr. (Rerum Naturalis Doctor)

title: User:Kychot/CV


text:
Talk:Ideas in Geometry/Instructive examples/Section 3.2 Number 28

I checked your math and it is correct. The only two suggestions I have for you would be to actually write out your equations so that you can see it when it is spread out and to explain how you took out the fractional parts and the whole parts and maybe why you take these parts out (to get the answer as accurate as possible). Otherwise, good job! [Katelin_Barker90]