text:
User:Egm6321.f09.team4/HW3

= Homework #3: Team 4 =

Problem #1:
(Lecture 13-1)

 * Given :
 * The equation $$\left ( x^ny^m \right )\left [ \sqrt{x}y^{\prime\prime}+2xy^{\prime}+3y \right ]=0$$

 * Find :
 * The values of m and n such that the equation is exact.

 * Solution :

 * The given equation can be rewritten in the following form (first condition of exactness):

 * $$ \begin{align}

f\left(x,y,p\right) &= \left(x^{n}y^{m}\right)x^{0.5} = x^{n+0.5}y^{m} \\ g\left(x,y,p\right) &= \left(x^{n}y^{m}\right)(2xp+3y) = 2x^{n+1}y^{m}p+3x^{n}y^{m+1} \\ \end{align} $$

 * To evaluate the second condition of exactness, various partial derivatives of $$f$$ and $$g$$ are required.

 * $$ \begin{align}

f_{xx} &= \frac{\partial}{\partial x}\left( \left(n+0.5\right)x^{n-0.5}y^{m}\right) = \left(n^{2}-0.25\right)x^{n-1.5}y^{m} \\ f_{xy} &= \frac{\partial}{\partial y}\left( \left(n+0.5\right)x^{n-0.5}y^{m}\right) = m\left(n+0.5\right)x^{n-0.5}y^{m-1} \\ f_y   &= mx^{n+0.5}y^{m-1} \\ f_{yy} &= m\left(m-1\right)x^{n+0.5}y^{m-2} \\ f_{xp} &= f_{yp} = 0.0 \\ \\  g_{xp} &= \frac{\partial}{\partial x}\left( 2x^{n+1}y^{m} \right) = 2\left(n+1\right)x^{n}y^{m} \\ g_{yp} &= \frac{\partial}{\partial y}\left( 2x^{n+1}y^{m} \right) = 2mx^{n+1}y^{m-1} \\ g_{pp} &= 0.0 \\ g_{y} &= 2mx^{n+1}y^{m-1}p + 3\left(m+1\right)x^{n}y^{m} \end{align} $$

 * The second condition of exactness requires satisfying two equations. The first equation (eq. 2A) is evaluated below:

 * $$ \begin{align}

&f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} \quad \text{(2a)} \end{align} $$

 * Plugging in the values obtained above for the various partial derivatives:

 * $$ \begin{align}

&\left(n^2-0.25\right)x^{n-1.5}y^{m} + 2pm\left(n+0.5\right)x^{n-0.5}y^{m-1} + p^{2}m\left(m-1\right)x^{n+0.5}y^{m-2} = 2\left(n+1\right)x^{n}y^{m} + \cancel{2mpx^{n+1}y^{m-1}} - \cancel{2mpx^{n+1}y^{m-1}}- 3\left(m+1\right)x^{n}y^{m} \\ &\left(n^2-0.25\right)x^{n-1.5}y^{m} + 2pm\left(n+0.5\right)x^{n-0.5}y^{m-1} + p^{2}m\left(m-1\right)x^{n+0.5}y^{m-2} = \left[ 2\left(n+1\right)-3\left(m+1\right)\right]x^{n}y^{m} \end{align} $$

 * Dividing both sides by $$x^{n}y^{m}$$:

 * $$ \begin{align}

&\left(n^2-0.25\right)x^{-1.5} + 2pm\left(n+0.5\right)x^{-0.5}y^{-1} + p^{2}m\left(m-1\right)x^{0.5}y^{-2} = 2n - 3m - 1 \\ \end{align} $$

 * The other equation for the second exactness condition (eq. 2B) is given as follows:

 * $$ \begin{align}

&f_{xp} + pf_{yp} + 2f_{y} = g_{pp} \\ &0.0 + p\left(0.0\right) + 2mx^{n+0.5}y^{m-1} = 0.0 \\ &2mx^{n+0.5}y^{m-1} = 0 \end{align} $$

 * The only way this equation can be satisfied for arbitrary $$x,y$$ is if $$m = 0$$. Plugging $$m=0$$ back into eq. 2A:

 * $$ \begin{align}

&\left(n^2-0.25\right)x^{-1.5} = 2n - 1 \end{align} $$

 * This equation can only be satisfied by $$n = 0.5$$. Therefore, the values of $$m,n$$ which make the original equation exact are:

 * $$ \begin{align}

m &= 0.0 \\ n &= 0.5 \end{align} $$  Good write up. Nice work. --Egm6321.f09.TA 04:13, 15 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 19:53, 7 October 2009 (UTC)
 * Egm6321.f09.Team4.meyerat 11:55, 7 October 2009 (UTC)

Problem #2:
(Lecture 13-1)

 * Given :
 * The equation $$\Phi(x,y,p)=xy'+(2x^{3/2}-1)y+k_1=k_2$$ where $$k_1$$ and $$k_2$$ are constants

 * Find :
 * Solve for y(x) by applying the integrating factor method

 * Solution :
 * This is a first-order ODE, which can be rearranged in the following form:

 * $$ \begin{align}

y' + p\left(x\right)y = g\left(x\right), \quad \text{where:} \quad p\left(x\right)&=2x^{0.5}-x^{-1} \\ g\left(x\right)&=x^{-1}\left(k_2-k_1\right) \end{align} $$

 * Let $$h\left(x\right)$$ be the integrating factor:

 * $$ \begin{align}

&h\left(x\right)y' + h'\left(x\right)y = h\left(x\right)g\left(x\right) \end{align} $$

 * Note: if we find $$h\left(x\right)$$ such that $$h\left(x\right)p\left(x\right)=h'\left(x\right)$$, the equation becomes the following:

 * $$ \begin{align}

&h\left(x\right)y' + h'\left(x\right)y = h\left(x\right)g\left(x\right) \\ &\frac{d}{dx}\left(h\left(x\right)y\right) = h\left(x\right)g\left(x\right) \quad \text{(product rule)} \\ &h\left(x\right)y+c = \int h\left(x\right)g\left(x\right)dx \\ &y\left(x\right) = \frac{\int h\left(x\right)g\left(x\right)dx - C_{1}}{h\left(x\right)} \end{align} $$

 * Then, to solve this, all we need is $$h\left(x\right)$$, which can be found by solving the constraint:

 * $$ \begin{align}

&h\left(x\right)p\left(x\right) = h'\left(x\right) \\ &h\left(x\right)\left(2x^{0.5}-x^{-1}\right) = h'\left(x\right) \\ &\frac{h'\left(x\right)}{h\left(x\right)} = 2x^{0.5} - x^{-1} \\ &\frac{d}{dx}\left\{\text{ln}\left[h\left(x\right)\right]\right\} = 2x^{0.5} - x^{-1} \\ &\text{ln}\left[h\left(x\right)\right] = \frac{4}{3}x^{1.5} - \text{ln}\left(x\right) + c \\ &h\left(x\right) = \text{exp}\left[ \frac{4}{3}x^{1.5}-\text{ln}\left(x\right)+c\right] \\ &h\left(x\right) = C_{2} x^{-1} \text{exp}\left(\frac{4}{3}x^{1.5}\right) \end{align} $$

 * Finally, $$y\left(x\right)$$ is found by plugging in $$h\left(x\right)$$:

 * $$ \begin{align}

y\left(x\right) &= \frac{\int\left[ C_{2}x^{-1}\text{ exp}\left(\frac{4}{3}x^{1.5}\right)\cdot x^{-1}\left(k_{2}-k_{1}\right)\right]dx - C_{1}}{C_{2}x^{-1}\text{ exp}\left(\frac{4}{3}x^{1.5}\right)} \\ y\left(x\right) &= x\text{ exp}\left(\frac{-4}{3}x^{1.5}\right)\left[ C_{2}\left(k_2-k_1\right)\int x^{-2}\text{ exp}\left(\frac{4}{3}x^{1.5}\right)dx - \frac{C_1}{C_2}\right] \end{align} $$

You have an extra power of $$x$$ in the denominator of the integrand. Egm6321.f09.TA 01:24, 28 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 16:55, 6 October 2009 (UTC)
 * Egm6321.f09.Team4.meyerat 11:58, 7 October 2009 (UTC)

Problem #3:
(Lecture 13-1)

 * Given :
 * A class of ODE's: linear, 2nd order ODE's with varying coefficients (L2_ODE_VC).

 * Find :
 * The mathematical structure of $$\Phi\left(x,y,p\right)$$ that yields the above class.

 * Solution :

 * Consider the following equation:

 * $$ \begin{align}

\Phi\left(x,y,p\right) = C, \quad \text{(where C is a constant)} \end{align} $$

 * Take the total derivative of this equation.

 * $$ \begin{align}

&\frac{d}{dx}\left[ \Phi\left(x,y,p\right) \right] = 0 \\ &\Phi_{x}dx + \Phi_{y}dy + \Phi_{p}dp = 0 \\ &\Phi_{x} + \Phi_{y}\frac{dy}{dx} + \Phi_{p}\frac{dp}{dx} = 0 \\ &\Phi_{x} + \Phi_{y}y' + \Phi_{p}y'' = 0 \quad \text{(since } p=y' \text{)} \end{align} $$

 * For any $$\Phi\left(x,y,p\right)$$, this equation will produce an exact, second-order ODE with varying coefficients. However, we were given the additional restriction that the ODE must be linear; i.e., the equation must be linear in both $$y$$ and $$y'$$, and there should be no terms containing $$y$$, $$y'$$, or $$y''$$ multiplied together. These requirements lead to three conditions on the partial derivatives of $$\Phi$$:
 * $$\Phi_x$$ cannot contain powers of $$y$$ or $$p$$ greater than 1. If both are present, they cannot be multiplied together.
 * $$\Phi_y$$ cannot contain $$y$$ or $$p$$.
 * $$\Phi_p$$ cannot contain $$y$$ or $$p$$.

 * To meet these conditions, we define $$\Phi$$ to have the following structure:

 * $$ \begin{align}

\Phi\left(x,y,p\right) = A\left(x\right) + B\left(x\right)y + C\left(x\right)p + D, \quad \text{(where D is a constant)} \end{align} $$

 * This yields the following values for the partial derivatives:

 * $$ \begin{align}

\Phi_x &= A'\left(x\right) + B'\left(x\right) y + C'\left(x\right) p \quad &\text{(condition 1 met)} \\ \Phi_y &= B\left(x\right) \quad &\text{(condition 2 met)}\\ \Phi_p &= C\left(x\right) \quad &\text{(condition 3 met)} \end{align} $$

It is not clear that you have arrived at the proper result. To finish this you need to equate your terms to those in the problem statement. Egm6321.f09.TA 02:01, 28 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 18:18, 6 October 2009 (UTC)
 * Egm6321.f09.Team4.meyerat 12:00, 7 October 2009 (UTC)

Problem #4:
(Lecture 13-3)

 * Given :
 * A nonlinear 1st order ODE $$F\left(x,y,y'\right)=0=\frac{d}{dx}\Phi\left(x,y\right)$$, and the exactness condition $$f_o-\frac{df_1}{dx}=0$$.

 * Find :
 * Show that this exactness condition is equivalent to the condition $$\Phi_{xy}=\Phi_{yx}$$, derived earlier in the course. Hint: $$f_1=\Phi_y$$.

 * Solution :

 * Start by expanding the definition of our ODE:

 * $$ \begin{align}

F\left(x,y,y'\right) &= \frac{d}{dx}\left[\Phi\left(x,y\right)\right] = 0 \\ &= \Phi_{x}\left(x,y\right)dx + \Phi_{y}\left(x,y\right) dy \\ &= \Phi_{x}\left(x,y\right) + \Phi_{y}\left(x,y\right)\frac{dy}{dx} \\ &= \Phi_{x}\left(x,y\right) + \Phi_{y}\left(x,y\right) p \quad \text{(where } p=y' \text{)} \end{align} $$

 * By definition, we know that:
 * $$ \begin{align}

f_{i} &= \frac{\partial F}{\partial y^{\left(i\right)}} \end{align} $$

 * Expanding this definition for $$i=0, i=1$$ yields:

 * $$ \begin{align}

f_{0} &= \frac{\partial F}{\partial y^{\left(0\right)}} = \frac{\partial F}{\partial y} \\ &= \Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p \\ \\ f_{1} &= \frac{\partial F}{\partial y^{\left(1\right)}} = \frac{\partial F}{\partial p} \\ &= \Phi_{y}\left(x,y\right) \end{align} $$

 * Inserting these two values into the exactness condition:

 * $$ \begin{align}

&f_{0} - \frac{d}{dx} f_{1} = 0 \\ &\Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p - \frac{d}{dx}\left[\Phi_{y}\left(x,y\right)\right] = 0 \\ &\Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p - \left[\Phi_{yx}\left(x,y\right)\frac{d}{dx}\left(x\right) + \Phi_{yy}\left(x,y\right)\frac{d}{dx}y\right] = 0 \\ &\Phi_{xy}\left(x,y\right) + \Phi_{yy}\left(x,y\right)p - \left[ \Phi_{yx}\left(x,y\right) + \Phi_{yy}p \right] = 0 \\ &\Phi_{xy}\left(x,y\right) = \Phi_{yx}\left(x,y\right) + \Phi_{yy}p - \Phi_{yy}p \\ &\Phi_{xy}\left(x,y\right) = \Phi_{yx}\left(x,y\right) \end{align} $$

 Good write up- nice work. This is very clear. --Egm6321.f09.TA 14:24, 15 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.sls 16:15, 7 October 2009 (UTC)

Problem #5:
(Lecture 14-1)
 * Given :
 * An exact nonlinear 2nd order ODE $$F\left(x,y,y',y''\right)=0=\frac{d}{dx}\Phi\left(x,y,y'\right)$$, and the exactness condition $$f_0-\frac{df_1}{dx}+\frac{d^{2}f_2}{dx^2}=0$$.

 * Find :
 * Show that $$f_1=\frac{d}{dx}f_2+\Phi_y$$
 * Show that $$\frac{d}{dx}(\Phi_y)=f_0$$
 * Show that $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$
 * Relate the condition $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$ to equations 4 and 5 on page 10-2.

 * Solution :

 * Expanding $$F(x,y,y',y'')$$ yields:
 * $$ \begin{align}

F &=\frac{d\Phi}{dx}(x,y,p) \qquad \text{(where } p=y' \text{)}\\ &=\Phi_x\frac{dx}{dx}+\Phi_y\frac{dy}{dx}+\Phi_p\frac{dp}{dx} \\ &=\Phi_x+\Phi_y p+\Phi_p p' \end{align} $$

 * By definition:
 * $$ \begin{align}

&f_i=\frac{\partial F}{\partial y^{(i)}} \end{align} $$

 * Expanding this definition for $$i=0,i=1,i=2$$ yields:
 * $$ \begin{align}

f_0&=\frac{\partial F}{\partial y} = \Phi_{xy} + \Phi_{yy}p + \Phi_{py}p' \\ \\ f_1&=\frac{\partial F}{\partial p} = \Phi_{xp} + \Phi_{yp}p + \Phi_{y} + \Phi_{pp}p' \\ \\ f_2&=\frac{\partial F}{\partial p'} = \Phi_p \end{align} $$

 * Part 1 :

 * $$ \begin{align}

f_{1} &= \frac{df_2}{dx} + \Phi_y \\ \Phi_{xp}+\Phi_{yp}p+\Phi_{y}+\Phi_{pp}p' &= \frac{d}{dx}\left(\Phi_p\right) + \Phi_{y} \\ \Phi_{xp}+\Phi_{yp}p+\Phi_{y}+\Phi_{pp}p' &= \Phi_{px} + \Phi_{py}p + \Phi_{pp}p' + \Phi_{y} \\ \Phi_{xp}+\Phi_{yp}p &= \Phi_{px} + \Phi_{py}p \\ 0 &= 0 \quad \text{Proof. (since } \Phi_{ab}=\Phi_{ba} \text{, for all a,b)} \end{align} $$

 * Part 2 :

 * $$ \begin{align}

\frac{d}{dx}\left(\Phi_y\right) &= f_{0} \\ \Phi_{yx}+\Phi_{yy}p+\Phi_{yp}p' &= \Phi_{xy} + \Phi_{yy}p + \Phi_{py}p' \\ \Phi_{yx}+\Phi_{yp}p' &= \Phi_{xy} + \Phi_{py}p' \\ 0 &= 0 \quad \text{Proof. (since } \Phi_{ab}=\Phi_{ba} \text{, for all a,b)} \end{align} $$

 * Part 3 :

 * $$ \begin{align}

f_0 - \frac{df_1}{dx} + \frac{d^{2}f_{2}}{dx^{2}} = 0 \end{align} $$

 * From part 1, we know that:

 * $$ \begin{align}

\frac{d}{dx}\left( f_{2} \right) = f_{1} - \Phi_{y} \end{align} $$

 * Differentiating both sides of this equation:

 * $$ \begin{align}

\frac{d^2}{dx^2}\left(f_{2}\right) &= \frac{d}{dx}\left(f_{1}\right) - \frac{d}{dx}\left(\Phi_{y}\right) \\ &= \frac{d}{dx}\left(f_{1}\right) - f_{0} \quad \text{(by result from Part 2)} \end{align} $$

 * Plugging this result back into the equation we're trying to prove:

 * $$ \begin{align}

f_{0} - \frac{df_1}{dx} + \frac{d^{2}f}{dx^{2}} &= 0 \\ f_{0} - \frac{df_1}{dx} + \left[ \frac{df_1}{dx} - f_{0} \right] &= 0 \\ f_{0} - f_{0} &= 0 \\ 0&=0 \quad \text{Proof.} \end{align} $$

 * Part 4 :

 * The old second-order exactness conditions which we're trying to derive are given as follows:

 * $$ \begin{align}

\quad &f_{xx} + 2pf_{xy} + p^{2}f_{yy} = g_{xp} + p g_{yp} - g_{y} \\ \quad &f_{xp} + pf_{yp} + 2f_{y} = g_{pp} \end{align} $$

 * Using the expansion of $$F$$ at the beginning of the solution, the functions $$f\left(x,y,p\right)$$ and $$ g\left(x,y,p\right)$$ are given as follows:

 * $$ \begin{align}

\text{(2a)} \qquad &f\left(x,y,p\right) = \Phi_p\left(x,y,p\right) \\ \text{(2b)} \qquad &g\left(x,y,p\right) = \Phi_x\left(x,y,p\right) + \Phi_y\left(x,y,p\right)p \end{align} $$

 * Taking the necessary partial derivatives:

 * $$ \begin{align}

f_{y} &= \Phi_{py}, \quad &f_{yy} = \Phi_{pyy}, \qquad &f_{xy} = \Phi_{pxy} \\ f_{xx} &= \Phi_{pxx}, \quad &f_{xp} = \Phi_{pxp}, \qquad &f_{yp} = \Phi_{pyp} \end{align} $$

 * $$ \begin{align}

g_{y} &= \Phi_{xy}+\Phi_{yy}p \\ g_{yp} &= \Phi_{xyp}+\Phi_{yy}+\Phi_{yyp}p \\ g_{xp} &= \Phi_{xxp} + \Phi_{yx} + \Phi_{yxp}p \\ g_{pp} &= \Phi_{xpp} + 2\Phi_{yp} + \Phi_{ypp}p \end{align} $$

 * Expand equation (2a):

 * $$ \begin{align}

f_{xx} + 2pf_{xy} + p^{2}f_{yy} &= g_{xp} + pg_{yp} -g_{y} \\ \Phi_{pxx} + 2p\Phi_{pxy} + p^{2}\Phi_{pyy} &= \Phi_{xxp} + \Phi_{px} + \Phi_{yxp}p + p\Phi_{xyp}+p\Phi_{yy}+p^{2}\Phi_{yyp}-\Phi_{xy}-\Phi_{yy}p \\ \\ \Phi_{xxp} + 2p\Phi_{xyp} + p^{2}\Phi_{yyp} &= \Phi_{xxp} + 2p\Phi_{xyp} + p^{2}\Phi_{yyp} \end{align} $$

 * Expand equation (2b):

 * $$ \begin{align}

f_{xp} + pf_{yp} + 2f_{y} &= g_{pp} \\ \\ \Phi_{xpp} + p\Phi_{ypp} + 2\Phi_{py} &= \Phi_{xpp} + 2\Phi_{py} + p\Phi_{ypp} \end{align} $$

 * Next, we'll expand the new exactness condition. First, note the following:

 * $$ \begin{align}

f_{0} &= \Phi_{xy} + \Phi_{yy}p + \Phi_{py}p' = \frac{d}{dx}\left(\Phi_y\right) \quad \text{(from part 2)} \\ \frac{df_{1}}{dx} &= \frac{d}{dx}\left[ \Phi_{xp} + \Phi_{yp}p + \Phi_{y} + \Phi_{pp}p' \right]\\ &= \left[\Phi_{xpx}+\Phi_{xpy}p+\Phi_{xpp}p'\right] + \left[p'\Phi_{yp}+\Phi_{ypx}p+\Phi_{ypy}p^{2}+\Phi_{ypp}pp'\right] + \left[\Phi_{ppx}p'+\Phi_{ppy}pp'+\Phi_{ypp}\left(p'\right)^{2}\right] +f_{0} \end{align} $$

 * Also, remember from the derivation of Part 3:

 * $$ \begin{align}

\frac{d^{2}f_{2}}{dx^{2}} = \frac{df_{1}}{dx} - f_{0} \end{align} $$

 * This allows us to rewrite the new exactness condition as follows:

 * $$ \begin{align}

&\frac{df_{1}}{dx} - f_{0} = \frac{df_{1}}{dx} - f_{0} \\ &\Phi_{xpx}+\Phi_{xpy}p+\Phi_{xpp}p' + p'\Phi_{yp}+\Phi_{ypx}p+\Phi_{ypy}p^{2}+\Phi_{ypp}pp' + \Phi_{ppx}p'+\Phi_{ppy}pp'+\Phi_{ypp}\left(p'\right)^{2}    = \cdots \quad \text{(both sides the same)}\\ \end{align} $$

 * We can then group the terms of this equation by powers of $$p'$$:

 * $$ \begin{align}

&\left[\Phi_{xpx}+\Phi_{xpy}p+\Phi_{ypx}p+\Phi_{ypy}p^{2}\right] + \left[\Phi_{xpp}+\Phi_{yp}+\Phi_{ypp}p+\Phi_{ppx}+\Phi_{ppy}p\right]p' + \left[\Phi_{ppp}\right]\left(p'\right)^{2} = \cdots \quad \text{(both sides the same)}\\ &\left[\Phi_{xxp}+2p\Phi_{xyp}+p^{2}\Phi_{yyp}\right] + \left[2\Phi_{xpp}+2p\Phi_{ypp}+\Phi_{yp}\right]p' + \left[\Phi_{ppp}\right]\left(p'\right)^{2} = \cdots \quad \text{(both sides the same)} \end{align} $$

 * Then, subtract $$\left[\Phi_{xpp}+p\Phi_{ypp}-\Phi_{yp}\right]p' + \Phi_{ppp}p^{2}$$ from both sides:

 * $$ \begin{align}

&\left[\Phi_{xxp}+2p\Phi_{xyp}+p^{2}\Phi_{yyp}\right] + \left[\Phi_{xpp}+p\Phi_{ypp}+2\Phi_{yp}\right]p' = \cdots \quad \text{(both sides the same)} \end{align} $$

 * Finally, we can split this single equality into two parts by equating the coefficients of each $$p'$$ term.

 * $$ \begin{align}

\text{(1)} \quad &\Phi_{xxp} + 2p\Phi_{xyp} +\Phi_{yyp}p^{2} = \Phi_{xxp} + 2p\Phi_{xyp} +\Phi_{yyp}p^{2}\\ \text{(2)} \quad &\Phi_{xpp} + p\Phi_{ypp} + 2\Phi_{yp} = \Phi_{xpp} + p\Phi_{ypp} + 2\Phi_{yp} \end{align} $$

 * These two equations are equal to the expansions of equations (2a) and (2b) that we performed earlier. Therefore, we've shown that the general exactness condition specialized to second-order equations is equivalent to the specific 2nd order conditions derived earlier in the course.

 Good work. Your solution could be more concise but it is complete. --Egm6321.f09.TA 01:36, 16 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.palubin.d 03:03, 7 October 2009 (UTC)
 * Egm6321.f09.team4.sls 19:48, 7 October 2009 (UTC)

Problem #6:
(Lecture 14-2)

 * Given :The form of the Legendre linear 2nd order ODE with varying coefficients:

$$F=(1-x^2)y^{\prime\prime}-2xy^{\prime}+n(n+1)y=0$$

 * Find :

1)Verify exactness using:

a)Equations 4 and 5 on p. 10-3

b)Equation 5 on p. 14-1

2)If not exact, see whether it can be made exact by using the integrating factor method with $$h(x,y)=x^my^n$$

Solution:

1a) Equations 4 and 5 on p. 10-3 are as follows:

 * $$ \begin{align}

\text{(a)} \quad & f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y} \\ \text{(b)} \quad & f_{xp}+pf_{yp}+2f_{y}=g_{pp} \end{align} $$

Where $$f=\Phi_p=(1-x^2)$$ and $$g=\Phi_x+\Phi_y p=n(n+1)y-2xp$$

Solving for the partial derivatives, we have:

$$f_x=-2x$$

$$f_y=0$$

$$f_{xx}=-2$$

$$f_{xy}=0$$

$$f_{yy}=0$$

$$f_{xp}=0$$

$$f_{yp}=0$$

$$g_x=-2p$$

$$g_y=n(n+1)$$

$$g_p=-2x$$

$$g_{xp}=-2$$

$$g_{yp}=0$$

$$g_{pp}=0$$

Substituting into the original expressions we get:

$$f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}\Rightarrow -2+0+0\neq -2+0+n(n+1)$$

and

$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}\Rightarrow 0+0+0=0$$

The first condition is not met, therefore the Legendre ODE is not exact.

1b) Equation 5 on p.14-1 is:

$$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

And by definition, $$f_i=\frac{\partial F}{\partial y^{(i)}}$$

From this we see that:

$$f_0=\frac{dF}{dy}=n(n+1)$$

$$f_1=\frac{dF}{dy'}=-2x$$

$$f_2=\frac{dF}{dy''}=(1-x^2)$$

Substituting into the exactness condition we get:

$$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0\Rightarrow n(n+1)-\frac{d}{dx}( -2x )+\frac{d^2}{dx^2}(1-x^2)=n(n+1)+2-2=n(n+1)\neq 0$$

Therefore the Legendre ODE is not exact.

2) To make the Legendre ODE exact, we attempt to use an integrating factor of the form $$x^m y^n$$, which yields:

$$F=x^My^N(1-x^2)y''-2x^My^Nxy'+x^My^Nn(n+1)y$$

This simplifies to:

$$(x^My^N-x^{M+2}y^N)y''-2x^{M+1}y^Ny'+x^My^{N+1}n(n+1)=0$$

Working in a similar fashion to part 1b, we get:

$$f_0=\frac{dF}{dy}=(N+1)x^My^Nn(n+1)$$

$$f_1=\frac{dF}{dy'}=-2x^{M+1}y^N$$

$$f_2=\frac{dF}{dy''}=x^My^N-x^{M+2}y^N$$

Substituting in to the original exactness condition we get:

$$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=(N+1)n(n+1)x^My^N-(-2(M+1)x^My^N)+\frac{d}{dx}(Mx^{M-1}y^N-(M+2)x^{M+1}y^N)$$

Solving for the last derivative gives:

$$(N+1)n(n+1)x^My^N+2(M+1)x^My^N+M(M-1)x^{M-2}y^N-(M+2)(M+1)x^My^N=0$$

We can simplify the problem by pulling out the term $$x^My^N$$ giving:

$$(N+1)n(n+1)+2(M+1)+M(M-1)x^{-2}-(M+2)(M+1)=0$$

We can see that in order to eliminate n(n+1) from this equation, N must equal -1. M can be solved for by trial and error. Attempting the solution where M=0 yields:

$$2(M+1)+M(M-1)x^{-2}-(M+2)(M+1)\Rightarrow 2(1)+0-(2)(1)=0$$

Therefore, the Legendre ODE can be made exact using the integrating factor $$h(y)=y^{-1}$$

Contributors:
 * Egm6321.f09.team4.palubin.d 16:24, 7 October 2009 (UTC)

Problem #7:
(Lecture 14-3)

 * Given :That an operator $$L(\cdot )$$ is linear if and only if:

1)For all u(x),v(x), $$L(u+v)=L(u)+L(v)$$

and

2)For all $$\lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)$$

 * Find : Show that the above conditions for linearity are equivalent to equation 3 on p.3-3:

$$\forall\alpha,\beta \in\mathbb{R}$$, $$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$$ where u and v are any functions of x

Solution:

Define A and B such that: $$A=\alpha u$$ and $$B=\beta v$$

A and B are therefore still both functions of x, and applying the first given relation (equation 1 on p.14-3) we can write:

$$L(A+B)=L(A)+L(B)$$ or $$L(\alpha u + \beta v)=L(\alpha u) + L(\beta v)$$

Applying the second relation (equation 2 on p.14-3) and considering each term individually, we get:

$$L(\alpha u) + L(\beta v)=\alpha L(u) + \beta L(v)$$

Therefore the combination of relations 1 and 2 on p.14-3 are equivalent to equation 3 on p.3-3,

$$L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)$$

 clear explanation. --Egm6321.f09.TA 04:04, 16 October 2009 (UTC)

Contributors:
 * Egm6321.f09.team4.palubin.d 23:21, 6 October 2009 (UTC)

Problem #8:
Lecture 15-2

 * Given :
 * Shape functions for $$N_{j}^1(x),N_j^2(x),N_{j+1}^1(x)$$

 * Find :
 * Plot the shape function for $$N_{j+1}^2(x)$$

 * Solution :

 * [[File:Egm6321_hw3_team4.PNG]]

correct. Egm6321.f09.TA 03:07, 28 October 2009 (UTC)

Contributors: :Egm6321.f09.team4.sls 17:32, 7 October 2009 (UTC) :Egm6321.f09.team4.palubin.d 17:32, 7 October 2009 (UTC)

Problem #9:
(Lecture 16-2)

 * Given : $$y(x)=y(x(t))$$ and $$x(t)=e^t$$

 * Find :Show that:

$$y_{xxx}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$$

and

$$y_{xxxx}=e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

Solution:

We start from equations 5 on p.16-1 and 7 on p.16-2:

$$y_x=e^{-t}y_t$$

$$y_{xx}=e^{-2t}(y_{tt}-y_t)$$

Solving for $$y_{xxx}$$ yields:

$$y_{xxx}=\left ( y_{xx} \right )_x=\frac{d}{dt}(e^{-2t}(y_{tt}-y_t))\frac{dt}{dx}$$ where $$\frac{dt}{dx}=e^{-t}$$

Applying the chain rule, we get:

$$\left [ \left ( -2e^{-2t} \right )\cdot \left ( y_{tt}-y_t \right )+(e^{-2t})\cdot (y_{ttt}-y_{tt}) \right ]e^{-t}=e^{-3t}(-2y_{tt}+2y_t+y_{ttt}-y_{tt})$$

Which equals the solution

$$y_{xxx}=e^{-3t}(y_{ttt}-3y_{tt}+2y_t)$$

Solving for $$y_{xxxx}$$ is similar. We start from:

$$y_{xxxx}=(y_{xxx})_x=\frac{d}{dt}(e^{-3t}(y_{ttt}-3y_{tt}+2y_t))\frac{dx}{dt}$$

Substituting in $$\frac{dt}{dx}=e^{-t}$$ and applying the chain rule we get:

$$\left [ (-3e^{-3t})\cdot (y_{ttt}-3y_{tt}+2y_t)+(e^{-3t})\cdot (y_{tttt}-3y_{ttt}+2y_{tt}) \right ]e^{-t}$$

which equals:

$$e^{-4t}(-3y_{ttt}+9y_{tt}-6y_t+y_{tttt}-3y_{ttt}+2y_{tt})$$

This simplifies down to the final solution:

$$y_{xxxx}=e^{-4t}(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

good - very legible. Egm6321.f09.TA 03:25, 28 October 2009 (UTC)

Contributors:

Egm6321.f09.team4.palubin.d 22:10, 6 October 2009 (UTC)

Problem #10:
(Lecture 16-4)

 * Given : The equation $$x^2 y^{\prime\prime} -2xy^{\prime}+2y=0$$

 * Find :
 * (a)Try to solve for this equation directly using the method of trial solutions with $$y=e^{rx}$$. For boundary conditions, set y(1)=3 and y(2)=4.
 * (b)Compare with the solution on p.16-3: $$y(x)=C_1 x^{r_1}+C_2 x^{r_2}$$
 * (c)Use Matlab to plot the solution.

 * Solution (a) :

 * $$y=e^{rx}$$
 * $$y^{\prime}=re^{rx}$$
 * $$y^{\prime\prime}=r^2e^{rx}$$

 * Substitute back in,

 * $$x^2(r^2e^{rx})-2xre^{rx}+2e^{rx}=0$$
 * $$e^{rx}(r^2x^2-2xr+2)=0$$
 * $$r^2x^2-2xr+2=0$$

 * Find roots,

This equation is quadratic in 2 variables. You cannot use the quadratic equation here. Egm6321.f09.TA 03:50, 28 October 2009 (UTC)

 * $$r_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
 * $$r_{1,2}=\frac{2\pm \sqrt{2^2-4(1)(2)}}{2(1)}$$
 * $$r_{1,2}=1\pm i$$

 * Solution has the form,

 * $$y=e^{1\pm i}$$
 * $$y=e^{x}e^{\pm i}$$
 * $$y=e^{x}(C_{1}cos(x)+C_{2}sin(x))$$

 * Apply boundary conditions,

 * $$3=e^{1}(C_{1}cos(1)+C_{2}sin(1))$$

 * and,

 * $$4=e^{2}(C_{1}cos(2)+C_{2}sin(2))$$

 * we find that,

 * $$C_{1}=\frac{3}{e^{1}cos(1)}-C_{2}tan(1)$$

 * and,

 * $$C_{2}=\frac{1}{tan(1)-tan(2)}\left ( \frac{3}{e^{1}cos(1)}-\frac{4}{e^{2}cos(2)} \right )$$

 * these are simply constants that correspond to $$C_{1}=0.6513$$ and $$C_{2}=0.8934$$

 * Solution:
 * $$y=e^{x}(C_{1}cos(x)+C_{2}sin(x))$$
 * $$C_{1}=\frac{3}{e^{1}cos(1)}-C_{2}tan(1)$$
 * $$C_{2}=\frac{1}{tan(1)-tan(2)}\left ( \frac{3}{e^{1}cos(1)}-\frac{4}{e^{2}cos(2)} \right )$$

 * Solution (b) :
 * We see that the solution has the same form as Equation (10) of 16-3; however, in this instance the polynomial solution is expressed in terms of an exponential function where $$x$$ has been substituted with $$e^{x}$$.

 * Solution (c) :
 * The plot of the solution is as follows:

 * Contributors:
 * Egm6321.f09.Team4.meyerat 11:23, 7 October 2009 (UTC)

Problem #11:
(Lecture 17-3)

 * Given :$$u_1(x)z^{\prime}+(a_1(x)u_1(x)+2u_1^{\prime}(x))z=0$$

 * Find :Solve for z using the integrating factor method

 * Solution :

 * $$u_1(x)z^{\prime}+(a_1(x)u_1(x)+2u_1^{\prime}(x))z=0$$
 * $$z^{\prime}+\frac{(a_1(x)u_1(x)+2u_1^{\prime}(x))}{u_1(x)}z=0$$
 * $$z^{\prime}+\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )z=0$$

 * Choose an integrating factor $$h(x)$$ and multiply through,

 * $$h(x)z^{\prime}+\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )h(x)z=0$$

 * And force $$h(x)$$ such that,

 * $$h^{\prime}(x)=\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )h(x)$$

 * $$\frac{h^{\prime}(x)}{h(x)}=\left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)} \right )$$

 * $$ln\left (h(x) \right )=\int \left (a_1(x)+\frac{2u_1^{\prime}}{u_1(x)}  \right )dx$$

 * $$ln\left (h(x) \right )=2ln(u_1(x))+\int a_1(x)dx+C$$

 * $$h(x)=e^{(2ln(u_1(x))+\int a_1(x)dx)+C}$$

 * $$h(x)=C_1(u_1(x))^{2}e^{\int a_1(x)dx}$$

 * Going back, recall we chose $$h(x)$$ such that,

 * $$h(x)z^{\prime}+h^{\prime}(x)z=0$$

 * Therefore,

 * $$\frac{\partial }{\partial x}\left (h(x)z) \right )=0$$

 * $$h(x)z=C_2$$

 * Plug in our expression for $$h(x)$$,

 * $$z=\frac{C_2}{C_1\left(u_1(x) \right )^{2}e^{\int a_1(x)dx}}$$

 * $$z=C_3(u_1(x))^{-2}e^{-\int a_1(x)dx}$$

 * (Answer)

very good. Egm6321.f09.TA 04:06, 28 October 2009 (UTC)

 * Contributors:
 * Egm6321.f09.Team4.meyerat 11:24, 7 October 2009 (UTC)

Problem #12:
(Lecture 18-1)

 * Given :

 * $$y^{\prime\prime}+a_1(x)y^{\prime}+a_0(x)y=0$$ and

 * a) $$y(x)= U(x)\pm u_1(x)$$

 * b) $$y(x)= \frac{U(x)}{u_1(x)}$$

 * c) $$y(x)= \frac{u_1(x)}{U(x)}$$

 * Find : Develop a reduction of order method 2 using algebraic operations (a), (b), and (c) above.

 * Solution :

 * Part (a)
 * $$y(x)= U(x)\pm u_1(x)$$
 * $$y^{\prime}(x)= U^{\prime}(x)\pm u_1^{\prime}(x)$$
 * $$y^{\prime\prime}(x)= U^{\prime\prime}(x)\pm u_1^{\prime\prime}(x)$$

 * Plug in:

 * $$U^{\prime\prime}(x)\pm u_1^{\prime\prime}(x)+a_1(x)\left (U^{\prime}(x)\pm u_1^{\prime}(x) \right )+a_0\left (U(x)\pm u_1(x)  \right )=0$$

 * Factoring out the $$u_1(x)$$ homogenous solution, we are left with:

 * $$U^{\prime\prime}(x)+a_1(x)U^{\prime}(x)+a_0U(x)=0$$

 * We have not reduced the order of the equation; we did not achieve an ODE with a "missing" dependent variable. Therefore, operation (a) cannot be used in reduction of order processes.

 * Part (b)
 * Drop "(x)" for convenience only
 * $$y= \frac{U}{u_1}$$
 * $$y^{\prime}= \frac{-U}{u_1^{2}}u_1^{\prime}+\frac{U^{\prime}}{u_1}$$
 * $$y^{\prime\prime}= \frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-\frac{U'u_1^{\prime}}{u_1^2}-\frac{u_1'^2}{u_1^2}+\frac{U''}{u_1}$$

 * Plug back into ODE:
 * $$y^{\prime\prime}+a_1y^{\prime}+a_0y=0$$

 * $$\left (\frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-\frac{U'u_1^{\prime}}{u_1^2}-\frac{u_1'^2}{u_1^2}+\frac{U''}{u_1} \right )+a_1\left (\frac{-Uu_1'}{u_1^{2}}+\frac{U^{\prime}}{u_1}  \right )+a_0\frac{U}{u_1}=0$$

 * Group terms,

 * $$\frac{U''}{u_1}+\left (a_1\frac{U^{\prime}}{u_1}-\frac{U'u_1^{\prime}}{u_1^2} \right )+\left (\frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-a_1\frac{Uu_1'}{u_1^{2}}+a_0\frac{U}{u_1}  \right )-\frac{u_1'^2}{u_1^2}=0$$

 * The only way to have a reduction in order is if

 * $$\left (\frac{2Uu_1'^{2}}{u_1^{3}}-\frac{Uu_1^{\prime\prime}}{u_1^2}-a_1\frac{Uu_1'}{u_1^{2}}+a_0\frac{U}{u_1} \right )=0$$

 * Working on this

 * $$U\left (\frac{2u_1'^{2}}{u_1^{3}}-\frac{u_1^{\prime\prime}}{u_1^2}-a_1\frac{u_1'}{u_1^{2}}+\frac{a_0}{u_1} \right )=0$$

 * $$a_0u_1^2+2u_1'^{2}-a_1u_1u_1'-u_1''u_1=0$$

 * $$u_1''+a_1u_1'-a_0u_1-\frac{2u_1'^{2}}{u_1}=0$$

 * However, because $$u_1$$ is a solution to the original such that

 * $$u_1''+a_1u_1'+a_0u_1=0 $$

 * We know that,

 * $$u_1''+a_1u_1'-a_0u_1-\frac{2u_1'^{2}}{u_1}\neq0$$

 * Answer.

 * Part (c)

 * Similar to Part (b), but here $$u_1$$ is replaced by $$U$$ and vice versa:

 * $$y= \frac{u_1}{U}$$
 * $$y^{\prime}= \frac{-u_1}{U^{2}}U^{\prime}+\frac{u_1^{\prime}}{U}$$
 * $$y^{\prime\prime}= \frac{2u_1U'^{2}}{U^{3}}-\frac{u_1U^{\prime\prime}}{U^2}-\frac{u_1'U^{\prime}}{U^2}-\frac{U'^2}{U^2}+\frac{u_1''}{U}$$

 * Plug back into ODE:
 * $$y^{\prime\prime}+a_1y^{\prime}+a_0y=0$$

 * $$\left (\frac{2u_1U'^{2}}{U^{3}}-\frac{u_1U^{\prime\prime}}{U^2}-\frac{u_1'U^{\prime}}{U^2}-\frac{U'^2}{U^2}+\frac{u_1''}{U} \right )+a_1\left (\frac{-u_1}{U^{2}}U^{\prime}+\frac{u_1^{\prime}}{U}  \right )+a_0\frac{u_1}{U}=0$$

 * Group terms,

 * $$-\frac{u_1U^{\prime\prime}}{U^2}+\left (\frac{2u_1U'^{2}}{U^{3}}-\frac{u_1'U^{\prime}}{U^2}-\frac{U'^2}{U^2}-a_1\frac{u_1U'}{U^{2}} \right )+a_1\frac{u_1^{\prime}}{U}+a_0\frac{u_1}{U}+\frac{u_1''}{U}=0$$

 * Simplify,

 * $$-u_1U^{\prime\prime}+\left (\frac{2u_1U'^{2}}{U}-u_1'U^{\prime}-U'^2-a_1u_1U' \right )+U\left (a_1u_1^{\prime}+a_0u_1+u_1''  \right )=0$$

 * $$U^{\prime\prime}+\left (\frac{-2U'^{2}}{U}+\frac{u_1'U^{\prime}}{u_1}+\frac{U'^2}{u_1}+a_1U' \right )-\frac{U}{u_1}\left (a_1u_1^{\prime}+a_0u_1+u_1''  \right )=0$$

 * And although the last term goes to zero, it does not eliminate the $$U$$ term in the second term. Therefore, we cannot perform a reduction of order for Part (c).

 * Answer

very good. Egm6321.f09.TA 04:16, 28 October 2009 (UTC)

 * Contributors:
 * Egm6321.f09.Team4.meyerat 11:26, 7 October 2009 (UTC)

Problem #13:
(Lecture 18-2)

 * Given :

 * Equation 1: $$(1-x^2)y''-2xy'+2y=0$$
 * a) $$y=ax^b$$
 * b) $$y=e^{rx}$$

 * Find : $$u_1$$ and $$u_2$$ of Equation (1) using trial solutions (a) and (b).

 * Solution :

 * Trial Solution (a)

 * $$y=ax^b$$
 * $$y'=bax^{b-1}$$
 * $$y''=b(b-1)ax^{b-2}$$

 * Plug into Equation 1,

 * $$(1-x^2)b(b-1)ax^{b-2}-2xbax^{b-1}+2ax^b=0$$
 * $$(1-x^2)b(b-1)x^{b-2}-2xbx^{b-1}+2x^b=0$$
 * $$(1-x^2)b(b-1)x^{b-2}-2(b-1)x^{b}=0$$

 * This is only true if b=1. Therefore,

 * $$y=ax$$

 * The value of a will be determined later.

 * $$u_1=ax$$

 * To find $$u_2$$, we then perform the following operation

 * $$y=Uu_1$$
 * $$y'=U'u_1+Uu_1'$$
 * $$y=Uu_1+2U'u_1'+u_1''U$$

 * Plug into Equation (1) and get,

 * $$(1-x^2)\left (Uu_1+2U'u_1'+u_1U \right )-2x(U'u_1+Uu_1')+2(Uu_1)=0$$

 * $$(1-x^2)Uu_1+U'\left ((1-x^2)2u_1'-2xu_1 \right )+U(-2xu_1'+2u_1+(1-x^2)u_1)=0$$

 * $$(1-x^2)Uu_1+U'\left ((1-x^2)2u_1'-2xu_1 \right )+\underset{0}{\underbrace{{}U(-2xu_1'+2u_1+(1-x^2)u_1)}}=0$$

 * $$U''+U'\left (\frac{2u_1'}{u_1}-\frac{2x}{(1-x^2)} \right )=0$$

 * With a change of variables $$U'=Z$$, we can show that,

 * $$U=\int Z=C_3u_1+C_2u_1\int \frac{1}{u_1^2}exp\left (\int\frac{2x}{1-x^2}dx \right ) dx$$

 * $$U=C_3u_1+C_2\underset{u_2}{\underbrace{u_1\int \frac{1}{u_1^2}exp\left (\int\frac{2x}{1-x^2}dx \right ) dx}}$$

 * Recall that $$y=Uu_1$$ and $$y=ku_1+k_2u_2$$. Therefore,

 * $$u_1=x$$
 * $$u_2=u_1\int \frac{1}{u_1^2}exp\left (\int\frac{2x}{1-x^2}dx\right ) dx$$

 * Which is evaluated as:

 * $$u_2=x\int \frac{1}{x^2(1-x^2)}dx$$

 * and

 * $$y(x)=C_3x+C_2u_1\int \frac{1}{u_1^2}exp\left (\int\frac{2x}{1-x^2}dx\right ) dx$$

 * ANSWER.

What about $$y=e^{rx}$$ ? Egm6321.f09.TA 05:34, 28 October 2009 (UTC)

 * Contributors:
 * Egm6321.f09.Team4.meyerat 13:04, 7 October 2009 (UTC)

title: User:Egm6321.f09.team4/HW3


text:
User:Kanit yartsangkad

My name is Mong Kanit Yartsangkad. Major :TIM Born in Bahrain,spend 5 years in Chiang mai. I love cars and wakeboarding.

CPU > Intel Core i7 Extreme 965 2.66 GHz > 38,500 Baht

Motherboard > ASUS P6T7 WS Supercom > 19,500 Baht

RAM > CORSAIR DOMINATOR DDR3 1866 MHz 6GB (Triple Channel) > 14,000 Baht

Graphic Card > ASUS GTX 295 1792MB GDDR3 (x3 3Way-SLI) > 59,700 Baht

Harddisk > G.SKILL SSD 256 GB FALCON SATA2 2.5" x6 RAID0 > 145,800 Baht

DVD-RW > LG 16x BE06LU Blu-Ray > 14,150 Baht  Sound Card > Creative Labs Sound Blaster X-Fi Elite Pro > 16,900 Baht

Power Supply > Enermax ERV1250EGT 1250W "Revolution" > 18,200 Baht

Floppy drive> no one use it anymore.

Case > THERMALTAKE Xpressar > 22,500 Baht

LCD > DELL Ultrasharp 24" LCD > 25,000 Baht

Speaker > Creative Gigaworks THX > 22,900 Baht

Mouse > Razer Mamba > 4,590 Baht

Keyboard > Razer Tarantula > 3,300 Baht

The differents between Twitter amd Facebook

Facebook is different from twitter in many ways. In twitter a user can choose follow to see another user's status,but for facebook, you have to add another user as a friend and wait for that user to accept youself first, then you came view another user's status updates and profile. Twitter is more easlier to play than facebook, but face has more features other than twitter has such as games, photo tags, etc.

Why is Twitter becoming a competitor with facebook ?

Twitter is becoming more competitive because it is easlier to get access to other user's without the other user's permission for access, which is different from "Facebook". With this advantage, Twitter is becoming more and more popular because a user like me can get access to any other user's status, updates,page easily.

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

title: User:Kanit yartsangkad


text:
User:KelleyP

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

I will plan an adventure hiking the Appalachian Trail for a future date.

 * Research Component

Research how to make proper arrangements to complete the hike.

 * week 1

What is a realistic goal for when I can take this journey? Approximately how long will I be away from home? Find info on the internet.

 * week 2

Are there groups/clubs I can join and hike with? Read 15 pages of "Walking the Appalachian".

 * week 3

How much money do I need to save to allow this much time off work/school and for supplies and gear? Read 15 pages of "Walking the Appalachian".
 * week 4

How will I ship supplies to towns I will visit? Who can I count on to send them? Read 15 pages of "Walking the Appalachian".

 * week 5

What supplies do I need from the start? What will I eat? Read 15 pages of "Walking the Appalachian".

 * week 6

Shopping/pricing gear and supplies. Route configuration with stops and supply list for each mail stop. Read 15 pages of "Walking the Appalachian". Training.
 * week 7

Set up post office pickups. Give family members plan. Read 15 pages of "Walking the Appalachian". Training.
 * week 8

Plan flight times and arrival at trail head. Read 15 pages of "Walking the Appalachian". Training.

Midterm Assignment 1: Taste & Aesthetics
Persephone Unbound stood paralyzed in the dirt under the dark clouds that showered her; no life there. This was the first encounter I had with her and I immediately disliked her. The diffused light failed to define her features, her figure was cold and stiff. I did not see anything in the sculpture before me.

As much as I tried to embrace this work I could not see any beauty in it. The material was dark and unmoving. Even the sheen from the rain did not provide any interest to me. It was the first sculpture I observed close up and I thought how ironic because I usually appreciate abstract art. I was surprised to find myself disliking this piece immediately. I moved on to the next sculpture, also created by Beverly Pepper, and it appealed to me as did most of the other sculptures in the park. I returned to Persephone after observing other pieces to make sure my dislike of her wasn’t caused by a need to “warm up” my brain. Not so; I did not like her still.

I wasn’t sure what I would find on Beverly Pepper’s website but I discovered the most amazing structures and sculptures I’ve ever seen. She’s designed amphitheaters and cool places to hang outside in public areas alongside her sculptures. The Land Art was my favorite of her work shown on the site. I was awestruck by the cast iron Palingenesis, it looked like a place I would go to get inspiration and recharge.

I felt bad for disliking Persephone after seeing what an amazing artist Beverly is. I began trying to associate characteristics of other pieces of hers I liked with the sculpture I didn’t like (to make myself like it). Next, I went to the Sculpture Park website to observe Persephone Unbound after my good experience with Beverly’s other works. First of all, the picture posted on the site was taken on a sunny day. That factor immediately changed the way the sculpture appeared to me. The sunlight created shadows where there were none on the cloudy day and gave depth to the piece. It created the lines of Persephone’s face where I didn’t see them before and the form of a woman began to appear in the sculpture.

Although I wasn’t completely successful in changing my own mind, I was able to appreciate the piece after learning more about the artist and observing the sculpture in different ways, and from different light-sources. I am not surprised by my results because I can usually find something to appreciate or a lesson in all my experiences. I can now see Persephone Unbound in the sculpture but it was still my least favorite in the park and of the work I have seen by Beverly Pepper .

Final Presentation

 * Topic (10%)

I have always wanted to plan a thru-hike on the Appalachian Trail, a 2,175 mile journey from Georgia to Maine.

 * Research (20%)

 * 1) Internet site:Home-Appalachian Trail Conservancy. Web. 07 Nov. 2009. . Appalachian Trail. Web. . Blazin' the Trail. Web. .
 * 2) Book:Luxenberg, Larry. Walking the Appalachian Trail. Mechanicsburg, PA: Stackpole Books, 1994. Print.
 * 3) Scholarly Journal:Ron Potts. ""Fellow travelers along the trail. "" The World & I 18.8 (2003): 267. Proquest Research Library. Web. 24 Nov. 2009.
 * 4) Armstrong, Brian, prod. ""Appalachian Trail"" America's Wild Spaces. Dir. Todd Hermann. National Geographic Channel. NGC, 10 Nov. 2009. Television.

 * Your accomplishment (20%)

I was able to plan my daily miles and where I will camp each night as well as what locations I will resupply and how many days of supplies I will need. I made my gear list and found out how to get to the trailhead and back after completion. I read my book and found out what trail guide is best to use while on the trail. I figured out about how much money I will need to cover bills and how much I will need to have on the trail. I didn't figure out how much I would spend on gear but I have most of it already.

 * Project Review (50%)
 * 1)What went well: I had an easy time finding information to help plan everything. I really enjoyed putting the powerpoint together because I saw so many pictures from the trail and read many blogs written by people who have hiked the trail. Reading what people say about the hike and also re-reading my original book got me excited to hike the trail.
 * 2)What went poorly: I had planned on doing the cardio training but didn't have time or extra motivation since I knew I wasn't actually hiking the trail anytime soon. I had also planned on figuring out what supplies I would mail drop to myself while hiking but didn't because I read that mail drops are not necessary.  I didn't find anyone who wants to do it with me but later realized it's probably better that way. I didn't plan an exact date to fly out or start but I know it will be around April 1.  I didn't price gear which is helpful to determine total cost of the adventure but reading other's information gave me a pretty good idea plus I have a lot of gear already.
 * 3)Future strategies: In the future when planning to achieve a goal I will probably not create a schedule because I don't use them. I didn't really look at mine during the process but that could partly be due to the fact that I keep all the things I have to get done in a mental list. I will try not to procrastinate in future endeavors because that is a character flaw of mine although I didn't do too bad on this one.  When I plan to do the Pacific Crest Trail I will try not to rely as much on planning that other people have posted to help me plan my own trip. It makes it easy to slack a bit on research because the info is given to you.

title: User:KelleyP


text:
User:Cscully

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

<>
 * Research Component

I will explain the difference between Type 1 and Type 2 diabetes.

<>
 * Start testing my blood sugar more regularly, preferably 4 times a day and most importantly in the morning and at night. I will also start to record my numbers more frequently and keep a self monitoring diary. I will use all the advantages of keeping notes about my day, such as, if I had a low blood sugar, if I ate in the middle of the night, or if I forgot to give insulin etc. Also in this six week time span I will be playing around with the dosage of my long acting insulin, for example should I take 17 units in the morning, or split it up to 10 units in the morning and at night. This will have a substantial effect on my A1C levels because I am still in the process of seeing which dosages work best for me. I was recently diagnosed with type 1 diabetes in December of 2007 since then I haven't done my best to keep track of my blood sugars and take my insulin at the correct times. Therefore making these adjustments to my life on the daily, instead of when "i feel like it" will be a rewarding experience that I am very ready to adjust to. Another challenge I might encounter during this project is feeling low when my blood sugar is actually normal.

<>
 * During this week I will still be taking 17 units of my long acting insulin in the mornings. My doctor told me to do this for about two to three weeks and if my blood sugars are still high in the morning, that means the insulin isn't staying in my body for 24 hours. Therefore I should split it up to 10 units in the morning, and ten at night. My biggest challenge with this is at night I tend to forget about the long acting. After this week if my numbers are still high in the morning I will make this adjustment. I should still be trying to test as much as possible. This will be a bit of a challenge for me because i tend to forget, especially before bed

<>
 * This week I will decide whether to change my long acting dosage. I will also make it a must to test 4 times a day. I will start to research the differences between Type 1 and Type 2 diabetes, I find this very interesting because people tend to have a one sided view on diabetes, that it only happens to overweight or old people. I want to teach more about juvenile diabetes which I feel tends to get pushed aside when people hear the word diabetes

<>
 * This week I will continue researching. Along with explaining the differences between Type 1 and Type 2, I also want to research on how far care for diabetes has come in the past 40 years and maybe look into what is in store for the future, such as a cure, or an easier way to manage it, and learn about the risks if it goes untreated for too long

<>
 * This will be a week before my doctors appointment so I want to make sure I am doing my best to record because with out recordings the doctors can't really determine much. By now I will have found which dosage of long acting insulin works best for me. Also once I learn more details about how the final project is going to be presented i will finish up researching and start to plan my final presentation.

<>
 * By now I will have had about 6 weeks since my last doctors appointment to make a significant change in my blood sugar levels and hopefully have brought my A1C level down. Although I'm shooting for 7 anything below ten will be good too. On Tuesday the 17th I will go to Renton for my follow up appointment, get my blood work done and within the next 7 days I should have my results.

<>
 * I should have received my blood work and will find out if i reached my goal! If I did, then my new routine I have adapted from this project shouldn't stop there. I will continue to do these things more seriously throughout my life and eventually have an A1C of 6. This week I will finish up my project and all of the formalities so I can be prepared to give my practice presentation week 8.

<>
 * I will give my practice presentation to my group in class. Once I receive feedback from them I will make the final adjustments to my project so I am ready for the presentation.

Midterm Assignment 1: Taste & Aesthetics
Sculptures have always been hard for me to fall in love with. One sculpture I found that I wasn’t fond of at the Olympic Park was a piece called Rivera by artist Anthony Caro. I had a problem with its abstract parts and no real end use. I wouldn’t say that I was able to manipulate myself into suddenly liking the piece but after some research I was able to gain a more open perspective to this style of work. After learning about the artists past, how the sculptures are made, and reviewing some of his other pieces I was able to appreciate things about this sculpture that other people might find appealing. So in a way I suppose that I was able to manipulate myself into changing my mind. I feel that if people are ignorant to a subject they can be quick to judge. But with more knowledge, stories, and or images about the subject it can be very easy for a person to change their mind without even realizing why they took the stance they did in the first place. I feel that something like this took place with me when I was reviewing the sculpture. At first glance I had already made up my mind it was ugly and I felt no emotion. But after gaining some facts and history I was now starting to have feeling associated with the sculpture which then changed the way I viewed it. I wasn’t just looking at it with my eyes now; I was looking at it with feelings, opinions, and knowledge. I am not depressed that my mind was able to change the way it did. I believe I can control my own brain but when new ideas are presented to me it allows me to question my old views.

Midterm Assignment 2: Manipulation
1. http://www.youtube.com/watch?v=tykkLaFpKok

This is an example of Dubious Authority because they are using the dermatologist in the ad to make claims about the product.

2. My second example stems from a conversation I had with my aunt, she made the statement "Barack Obama is Muslim". This is an example of Alness because it is a statement made without evidence to back it up.

Final Presentation

 * Topic (10%)

<>

 * Research (20%)

<>

 * 1) Internet site: American Diabetes Association. "Type 2." Diabtetes.org. Web. 13 Dec. 2009.

MedicineNet. "Hemoglobin A1C definition - Medical Dictionary definitions of popular medical terms easily defined on MedTerms." Web. 14 Dec. 2009.

Nutrition Click®. "Type 1 Diabetes Type 2 Diabetes." Diabetes Planner - Diabetes Menus Grocery Lists Diabetes Recipes & Nutrition Facts. Web. 14 Dec. 2009.

 * 2) Book: N/A

 * 3) Scholarly Journal: N/A

 * 4) Other (interview, TV show,etc): Http://www.encognitive.com. "Diabetes, Type 1, Type 2, Glucose, Insulin." Video blog post. Youtube.com. Web. 13 Dec. 2009.

 * Your accomplishment (20%)

I believe I am definitely on the right track to reaching my final goal. I know that if I continue to practice what I acclimatized throughout my project I will have no problem successfully lowering my A1C in the future. Besides from the major goal of the project, I feel I have also reached tinier ones that I set up in order to create the final goal. I have productively accomplished the smaller ones. Therefore I may have not reached my goal in 8 weeks, but I am on the appropriate path towards achievement.

 * Project Review (50%)
 * 1) There were a few things that went really well for me. One of those was getting back on a daily insulin routine. Knowing about this project and having a goal really helped me take action. I remembered doing daily tasks that I had set aside in the past. Since the beginning of the project I have successfully incorporated things such as carb counting and taking my insulin at appropriate times. All these things are key steps that I needed to perform in order to get to my end result. I also discussed in my plan something about changing my insulin regimen half way through. I did in fact change it and it went over smoothly and works about better as opposed to the dosage I was taking when I began the project. So basically adjusting to an insulin regimen, carb counting, and testing all went well for me. In the end the doctor wasn’t able to test my A1C at that particular appointment. She wanted me to wait two more months, but from the numbers I did have recorded the drop in high blood sugars is obvious and I am headed down the right path.

 * 2) Even though I now have my blood sugars under control there were a couple things that went poorly or didn't go quite as intended. One being that I still don’t test up to four times a day. I tested barely enough to provide information on how my body was reacting to the insulin, but still not as much as I should. The more I test the better I will know my body. Even though they can draw my blood to see where my blood sugar levels are at, basically an A1C, it is helpful to have numbers written down though. This was the major thing I struggled with throughout the whole project. I also planned to have my A1C tested at my doctor’s appointment, but unfortunately that didn't go as planned. This kind of upset me because the timeline of the date when the project was given and the date of my doctors appointment matched up so well. It was one of the main reasons I had chosen this as my topic in the first place. But I have to wait two more months. So I can’t say that I reached my goal exactly, but I am taking the appropriate steps.

 * 3) This whole project was beneficial in many ways. Knowing that in the end I would have to share this with the class really helped with my motivation. I also feel a lot healthier. Since these tasks are now second nature to me I have no problem with forgetting things. Since I will have diabetes for the rest of my life everything I have learned and practiced in this project will continue to be utilized every day. It was really nice getting back on track and organized. My newly adapted, but long overdue, routine will not stop here. I will continue to practice and set goals such as lowering my A1C, throughout my life, to ensure a future of health and happiness.

Self-Assessment
General Education was actually one of my classes that I really enjoyed. It was a really good class to have during my first quarter at the Art Institute because it gave a preview of what to expect but comfortably allowed us to ease in. I really enjoyed all of the information that you provided about the library, free websites to help with school, and other various information about the city. All of this has helped me with other subjects as well. I don’t feel that I could easily find this information anywhere else and I appreciate you taking the time to show all of your students step by step. I feel the class accurately summarized what to look for when researching and how to distinguish a good source from a bad source. The final project I really liked and I also benefited personally from it. I feel this class has prepared me for future classes. As a student I feel that I could have participated more and this is something I will continue to work on. I believe if I had participated more then I might have contributed to the experiences of another student in the class, but as of now I do not feel that I have. Looking ahead I know that I have a lot left to learn, but I definitely feel more readily prepared.

title: User:Cscully


text:
User:Kaseyc~enwikiversity

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

Completely read Dante Alighieri's Divine Comedy.
 * Research Component

Look up information Alighieri, where he was in his life while wrighting the epic, if there were any parts of his life that affected his writing of the epic, and if he used a certain writing process.
 * week 1

 * Read to page 60

 * week 2

 * Read to page 160

 * week 3

 * Read to page 220
 * Conclude research on Alighieri's overall biography (born, lived, major events in life, etc)

 * week 4

 * Read to page 280

 * week 5

 * Read to page 340
 * Conclude research on whether there were any moments in Dante's life that affected his writing of the book, and if he used a certain writing style

 * week 6

 * Read to page 400
 * Make sure all research is completed, and begin creating presentation

 * week 7

 * Read to page 460

 * week 8

 * Finish book
 * Finish presentation and practice

Midterm Assignment 1: Taste & Aesthetics
The sculpture I decided to focus on was the Love & Loss sculpture by Roy McMakin. I think the sculpture is great, except for one thing: the rotating “&” symbol. I really like the concept behind the sculpture and how it is interactive and looks great, but the “&” really ruined it for me. I don’t feel like it fits very well with the sculpture at all. The sculpture is mostly concrete, aside from the “V” tree and you can interact with it in various ways. For example: you can sit on the “e” and both the “L”s. Everything also just flows so well together. But then there’s just this “&” symbol that doesn’t really fit with everything else. After doing research on the sculpture, I learned more about the idea behind it. McMakin says
 * “My sculptures are functional but they also have content that you discover through participation. You need to put the meaning together by sitting on it, 	walking through it, and then recomposing what you have seen and experienced in your mind’s eye. Only then will the theme—the process of love and loss—be revealed for you to sit here and contemplate.”

I really like this concept; how everything is interactive and works together; the “V” tree also blossoms seasonally, a cycle reminiscent of the human experience. At the same time though, all this makes the “&” even more pointless. To me, it doesn’t seem to serve any function interactively or an idea that we can relate to. It also doesn’t go along with the kind of concrete them that the rest of the sculpture has (besides the “V” tree). So in the end, I have decided that it’s a great sculpture, but the whole “&” symbol still kind of ruins it for me.

Final Presentation

 * Topic (10%)

Dante's poem "The Divine Comedy"

 * Research (20%)

<>

 * 1) Internet site:

"The Princeton Dante Project (2.0)." Almagest - A multimedia database for teaching & learning. Web. 22 Nov. 2009.
 * 2) Book:

Alighieri, Dante. The Divine Comedy. Trans. C. H. Sisson. New York: Oxford UP, 2008. Print.

 * Your accomplishment (20%)

I was only able to read through Inferno and Purgatorio, not Paradiso. I wasn't able to keep up with all the reading that I needed to get done, and should have set some more realistic small-goals. Presentation: read from Inferno, Canto III, lines 22-33

 * Project Review (50%)
 * 1)What went well

I was able to get through most of my project and get all of my research done. I enjoyed reading the poem and learning about all its intricate complexities.
 * 2)What went poorly

I wasn't able to keep up with the reading goals I had set, and thus wasn't able to complete me overall goal of reading all three sections of the poem.
 * 3)Future strategies

If there's a next time, I'll have to set a more realistic weekly reading goal. Instead of planning on reading every night, I'll have to break it up into days. I'll also have to not plan on reading over the weekend or any breaks that I'll have (such as Thanksgiving break). This way I can figure out a more realistic plan of action, and how much I'll actually be able to read.

title: User:Kaseyc~enwikiversity


text:
User:Jacquelinec~enwikiversity

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

The Art of Tattooing
 * Research Component

I will be researching how to tattoo and become a tattoo artist.
 * week 1

Research the history of tattooing and what it symbolized. How has it changed into modern days.
 * week 2

Research modern day culture and tattoos. Why is it popular?
 * week 3

What tools do I need for tattooing, and where can I get them? What training/schooling is needed to become a tattoo artist?
 * week 4

Get tattoo equipment and order "skin" sheets. Read tattoo magazines and start drawing flash.
 * week 5

Interview a tattoo artist/shop owner.
 * week 6

Start practicing tattooing. Start putting research information together.
 * week 7

Practice. Try to find a place to be an apprentice at or work at.
 * week 8

Put presentation together. State out future goals.

Final Presentation

 * Topic (10%)

< >

 * Research (20%)

<>

 * 1) Internet site:
 * 2) Book:
 * 3) Scholarly Journal:
 * 4) Other (interview, TV show,etc):

 * Your accomplishment (20%)

<>

 * Project Review (50%)
 * 1)What went well
 * 2)What went poorly
 * 3)Future strategies

title: User:Jacquelinec~enwikiversity


text:
User talk:81.157.237.157/How was the first horse shoe discover

How was the first horse shoe discover and made by whom?


text:
User:JoelSimler

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

I will study through all sections of "Guitar Bible". <> What is the historical significance of Gibson guitars, and what are the different types of guitars does Gibson make?
 * week 1

Go over "The Basics" chapter of book. Also I must procure my guitar from my home in Bellingham
 * week 2

Go over "Blues & Rock 'n' Roll" chapter. I will also need to review any basics that i may need in order to learn this next section. Research Gibson, and Gibson guitars for general information.
 * week 3

Go over "Classic Rock" chapter. Also, I will find songs to practice jamming along to for the previous section. Continue research
 * week 4

Begin studying "Alt. Indie & Power Pop" chapter. Review last chapter for added clarity. Continue research, finding various sources to meet requirements.
 * week 5

Move onto "Metal, Punk & Grunge" Chapter. Continue to practice previous sections of book. Begin planning for presentation
 * week 6

Read the "Funk, Soul & Reggae" Chapter. Review previous chapters. Continue presentation planning, paper in process of being written.
 * week 7

Move onto final chapter of book "Americana". Practice, practice, practice. Finish paper, and finish planning of presentation.
 * week 8

Review every single chapter and make sure that all makes sense. Review paper, and presentation to make sure its all good to go.

Midterm Assignment 1: Taste & Aesthetics
Whilst wandering amidst the Olympic Sculpture Park, I came across, as should be expected, a lot of interesting sculptures. I am not a connoisseur of sculpture art, or much of any visual art really, but I can appreciate the interesting shapes that inhabit this park. On my stroll through I came across one particular sculpture(s) that I actually thought was garbage, and a bench, literally garbage. It was Roy McMakin’s Untitled (concrete bench, metal lawn chair, and cardboard box). At first I believed someone had left a little package next to a bench, and a lawn chair had been placed close by it. It wasn’t until I gave he box a tap with my foot that I realized it was metal, and looked over to my right to see a description of the three chairs. Initially I was thinking “wow, that’s really stupid. . .” and I actually continued to think that all week, practically until I began writing this paper! Even after I did some basic research, I didn’t think much Of Roy McMakin. Actually, I probably began to dislike the piece even more. So how do I feel now? Well, I’m still not a fan of this particular piece of work, I at the very least don’t dislike it. What changed my mind was looking at a collection of his work. He mostly does furniture, much of which is useful. When I began looking more closely at his work I started to notice details and nuances that made me want to stare at them longer. With his other work in mind I was able to see how it related to his other works that I did like. Though I wonder if I really changed my mind, or if I’m just telling myself that so I don’t get depressed, knowing I can’t control my own brain! Which also makes me wonder what it is that controls my brain, if my brain is not me. . . Off topic, but still interesting.

Final Presentation

 * Topic (10%)

<>

 * Research (20%)

<>

 * 1) Internet site: Unknown. Google Images. Web. 12 Dec. 2009. .
 * 2) Book: Hunter, Dave, and Deirdre Cartwright. Totally Interactive Guitar Bible. New York: Thunder Bay, 2006. Print.
 * 3) Scholarly Journal: Martin, Darryl. "Innovation and the Development of the Modern Six-String Guitar." Galpin Society (1998): 86-86. Print.
 * 4) Other (interview, TV show,etc):

 * Your accomplishment (20%)

<<Technically I was very close to accomplishing my goal. I only missed it by about 1 and a half chapters. However, in actuality I feel that I was far from my goal. A major problem was that I shot myself in the foot right off the bat. I planned the project to start on week 2 or 3, not realizing that the project technically begins week 1. So, as soon as i knew what I was going to do for my project, I was behind about 2 chapters! Which meant I was suddenly rushing to get through the book. This was very bad for I feel I didn't get everything out of the book that i should, since I was in such a hurry. So, yes, on the surface I was basically succesfull, but i still feel unsuccesful in my project, but I have only me to blame for it. However, on a positive note, I did manage to find out what I want to research on my own to continue this project. I will begin to focus on playing the blues. >>

 * Project Review (50%)
 * 1)What went well

What went well? Well, I would have to say that the project itself went fairly well. I ended up learning more about guitar than I knew beforehand, and feel like, at the very least, a slightly more capable guitar player. Practice was the part of the project that went best, when I actually did practice that is. More specifically, the Blues section I feel was the best and most informative out of all the chapters I made it to.
 * 2)What went poorly

What went poorly was my motivation in the later parts of the project. As time moved on I got busier and busier, which led to me not practicing everyday like I was supposed to. More specifically, the last chapter and a half went terribly awry, being as I didn't even get to them!
 * 3)Future strategies

For a future strategy, I would like to narrow down my idea for learning guitar. Instead of learning to play several different styles half-assedly, I would like to work on just one style of guitar. Blues is what I found I would like to focus on. Also, I would very much like to find a way to motivate myself to practice 10 minutes a day. Often while practicing I would work for an hour, and the not play for a few days. i feel it would be a better strategy to practice a little bit, but way more often.

Self-Assessment
I can definitely see why the art institute would have this be a mandatory class for freshman. It helps to reach you research strategies, self motivation, and it is a good general way to step into a college situation. Personally though I am slightly frustrated that I had to pay for this class. Not that I didn't learn anything from this class, but I have been in college for a number of years now, and feel that I didn't really need the class. I did enjoy the class, and it was especially fun because I enjoy your teaching style, but I feel that the money I paid to take the class was not entirely worth it (it would have been nice to save that money for another class related more to my field). Financial matters aside, I did like the class a lot. The assignments were great because I could really see the purpose for them. Such as the sculpture assignment, where we learn what/who can manipulate ur own thoughts patterns. The "always wanted too. . ." assignment was one of the best assignments I've had, because the guidelines were so vague, that there was little hinderance for creativity. I feel that I was successful in this class because I was able to see the purpose in the assignments.

title: User:JoelSimler


text:
User:KipYellowjacket~enwikiversity

Kip Boahn On-line Alias: Kip Yellowjacket On-line / Off-line Language Learning Facilitator eLearning Enthusiast & Developer EFL/ESL Content Creator Creator of Second Life English & Other On-line Communities Co-owner of Oxford School for English

Oxford School for English Englisch lernen? No problem! E-mail: info@oxfordschool.de Website: www.oxfordschool.de

Second Life English Welcome to the r(e)volution... Do YOU speak SLEnglish? E-mail: secondlifeenglish@gmail.com Website: secondlifeenglish.com Teleport Link: www.tinyurl.com/2987yw

Other Contact Info... Google Talk: secondlifeenglish Skype: secondlifeenglish Tokbox: KipYellowjacket Twitter: www.twitter.com/kipyellowjacket

I do eLearning... www.secondlifeenglish.com www.secondlifeactivities.com www.secondlifescripts.com www.teachyouteachme.ning.com www.virtualworldholodecks.com etc.

title: User:KipYellowjacket~enwikiversity


text:
User:Aneesa~enwikiversity

hi... aneesae how r u 'fahadsaeed@hotmail.com'

title: User:Aneesa~enwikiversity


text:
User:RGHwiki

This is my first wiki page. I want to share a number of areas through it. #(How does the list text work?) Some DT021_4DSP2 observations. DT021_3 Controls Course. Use of webcourses.

DSP2. Currently going through convolution using the DFT. It is required to be able to write programs in MATLAB to implement circular convolution and linear convolution

title: User:RGHwiki


text:
User:Egm6321.f09.team7/HW3

Homework Three: Problem 1
Problem 1 Find $$(m,n)$$ such that eqn. 1 on is exact. A first integral is $$ \Phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y+k_1=k_2 $$ where $$k_1,k_2$$ are constants.

Given: Equation to be solved: $$x^\frac12y'' + 2xy' + 3y = 0$$

Problem solution: First exactness condition is met with the form of the equation in that y'' is unmodified Second exactness condition:

$$Fxx+2P*Fxy+P^2*Fyy=Gxp+P*Gyp-Gy$$

$$Fxp+P*Fyp+2Fy=Gpp$$

First step is to multiply through by the h term in the form $$h=x^m*y^n$$

giving us : $$ (x^{\frac12+m}y^n)y''+2x^{m+1}y^np+3x^my^{n+1}$$ then we can identify F and G for the equation in the form $$F*y''+G$$: F is given as: $$x^{\frac12+m}y^n$$ G is given as: $$2x^{m+1}y^np+3x^my^{n+1}$$

The next step will be to identify the necessary partial derivatives to fulfill the second exactness condition:

$$Fx=(m+0.5)x^{(m-0.5)}y^n$$ $$Fxp=0$$ $$Fxx=(m^2-0.25)x^{m-1.5}y^n$$ $$Fxy=(m+0.5)nx^{m-0.5}y^{n-1}$$ $$Fy=nx^{m+0.5}y^{n-1}$$ $$Fyy=n(n-1)x^{m+0.5}y^{n-2}$$ $$Fyp=0$$ $$Gp=2x^{m+1}y^n$$ $$Gx=2p(m+1)x^my^n+3mx^{m-1}y^{n+1}$$ $$Gy=2pnx^{m+1}y^{n-1}+3(n+1)x^my^n$$ $$Gpp=0$$ $$Gxp=2(m+1)x^my^n$$ $$Gyp=2nx^{m+1}y^{n-1}$$

Finally we see the above partials should be inserted into the second exactness condition:

condition 2 part 2: $$0+p(0)+2nx^{m+0.5}y^{n-1}=0\rightarrow n=0$$

this condition can be used to simplify the arguments for the first part of the second exactness condition

condition 2 part 1: $$(m^2-0.25)x^{m-1.5}=2(m+1)x^m-3x^m\rightarrow m=0.5$$ note:m=0.5 found through trial and error and recognizing that this forces the LHS of the above equation to zero meaning that $$2(m+1)x^m=3x^m$$which is true for 0.5  You do not have to use trial and error. Start with the shorter of the two equations for the second exactness condition and you will find that $$n=0$$. Then use this result in the other equation and it will naturally fall out that $$m=1/2$$. --Egm6321.f09.TA 04:35, 15 October 2009 (UTC)

thus using the found m and n to modify the original equation and make it exact we can find the exact form of this equation is: $$(x^my^n)*[x^\frac12y'' + 2xy' + 3y] = 0$$ Meeting both conditions of exactness

 Express the final form with the values you found for m,n: i.e., $$xy''+2x\sqrt{x}y'+3\sqrt{x}y=0$$.--Egm6321.f09.TA 04:35, 15 October 2009 (UTC)

Homework Three: Problem 2
Problem Statement and Given:  Solve eqn. 2 on for $$y(x)$$. $$ xy' + (2x^{\frac{3}{2}}-1)y = k_2 - k_1 $$

Solution:

Divide by x to put the ODE into a general form: $$ y' + p(x)y = q(x) $$ Where: $$ p(x)=\frac{(2x^{\frac{3}{2}}-1)}{x}=2x^{\frac{1}{2}}-x^{-1}$$ $$ q(x)=\frac{k_2 - k_1}{x}$$ The ODE can be solved by finding an integrating factor $$u(x)$$ such that: $$\frac{d(u(x)y)}{dx}=u(x)q(x)$$ Dividing by $$u(x)y$$ gives: $$\frac{1}{y}\frac{dy}{dx}+\frac{1}{u(x)}\frac{du(x)}{dx}=\frac{q(x)}{y}$$ $$p(x)=\frac{1}{u(x)}\frac{du(x)}{dx}$$ so: $$\frac{1}{y}\frac{dy}{dx}+p(x)=\frac{q(x)}{y}$$ Integrate both sides to obtain: $$\int p(x)dx=\int \frac{du(x)}{u(x)}=ln(u(x))+c$$ This gives: $$u(x)=exp(\int p(x)dx)$$ Integrating $$\frac{d(u(x)y)}{dx}=u(x)q(x)$$ gives: $$u(x)y=\int u(x)q(x)dx+c $$ Solving for $$y$$ and substituting in $$u(x)=exp(\int p(x)dx)$$ gives: $$y=\frac{\int exp(\int dxp(x))q(x)+c}{exp(\int p(x)dx)}$$ $$\int p(x)dx=\int 2x^{\frac{1}{2}}-x^{-1}=\frac{4}{3}x^{\frac{3}{2}}-ln(x)+C$$ The final solution becomes: $$y=\frac{\int C_2x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})\frac{k_2 - k_1}{x}+C_1}{ C_2x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})}$$

Good work - nice solution. Egm6321.f09.TA 01:33, 28 October 2009 (UTC)

Homework Three: Problem 3
Problem Statement and Given:  From, find the mathematical structure of $$\Phi$$ that yields the above class of ODE.

Solution:

'''See attached file contains the solution! '''

You are using the result for the derivation. This is circular. Egm6321.f09.TA 02:13, 28 October 2009 (UTC)

Homework Three: Problem 4
Problem 4 From, for the case $$n=1$$ (N1_ODE) $$F(x,y,y')=0=\frac{d\Phi}{dx}(x,y)$$. Show that $$f_0-\frac{df_1}{dx}=0 \Leftrightarrow\Phi_{xy}=\Phi_{yx}$$. Hint: Use $$f_1=\Phi_y$$. Specifically: 4.1) Find $$f_0$$ in terms of $$\Phi$$ 4.2) Find $$f_1$$ in terms of $$\Phi$$($$f_1=\Phi_y$$) 4.3) Show that $$ f_0-\frac{df_1}{dx}=0\Leftrightarrow \Phi_{xy}=\Phi_{yx}$$.

Problem solution: For the universal solution given in class the following statements apply $$F=\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

$$f_{i}=\frac{dF}{dy^{(i)}}$$

since this problem is for a first order equation n=1 and therefore:

$$ F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}\rightarrow F= \Phi_{x}+\Phi_{y}y'$$

So the first function to find is $$f_0$$

$$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy}$$  This should be a partial differential with respect to $$y$$. --Egm6321.f09.TA 14:41, 15 October 2009 (UTC) $$f_0=\Phi_{xy}+\Phi_{yy}P$$

$$f_1=\frac{dF}{dy'}=\frac{d(\Phi_{x}+\Phi_{y}y')}{dy'}$$

Since Phi is only a function of x and y

$$f_1=\Phi_{y}$$

now to take the full derivative of f_1 W.R.T. x

$$\frac{df_1}{dx}\rightarrow \frac{d\Phi_{y}}{dx}$$ Phi being a function of x and y

$$=\frac{d\Phi_{y}}{dx}\frac{dx}{dx}+\frac{d\Phi_{y}}{dy}\frac{y}{dx}\rightarrow \Phi_{yx}+\Phi_{yy}P$$  This first term should be $$\frac{\partial \Phi_y}{\partial x}\frac{dx}{dx}$$. The chain rule uses partials with respect to the first variable of differentiation, and then total derivatives for the implicit dependencies. Same thing with $$\frac{\partial\phi_y}{\partial x}\frac{dy}{dx}$$ --Egm6321.f09.TA 14:41, 15 October 2009 (UTC) $$ f_0-\frac{df_1}{dx}=\Phi_{xy}+\Phi_{yy}P-\Phi_{yx}+\Phi_{yy}P=0$$

Finally cancelling terms leaves us with:

$$\Phi_{yx}-\Phi_{xy}=0$$ or $$\Phi_{yx}=\Phi_{xy}$$

Homework Three: Problem 5
Problem 5 From (p.13-3), for the case n = 2 (N2_ODE) show:

5.1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$

5.2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$

5.3) $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

5.4) Relate eqn. 5 to eqs. 4&5 from p.10-2.

Problem solution: For the universal solution given in class the following statements apply $$F=\Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+...+\Phi_{y^{(n-1)}}y^{(n)} $$

$$f_{i}=\frac{dF}{dy^{(i)}}$$

since this problem is for a first order equation n=1 and therefore:

$$ F= \Phi_{x}+\Phi_{y^{(0)}}y^{(1)}+\Phi_{y^{(1)}}y^{(2)}\rightarrow F= \Phi_{x}+\Phi_{y}y'+\Phi_{y'}y''$$

So the first step is to find the values for the functions $$f_0,f_1,f_2$$

$$f_0=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y'')}{dy}$$

$$f_0=\Phi_{xy}+\Phi_{yy}P+\Phi_{py}y''$$

$$f_1=\frac{dF}{dp}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y'')}{dp}$$

$$f_1=\Phi_{xp}+\Phi_{yp}P+\Phi_{y}+\Phi_{pp}y''$$

$$f_2=\frac{dF}{dy}=\frac{d(\Phi_{x}+\Phi_{y}y'+\Phi_{y'}y)}{dy''}$$

$$f_2=\Phi_{p}$$

Starting with the first problem statement:

5.1) Show $$f_1=\frac{df_2}{dx}+\Phi_y$$

$$f_1 and f_2$$ are shown above so now we will solve $$\frac{df_2}{dx}+\Phi_y$$

the full derivative is: $$=\frac{d\Phi_{p}}{dx}\frac{dx}{dx}+\frac{d\Phi_{p}}{dy}\frac{y}{dx}+\frac{d\Phi_{p}}{dp}\frac{dp}{dx}\rightarrow \Phi_{px}+\Phi_{py}P+\Phi_{pp}y''$$

adding this to the problem statement and description of the first f function gives the answer:

$$\Phi_{px}+\Phi_{py}P+\Phi_{pp}y+\Phi_{y}=\Phi_{px}+\Phi_{py}P+\Phi_{pp}y+\Phi_{y}$$

5.2) Show $$\frac{d}{dx}(\Phi_y)=f_0$$

$$f_o$$ es show above so now we will solve $$\frac{d\phi_y}{dx}$$

the full derivative is: $$ = \frac{d\Phi_y}{dx}\frac{dx}{dx}+\frac{d\Phi_{y}}{dy}\frac{y}{dx}+\frac{d\Phi_{y}}{dp}\frac{dp}{dx}$$

$$= \Phi_{yx}+\Phi_{yy}P+\Phi_{yp}y''$$

This is exactly the same as the statement for $$f_0$$ and satisfies the problem statement

5.3) $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

As was shown in the first problem $$f_1=\frac{df_2}{dx}+\Phi_y$$ so when subtracted out these leave only the $$\Phi_y$$ term which reduces the problem statement to:

$$f_0-\frac{\Phi_y}{dx}=0$$

since $$f_0$$ is known from above the only task is to find $$\frac{\Phi_y}{dx}=0$$:

$$\frac{\Phi_y}{dx}+\frac{\Phi_y}{dy}\frac{dy}{dx}+\frac{\Phi_y}{dp}\frac{dp}{dx} \rightarrow \Phi_{yx}+\Phi_{yy}P+\Phi_{yp}y''$$

This is the same function as $$f_0$$ and since they are subtracted = 0 (answer could have also been inferred from a combination of the first and second parts which are also related to the two exactness condition mentioned in 5.4)

 Your results for this problem are all correct - nice write up. You want to restate the expression that you have proven (and possibly box it) so that it is evident you have arrived at the expected result. --Egm6321.f09.TA 02:00, 16 October 2009 (UTC)

Homework Three: Problem 6
Problem Statement and Given:  From, for the Legendre differential equation $$F=(1-x^2)y''-2xy'+n(n+1)y=0$$, 6.1 Verify exactness of this equation using two methods: 6.1a.), Equations 4&5. 6.1b.) , Equation 5. 6.2 If it is not exact, see whether it can be made exact using the integrating factor with $$h(x,y)=x^my^n$$.

Solution 6.1a:

The general form is: $$F=f(x,y,p)y''+g(x,y,p)$$ Where: $$p=y' $$ $$f(x,y,p)=1-x^2$$ $$g(x,y,p)=-2xp+n(n-1)y$$

, Equations 4&5. i)$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ ii)$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

The partial derivatives are: $$f_{xx}=-2$$ $$f_{xy}=0$$ $$f_{yy}=0$$ $$f_{xp}=0$$ $$f_{yp}=0$$ $$f_{y}=0$$ $$g_{xp}=-2$$ $$g_{yp}=0$$ $$g_{y}=n(n-1)$$ $$g_{pp}=0$$

Substituting in the partial derivatives: i) $$-2+2p0+p^20=-2+p0-n(n-1)$$ i) $$-2\neq-2-n(n-1)$$ and ii) $$0+p0+20=0$$ ii) $$0=0$$ Since i) does not equal, this ODE is not exact.

Solution 6.1b:

, Equation 5. $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$

Where: $$f_0=\frac{dF}{dy}=n(n+1)$$ $$f_1=\frac{dF}{dy'}=-2x$$ $$\frac{f_1}{dx}=-2$$ $$f_2=\frac{dF}{dy''}=1-x^2$$ $$\frac{d^2f_2}{dx^2}=-2$$

Substituting into $$f_0-\frac{df_1}{dx}+\frac{d^2f_2}{dx^2}=0$$ gives: $$n(n+1)+2-2\neq0$$ so the ODE is not exact.

Solution 6.1c:

With an integrating factor $$h(x,y)=x^my^n$$, the ODE becomes: $$F=(x^my^n)(1-x^2)y''-(x^my^n)2xy'+(x^my^n)n(n+1)y=0$$ Where: $$p=y' $$ $$f(x,y,p)=x^my^n-x^{m+2}y^n$$ $$g(x,y,p)=-2x^{m+1}y^np+n(n-1)x^my^{n+1}$$

, Equations 4&5. i)$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ ii)$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

The partial derivatives are: $$f_{xx}=m(m-1)x^{m-2}y^n-(m+2)(m+1)x^my^n$$ $$f_{xy}=mnx^{m-1}y^{n-1}-n(m+2)x^{m+1}y^{n-1}$$ $$f_{yy}=n(n-1)x^my^{n-2}-n(n-1)x^{m+2}y^{n-2}$$ $$f_{xp}=0$$ $$f_{yp}=0$$ $$f_{y}=nx^my^{n-1}-nx^{m+2}y^{n-1}$$ $$g_{xp}=-2(m+1)x^my^n$$ $$g_{yp}=-2nx^{m+1}y^{n-1}$$ $$g_{y}=-2nx^{m+1}y^{n-1}p+a(a-1)x^m(n+1)y^n$$ $$g_{pp}=0$$

Substituting in the partial derivatives: i)$$m(m-1)x^{m-2}y^n-(m+2)(m+1)x^my^n+2pmnx^{m-1}y^{n-1}-n(m+2)x^{m+1}y^{n-1}+p^2n(n-1)x^my^{n-2}-n(n-1)x^{m+2}y^{n-2}=-2(m+1)x^my^n+p-2nx^{m+1}y^{n-1}+2nx^{m+1}y^{n-1}p+a(a+1)x^m(n+1)y^n$$ ii)$$0+p0+2nx^my^{n-1}-nx^{m+2}y^{n-1}=0$$

ii)$$2nx^my^{n-1}-nx^{m+2}y^{n-1}=0$$

From equation ii, n=0. This gives i as: i)$$m(m-1)x^{m-2}-(m+2)(m+1)x^m=-2(m+1)x^m+a(a+1)x^m$$ i)$$m(m-1)x^{-2}-(m+2)(m+1)=-2(m+1)+a(a+1)$$

Homework Three: Problem 7
Problem Statement and Given:  From, Show that equations 1 and 2, namely 7.1 $$\forall u,v$$ functions of $$x$$, $$L(u+v)=L(u)+L(v)$$. and 7.2 $$\forall \lambda\in\mathbb{R},L(\lambda u)=\lambda L(u)\forall$$ functions of $$ x$$. are equivalent to equation 3 on p.3-3.

Solution:

Alternate definition of Linearity:  L(.) is linear if and only if: 1) $$L(u+v)  =  L(u) + L(v) $$ 2)  $$L(\lambda u) = \lambda L(u)$$

Let, $$\bar u = u \lambda_1 $$ $$\bar v = v \lambda_2 $$

Use property 1 for the above definitions: $$L(\bar u + \bar v) =  L(\bar u) + L(\bar v) $$

Rewrite in terms of u: $$L(u \lambda_1 + v \lambda_2) =  L(u \lambda_1) + L(v \lambda_2) $$

Use the second property twice to get: $$L(u \lambda_1) =  \lambda_1 L(u)$$ $$L(v \lambda_2) =  \lambda_2 L(v)$$

Substitute the above expressions: $$L(u \lambda_1 + v \lambda_2) =   \lambda_1 L(u) + \lambda_2 L(v) $$

The result is the former definition of linearity in 3-3 but different nomenclature.

 Good explanation. Why is your font so small? --Egm6321.f09.TA 04:08, 16 October 2009 (UTC)

Homework Three: Problem 8
Problem Statement and Given:  From, plot the shape function $$N_{j+1}^{2}(x)$$.

Solution:

correct. Egm6321.f09.TA 03:10, 28 October 2009 (UTC)

Homework Three: Problem 9
Problem Statement and Given:  (| p.16-2), show that $$y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_t\right)$$ $$y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t\right)$$ Solution:

 Part 1 $$y_{xxx} = \frac{d}{dx}(y_{xx}) = \frac{dt}{dx} \frac{d}{dt}(y_{xx})$$

$$y_{xx} = \frac{d}{dx}(\frac{d}{dx}(y))= \frac{dt}{dx} \frac{d}{dt}(\frac{dt}{dx} \frac{d}{dt}(y))$$

$$ y_{xxx} = e^{-t} \frac{d}{dt} (e^{-2t}(\ddot y - \dot y))$$

Define the following because Wiki Latex won’t allow a triple dot accent: $$\frac{d}{dt}(\ddot y ) = \check{Y}$$

$$ y_{xxx} = e^{-t} (e^{-2t}(\check{Y}  - \ddot y  - 2 e^{-2t}(\ddot y  - \dot y)))$$

$$ y_{xxx} = e^{-3t} (\check{Y} - 3 \ddot y +2  \dot y))$$

The above equation matches!

 Part 2 $$ y_{xxxx} =\frac{d}{dx}(y_{xxx})= e^{-t} \frac{d}{dt}( e^{-3t} (\check{Y} - 3 \ddot y  +2  \dot y))$$

Define the following because Wiki Latex won’t allow a quadruple dot accent: $$ \frac{d}{dt}(\check{Y}) = \dot \check{Y}$$

$$ y_{xxxx} = e^{-t} [ -3 e^{-3t} (\check{Y} - \ddot y - 3 \ddot y  -2  \dot y) + e^{-3t} (\dot \check{Y}- 3 \check{Y}  +2  \ddot y)]$$

$$ y_{xxxx} = e^{-4t} [\dot \check{Y} - 6 \check{Y} + 11 \ddot y - 6 \dot y]$$

The above equation matches!

Very good. Boxing your answers would make the results more visible. Nice concise approach. Egm6321.f09.TA 03:30, 28 October 2009 (UTC)

Homework Three: Problem 10
Problem Statment and Given:  From (| p.16-4) Solve equation 1 on p.16-1, $$ x^2y''-2xy'+2y=0 $$ using the method of trial solution $$ y=e^{rx}$$ or $$ y=x^{r}$$ directly for the boundary conditions $$\left\{ \begin{array}{rl} y(1)=&3\\ y(2)=&4\\ \end{array}\right.$$ Compare the solution with equation 10 on p.16-3. Use matlab to plot the solutions.

Solution:  First Try: $$ y=e^{rx}$$

$$y'=re^{rx}$$

$$y''=r^{2}e^{rx}$$

Plug into the ODE and you will notice that $$ e^{rx}$$ can factored out:

$$x^2 r^2 -2xr + 2 = 0 $$

The quadratic equation can be used to get $$r_1(x)$$ and $$r_2(x)$$ ; however, this contradicts the implicit assumption made in calculating $$y'$$ and $$y''$$. We assumed r was constant therefore obtaining r(x) is in direct violation of our assumption.

$$x^2y''-2xy'+2y=0$$

The ODE's solution is not of the form $$y = e^{rx}$$ !

Instead, the solution takes the form: $$y = x^{r}$$  $$y'=rx^{r-1}$$

$$y''=r(r-1)x^{r-2}$$

Plug into the ODE and you will notice that $$ x^{r}$$ can factored out: $$x^r(r(r-1)-2r+2)=0$$

$$r^2-3r+2=0$$

$$ (r-2)(r-1)=0$$

$$r_1=2$$ and $$r_2=1$$

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

y(x) = c_1 x^2 + c_2 x$$
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
 * }
 * }

The above solution to the ODE identical to that in equation 10 on p.16-3!

Use the BVC to get the constants:

$$3 = c_1 + c_2$$

$$4 = 4 c_1 + 2 c_2$$

$$ c_1 = -1$$ and $$c_2 = 4$$

$$ y(x) = -x^2 + 4x$$

The solution is plotted:

very good. Egm6321.f09.TA 03:56, 28 October 2009 (UTC)

Homework Three: Problem 11
Problem Statement and Given:

From (| p.17-4) obtain equation 2 from p.17-3

$$Z(x)=\frac{c}{u_{1}^2}\exp\left(-\int^x a_1(s)ds\right)$$

using the integrator factor method.

Problem solution:

Starting from equation 1 from p.17-3.

$$u_{1}(x) Z'+\left[a_{1}u_{1}(x)+2u_{1}' (x) \right]Z=0 $$

$$\mu(x)\left[Z'+\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]Z=0 \right]$$

$$\mu(x)Z'+\mu(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]Z=0 $$

$$\frac{d}{dx}\left[\mu(x)Z(x) \right]=\mu(x)Z'(x)+\mu'(x)Z$$

Let $$\mu'(x)= \mu(x)\left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

$$\frac{\mu'(x)}{\mu(x)}= \left[a_{1}+\frac{2u_{1}' (x)}{u_{1}(x)} \right]$$

Integrating this will give:

$$\mu(x)=B exp \left[\int_{}^{x}a_{1}(s)ds+ln\left|u_{1} \right|^{2} \right]$$

$$\mu(x)Z'+\mu'(x)Z=0$$

$$\int_{}^{}\mu(x)*Z(x)=\int_{}^{}0$$

$$\mu(x)Z(x)=K$$

$$ B exp \left[\int_{}^{x}a_{1}(s)ds+ln\left|u_{1} \right|^{2} \right]Z(x)=K $$

$$Z(x)= C exp \left[-\int_{}^{x}a_{1}(s)ds-ln\left|u_{1} \right|^{2} \right]$$

$$Z(x)= \frac{C}{u_{1}^{2}} exp \left[-\int_{}^{x}a_{1}(s)ds \right]$$

very good. Egm6321.f09.TA 04:10, 28 October 2009 (UTC)

Homework Three: Problem 12
Problem Statement and Given:

From (| p.18-1), develop reduction of order method using the following algebraic options

$$y(x)=U(x)\pm u_1 (x)$$

$$y(x)=\frac{U(x)}{u_1 (x)}$$

$$y(x)=\frac{u_1 (x)}{U(x)}$$

Problem solution:

Part a:

$$y(x)=U(x)\pm u_1 (x)$$

$$y' (x)=U' (x)\pm u_1 ' (x)$$

$$y (x)=U (x)\pm u_1 '' (x)$$

$$ (U (x)\pm u_1  (x))+a_1(U' (x)\pm u_1 ' (x))+a_0(U(x)\pm u_1 (x))=0$$

$$ (U (x)+a_1 U'(x) + a_0 U(x))\pm ((u_1(x)+a_1 u_1'(x)+a_0 u_1(x))=0$$

The reduction of order method can not be implemented since U is not missing.

Part b:

$$y(x)=\frac{U(x)}{u_1 (x)}$$

$$y' (x)= -\frac{U(x)}{u_{1}^{2}(x)}u_{1}'(x)+\frac{U'(x)}{u_{1}(x)}$$

$$y (x)= U(x)\left[-\frac{u_{1}(x)}{u_{1}(x)^{2}}+2\frac{u_{1}'(x)^{2}}{u_{1}(x)^{3}} \right]-2\frac{U'(x)u_{1}'(x)}{u_{1}(x)^{2}}+\frac{U''(x)}{u_{1}(x)}$$

$$\left[U(x)\left[-\frac{u_{1}(x)}{u_{1}(x)^{2}}+2\frac{u_{1}'(x)^{2}}{u_{1}(x)^{3}} \right]-2\frac{U'(x)u_{1}'(x)}{u_{1}(x)^{2}}+\frac{U(x)}{u_{1}(x)} \right]+a_{1}\left[-\frac{U(x)}{u_{1}^{2}(x)}u_{1}'(x)+\frac{U'(x)}{u_{1}(x)}\right]+a_{0}\left[\frac{U(x)}{u_{1}(x)} \right]=0$$

The reduction of order method can not be implemented since U is not missing.

Part c:

$$y(x)= \frac{ u_1 (x)}{ U (x) } $$

$$y' (x)= -\frac{u_1(x)}{U(x)^{2}}U'(x)+\frac{u_1'(x)}{U(x)}$$

$$y (x)= u_1(x)\left[-\frac{U(x)}{U(x)^{2}}+2\frac{U'(x)^{2}}{U(x)^{3}} \right]-2\frac{u_1'(x)U'(x)}{U(x)^{2}}+\frac{u_1''(x)}{U(x)}$$

$$\left[u_1(x)\left[-\frac{U(x)}{U(x)^{2}}+2\frac{U'(x)^{2}}{U(x)^{3}} \right]-2\frac{u_1'(x)U'(x)}{U(x)^{2}}+\frac{u_1(x)}{U(x)} \right]+a_{1}\left[-\frac{u_1(x)}{U(x)^{2}}U'(x)+\frac{u_1'(x)}{U(x)}\right]+a_{0}\left[\frac{ u_1 (x)}{ U (x) } \right]=0$$

The reduction of order method can not be implemented since U is not missing.

very good. Egm6321.f09.TA 04:18, 28 October 2009 (UTC)

Homework Three: Problem 13
Problem Statement and Given:

From (| p.18-1), Find $$u_{1}(x)$$ and $$ u_{2}(x)$$ of equation 1 on p.18-1 using 2 trial solutions:

$$ y=ax^b$$

$$ y=e^{rx}$$

Compare the two solutions using boundary conditions $$y(0)=1$$ and $$ y(1)=2$$ and compare to the solution by reduction of order method 2. Plot the solutions in Matlab.

Problem solution:

Part 1:

Starting with the equation:

$$(1-x^{2})y''-2xy'+2y=0 $$

Using the trial solution given:

$$ y=ax^{b}$$

$$ y'=abx^{b-1}$$

$$ y''=a(b-1)x^{b-2}$$

The equation becomes:

$$ (1-x^{2})b(b-1)ax^{b-2}-2xbax^{b-1}+2ax^{b}=0$$

Now divide by: $$ ax^{b}$$

$$ (b^{2}-b)(x^{-2}-1)-2(b-1)=0$$

It is obvious that at x=0 the equation would be undefined unless b^2-b=0 which would make b=1 and at x=1 it is obvious too that b=1.

$$ y(x)=ax$$

The Boundary Conditions for this problem are y(0)=1 and y(1)=2 which only prove that $$ y=ax^{b}$$ cannot be a solution to this problem.

Part 2:

Starting with the equation:

$$(1-x^{2})y''-2xy'+2y=0 $$

Using the trial solution given:

$$ y= exp(rx)$$

$$ y'=r exp(rx)$$

$$ y''=r^{2}exp(rx) $$

The equation becomes:

$$ (1-x^{2}) r^{2}exp(rx)-2x r exp(rx)+2 exp(rx)=0$$

Now divide by: $$ exp(rx)$$

$$ (1-x^{2}) r^{2} -2(x r -2) =0$$

It is obvious that at x=0 the value of r=(-2)^(1/2) and at x=1 the value of r= 1.

$$ y(x)= exp(x)+ exp(xi\sqrt[]{2})$$

The Boundary Conditions for this problem are y(0)=1 and y(1)=2 which only prove that $$ y= exp(rx)$$ cannot be a solution to this problem since the result should only be real numbers.

Where is your plot? Egm6321.f09.TA 06:00, 28 October 2009 (UTC)

Contributing Authors
Egm6321.f09.team7.mm 03:16, 6 October 2009 (UTC) Egm6321.f09.team7.benedict 01:10, 7 October 2009 (UTC) Egm6321.f09.team7.hua 04:05, 7 October 2009 (UTC) User:Emg6321.f09.blanco 11:17, 7 October 2009 (UTC)

title: User:Egm6321.f09.team7/HW3


text:
User:Drjwsimmons

I shall update this page as I learn more. This is certainly an interesting concept.

title: User:Drjwsimmons


text:
User talk:Drjwsimmons

Well, I guess I can send a message to myself. I'm still learning. --Drjwsimmons 22:27, 5 October 2009 (UTC)

title: User talk:Drjwsimmons


text:
User:ShuyuL

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

I have always wanted to go to scuba diving Research Component Scuba diving
 * week 1

Find out what training I will need to do to get a basic diving license.
 * week 2

Because I will be back in Taiwan during the summer vocation, I want to complete all my training and get my license in Taiwan. Thus I will search the available scuba diving school in Taiwan. Possible sources: Ask my high school’s biology teacher who has had a lot of diving experience.
 * week 3

Collecting all the information I might need to make a plan on how I will spend my time to get my license during the summer vocation.
 * week 4

Make a detailed plan on how I will spend my time to get my license. Including which training school I will go, how much time do I need to spend on practicing, where am I going to practice, where am I going to get the license, how much money I might need..etc.
 * week 5

Convince my parent with the plan I had to let me go get the scuba diving license. (I know my father pretty well; I think he will ask me to reach a goal to get his permission. If he do so, complete the goal as well.)
 * week 6

Do research on the scuba diving history.
 * week 7

Collect the research information, and make a slide show of it.
 * week 8

Prepare for the presentation.

Midterm Assignment 1: Taste & Aesthetics
The first time I saw the sculpture that I don’t like was when I was in the car to somewhere. Although I only had a glimpse of it, I decided “I don’t like that sculpture.” And then forgot about it after I made that decision, left it with those things I don’t like behind my mind. Because of this assignment, I went to the sculpture park at Tuesday November 10; it was a bit rain when I got there. I saw the sculpture that I decided I don’t like from far away. The shape of the line of those blue strips is still staying the way that I couldn’t agree with. Because of the rain, the water on the silver part of the sculpture reflected tiny light to my eyes created a glowing illusion to my eyes. Typewrite Eraser, Scale X is the name of that art piece. Basically the red with the silver bottom didn’t bother me. It was all about those blues strip flying in the air. I thought it was because the strange angles and the way those strip were placed causing dislike of that piece. I looked up some information and learned about this sculpture and the artist, Claes Oldenburg who created it. I found out that most of Clase Oldenburg’s art are sculptures of something from our real life but in huge sizes. I found that I actually like some of his art work like Dropped Cone (in Cologne, Germany) and Spoonbridge and Cherry (in Minneapolis, MN) and others. I tried to manipulate myself into changing my mind, but I failed. I’m not depressing with that I couldn’t change my mind about those annoying blue strips, because if I force myself to change my mind, that person might won’t be me anymore, might be some other person who are effect by the assignment’s explanation that her instructor gave. (Although on the other hand, I might still be the same person because I myself am the one who decided to change my mind.) I still have problems with those weird angles of that blue strips.

Final Presentation

 * Topic (10%)

<>

 * Research (20%)

<> -Bellis, Mary. "History of Scuba Diving." About.com. The New York Times Company, 2009. Web. 10 Dec. 2009. .

-Epp, Karl H., comp. "SSI free online open water course." SSI. Web. 10 Dec. 2009. .
 * Your accomplishment (20%)

<<98%>>

 * Project Review (50%)
 * 1)What went well
 * 2)What went poorly
 * 3)Future strategies

Self-Assessment
In this class, I learned to see and to think of one thing from different perspectives. Learning from different people’s experience, I noticed that many things I thought as a common sense or some natural results of certain things is unreasonable in others’ mind. When we were talking about the “dark side” of the critical thinking, it was the first time that I realized that I had been doing or I think I had been doing this critical thinking without knowing exactly how to give this term a clear definition. Because critical thinking is what school keeps telling us to do, I keep doing it without second thoughts; maybe that’s the reason why it was so hard when facing more complicated situation. I think what I did well in this class is that I tried my best to understand what my teacher was telling us by giving different examples in his or his friend’s life. Because of my cultural background, I am not used to express my opinion during class period. What I could have done better is to participate more in the class. I was expecting a boring class when I saw this class on the schedule and now I am glad that I was wrong. I like this class and the way our instructor allows us to think in our way and not just point out what the fact should be.

title: User:ShuyuL


text:
User:KatherineGibson

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

Learn more about the country of Morocco

 * Research ComponentI will be learning more about the country of Morocco. I have always been interested in the different culture. I will be learning the history of current and ancient Morocco. I plan on going there one day, but for now I want to learn more about it before I embark on the journey.

 * week 1

Read about the history of Morocco. How it became an independent country. Learn about how Morocco is like today and how it has changed throughout the years.

 * week 2

Study the laws and the people. How they interact with each other. Learn more about modern times.

 * week 3

Research the landmarks and learn how they are significant. Study the clothes, lifestyle, food, and games.

 * week 4

Find Moroccan inspired items or actual Moroccan items that I can look at to get the better idea of what they used in their daily lives.

 * week 5

Pick a recipe that is used in Morocco and that is still made to this day in the country.

 * week 6

Practice on making the recipe and figure out the flaws and what I can do to make it authentic.

 * week 7

Make a power point of all the information I've learned. Remake the recipe if I didn't do it accurately.

 * week 8

Organize and get myself ready for the presentation.

Final Presentation

 * Topic (10%)

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 * Research (20%)

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 * 1) Internet site:
 * 2) Book:
 * 3) Scholarly Journal:
 * 4) Other (interview, TV show,etc):

 * Your accomplishment (20%)

<>

 * Project Review (50%)
 * 1)What went well
 * 2)What went poorly
 * 3)Future strategies

title: User:KatherineGibson


text:
Death

This resource deals with Medical Ethics.

Death and Futility Outline Fall 2009

Death and Futility Cases Fall 2009

Audio of Lecture on Death Fall 2009

Brain Death Test AAN detailed report on how to determine brain death in a patient. This raises the question: If you perform the test on yourself and satisfy all of the conditions, are you dead or not?

Locked in Syndrome A man in Belgium is believed to be in PVS, but is really in Locked in Syndrome

title: Death


text:
User:Carissabauman

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

< >
 * Research Component

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 * week 1

 * week 2

 * week 3

 * week 4

 * week 5

 * week 6

 * week 7

 * week 8

Final Presentation

 * Topic (10%)

< >

 * Research (20%)

<>

 * 1) Internet site:
 * 2) Book:
 * 3) Scholarly Journal:
 * 4) Other (interview, TV show,etc):

 * Your accomplishment (20%)

<>

 * Project Review (50%)
 * 1)What went well
 * 2)What went poorly
 * 3)Future strategies

title: User:Carissabauman


text:
User:AshtonR

=Freshman Seminar=

Always Wanted To Schedule

 * Topic

<>
 * Research Component

<>
 * week 1

My goal is to smoke ten cigarrettes to cut back from a pack a day. I will research heart disease and problems with smoking
 * week 2

My goal is to smoke eight cigarrettes a day. I will research lung cancer and the effects smoking has on the lungs and cardio and respitory.
 * week 3

My goal is to smoke six cigarrettes a day. I will research a helpline to quit smoking.
 * week 4

My goal is to smoke five cigarrettes a day. I will hopefully make a conference call to the smoking helpline. I will research the techniques the helpline gives me.
 * week 5

My goal is to smoke four cigarrettes a day. I will research ingredients in cigarettes.
 * week 6

My goal is to smoke three cigarrettes a day. I will research healthy habits on quitting.
 * week 7

My goal is to smoke two cigarrettes a day. I will research pictures and stories of people who have quit smoking or still smoke and the effects it had on them.
 * week 8

My goal is to smoke one cigarette that day and it will hopefully be the last cigarette I will ever smoke.

Midterm Assignment 2: Manipulation
Upon reaching the Olympic Sculpture Park I did not find myself in awe of the modern forms of art, in fact I found myself dissapointed with more than half the pieces. Also the pieces I found interesting were still the least bit visual stimulating. Yet, I digress. The purpose of this particular assignment was to find a sculpture that I utterly hated and try to prove myself to believe the exact opposite. I simply failed.

While walking through the park i stumbled upon the most ridiculous sculpture I had seen thus far and would see the remainder of the time I was there. The sculture was Bunyon's Chess. The sculpture looked as if i had been made from a few pieces beach wood and strapped in the air. It immediatly seemed to me that either the artist is not very good at all, or maybe he was at one time a brilliant artist whom had become vain with his own creations, after which he came to believe it would be perfectly fine to create any old thing and have it be as good as gold.

Now i had to try to make myself like Bunyon's Chess? "hmmm...simpple enough" i thought to myself. Wrong! I spent a few hours researching the piece Mark di Suvero, and the pieces creation in 1985. Though there was alot information on the creator and the piece, i could not find any unkown traits that would change my mind about having distaste for the piece. i tried.

Final Presentation

 * Topic (10%)

<>

 * Research (20%)

<>

 * 1) Internet site:"Smoking Cigarettes and Cardiovascular Disease." American Heart Association. 10 12 2009. American Heart Association, Web. 10 Dec 2009.
 * 2) Book:Rabin, Robert. "Regulating Tobacco". 1st. New York: Ocford University Press, 2001. Print.
 * 3) Scholarly Journal:Raymond Pearl

Science, New Series, Vol. 87, No. 2253. (Mar. 4, 1938), pp. 216-217.
 * 4) Other (interview, TV show,etc): "TRUTH" commercials

 * Your accomplishment (20%)

I did not accomplish my goal of quitting smoking. I still smoke around a half a pack a day.

 * Project Review (50%)
 * 1) My perseverance through the first few weeks was unyielding.
 * 2) After about the 6th week I found myself becoming irritable and easily distracted after lowering the amount I smoked a day.
 * 3) In the fute I plan to uphold goals io set in place for myself. I also plan to not let my meager attempts at hinder certain habits be enough to be called satisfactory. "will I try to quit smoking again?" my answer would most likely be no.

title: User:AshtonR