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University of Florida/Egm4313/s12.team11.R3

Intermediate Engineering Analysis Section 7566 Team 11 Due date: February 22, 2012.

Problem 3.1
Solved by Luca Imponenti

Problem Statement
Find the solution to the following L2-ODE-CC: $$y''-10y'+25y=r(x)\!$$

With the following excitation: $$r(x)=7e^{5x}-2x^2\!$$

And the following initial conditions: $$y(0)=4, y'(0)=-5\!$$

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

Homogeneous Solution
To find the homogeneous solution we need to find the roots of our equation

$$\lambda^2-10\lambda+25=0\!$$ $$(\lambda-5)(\lambda-5)=0\!$$ $$\lambda=5\!$$

We know the homogeneous solution for the case of a real double root with $$\lambda=5\!$$ to be

$$y_h=c_1e^{5x}+c_2xe^{5x}\!$$

Particular Solution
For the given excitation we must use the Sum Rule to the particular solution as follows

$$y_p=y_{p1}+y_{p2}\!$$ where $$y_{p1}\!$$ and $$y_{p2}\!$$ are the solutions to $$r_1(x)=7e^{5x}\!$$ and $$r_2(x)=-2x^2\!$$, respectively

First Particular Solution
$$r_1(x)=7e^{5x}\!$$,

from table 2.1, K 2011, pg. 82 we have

$$y_{p1}=Ce^{5x}\!$$

but this corresponds to one of our homogeneous solutions so we must use the modification rule to get

$$y_{p1}=Cx^2e^{5x}\!$$

Plugging this into the original L2-ODE-CC then substituting;

$$y_{p1}''-10y_{p1}'+25y_{p1}=r_1(x)\!$$

$$(Cx^2e^{5x})''-10(Cx^2e^{5x})'+25(Cx^2e^{5x})=r_1(x)\!$$

$$25Cx^2e^{5x}+10Cxe^{5x}+10Cxe^{5x}+2Ce^{5x}-10(5Cx^2e^{5x}+2Cxe^{5x})+25Cx^2e^{5x}=7e^{5x}\!$$

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

$$e^{5x}2C=7e^{5x}\!$$

so $$C=\frac{7}{2}\!$$ and the first particular solution is,
 * $$y_{p1}=\frac{7}{2}x^2e^{5x}\!$$

Second Particular Solution
$$r_2(x)=-2x^2\!$$,

from table 2.1, K 2011, pg. 82 we have

$$y_{p2}=a_2x^2+a_1x+a_0\!$$

Plugging this into the original L2-ODE-CC then substituting;

$$y_{p2}''-10y_{p2}'+25y_{p2}=r_2(x)\!$$

$$(a_2x^2+a_1x+a_0)''-10(a_2x^2+a_1x+a_0)'+25(a_2x^2+a_1x+a_0)=-2x^2\!$$

$$2a_2-10(2a_2x+a_1)+25(a_2x^2+a_1x+a_0)=-2x^2\!$$

grouping like terms we get three equations to solve for the three unknowns, these are written in matrix form

$$25a_2x^2+(25a_1-20a_2)x+(25a_0-10a_1+2a_2)=-2x^2\!$$

 * $$ \begin{bmatrix}

2 & -10 & 25\\ -20 & 25 & 0\\  25 & 0 & 0\end{bmatrix}

\begin{bmatrix} a_2\\ a_1\\ a_0\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ -2\end{bmatrix}\!$$

solving by back subsitution leads to $$a_2=-\frac{2}{25}, a_1=\frac{8}{125}, a_0=\frac{4}{125}\!$$

so the second particular solution is,
 * $$y_{p2}=-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

General Solution
The general solution is the summation of the homogeneous and particular solutions

$$y=y_h+y_{p1}+y_{p2}\!$$

$$y=c_1e^{5x}+c_2xe^{5x}+\frac{7}{2}x^2e^{5x}-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

$$y=e^{5x}(c_1+c_2x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

Applying the first initial condition $$y(0)=4\!$$

$$y(0)=c_1+\frac{4}{125}=4\!$$

$$c_1=\frac{496}{125}\!$$

Second initial condition $$y'(0)=-5\!$$

$$y'=\frac{d}{dx}y=e^{5x}[5(c_1+c_2x+\frac{7}{2}x^2)+c_2+7x]-\frac{4}{25}x+\frac{8}{125}\!$$

$$y'=e^{5x}[\frac{35}{2}x^2+(5c_2+7)x+5c_1+c_2]-\frac{4}{25}x+\frac{8}{125}\!$$

$$y'(0)=5c_1+c_2+\frac{8}{125}=-5\!$$

$$c_2=-\frac{3113}{125}\!$$

The general solution to the differential equation is therefore

$$y(x)=e^{5x}(\frac{496}{125}-\frac{3113}{125}x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

Plot
Below is a plot of this solution and the solution to in the example on p.7-3

our solution $$y(x)=e^{5x}(\frac{496}{125}-\frac{3113}{125}x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$ (shown in red)

example on p.7-3 $$y(x)=e^{5x}(4-25*x+*\frac{7}{2}x^2)\!$$ (shown in blue)

Egm4313.s12.team11.imponenti 05:55, 20 February 2012 (UTC)

Problem 3.2
Solved by Gonzalo Perez

Problem Statement
Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.

Given
Consider two distinct roots of the form:

$$ \lambda_1 = x \! $$ and $$ \lambda_2 = x + \epsilon \! $$

(where $$ \epsilon \! $$ is perturbation).

Given
Find the homogeneous L2-ODE-CC having the above distinct roots.

Solution
$$ (\lambda - \lambda_1)(\lambda - (\lambda_1))=0 \! $$

$$ (\lambda - x)(\lambda - (x + \epsilon))=0 \! $$

$$ \lambda^2 - \lambda x - \lambda \epsilon - \lambda x + x^2 + x \epsilon =0 \! $$

$$ \lambda^2 - \lambda (2x+\epsilon) + x (x+\epsilon )=0 \! $$

$$ \therefore y'' - y'(2\lambda +\epsilon) + y \lambda (\lambda + \epsilon) = 0 \! $$ (1)

Given
Show that $$ \frac{e^{(\lambda +\epsilon) x} - e^{\lambda x}}{\epsilon } \! $$ is a homogeneous solution. (2)

Solution
Let's find the corresponding derivatives:

$$ y = \frac{e^{(\lambda+\epsilon) x} - e^{\lambda x}}{\epsilon} \! $$

$$ y' = \frac{(\lambda +\epsilon) e^{\lambda + \epsilon x}-\lambda e^{\lambda x}}{\epsilon } \! $$

$$ y'' = \frac{(\lambda +\epsilon)^2 e^{(\lambda + \epsilon) x}-\lambda^2 e^{\lambda x}}{\epsilon} \! $$

If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:

$$ e^{(\lambda + \epsilon)x} (\lambda^2+2\lambda \epsilon + \epsilon^2 - 2\lambda^2-2\lambda \epsilon - \lambda \epsilon -\epsilon^2 + \lambda^2 + \epsilon \lambda)+e^{\lambda x}(- \lambda^2+2\lambda^2+\lambda \epsilon - \lambda ^2 - \lambda \epsilon) = 0 \! $$

$$ \therefore 0 \equiv 0. \! $$

Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.

Given
Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).

Solution
Using l'Hopital's Rule,

$$ \lim_{\epsilon \rightarrow 0}\frac{e^{(\lambda +\epsilon) x }-e^{\lambda x}}{\epsilon} = \frac{0}{0} \! $$

(this is an indeterminate form).

L'Hopital's Rule states that we can divide this function into two functions, $$ f(\epsilon) \! $$ and $$ g(\epsilon) \! $$, and then find their derivatives and attempt to find the limit of $$ \frac{f'(\epsilon)}{g'(\epsilon)} \! $$. If a limit exists for this, then a limit exists for our original function.

$$ \lim_{\epsilon \rightarrow 0} \frac{f'(\epsilon)}{g'(\epsilon)} = \lim_{\epsilon \rightarrow 0} \frac{xe^{x(\lambda + \epsilon)}}{1}\! $$

$$ = \frac{x e^{(\lambda+0)x}}{1} \! $$

$$ = xe^{x \lambda} \! $$

Given
Take the derivative of $$ e^{\lambda x} $$ with respect to lambda.

Solution
Taking the derivative with respect to lambda, we find that:

$$ \frac{d(e^{\lambda x})}{d\lambda} = xe^{\lambda x} $$.

It is important to remember that we must hold $$ x \! $$ as a constant when finding this derivative.

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

Solution
Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:

$$ \frac{d(e^{\lambda x})}{d\lambda }=\lim_{e\rightarrow 0}\frac{e^{(\lambda +\epsilon )x} - e^{\lambda x}} {\epsilon} = xe^{\lambda x} \! $$

Given
Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001, and compare to the value obtained from the exact second homogeneous solution.

Solution
After performing these calculations, from (2) we get 148.478.

And from the exact second homogeneous solution, we get 200.05.

Egm4313.s12.team11.perez.gp 20:32, 22 February 2012 (UTC)

Problem 3.3
Solved by Jonathan Sheider

Problem Statement
Find the complete solution to the ODE with the given initial conditions. Plot the solution y(x).

Given
$${y}''-3{y}'+2y = 4x^2\!$$ $$y(0) = 0, {y}'(0) = 0\!$$

Solution
First let us analyze the homogenous solution to the given ODE: $${y}''-3{y}'+2y = 0\!$$ The characteristic equation for this ODE is therefore: $$\lambda^2 - 3\lambda + 2 = 0\!$$ Solving for $$\lambda\!$$: $$(\lambda-2)(\lambda-1) = 0\!$$ $$\lambda = {1,2}\!$$ $$\lambda_{1} = 1\!$$ and $$\lambda_{2} = 2\!$$ Therefore the equation has two real roots and a homogenous solution of the following form (from Kreyszig 2011, p.54-57): $$ y_{h} = c_{1}e^{\lambda_{1}x} + c_{2}e^{\lambda_{2}x}\!$$ And finally we find the general homogenous solution: $$ y_{h} = c_{1}e^{x} + c_{2}e^{2x}\!$$ Next let us evaluate the particular solution to the given ODE: $${y}''-3{y}'+2y = 4x^2\!$$ The function on the right hand side of this equation implies that the particular solution has the following form based on Table 2.1 (from Kreyszig 2011, p.82): $$ y_{p} = K_{n}x^n + K_{n-1}x^{n-1} + ... + K_{1}x + K_{0} \!$$ Therefore for this ODE we have: $$ y_{p} = K_{2}x^2+K_{1}x+K_{0}\!$$ Differentiating $$ y_{p}\!$$ to obtain $${y_{p}}'\!$$ and $${y_{p}}''\!$$ respectively: $$ {y}' = 2K_{2}x+K_{1}\!$$ $$ {y}'' = 2K_{2}\!$$ Substituting these equations into the original ODE yields: $$ (2K_{2}) -3(2K_{2}x+K_{1}) + 2(K_{2}x^2+K_{1}x+K_{0}) = 4x^2\!$$ $$ 2K_{2} -6K_{2}x-3K_{1} + 2K_{2}x^2+2K_{1}x+2K_{0} = 4x^2\!$$ And we know that the coefficients of the variables on each side of the equation must be equal: $$ (2K_{2})x^2 + (2K_{1}-6K_{2})x + (2K_{2} - 3K_{1} + 2K_{0}) = 4x^2\!$$ Therefore we find: $$ 2K_{2} = 4\!$$ $$ K_{2} = 2 \!$$ Now, solving for $$ K_{1}\!$$ and $$ K_{0}\!$$: $$ 2K_{1} - 6K_{2} = 0 \!$$ $$ 2K_{1} = 6(2) \!$$ $$ K_{1} = 6\!$$ And also: $$ 2K_{2} - 3K_{1} + 2K_{0} = 0 \!$$ $$ 2K_{0} = 3(6) - 2(2)\!$$ $$ K_{0} = 7\!$$ Finally we arrive at the particular solution: $$ y_{p} = 2x^2 + 6x + 7\!$$ By superposition, we can find the complete general solution: $$ y = y_{h} + y_{p}\!$$ $$ y = c_{1}e^{x} + c_{2}e^{2x} + 2x^2 + 6x + 7\!$$ Using the given initial conditions, we can solve for $$ c_{1}\!$$ and $$c_{2}\!$$, first by using the initial conditions for $$y\!$$: $$ y(0) = c_{1} + c_{2} + 7 = 0\!$$ Differentiating the complete general solution and using the given initial condition for $${y}'\!$$: $$ {y}' = c_{1}e^{x} + 2c_{2}e^{2x} + 4x + 6\!$$ $$ {y}'(0) = c_{1} + 2c_{2} + 6 = 0\!$$ Solving this system of equations, by solving the first equation for $$c_{1}\!$$: $$c_{1} = -7 - c_{2}\!$$ Plugging this into the second equation yields: $$ (-7 - c_{2}) +2c_{2} + 6 = -1 + c_{2} = 0\!$$ $$ c_{2} = 1\!$$ And therefore: $$ c_{1} = -7 - (1)\!$$ $$ c_{1} = -8\!$$ Finally we have the complete general solution that evaluates the given initial conditions: $$ y(x) = -8e^{x} + e^{2x} +2x^2+6x+7\!$$ A plot of $$y(x)\!$$ from $$x = -3\!$$ to $$x = 3\!$$ is shown using MatLAB:

Checking
Differentiating the solution to find $$ {y}'\!$$ and $$ {y}''\!$$ respectively: $$ y = -8e^{x} + e^{2x} +2x^2+6x+7\!$$ $$ {y}' = -8e^{x} + 2e^{2x} +4x+6\!$$ $$ {y}'' = -8e^{x} + 4e^{2x} +4\!$$ Substituting into the original ODE yields: $$ {y}'' - 3{y}'+2y = 4x^2\!$$ $$ (-8e^{x} + 4e^{2x} +4) -3(-8e^{x} + 2e^{2x} +4x+6) +2(-8e^{x} + e^{2x} +2x^2+6x+7) = 4x^2\!$$ $$ -8e^{x} + 4e^{2x} +4 +24e^{x} - 6e^{2x} -12x-18 -16e^{x} + 2e^{2x} +4x^2+12x+14 = 4x^2\!$$ $$ (-8+24-16)e^{x} + (4-6+2)e^{2x} + (-12+12)x + (-18+4+14) +4x^2 = 4x^2\!$$ $$ 4x^2 = 4x^2\!$$ Therefore this solution is correct. --Egm4313.s12.team11.sheider 21:30, 21 February 2012 (UTC)

Problem 3.4
Solved by Daniel Suh

Problem Statement
Use Basic Rule (1) and Sum Rule (3) to show that the appropriate particular solution for $$y''-3y'+2y = 4x^{2}-6x^{5}\!$$ is of the form $$\sum_{j=0}^{n}c_{j}x^{j}\!$$, with n = 5.

Finding the Particular Solution with Basic Rule and Sum Rule
We know that in standard form, for a particular solution, $$y''_{p}+ay'_{p}+by_{p}=r(x)\!$$ $$y_{p}=y_{p1}+y_{p2}\!$$

Using Basic Rule (1) and Sum Rule (3), we know that $$y''_{p}+ay'_{p}+by_{p}=\sum_{i}r_i(x)\!$$

Additionally, we choose $$r_1(x)\rightarrow y_{p1}(x)\!$$ $$r_2(x)\rightarrow y_{p2}(x)\!$$

Solving for Particular Solution 1
Using the Method of Undetermined Coefficients, we find that $$y_{p1}=C_{2}x^{2}+C_{1}x+C_{0}\!$$ $$y'_{p1}=2C_{2}x+C_{1}\!$$ $$y''_{p1}=2C_{2}\!$$

Plugging $$y_{p1}\!$$ into $$y''-3y'+2y=4x^{2}\!$$ gives $$(2C_{2})-3(2C_{2}x+C_{1})+2(C_{2}x^{2}+C_{1}x+C_{0})=4x^{2}\!$$

Solving for coefficients $$2C_{2}=4\!$$ $$-6C_{2}+2C_{1}=0\!$$ $$-3C_{1}+2C_{0}+2C_{2}=0\!$$

Results in $$C_{2}=2\!$$ $$C_{1}=6\!$$ $$C_{0}=7\!$$

$$y_{p1}=2x^{2}+6x+7\!$$

Solving for Particular Solution 2
Using the Method of Undetermined Coefficients, we find that $$y_{p2}=K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}\!$$ $$y'_{p2}=5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1}\!$$ $$y''_{p2}=20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\!$$

Plugging $$y_{p2}\!$$ into $$y''-3y'+2y=-6x^{5}\!$$ gives $$(20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2})-3(5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1})+2(K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0})=-6x^{5}\!$$

Solving for coefficients results in $$K_{5}=-3\!$$ $$K_{4}=-22.5\!$$ $$K_{3}=-105\!$$ $$K_{2}=-337.5\!$$ $$K_{1}=-697.5\!$$ $$K_{0}=-708.75\!$$

$$y_{p2}=-3x^{5}-22.5x^{4}-105x^{3}-337.5x^{2}-697.5x-708.75\!$$

Combining Particular Solutions
Plugging $$y_{p1}\!$$ and $$y_{p2}\!$$ into $$y_{p}=y_{p1}+y_{p2}\!$$ gives

$$y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$$

Comparing to Summation Form
The particular solution in the summation form is $$y_{p}=\sum_{j=0}^{n}c_{j}x^j\!$$

if n=5, then $$y_{p}=\sum_{j=0}^{5}c_{j}x^j\!$$ $$y'_{p}=\sum_{j=0}^{5}c_{j}*j*x^{(j-1)}\!$$ $$y''_{p}=\sum_{j=0}^{5}c_{j}*j*(j-1)*x^{(j-2)}\!$$

Plugging the particular solution into $$y''-3y'+2y=-6x^{5}\!$$, the final solution gives $$y_{p}=-3x^{5}-22.5x^{4}-105x^{3}-335.5x^{2}-691.5x-701.75\!$$ (check Problem 3.5 for explanation on how to get this answer)

Result
As you can see, the two particular solutions of $$y_{p}\!$$ are equal. Thus, using the Basic Rule (1) and Sum rule (3) does give you the correct particular solution for $$y''-3y'+2y=4x^{2}-6x^{5}\!$$, in the form of $$y_{p}=\sum_{j=0}^{n}c_{j}x^j\!$$

Created by [| Daniel Suh] 20:57, 21 February 2012 (UTC)

Problem 3.5
Solved by: Andrea Vargas

Problem Statement
Given $$y''-3y'+2y=4x^2-6x^5\!$$

1. Obtain the coefficients of $$ x,x^2,x^3,x^5 \!$$ 2.Verify all equations by long-hand expansion. Use the series before adjusting the indexes.

3. Put the system of equations in an upper triangular matrix.

4. Solve for $$c_0,c_1,c_2,c_3,c_4,c_5\!$$ by using back substitution.

5. Using the initial conditions $$ y(0)=1, y'(0)=0 \!$$ find $$y(x)\!$$ and plot it

Solution
1. To obtain the equations for the coefficients we use the following equation from (1) p7-11:

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

Finding the coefficients of $$x\!$$ where $$j=1\!$$:

$$[c_3(3)(2)-3(c_2)(2)+2c_1]x^1=[6c_3-6c_2+2c_1]x\!$$

Finding the coefficients of $$x^2\!$$ where $$j=2\!$$:

$$[c_4(4)(3)-3(c_3)(3)+2c_2]x^2=[12c_4-9c_3+2c_2]x^2\!$$

Finding the coefficients of $$x^3\!$$ where $$j=3\!$$:

$$[c_5(5)(4)-3(c_4)(4)+2c_3]x^3=[20c_4-12c_3+2c_2]x^3\!$$

Finding the coefficients of $$x^5\!$$ where $$j=5\!$$:

$$c_5x^5\!$$

2. To verify all the above equations by long hand expansion we use the following equation from (4) p7-12: $$\sum_{j=2}^{5}c_j\times j\times(j-1)\times x^{j-2}-3\sum_{j=1}^5c_j\times j\times x^{j-1}+2\sum_{j=0}^5c_j\times x^j=4x^2-6x^5\!$$

Finding the coefficients when $$j=0\!$$:

$$2c_0\!$$

Finding the coefficients when $$j=1\!$$:

$$-3c_1+2c_1x_1\!$$

Finding the coefficients when $$j=2\!$$:

$$2c_2-6c_2x+2c_2x^2\!$$

Finding the coefficients when $$j=3\!$$:

$$6c_3x-9c_3x^2+2c_3x^3\!$$

Finding the coefficients when $$j=5\!$$:

$$20c_5x^3-15c_5x^4+2c_5x^5\!$$

By collecting these terms we can compare them to the equations of part 1.

Coefficients of $$x\!$$:

$$[c_3(3)(2)-3(c_2)(2)+2c_1]x^1=[6c_3-6c_2+2c_1]x\!$$

Coefficients of $$x^2\!$$:

$$[c_4(4)(3)-3(c_3)(3)+2c_2]x^2=[12c_4-9c_3+2c_2]x^2\!$$

Coefficients of $$x^3\!$$:

$$[c_5(5)(4)-3(c_4)(4)+2c_3]x^3=[20c_4-12c_3+2c_2]x^3\!$$

Coefficients of $$x^5\!$$:

$$c_5x^5\!$$

We can see that we obtain the same system of equations to solve for the coefficients with both methods.

3.Constructing the coefficients matrix: $$\begin{bmatrix} 2&-3&2&0&0&0 \\ 0&2&-6&6&0&0 \\  0&0&2&-9&12&0 \\  0&0&0&2&-12&20 \\  0&0&0&0&2&-15 \\  0&0&0&0&0&2 \end{bmatrix}\!$$

Then, the system becomes:

$$\begin{bmatrix} 2&-3&2&0&0&0 \\ 0&2&-6&6&0&0 \\  0&0&2&-9&12&0 \\  0&0&0&2&-12&20 \\  0&0&0&0&2&-15 \\  0&0&0&0&0&2 \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 4 \\ 0 \\ 0 \\ -6 \\ \end{bmatrix}\!$$ 4. Solving for the coefficients: $$2c_5=-6 \rightarrow c_5=-3\!$$ $$ 2c_4-(15)(-3)=0\rightarrow c_4=\frac{-45}{2}\!$$ $$ 2c_3-12(-\frac{45}{2})+20(-3)=0 \rightarrow c_3=-105 \!$$ $$ 2c_2-9(-105)+12(-\frac{45}{2})=4 \rightarrow c_2=-\frac{671}{2}\!$$ $$ 2c_1-6(-\frac{671}{2})+6(-105)=0 \rightarrow c_1=-\frac{1383}{2} \!$$ $$ 2c_0-3(-\frac{1383}{2})+2(-\frac{671}{2})=0 \rightarrow c_0=-\frac{2807}{4} \!$$

Particular solution
This yields the particular solution: $$y_p(x)=-3x^5-\frac{45}{2}x^4-105x^3-\frac{671}{2}x^2-\frac{1383}{2}x-\frac{2807}{4}\!$$

Homogeneous Solution
$$y{}''_h-y{}'_h+2y_h=0\!$$ $$\lambda^2-3\lambda+2=0\!$$ Then, we can find the characteristic equation: $$(\lambda-2)(\lambda-1)\!$$ $$\lambda=2,1\!$$ Then the solution for the homogeneous equation becomes: $$y_h(x)=C_1e^{2x}+C_2e^{x}\!$$

General Solution
Using the given initial conditions $$ y(0)=1, y'(0)=0 \!$$ we find the overall solution:

$$ y(x)=y_h+y_p \!$$ $$ y(x)=C_1e^{2x}+C_2e^x-3x^5-\frac{45}{2}x^4-105x^3-\frac{671}{2}x^2-\frac{1383}{2}x-\frac{2807}{4} \!$$ $$ y'(x)=2C_1e^{2x}+C_2e^x-15x^4-90x^3-315x^2-671x-\frac{1383}{2}\!$$ Using the initial conditions to solve for $$ C_1 \!$$ and $$ C_2 \!$$ $$ 1=C_1+C_2-\frac{2807}{4} \!$$ $$ 0=C_1+C_2-\frac{1383}{2} \!$$ $$ C_1=-\frac{45}{4} \; \; \; C_2=714 \!$$ The general solution becomes

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

Plot
Below is a plot of the solution:

--Egm4313.s12.team11.vargas.aa 05:56, 21 February 2012 (UTC)

Problem 3.6
Solved by Francisco Arrieta

Problem Statement

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

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

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

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

Given

 * First particular solution

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

 * Initial Conditions

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

Particular Solution

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

 * Then

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

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

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

 * Using coefficient matching of left hand side to the right hand side of the equation, the following equations are obtained

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

 * Which is equivalent to the following matrix

 * $$ \begin{bmatrix}

2&-3 & 2 & 0 & 0 & 0\\ 0& 2 &-6 & 6 & 0 & 0\\  0& 0 & 2 &-9 &12 & 0\\  0& 0 & 0 & 2 &-12& 20\\  0& 0 & 0 & 0 & 2 & -15\\  0& 0 & 0 & 0 & 0 & 2 \end{bmatrix}

\begin{bmatrix} c_0\\ c_1\\ c_2\\ c_3\\ c_4\\ c_5 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ -6 \end{bmatrix} \!$$

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

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

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

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

 * Since these equations are L2-ODE-CC, the superposition principle applies and $$ y_p=y_{p,1}+y_{p,2} \!$$ and the general particular solution becomes

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

Homogeneous Solution

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

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

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

 * With solutions

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

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

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

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

 * Then the solutions for each distinct linearly independent homogeneous equation becomes

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

 * Because of linearity, the linear combination of the previous two equations times two constants that satisfy 2 initial conditions, is also a solution

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

General Solution

 * The overall solution for the L2-ODE-CC

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

 * Using the initial conditions to solve for $$ C_1 \!$$ and $$ C_2 \!$$

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

 * The general solution becomes

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

--Egm4313.s12.team11.arrieta 20:26, 19 February 2012 (UTC)

Problem 3.7
Solved By Kyle Gooding

Problem Statement
Expand the series on both sides of (1),(2) pg. 7-12b to verify these equalities. (1)

Given
$$\sum_{j=2}^{5}C_j*j(j-1)x^{j-2}= \sum_{j=0}^{3}C_{j+2}*(j+2)(j+1)x^{j} $$

(2) $$ \sum_{j=1}^{5}C_j*jx^{j-1}= \sum_{j=0}^{4}C_{j+1}*(j+1)x^{j}$$

Solutions
Expanding both sides of (1) results in: $$\sum_{j=2}^{5}C_j*j(j-1)x^{j-2}=C_2(2)(2-1)x^0+ C_3(3)(3-1)x^1+C_4(4)(4-1)x^2 + C_5(5)(5-1)x^3 $$ $$\sum_{j=0}^{3}C_j(j+2)(j+1)x^{j}=C_{0+2}(0+2)(0+1)x^0+ C_{1+2}(1+2)(1+1)x^1+C_{2+2}(2+2)(2+1)x^2 + C_{3+2}(3+2)(3+1)x^3 $$

Simplifying: $$\sum_{j=2}^{5}C_j*j(j-1)x^{j-2}=C_2(2)+ C_36x+C_412x^2 + C_520x^3 $$ $$\sum_{j=0}^{3}C_j(j+2)(j+1)x^{j}=C_2(2)+ C_36x+C_412x^2 + C_520x^3 $$

The two sums are equal. Expanding both sides of (2) results in: $$ \sum_{j=1}^{5}C_j*jx^{j-1}= C_1(1)x^0+ C_2(2)x^1+C_3(3)x^2+C_4(4)x^3 + C_5(5)x^4 $$ $$ \sum_{j=0}^{4}C_{j+1}*(j+1)x^{j}= C_{0+1}(0+1)x^0+ C_{1+1}(1+1)x^1+C_{1+2}(2+1)x^2+C_{2+2}(3+1)x^3 + C_{3+2}(4+1)x^4 $$

Simplifying: $$ \sum_{j=1}^{5}C_j*jx^{j-1}= C_1(1)+ C_2(2)x^1+C_3(3)x^2+C_4(4)x^3 + C_5(5)x^4 $$ $$ \sum_{j=1}^{5}C_j*jx^{j-1}= C_1(1)+ C_2(2)x^1+C_3(3)x^2+C_4(4)x^3 + C_5(5)x^4 $$ The two sums are equal.

Egm4313.s12.team11.gooding 23:49, 19 February 2012 (UTC)

Problem 3.8
solved by Luca Imponenti

Problem Statement
Find a (real) general solution. State which rule you are using. Show each step of your work.
 * $$y''+4y'+4y=e^{-x}cos(x)\!$$

Homogeneous Solution
To find the homogeneous solution, $$y_h\!$$, we must find the roots of the equation

$$\lambda^2+4\lambda+4=0\!$$

$$(\lambda+2)(\lambda+2)=0\!$$

$$\lambda=-2\!$$

We know the homogeneous solution for the case of a double root to be

$$y_h=c_1e^{\lambda x}+c_2xe^{\lambda x}\!$$

 * $$y_h=c_1e^{-2x}+c_2xe^{-2x}\!$$

Particular Solution
We have the following excitation
 * $$r(x)=e^{-x}cos(x)\!$$

From table 2.1, K 2011, pg. 82, we have
 * $$y_p(x)=e^{-x}[Kcos(x)+Msin(x)]\!$$

Since this does not correspond to our homogeneous solution we can use the '''Basic Rule (a), K 2011, pg. 81''' to solve for the particular solution

$$y_p''+4y_p'+4y_p=r(x)\!$$

where

$$y_p'=e^{-x}[-Ksin(x)+Mcos(x)]-e^{-x}[Kcos(x)+Msin(x)]\!$$

$$y_p'=e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$$

and

$$y_p''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)]-e^{-x}[Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]\!$$

$$y_p''=e^{-x}[-Kcos(x)-Msin(x)+Ksin(x)-Mcos(x)+Ksin(x)-Mcos(x)+Kcos(x)+Msin(x)]\!$$

$$y_p''=e^{-x}[2Ksin(x)-2Mcos(x)]\!$$

Plugging these equations back into the differential equation

$$y_p''+4y_p'+4y_p=r(x)\!$$

$$e^{-x}[2Ksin(x)-2Mcos(x)]+4e^{-x}[-Ksin(x)+Mcos(x)-Kcos(x)-Msin(x)]+4e^{-x}[Kcos(x)+Msin(x)]=e^{-x}cos(x)\!$$

$$e^{-x}[2Ksin(x)-2Mcos(x)-4Ksin(x)+4Mcos(x)-4Kcos(x)-4Msin(x)+4Kcos(x)+4Msin(x)]=e^{-x}cos(x)\!$$

$$2Mcos(x)-2Ksin(x)=cos(x)\!$$

from the above equation it is obvious that $$K=0\!$$ and $$M=\frac{1}{2}\!$$

therefore the particular solution to the differential equation is

 * $$y_p(x)=\frac{1}{2}e^{-x}sin(x)\!$$

General Solution
The general solution will be the summation of the homogeneous and particular solutions

$$y(x)=y_h(x)+y_p(x)\!$$

$$y(x)=c_1e^{-2x}+c_2xe^{-2x}+\frac{1}{2}e^{-x}sin(x)\!$$

$$y(x)=e^{-2x}(c_1+c_2x)+\frac{1}{2}e^{-x}sin(x)\!$$

The coefficients $$c_1\!$$ and $$c_2\!$$ can be readily solved for given either initial conditions or boundary value conditions.

Problem Statement
Find a (real) general solution. State which rule you are using. Show each step of your work.
 * $$y''+y'+(\pi ^2+\frac{1}{4})y=e^{-\frac{x}{2}}sin(\pi x)\!$$

Homogeneous Solution
To find the homogeneous solution, $$y_h\!$$, we must find the roots of the equation

$$\lambda^2+\lambda+(\pi ^2+\frac{1}{4})=0\!$$

$$\lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\!$$ with $$a=1, b=1,  c=(\pi ^2+\frac{1}{4})\!$$

$$\lambda = \frac{-1 \pm \sqrt{1^2-4*1*(\pi ^2+\frac{1}{4})}}{2*1}\!$$

$$\lambda =-\alpha \pm i\omega=-\frac{1}{2} \pm \pi i\!$$

We know the homogeneous solution for the case of a double root to be

$$y_h=e^{-\alpha x}[Acos(\omega x)+Bsin(\omega x)]\!$$

 * $$y_h=e^{-\frac{x}{2}}[Acos(\pi x)+Bsin(\pi x)]\!$$

Particular Solution
We have the following excitation
 * $$r(x)=e^{-\frac{x}{2}}sin(\pi x)\!$$

From table 2.1, K 2011, pg. 82, we have
 * $$y_p(x)=e^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]\!$$

Since this corresponds to our homogeneous solution we must use the '''Modification Rule (b), K 2011, pg. 81''' to solve for the particular solution

so $$y_p(x)=xe^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]\!$$

differentiating

$$y_p'=\pi xe^{-\frac{x}{2}}[-Ksin(\pi x)+Mcos(\pi x)]+(e^{\frac{x}{2}}-\frac{1}{2}xe^{-\frac{x}{2}})[Kcos(\pi x)+Msin(\pi x)]\!$$

$$y_p'=e^{-\frac{x}{2}}(\pi x[-Ksin(\pi x)+Mcos(\pi x)]+(1-\frac{1}{2}x)[Kcos(\pi x)+Msin(\pi x)])\!$$

$$y_p'=e^{-\frac{x}{2}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-\frac{1}{2}xKcos(\pi x)-\frac{1}{2}xMsin(\pi x)]\!$$

and

$$y_p''=e^{-\frac{x}{2}}[-\pi ^2xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^2xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+\frac{1}{2}\pi xKsin(\pi x)-\frac{1}{2}Kcos(\pi x)-\frac{1}{2}\pi xMcos(\pi x)-\frac{1}{2}Msin(\pi x)]-\frac{1}{2}e^{-\frac{x}{2}}[-\pi xKsin(\pi x)+\pi xMcos(\pi x)+Kcos(\pi x)+Msin(\pi x)-\frac{1}{2}xKcos(\pi x)-\frac{1}{2}xMsin(\pi x)]\!$$

$$y_p''=e^{-\frac{x}{2}}[-\pi ^2xKcos(\pi x)-\pi Ksin(\pi x)-\pi ^2xMsin(\pi x)+\pi Mcos(\pi x)-\pi Ksin(\pi x)+\pi Mcos(\pi x)+\frac{1}{2}\pi xKsin(\pi x)-\frac{1}{2}Kcos(\pi x)-\frac{1}{2}\pi xMcos(\pi x)-\frac{1}{2}Msin(\pi x)+\frac{1}{2}\pi xKsin(\pi x)-\frac{1}{2}\pi xMcos(\pi x)-\frac{1}{2}Kcos(\pi x)-\frac{1}{2}Msin(\pi x)+\frac{1}{4}xKcos(\pi x)+\frac{1}{4}xMsin(\pi x)]\!$$

grouping cosine and sine terms we get

$$y_p''=e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K)sin(\pi x)]\!$$

and

$$y_p'=e^{-\frac{x}{2}}[(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK +M-\frac{1}{2}xM)sin(\pi x)]\!$$

next we substitute the above equations into the ODE

$$y_p''+y_p'+(\pi ^2+\frac{1}{4})y_p=r(x)\!$$

$$e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K)sin(\pi x)]+e^{-\frac{x}{2}}[(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK +M-\frac{1}{2}xM)sin(\pi x)]+(\pi ^2+\frac{1}{4})xe^{-\frac{x}{2}}[Kcos(\pi x)+Msin(\pi x)]=e^{-\frac{x}{2}}sin(\pi x)\!$$

$$e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K)sin(\pi x)+(\pi xM+K-\frac{1}{2}xK)cos(\pi x)+(-\pi xK +M-\frac{1}{2}xM)sin(\pi x)+(\pi ^2+\frac{1}{4})x(Kcos(\pi x)+Msin(\pi x))]=e^{-\frac{x}{2}}sin(\pi x)\!$$

$$e^{-\frac{x}{2}}[(2\pi M- \pi xM+(\frac{1}{4}-\pi ^2)xK-K+\pi xM+K-\frac{1}{2}xK+(\pi ^2+\frac{1}{4})xK)cos(\pi x)+((\frac{1}{4}-\pi ^2)xM-M+\pi xK-2\pi K-\pi xK +M-\frac{1}{2}xM+(\pi ^2+\frac{1}{4})xM)sin(\pi x)]=e^{-\frac{x}{2}}sin(\pi x)\!$$

$$2\pi Mcos(\pi x)-2\pi Ksin(\pi x)=sin(\pi x)\!$$

after cancelling terms; we can equate cosine and sine coefficients to get two equations

$$2\pi M=0\!$$

$$-2\pi K=1\!$$

so $$M=0\!$$ and $$K=-\frac{1}{2\pi}\!$$

and the particular solution to the ODE is

 * $$y_p(x)=-\frac{1}{2\pi}xe^{-\frac{x}{2}}cos(\pi x)\!$$

General Solution
The general solution will be the summation of the homogeneous and particular solutions

$$y=y_h+y_p\!$$

$$y=e^{-\frac{x}{2}}[Acos(\pi x)+Bsin(\pi x)]-\frac{1}{2\pi}xe^{-\frac{x}{2}}cos(\pi x)\!$$

$$y=e^{-\frac{x}{2}}[Acos(\pi x)+Bsin(\pi x)-\frac{1}{2\pi}xcos(\pi x)]\!$$

Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)

Problem 3.9
Solved by Gonzalo Perez

Problem Statement
Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.

(K 2011 pg.85 #13)
$$ 8y'' - 6y' + y = 6 \cosh x \! $$              (1)

Initial conditions are:

$$ y(0) = 0.2, y'(0) = 0.05 \! $$

Solution
The general solution of the homogeneous ordinary differential equation is

$$ 8y'' - 6y' + y = 0 \! $$

We can use this information to determine the characteristic equation:

$$ 8 \lambda^2 - 6 \lambda + 1 = 0 \! $$

And proceeding to find the roots,

$$ 4\lambda(2 \lambda - 1) - 1(2 \lambda - 1) = 0 \! $$

Thus, $$ (4 \lambda - 1)(2 \lambda - 1) = 0 \! $$.

Solving for the roots, we find that $$ \lambda = \frac{1}{4}, \frac{1}{2}, \! $$

where the general solution is

$$ y_k = c_1 e^{\frac{1}{4}x} + c_2 e^{\frac{1}{2}x} \! $$.

The solution of $$ y_p \! $$ of the non-homogeneous ordinary differential equation is

$$ x = \frac{e^x+e^{-x}}{2} \! $$.

Using the Sum rule as described in Section 2.7, the above function translates into the following:

$$ y_p = y_{p1} + y_{p2} \! $$, where Table 2.1 tells us that:

$$ y_{p1} = Ae^x \! $$ and $$ y_{p2} = Be^x \! $$.

Therefore, $$ y_{p} = Ae^x + Be^{-x} \! $$.

Now, we can substitute the values ($$ y_p, y_p', y_p'' \! $$) into (1) to get:

$$ 8(Ae^x+Be^{-x}) - 6(Ae^x - Be^{-x}) + Ae^x + Be^{-x} = 6(\frac{e^x + e^{-x}}{2}) \! $$

$$ = 3Ae^x + 15Be^{-x} \! $$

$$ = 3(e^x + e^{-x}) \! $$

Now that we have this equation, we can equate coefficients to find that:

$$ 3A = 3 \! $$

$$ \therefore A = 1 \! $$

$$ B = \frac{3}{15} = \frac{1}{5} \ $$

and thus, $$ y_p = e^x + \frac{1}{5}e^{-x} \! $$

We find that the general solution is in fact:

$$ y = y_k + y_p \! $$

$$ y = c_1 e^{\frac{1}{4}}x + c_2 e^{\frac{1}{2}}x + e^x + 3e^{-x} \! $$

whereas the general solution of the given ordinary differential equation is actually:

$$ y = c_1 e^{\frac{1}{4}}x + c_2 e^{\frac{1}{2}}x + e^x + \frac{1}{5} e^{-x} \! $$

Solving for the initial conditions given and first plugging in $$ y(0) = 0.2 \! $$, we get that:

$$ 0.2 = c_1 e^{\frac{1}{4}(0)} + c_2 e^{\frac{1}{2}(0)} + e^0 + 3e^{(0)} \! $$

$$ 0.2 = c_1 e^{(0)} + c_2 e^{(0)} + e^{(0)} + 3e^{(0)} \! $$

$$ 0.2 = c_1 + c_2 + 1 + \frac{1}{5} \! $$

$$ \therefore c_1 + c_2 = - 1 \! $$.             (2)

And now we can determine the first order ODE :

$$ y' = \frac{1}{4}c_1 e^{\frac{1}{4}(x)} + \frac{1}{2}c_2 e^{\frac{1}{2}(x)} + e^{x} - \frac{1}{5}e^{x} \! $$

The second initial condition that was given to us, $$ y'(0) = 0.05 \! $$ can now be plugged in:

$$ 0.05 = \frac{1}{4}c_1 e^{\frac{1}{4}(0)} + \frac{1}{2}c_2 e^{\frac{1}{2}(0)} + e^{(0)} - \frac{1}{5}e^{(0)} \! $$

$$ 0.05 = \frac{1}{4}c_1 e^{(0)} + \frac{1}{2}c_2 e^{(0)} + e^{(0)} - \frac{1}{5}e^{(0)} \! $$

$$ 0.05 = \frac{1}{4}c_1 + \frac{1}{2}c_2 + 1 - \frac{1}{5} \! $$

$$ \frac{1}{4}c_1 + \frac{1}{2}c_2 = -0.75 \! $$

$$ \therefore c_1 + 2 c_2 = -3 \! $$             (3)

Once we solve (2) and (3), we can get the values:

$$ c_1 = 1, c_2 = - 2 \! $$.

And once we substitute these values, we get the following solution for this IVP:

$$ y = e^{\frac{1}{4}x} - 2 e^{\frac{1}{2}x} + e^{x} + \frac{1}{5} e^{-x} \! $$

(K 2011 pg.85 #14)
$$ y'' + 4y' + 4y = e^{-2x} sin2x \! $$              (1)

Initial conditions are:

$$ y(0)=1, y'(0) = -1.5 \! $$

Solution
The general solution of the homogeneous ordinary differential equation is

$$ y'' + 4y' + 4y = 0 \! $$

We can use this information to determine the characteristic equation:

$$ \lambda^2 + 4 \lambda + 4 = 0 \! $$

And proceeding to find the roots,

$$ (\lambda + 2) (\lambda + 2) = 0 \! $$

Solving for the roots, we find that $$ \lambda = -2, -2 \! $$

where the general solution is:

$$ y_k = c_1 e^{-2x} + c_2 e^{-2x} x \! $$, or:

$$ y_k = (c_1 + c_2 x) e^{-2x} \! $$

Now, according to the Modification Rule and Table 2.1 in Section 2.7, we know that we have to multiply by x to get:

$$ y_p = e^{-2x}(K x cos2x + Mx sin2x) \! $$, since the solution of $$ y_k \! $$ is a double root of the characteristic equation.

We can then derive to get $$ y_p' \! $$:

$$ y_p' = -2 e ^{-2x} (K x cos2x + M x sin2x) + e^{-2x} (Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x) \! $$

$$ y_p' = (-2K+2M) e^{-2x} x cos2x + (-2M-2K) e^-2x xsin2x + Ke^{-2x} cos2x + Me^{-2x} sinx \! $$

Deriving once again to solve for $$ y_p'' \! $$, we get the following:

$$ y_p'' = 4e^{-2x} (Kxcos2x + Mxsin2x) - 2e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) - 2e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) \! $$

$$ y_p'' = 4e^{-2x} (Kxcos2x + Mxsin2x) - 4e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) + e^{-2x} (-4Ksin2x - 4Kxcos2x + 4Mcos2x - 4Mxsin2x) \! $$

$$ y_p'' = (-4K + 4M) e^{-2x} cos2x + (-4K - 4M)e^{-2x} sin2x \ $$

Now, we can substitute the values ($$ y_p, y_p', y_p'' \! $$) into (1) to get:

$$ (-4K + 4M) e^{-2x}cos2x+(-4M-4K)e^{-2x}sin2x+4((-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-2x}xsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx) + 4(e^{-2x}(Kxcos2x + Mxsin2x)) = e^{2x}sin2x \! $$

$$ \therefore (-3K+4M)e^{-2x}cos2x + (-3M-4K)e^{-2x}sin2x = e^{-2x}sin2x \! $$

Now that we have this equation, we can equate coefficients to find that:

$$ -3K + 4M =0 \! $$ and $$ -4K - 3M = 1 \! $$

and finally discover that:

$$ M = - \frac{3}{25} \! $$ and $$ K = - \frac{4}{25} \! $$.

Plugging in these values in $$ y_p \! $$, we find that:

$$ y_p = e^{-2x}( - \frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$

And finally, we arrive at the general solution of the given ordinary differential equation:

$$ y = y_k + y_p \! $$

$$ y = (c_1 + c_2 x) e^{-2x}+e^{-2x} (-\frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$

Solving for the initial conditions given and first plugging in $$ y(0) = 1 \! $$, we get that:

$$ 1 = (c_1 + c_2 (0)) e^{-2(0)}+e^{-2(0)} (-\frac{4}{25}(0)cos2(0) - \frac{3}{25}(0)sin2(0)) \! $$

$$ 1 = (c_1 + c_2(0))e^0 \! $$

$$ \therefore c_1 = 1 \! $$

The second initial condition that was given to us, $$ y'(0) = -1.5 \! $$ can now be plugged in:

$$ y' = - \frac{1}{5} e^-2x (10c_1 + 10c_2x - 5c_2 + (3-14x)sin2x+(4-2x)cos(2x)) \! $$

$$ -1.5 = -\frac{1}{5}(10c_1 - 5c_2 + 4) \! $$

$$ \therefore c_2 = -3.5 \! $$

And once we substitute these values, we get the following solution for this IVP:

$$ y = (1 - 3.5 x) e^{-2x}+e^{-2x} (-\frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$

Egm4313.s12.team11.perez.gp 20:34, 22 February 2012 (UTC)


text:
User:Egm4313.s12.team1.wyattling/r3

=Problems for Report 3=

Statement
Given the double root $$ \lambda = 5\!$$ and the excitation $$ r(x)=7e^{5 x}-2 x^2 \!$$, with the initial conditions $$ y(0) = 4, y'(0) = 5 \!$$ find the solution $$ y(x)\!$$.

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$.

Using the given double root, we can find the equation for the characteristic equation which leads to the homogeneous solution:

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$$\displaystyle{(\lambda - 5)^2 = \lambda^2 - 10\lambda + 25 = 0}$$ (1.0)
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Homogeneous solution:
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$$\displaystyle{y'' - 10y' + 25y = r(x)}$$ (1.1)
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$$\displaystyle{y'' - 10y' + 25y = 7e^{5x} - 2x^2}$$ (1.2)
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First, we find that the general equation associated with the given double root is:

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$$\displaystyle{y_g(x) = C_1e^{5x} + C_2xe^{5x}}$$ (1.3)
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We need to find the particular solution to the excitation $$ r(x) = 7e^{5x} - 2x^2\!$$. In analyzing the excitation, it is found that the particular solution looks like this:

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$$\displaystyle{y_p(x) = Cx^2e^{5x} - [K_2x^2 + K_1x + K_0]}$$ (1.4)
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Now, we need to find the values for the constants $$ C, K_2, K_1, K_0\!$$, by taking the first and second derivatives of the particular solution:

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$$\displaystyle{y_p(x) = Cx^2e^{5x} - [K_2x^2 + K_1x + K_0]}$$ (1.5)
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$$\displaystyle{y_p(x)' = 2Cxe^{5x} + 5Cx^2e^{5x} - [2K_2x + K_1]}$$ (1.6)
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$$\displaystyle{y_p(x)'' = 2Ce^{5x} + 20Cxe^{5x} + 25Cx^2e^{5x} - [2K_2]}$$ (1.7)
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Now, plug these into the homogeneous solution (1.2) and simplify:

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$$\displaystyle{2Ce^{5x} - 2K_2 + 20K_2x + 10K_1 - 25K_2x^2 - 25K_1x - 25K_0 = 7e^{5x} - 2x^2}$$ (1.8)
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Comparing the coefficients for $$ e^{5x}, x^2, and x\!$$ allows us to solve for the unknown coefficients:

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$$\displaystyle{e^{5x}: 2C = 7 \rightarrow C = 7/2}$$ (1.9)
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$$\displaystyle{x^2: -25K_2 = -2 \rightarrow K_2 = 0.08}$$ (1.10)
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$$\displaystyle{x: 20K_2 - 25K_1 = 0 \rightarrow K_1 = 0.064}$$ (1.11)
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$$\displaystyle{-2K_2 + 10K_1 - 25K_0 = 0 \rightarrow K_0 = 0.0192}$$ (1.12)
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Plug into $$ y_p(x)\!$$:

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$$\displaystyle{y_p(x) = \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.13)
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Now that we have the particular solution, we can use the initial conditions to solve for the unknown constants in the general solution, $$ y_g(x)\!$$, and then we will be able to solve for the final solution $$ y(x)\!$$:

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$$\displaystyle{y(x) = C_1e^{5x} + C_2xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192]}$$ (1.14)
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Take derivative:

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$$\displaystyle{y(x)' = 5C_1e^{5x} + C_2e^{5x} +5C_2xe^{5x} + 7xe^{5x} + \frac{35}{2}x^2e^{5x} - [0.16x + 0.064]}$$ (1.15)
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Using the given initial conditions:

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$$\displaystyle{y(0) = 4 \rightarrow 4 = C_1 - 0.0192 \rightarrow C_1 = 4.0192}$$ (1.16)
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$$\displaystyle{y'(0) = 5 \rightarrow 5 = 5C_1 + C_2 - 0.064 \rightarrow C_2 = -15.032}$$ (1.17)
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Plug these constants back into the solution to obtain the final solution:

$$  \displaystyle y(x) = 4.0192e^{5x} - 15.032xe^{5x} + \frac{7}{2}x^2e^{5x} - [0.08x^2 + 0.064x + 0.0192] $$     (1.18)
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Author
Solved and Typed By -

Reviewed By -

Statement
Find a general solution for the following two problems:

Given
Homogeneous solution: $$ y'' + 4y' + 4y = e^{-x}cosx\!$$

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$.

First, we will find the general solution, $$ y_g(x) \!$$, by finding the roots of the characteristic equation:

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$$\displaystyle{\lambda^2 + 4\lambda + 4\lambda = 0 \rightarrow (\lambda + 2)^2}$$ (8.0)
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Which means that the characteristic equation has a double root of $$ \lambda = -2 \!$$.

Based on the double root, then, the general solution is:

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$$\displaystyle{y_g(x) = (C_1 + C_2x)e^{-\frac{ax}{2}} \rightarrow y_g(x) = (C_1 + C_2x)e^{-2x}}$$ (8.1)
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Next, we need to find the particular solution, and based on analysis of the excitation, $$ r(x) = e^{-x}cosx \!$$, we find that the particular solution is the following:

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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
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Now, we need to find the values for the constants $$ K, M\!$$, by taking the first and second derivatives of the particular solution (8.2):

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$$\displaystyle{y_p(x) = e^{-x}Kcosx + e^{-x}Msinx}$$ (8.2)
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$$\displaystyle{y_p(x)' = -e^{-x}Ksinx - e^{-x}Kcosx + e^{-x}Mcosx - e^{-x}Msinx}$$ (8.3)
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$$\displaystyle{y_p(x)'' = -e^{-x}Kcosx + e^{-x}Ksinx + e^{-x}Ksinx + e^{-x}Kcosx - e^{-x}Msinx - e^{-x}Mcosx - e^{-x}Mcosx + e^{-x}Msinx}$$ (8.4)
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Now, plug these into the homogeneous solution and simplify:

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$$\displaystyle{-e^{-x}(Kcosx - Ksinx - Ksinx - Kcosx + Msinx + Mcosx + Mcosx - Msinx) + 4[-e^{-x}(Ksinx + Kcosx - Mcosx + Msinx)] + 4[e^{-x}(Kcosx + Msinx)] = e^{-x}cosx}$$ (8.5)
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$$\displaystyle{-e^{-x}[-2Ksinx + 2Mcosx] - e^{-x}[4Ksinx + 4Kcosx - 4Mcosx + 4Msinx] + e^{-x}[4Kcosx + 4Msinx] = e^{-x}cosx}$$ (8.6)
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Pull $$ e^{-x} \! $$ out of equation (8.6) and simplify:

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$$\displaystyle{e^{-x}[4Kcosx + 4Msinx - 4Ksinx - 4Kcosx + 4Mcosx - 4Msinx + 2Ksinx - 2Mcosx] = e^{-x}cosx}$$ (8.7)
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$$\displaystyle{e^{-x}[-2Ksinx + 2Mcosx] = e^{-x}cosx}$$ (8.8)
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Now, solve for the unknown coefficients:

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$$\displaystyle{2M = 1 \rightarrow M = \frac{1}{2}}$$ (8.9)
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$$\displaystyle{-2K = 0 \rightarrow K = 0}$$ (8.10)
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Plug into $$ y_p(x)\!$$ (8.2):

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$$\displaystyle{y_p(x) = \frac{1}{2}e^{-x}sinx}$$ (8.11)
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Now, we have both the particular solution and the general solution, which allows for us to solve for the final solution, $$ y(x) \! $$

$$  \displaystyle y(x) = (C_1 + C_2x)e^{-2x} + \frac{1}{2}e^{-x}sinx $$     (8.12)
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Given
Homogeneous solution: $$ y'' + y' (\pi^2 + \frac{1}{4}) = e^{-\frac{x}{2}}sin{\pi x}\!$$

Solution
The solution $$ y(x)\! $$ is composed of a general solution and a particular solution so that $$ y(x) = y_g(x) + y_p(x)\!$$.

First, we will find the general solution, $$ y_g(x) \!$$, by finding the roots of the characteristic equation, using the quadratic equation:

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$$\displaystyle{\lambda^2 + \lambda + (\pi^2 + \frac{1}{4}) = 0 \rightarrow \lambda = \frac{-1 \pm \sqrt{1-4(1)(\pi^2 + \frac{1}{4})}}{2}}$$ (8.13)
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$$\displaystyle{\lambda = -\frac{1}{2} + i\pi}$$ (8.14)
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The roots of the characteristic equation are complex conjugates, meaning that the general solution, $$ y_g(x) \!$$, is the following:

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$$\displaystyle{y_g(x) = e^{-\frac{x}{2}}(Acos{\pi x} + Bsin{\pi x})}$$ (8.15)
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Next, we need to find the particular solution, and based on analysis of the excitation, $$ r(x) = e^{-\frac{x}{2}}sin{\pi x} \!$$, we find that the particular solution is the following:

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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-\frac{x}{2}}Kcos{\pi x} + e^{-\frac{x}{2}}Msin{\pi x}}$$ (8.16)
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Now, we need to find the values for the constants $$ K, M \!$$, by taking the first and second derivatives of the particular solution (8.16):

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$$\displaystyle{y_p(x) = e^{\alpha x}(Kcos\omega x + Msin\omega x) \rightarrow y_p(x) = e^{-\frac{x}{2}}Kcos{\pi x} + e^{-\frac{x}{2}}Msin{\pi x}}$$ (8.16)
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$$\displaystyle{y_p(x)' = {-\pi e^{\frac{-x}{2}}Ksin{\pi x} - \frac{{}e^{\frac{-x}{2}}}{2}Kcos{\pi x}} + {\pi e^{\frac{-x}{2}}Mcos{\pi x} + \frac{{}e^{\frac{-x}{2}}}{2}Msin{\pi x}}}$$ (8.17)
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$$\displaystyle{y_p(x)'' = {-\pi^2 e^{\frac{-x}{2}}Kcos{\pi x} +  \pi \frac{e^{\frac{-x}{2}}}{2}Ksin{\pi x}}  +  \pi \frac{e^{\frac{-x}{2}}}{2}Ksin{\pi x}  + \frac{e^{\frac{-x}{2}}}{4}Kcos{\pi x}  - \pi^2 {e^{\frac{-x}{2}}}Msin{\pi x} - \pi \frac{e^{\frac{-x}{2}}}{2}{}Mcos{\pi x} - \pi \frac{e^{\frac{-x}{2}}}{2}{}Mcos{\pi x} + \frac{e^{\frac{-x}{2}}}{4}{}Msin{\pi x}}$$ (8.18)
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Now, we plug these into the homogeneous solution and simplify but we find that everything cancels out, meaning the unknown constants, $$ M, K \!$$ are both equal to $$ 0 \! $$. This means that the final solution is equal to just the general solution:

$$  \displaystyle y(x) = e^{\frac{-x}{2}}(Acos{\pi x} + Bsin{\pi x}) $$     (8.19)
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Author
Solved and Typed By - Reviewed By -

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

title: User:Egm4313.s12.team1.wyattling/r3


text:
Engineering Experience 4: Design a Small Solar Vehicle/Nl/2012: Team PM8

Inleiding
Deze engineering experience draait om “make stuff work”, meer bepaald om het zelf ontwerpen en ontwikkelen van een kleine zonnewagen. Deze zal het dan op het einde van het project moeten opnemen in een race tegen de andere teams. Het bouwen van zo’n zonnewagen zal bekeken en besproken worden vanuit drie invalshoeken. “Engineering”, waarbij onder andere de elektronische en mechanische werking en de sterkte van de materialen zal worden besproken en gesimuleerd met Simulink, alvorens effectief tot het werkelijk testen van de SSV wordt overgegaan. “Enterprising”, waarin de kost van het bouwen van een zonnewagen zal worden berekend en een marketingonderzoek zal worden gedaan. En als derde een gedeelte “Educating”, dat het opstellen van een rapport omvat.

Opdrachtgever
Dit EE4-project wordt als onderdeel van de opleiding Industriële Wetenschappen in de 2de bachelorfase van Groep T gegeven. Dit project is ook sterk verbonden aan het Umicore Solarteam.

1.1. Afspraken
In normale omstandigheden wordt het werk evenredig verdeeld in samenspraak met de groep. Iedereen wordt op elke vergadering verwacht. In het geval dat de afwezigheid van een teamlid problematisch is, wordt daarover gepraat binnen de groep en met het teamlid om tot een oplossing te komen die voor iedereen past. In geval van onkunde of onwil wordt de persoon aangesproken en erop toegewezen. Indien er niets veranderd aan de probleemsituatie na samenspraak en overleg, wordt dit gemeld aan de vakverantwoordelijke.

1.2. Verantwoordelijkheden
Meer gespecifieerde taakverdeling is terug te vinden in de gantt chart.

1.3. Wekelijkse vergadering
Wekelijkse meetings zullen plaatsvinden op dinsdagnamiddag in de cafetaria van GroepT tijdens of na het geplande EE4-moment, afhankelijk of er les wordt gegeven. Iedereen wordt verwacht op elke meeting, indien afwezig moet de teamleider verwittigd worden. Inhalen gebeurt op eigen houtje en voor de volgende vergadering. De persoon die zonder geldige reden (bv. overslapen is geen geldige reden) een vergadering mist, zal een extra workload toegewezen krijgen gelijk aan de workload dat de overige teamleden aan de vergadering hebben besteed. De vergadering is een moment om samen te werken, problemen te voorkomen en te bespreken. Voorzitter en notulist van de vergaderingen zijn geen vaste functies, maar veranderen naargelang het onderwerp van de vergadering. Voorzitter en notulist worden besproken in de vergaderverslagen.

1.4. Communicatie
Het merendeel van de communicatie verloopt binnen de vergaderingen en EE4-momenten. Andere communicatie buiten deze momenten verlopen via bovenvermelde mailadressen of via de facebookgroep.

1.5. Kosten
De maximale kost van dit project bedraagt 200,00 €. Deze kost wordt evenredig gedragen door de 9 teamleden.

2.1. Beginsituatie
De opdracht bestaat dus uit het ontwerpen en ontwikkelen van een SSV, een small solar vehicle of kleine zonnewagen. Hiervoor krijgen de studenten reeds een zonnepaneeltje en een motor van hun opdrachtgevers. De rest van het benodigde materiaal moeten ze zelf voorzien, maar met een beperking van €200. De bedoeling is dan deze zonnewagen te optimaliseren voor een race die dan op het einde van het project wordt gehouden. Daarnaast worden de wagentjes ook beoordeeld op innovativiteit en design. Om de wagen te optimaliseren dienen de nodige berekeningen te worden gemaakt qua elektronische en mechanische werking, sterkte van de materialen, … En deze worden dan ook getest met behulp van Simulink. Daarnaast wordt ook het economische gedeelte in rekening gebracht, hiervoor wordt de kostprijs afgeleid voor het bouwen van zo’n zonnewagentje en een marketingonderzoek uitgevoerd. Dit alles wordt dan uiteindelijk schriftelijk samengebracht in een eindrapport.

2.2 Doelen
De doelen worden besproken aan de hand van de 3 onderdelen van het project.

2.2.1. Engineering
Alvorens tot het einddoel te komen worden er 3 stappen of cases doorlopen.
 * De eerste case bestaat uit het ontwerpen en bouwen van de zonnewagen. Het doel hierbij is om het design van de zonnewagen te optimaliseren door de nodige berekeningen te doen.
 * Daarna komt de Simulink Case. Waar de optimalisatie verder wordt te verwezenlijkt door het gebruik van simulatie software “Matlab Simulink”.
 * Om vervolgens over te gaan naar de derde case waarbij de SSV wordt getest en uitvoerig besproken en berekend. Daarnaast wordt ook een technische tekening in 3D gemaakt van de zonnewagen.

Eenmaal deze cases doorlopen zijn wordt er overgegaan naar de race met als belangrijkste einddoel het winnen van de race.

2.2.2. Enterprising
Hierbij zal naar 2 zaken worden toegewerkt opdat de zonnewagen succesvol kan worden verkocht wanneer dit de bedoeling zou zijn.
 * Daarvoor is ten eerste de kostprijs belangrijk. Deze wordt op voorhand geschat en berekend. Er wordt hier dan bij op toegezien dat deze onder de €200 blijft. Dit is belangrijk om de productie winstgevend te maken.
 * Daarnaast wordt ook de nodige marketing toegepast. Dit omvat dan het maken van een businessplan, een marktonderzoek, een website en een herkenbaar logo voor op de zonnewagen.

2.2.3. Educating
Dit omvat het opstellen van een goed geschreven en overzichtelijk eindrapport. En elke week wordt op een blog een logboek bijgehouden.

2.3. Problemen
Natuurlijk komen er bij het uitvoeren van een project ook vaak heel wat problemen kijken. Hierbij is het dan best om op voorhand al in te schatten welke deze zouden kunnen zijn. Problemen zullen zich echter steeds voordoen en dan is het belangrijk om vooral niet bij de pakken te blijven zitten en de motivatie en sfeer zo hoog mogelijk te houden.
 * Een eerste probleem is dat niet iedereen mekaar goed kent omdat de teams ook groepsoverschrijdend zijn. Hierdoor is het onder andere moeilijk om elkaars sterktes en zwaktes te kunnen inschatten wat een probleem kan vormen bij de taakverdeling. Zeker in het begin zal de samenwerking nog wat stroef verlopen. Maar als echte ingenieurs mag dit uiteindelijk geen al te groot probleem worden.
 * Een ander mogelijk probleem is het wegvallen van een teamlid door ziekte. Hierbij zal de ernst van de ziekte moeten worden ingeschat en hierbij dan kijken of de zieke zijn werk zelfstandig kan afmaken. Indien dit niet het geval is zal er een kleine verschuiving moeten gebeuren in de werkverdeling.
 * Alsook tijdsnood kan een mogelijk probleem vormen. Daarom is het belangrijk om op voorhand een goede planning en werkverdeling te maken en deze dan ook effectief op te volgen. Alleen dan kan een tekort aan tijd vermeden worden. Wanneer er een bepaalde fout in de planning is gemaakt, zoals meer werkuren dan eerst gepland, dienen de nodige verschuivingen te gebeuren en zal er een extra tandje moeten worden bijgestoken.
 * Dan kan het ook mogelijk zijn dat er iets van de wagen stuk gaat kort voor de race. Daarom is het belangrijk om al zeker de sterkte van de materialen goed te berekenen en voorzichtig met het materiaal om te springen. Wanneer dan toch nog iets zou breken is het best om het nodige herstelmateriaal te hebben voorzien.

3. WBS
Omdat dit te grote tabellen zijn om hier te zetten kunt u de WBS vinden via deze link: WBS the 8 team

4. Gantt chart
Net zoals de WBS is de gantt chart een groot bestand dus zullen we hier enkel de link posten: Gantt chart the 8 team

Week 1
Het eerste seminarie van EE4 was een introductie. Na deze introductie hebben we samengezeten met ons team en even tijd genomen om kennis te maken. We hebben de taken veldeeld, afspraken gemaakt en contactgegevens uitgewisseld. Ook hebben we besloten een facebookgroep op te richten en een dropboxmap aan te maken, hierdoor kan iedereen op elk moment aan de documenten.

Week 2
Deze week hebben we een seminarie gehad over de dc motor en het zonnepaneel dat we zullen gebruiken ons wagentje te bouwen. We hebben over dit seminarie ook een testje moeten afleggen via toledo. Na het seminarie zijn we een aantal metingen gaan doen met het zonnepaneel. Deze week hebben we ook feedback gekregen op de documenten die we vorige week hebben moeten indienen.

Week 3
Het seminarie ging in week 3 over: Na het seminarie hebben we samen gezeten met het team en hebben we een schets opgesteld voor het frame van onze wagen. We hebben ook nagedacht over een aantal voorstellen voor het project een beetje innovatiever te maken.
 * De belangrijke krachten zoals
 * De luchweerstand
 * De rolweerstand
 * De weerstand door de helling
 * De racecondities
 * De transmissie ratio (2 types)
 * Hoe de optimale gear ratio te berekenen
 * Introductie van matlab

Week 4
We hebben deze week een inleiding gekregen over het programma simunlink en mathlab. Hoe we deze moeten gebruiken wordt uitgelegd in video's die op toledo staan. We moeten van deze software gebruik maken om de race en onze wagen te simuleren, hierdoor zullen we een beeld krijgen van de prestaties van onze wagen zonder dat we hem werkelijk moeten testen. Voor deze simulaties zullen we een programma moeten maken in simulink dit zal bestaan uit delen die we hebben gekregen en delen die we zelf zullen moeten maken.

Deze week zijn we ook gestart met de schetsen van het frame in 3D zodat het ontwerp zeker klaar raakt in week 6.. De tekeingen zullen gemaakt worden in solid edge zodat er vanuit de 3D tekeningen makkelijk en correct 2D-tekingen kunnen gehaald worden. Als we de tekeningen hebben zou het bouwen van het wagentje op zich een minder moeilijke taak moeten zijn.

Week 5
Het waren deze week twee aparte onderdelen van het seminarie namelijk:
 * Umicore solar team: Hier heeft een lid van het solar team een presentatie gegeven over wat de belangrijke dingen zijn voor een zonnewagen. Een beetje extra uitleg over het team hebben we ook gekregen.

 * Het fablab in leuven: Deze presentatie ging over de verschillende machines die in het lab staan. Ook de openingsuren en de locatie van het lab zijn we te weten gekomen.

Na deze 2 seminaries zijn we met een deel van het team begonnen aan het simulink programma en de handmatige berekingen. Het simulink programma zorgde voor een aantal problemen en hier zijn we dan ook lang mee bezig geweest. Het ontwerp van de wagen in 3D is deze week afgewerkt zodat we kunnen beginnen hier uit de 2D tekingen te halen.

Week 6
Deze week zal er geen seminarie plaatsvinden. We hebben deze week de 2de deadline van dit project, voor deze deadline moeten we het volgende indienen:

 * Binnenleveren Case SSV deel 1 en Simulink op Toledo.
 * Eerste versie Proces Report online.
 * (Peer assesment 1 moeten we deze week ook invullen)

We hebben de bouw deze week afgerond. Het simulink programma zorgt voor een aantal moeilijkheden maar we hopen dat deze zullen opgelost zijn tegen de deadline.

Dinsdag van deze week zal er wel een vergadering met de coach plaatsvinden waarin we ons kunnen informeren hoe we het simulink programma in orde kunnen krijgen. Voor deze vergadering met onze coach zullen we met ons team nog even samen zitten om te zien wat er nog moet er nog gebeuren voor de deadline van deze vrijdag.

Week 7
Er was deze week weer geen seminarie. We hadden wel een afspraak gemaakt met onze coach. We hebben feedback gekregen op het ingeleverde rapport van vorige vrijdag.

Aangezien er nog een aantal aanpassingen moeten gebeuren aan het rapport zullen we ons hier deze week mee bezig houden. Iedereen zal het deel dat hij geschreven heeft voor het rapport verbeteren en dan zullen we deze verbeterde versies samen zetten in een nieuw rapport. De werkelijke bouw van het volledige wagentje zal ook nog verder gebeuren in de paasvakantie.

Na de vakantie kunnen we dan beginnen met SSV Part 2.

Week 8
Deze week was het de eerste echte test van het project. De test had als onderwerp: Ter voorbereiding van deze test hebben we ons rapport geupdate zodat alle fouten er uit zijn gehaald. Uitzonderlijk zullen we in deze week nog vergaderen op woensdag zodat we volgende week zeker klaar raken met de bouw van het wagentje. Dit is nodig omdat we normaal gezien volgende week onze SSV al zouden moeten testen op een aangelegde testbaan.
 * SSV Case 1
 * Simulink

Week 9
Het was deze week een week waar we de wagentjes al eens konden testen op een parcour dat was opgesteld in de kelder van groept. Aangezien hier geen zon is kon men de wagentjes testen m.b.v. een batterij. Wij hebben echter niet de mogelijkheid gehad om ons wagentje te testen omdat onze bestelde tandwielen,... nog niet waren toegekomen. We hadden begin deze week wel al de wieltjes van onze wagen gaan snijden in fablab.

We hebben uiteindelijk op vrijdag toch nog de bestelde onderdelen gekregen en het wagentje in elkaar gestoken. Normaal gezien is het wagentje nu dus af. We hebben de voorbije weken een aantal testen uitgevoerd voor een koeling aan te brengen op onze wagen, dit is dus nog het enige wat er moet aangebracht worden op het wagentje.

Week 10
Aangezien het deze week een korte week was zijn we niet samen gekomen met het hele team. We hebben wel nog afzonderlijk gewerkt aan de zaken voor de deadline in week 11.

Het wagentje is deze week ook getest geweest. Aangezien het parcour er nog stond hebben we ons wagentje hierop kunnen testen. De test ging vlot en we gaan dus ook met vertrouwen naar een mogelijke racve volgende week.

We hebben deze week ook wel gezien dat ons zonnepaneel niet werkte zoals verwacht. Er zal hier dus nog een oplossing voor moeten gevonden worden.

Week 11
De race met het wagentje zal deze week niet doorgaan omdat het weer dit niet wil toelaten. We hebben deze week wel een finale deadline voor het project. De technische tekeningen en de eindrapporten moeten worden ingeleverd. Ook moeten we deze week onze 2de peer assesment invullen.

Voor de race is het dus wachten op volgende week!


text:
University of Florida/Egm4313/s12.team11.gooding/R3

Problem 3.7
Solved By Kyle Gooding

Problem Statement
Expand the series on both sides of (1),(2) pg. 7-12b to verify these equalities. (1)

Given
$$\sum_{j=2}^{5}C_j*j(j-1)x^{j-2}= \sum_{j=0}^{3}C_{j+2}*(j+2)(j+1)x^{j} $$

(2) $$ \sum_{j=1}^{5}C_j*jx^{j-1}= \sum_{j=0}^{4}C_{j+1}*(j+1)x^{j}$$

Solutions
Expanding both sides of (1) results in: $$\sum_{j=2}^{5}C_j*j(j-1)x^{j-2}=C_2(2)(2-1)x^0+ C_3(3)(3-1)x^1+C_4(4)(4-1)x^2 + C_5(5)(5-1)x^3 $$ $$\sum_{j=0}^{3}C_j(j+2)(j+1)x^{j}=C_{0+2}(0+2)(0+1)x^0+ C_{1+2}(1+2)(1+1)x^1+C_{2+2}(2+2)(2+1)x^2 + C_{3+2}(3+2)(3+1)x^3 $$

Simplifying: $$\sum_{j=2}^{5}C_j*j(j-1)x^{j-2}=C_2(2)+ C_36x+C_412x^2 + C_520x^3 $$ $$\sum_{j=0}^{3}C_j(j+2)(j+1)x^{j}=C_2(2)+ C_36x+C_412x^2 + C_520x^3 $$

The two sums are equal. Expanding both sides of (2) results in: $$ \sum_{j=1}^{5}C_j*jx^{j-1}= C_1(1)x^0+ C_2(2)x^1+C_3(3)x^2+C_4(4)x^3 + C_5(5)x^4 $$ $$ \sum_{j=0}^{4}C_{j+1}*(j+1)x^{j}= C_{0+1}(0+1)x^0+ C_{1+1}(1+1)x^1+C_{1+2}(2+1)x^2+C_{2+2}(3+1)x^3 + C_{3+2}(4+1)x^4 $$

Simplifying: $$ \sum_{j=1}^{5}C_j*jx^{j-1}= C_1(1)+ C_2(2)x^1+C_3(3)x^2+C_4(4)x^3 + C_5(5)x^4 $$ $$ \sum_{j=1}^{5}C_j*jx^{j-1}= C_1(1)+ C_2(2)x^1+C_3(3)x^2+C_4(4)x^3 + C_5(5)x^4 $$ The two sums are equal.

Egm4313.s12.team11.gooding 23:49, 19 February 2012 (UTC)


text:
User:Lovysinghal

Greetings! Please contact me at my main user talk page on the Hindi Wikipedia.

title: User:Lovysinghal


text: Template:What Matters

title: Template:What Matters


text:
Dominant group/Music

As with other humanities, music may have dominant groups of music itself or in some way associated with music and the artists that perform it.

Dominant group may be an artifact of human endeavor or may have preceded humanity.

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

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

Music theory
"This grouping is shared with the major set, however, the pre-dominant group does not include major chords on II."

"With astonishing abruptness the bridge begins in G major (m. 9); it leads, through the V of B major, to a dominant group containing two ideas, a "second theme" (mm. 14-18) and a codetta, (mm. ... A striking augmentation marks the climax of the harmonic transition from B minor to A major (Ex. 16c). A diminution begins the first strain of the dominant group (Ex. 16d); later on the motive is transposed to AG# -F# -E (Ex. 16e)."

Music analysis
“The new 'key' would then dissolve into an augmented sixth leading to the introductory V of the dominant group.”

"Music theorists perpetuate this understanding by juxtaposing description and analysis: Music theory pedagogue Michael Rogers writes that “the goal of description…is to collect information,” whereas “analysis seeks to answer ‘how’ and ‘why’ questions” (2004:74-75). Rogers sees description as a necessary, but separate prerequisite to analysis. As he sees it, it is only once we begin to analyze, to move beyond mere description, that we show our intellectual, stylistic, and political selves."

"Heteroglossia is not often recognized by speakers. This, Bakhtin suggests, is no accident: Given that dominant groups often seek to impose a particular notion of reality on others, their rule is threatened by the existence of multiple worldviews. Thus by highlighting one linguistic genre as primary dominant groups elevate a single worldview—one that likely reinforces their position of power."

The dominants
"Traditionally, the harmony moves from subdominant to dominant group."

In German, the word “Dominanten” may mean a dominant or dominants. The phrase “den Dominanten” occurs in 1774 usually referring to “die Dominante” in music. The plural is “die Dominanten” (the dominants). But, “the dominant” translates to “die dominierende”.

A teaching textbook for piano from 1826 contains “Sind die Grundtöne der Dur- und Moll-Tonarten, jedoch ohne die mindeste Beziehung auf ihre Dominanten, durch den Zirkel angesetzt, so werden die Dominanten dann hinterher beigefügt.”

Texts for musical composition contain "the dominants", such as “Doch leide« dieses Berkoth eine Ausnahme, wenn die Dominanten häufig von kommen.” from 1781.

The earliest occurrence so far for "the dominants" comes from 1757, “so sind solche entweder tonische Noten, Dominanten, oder Unterdominanren, und die Dominanten können simple oder tonische Dominanten seyn.”

Criticism
"These are the "insight" critics, who Atkins believes to be the dominant group of his period. ... What all this comes back to in the end is the position Atkins has been staking out in his notes, reviews, poems, and music theory all along, an emphasis upon conspicuous technique and "designed" imagination."

Music education
"The word 'different' to refer to world music occurs 25 times within the 30 minute interview. The centre to which this 'difference' is compared however, is only implied. Moon (1992) notes that the dominant group, the seeming foundation to which all other styles are likened, is often a hidden and implicit construct, which functions to naturalise its position of power."

"Functionalism’s approach to the transmission of knowledge also supports the reproduction of the social status quo—in particular, of the knowledge, ideologies, and practices of dominant social groups that have a vested interest in preventing or minimizing change and, thus, of maintaining their advantages over other groups. ... These theories stress the problems (conflicts) associated with schools that serve mainly (or only) the purposes of dominant groups that have vested interests in maintaining social, economic, gender, and ethnic inequities. ... Viewed as a social practice, it is not a question of whose music, which musics (etc.) to teach, but of stressing the personal and social agency of music—in general and across various groups (dominant or not, in conflict or not). ... Teachers will not, as a result, presume to impose the music of a dominant group, nor will they seek to reproduce society (or music in society) in status quo terms. ... The question of “whose” music to teach recognizes music as social practice from the very first. It can refer to the musics of different ethnic groups in a multicultural community, or to the music imposed by dominant groups on the rest of society. Dominance (power) is not simply an economic or ‘class’ issue (although whose musics get government subsidy does highlight economic and class inequalities): it can involve the dominance of an ethnic majority, or the power (authority) granted to music teachers over their students and, thus, the power to impose one music or a narrow range of musics on students to the exclusion of others."

"Pierre Bourdieu and Jean-Claude Passeron (1977) have written extensively about how the relatively autonomous character of a school conceals its function as the most effective means for the cultural reproduction of the privileges of the dominant class. ... According to Young (1990) cultural imperialism involves the universalisation of a dominant group's experience and culture, and its establishment as the norm. Because the values and perspectives of dominant groups permeate cultural and institutional norms, members of oppressed groups have their lives interpreted through the lens of the dominant group, defined as ‘common sense’. Furthermore, oppressed groups often internalise the negative stereotypes to which their group is subjected (Baker, et al., 2004; Bell, 1997; Freire 1997a; 1997b). The focus of the music circles is on listening to what would be regarded by the dominant groups as the pinnacles of musical achievement, and to act as a culture study (Giroux, 1986). ... The procedure followed within these circles tries to change the meaning of those cultural products which express the dominant group's perspective on and interpretation of events and elements in society, including other groups in society, insofar as they attain cultural status at all (Young, 1990)."

Music professionalism
"Gramsci maintains that for a sociopolitically dominant group to sustain power, it must forge alliances that establish comprehensive social authority by winning the consent of subordinate groups, thus making this authority seem a natural and legitimate construction of social reality -- a reality taken for granted by culture members as the truth of common sense".

Latin American music
“These comprise: 1. The "spontaneous" consent given by the great masses of the population to the general direction imposed on social life by the dominant fundamental group; this consent is "historically" caused by the prestige (and consequent confidence) which the dominant group enjoys because of its position and function in the world of production.”

Modernist music
"In the first analysis, Crawford's String Quartet (1931) movement iii, Hisama weaves several analytical threads: composer/theorist Robert Morris's innovative analysis of "musical contour"; literary theorist Elaine Showalter's term "double-voiced discourse" and related literary concepts of Showalter's; anthropologist Edwin Ardener's position that "women in a society operate both within a dominant group and in a space outside it [and consequently] express their views of the world".

Popular music
“Within this framework, popular music represents the legitimate expression of members of the subdominant group who resist and oppose oppression by members of the dominant group.”

"It can be used by a dominant group to arouse loyalty to the nation and to those in power, and this is visible in the patriotic songs which are found in many countries."

Rap
"Dominant group perception of rap as violent and loud emerges, for Rose, out of a fear of black resistance and defiance."

"Sterling Stuckey (1987) and James Scott (1990) recognize African American music as critical and resistant to the dominant group."

Religious music
Notation: let the symbol CCM stand for "Contemporary Christian music".

“Avant garde artists are not the dominant group in terms of record sales and religious radio airplay, but they command a significant following among Christian youth.” "In challenging these basic premises of the dominant group's hegemony, CCM provides a space for potentially overcoming that dominance."

Rhythm and blues
"The concept of folk music was created by academic elites, but remained unfamiliar to most people until the organized left took it on as a cultural project in the late 1930s and 1940s. Both academic elites and political activists constructed the genre as an alternative to the racialized genres that the commercial recording industry had dubbed “race records” and “hillbilly music.”"

"As the music of “a people” or “a folk” it conforms to the principle of homology. But, in fact, folk music is typically the appropriation by one group, usually a dominant group, of someone else's music, fortifying social boundaries by breaching the principle of homology. ... Rhythm and blues has been so explicitly coded as black that it was first recorded on “race records” (Oliver 1984)."

"Unlike the proletarian, slave or peasant, the teenager through the passage of time becomes part of the gerontocratically dominant group. ... However, the influences here were the Donald Wood rhythm and blues hit “Death of an Angel” (c. 1955) with a touch of “Endless Sleep.""

Hypotheses

 * 1) Dominant group is more than musical notes in music.

For music the proof of concept initially is that dominant group is involved at all.

title: Dominant group/Music


text:
Volcanoes, list/Pinatubo

Location and Characteristics
The Mount Pinatubo volcano is located on the island of Luzon,Philippines and is one volcano of a chain of compsite volcanoes called the Luzon volcanic arc. This arc goes along the west coast of the island because of the subduction of Manila Trench which is off to the west.It is located on a convergent boundary where two tectonic plates collide (Tectonic plates are pieces of the Earth's lithosphere that are on the Earth's asthenosphere.).

What Happens at This Type of Boundary
One of these plates is made of oceanic crust while the other is made of continental crust. The oceanic plate gets subducted(one tectonic plate moving under the other) and magma forms after processes within the Earth. This magma evenutually rises to the Earth's surface and causes a volcanic eruption of explosive nature because of the silica-rich magma.

Characteristics
Mt.Pinatubo is a composite volcano because of its broad base and because the slopes of the volcano get steeper the closer you get to the summit. Composite volcanoes are made with a combination of pyroclastic material and lava. Composite volcanoes are also referred to as stratovolcanoes.

The Reawakening
Mt.Pinatubo had been dormant for about 500 years. And then a major earthquake with a magnitude of 7.8(quite similar to the San Francisco earthquake of 1906)rocked the area about 60 miles(100 kilometers)northeast of the Pinatubo volcano. This is what people think awoke the volcano from its long sleep. Apart from landslides,several minor local earthquakes and steam emissions, the Pinatubo volcano seemed undisturbed.However,scientists were wrong because in March and April 1991,magma started rising from more than 20 miles(32 kilometers) below Pinatubo leading to small earthquakes and steam explosions which led to the blasting of 3 craters. In April,May,and June 1991 many minor earthquakes occurred and noxious sulphur dioxide was released.

The Eruption
On the days of June 7-11 of 1991 the magma began to ooze out of the central vent but no real danger was presented due to lost pressure. On June 12th the first explosive eruption occurred and then the big one erupted on June 15th. The volcano exploded woth massive power,releasing 5 cubic kilometers of volcanic material and releasing ash clouds that rose 22 miles(35 kilometers)into the air. Fine ash travelled to the Indian Ocean and satellites tracked the ash cloud a handful of times around the globe. Very hot ash,gas,and pumice bits and fragments(pyroclastic flows)blanketed deep valleys with deposits as deep as 660 feet(200 meters)thick.The eruption blasted out a great deal of magma and rock from below the volcano,causing the summit to collapse and form a caldera(depression)a whopping 1.6 miles(2.5 kilometers)across. Typhoon Yunya was nearby when the volcano erupted so the ash from the eruption and the water from the typhoon mixed together to form tephra with a high water content. Ash and water also mixed on the ground forming lahars(mud flows).

Damage/Effects and a Lasting Impact
The eruption affected many people and their lives, but much more would've been affected if scientists hadn't been able to predict the volcano's eruption.If it weren't for the Philippines Institute of Volcanology and Seismology(PHILVOLCS) and the United States Geological Survey(USGS),many more lives would've been affected.Commmercial planes and other aircraft were warned of the rising ash cloud and most were able to avoid it.The tephra that rained down caused the roofs of many homes near the eruption site to collapse when it crushed them with massive weight and force. Lahars(destructive mud flows)also posed a threat to lives and property. The lahars completely submerged houses with a mud-like substance.The ash clouds also affected two nearby U.S. military bases-the Clark Air Base and the Subic Bay Naval Station. About 20million tons of sulphur dioxide was blasted into the Earth's stratosphere. This cloud spreaded around the world and temporarily caused global temperatures to drop half a degree Celsius even a few years after the eruption. The death toll was 300-800 people. Native people who lived on the slopes of Pinatubo called the Aeta Highlanders(20,000 people)were forced to move and many of them today still live in resettlement camps awaiting the day they can return to their home. The other 200,000(about)people who were evacuated prior and to the big eruption have returned but they still face threats of lahars burying their homes.

Other Eruptions/Eruption History
There have been seven other eruptions.The first ever eruption was around 7460 B.C. and it was an explosive eruption that had pyroclastic flows. The most recent eruption was in 1993 and there was some damage to property,there were evacuations,and there were some deaths. It was not as effective as the 1991 eruption,though.

References/Bibliography
http://www.georesources.co.uk/pinatubo.htm

http://www.volcano.si.edu/world/volcano.cfm?vnum=0703-083&volpage=erupt

http://vulcan.wr.usgs.gov/Volcanoes/Philippines/Pinatubo/Publications/Wolfe/wolfe_report.html

Berry,Kathleen M.,et al.Earth Science

title: Volcanoes, list/Pinatubo


text:
University of Florida/Egm4313/s12.team11.imponenti/R3.1

Problem Statement
Find the solution to the following L2-ODE-CC: $$y''-10y'+25y=r(x)\!$$

With the following excitation: $$r(x)=7e^{5x}-2x^2\!$$

And the following initial conditions: $$y(0)=4, y'(0)=-5\!$$

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

Homogeneous Solution
To find the homogeneous solution we need to find the roots of our equation

$$\lambda^2-10\lambda+25=0\!$$ $$(\lambda-5)(\lambda-5)=0\!$$ $$\lambda=5\!$$

We know the homogeneous solution for the case of a real double root with $$\lambda=5\!$$ to be

$$y_h=c_1e^{5x}+c_2xe^{5x}\!$$

Particular Solution
For the given excitation we must use the Sum Rule to the particular solution as follows

$$y_p=y_{p1}+y_{p2}\!$$ where $$y_{p1}\!$$ and $$y_{p2}\!$$ are the solutions to $$r_1(x)=7e^{5x}\!$$ and $$r_2(x)=-2x^2\!$$, respectively

First Particular Solution
$$r_1(x)=7e^{5x}\!$$,

from table 2.1, K 2011, pg. 82 we have

$$y_{p1}=Ce^{5x}\!$$

but this corresponds to one of our homogeneous solutions so we must use the modification rule to get

$$y_{p1}=Cx^2e^{5x}\!$$

Plugging this into the original L2-ODE-CC then substituting;

$$y_{p1}''-10y_{p1}'+25y_{p1}=r_1(x)\!$$

$$(Cx^2e^{5x})''-10(Cx^2e^{5x})'+25(Cx^2e^{5x})=r_1(x)\!$$

$$25Cx^2e^{5x}+10Cxe^{5x}+10Cxe^{5x}+2Ce^{5x}-10(5Cx^2e^{5x}+2Cxe^{5x})+25Cx^2e^{5x}=7e^{5x}\!$$

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

$$e^{5x}2C=7e^{5x}\!$$

so $$C=\frac{7}{2}\!$$ and the first particular solution is,
 * $$y_{p1}=\frac{7}{2}x^2e^{5x}\!$$

Second Particular Solution
$$r_2(x)=-2x^2\!$$,

from table 2.1, K 2011, pg. 82 we have

$$y_{p2}=a_2x^2+a_1x+a_0\!$$

Plugging this into the original L2-ODE-CC then substituting;

$$y_{p2}''-10y_{p2}'+25y_{p2}=r_2(x)\!$$

$$(a_2x^2+a_1x+a_0)''-10(a_2x^2+a_1x+a_0)'+25(a_2x^2+a_1x+a_0)=-2x^2\!$$

$$2a_2-10(2a_2x+a_1)+25(a_2x^2+a_1x+a_0)=-2x^2\!$$

grouping like terms we get three equations to solve for the three unknowns, these are written in matrix form

$$25a_2x^2+(25a_1-20a_2)x+(25a_0-10a_1+2a_2)=-2x^2\!$$

 * $$ \begin{bmatrix}

2 & -10 & 25\\ -20 & 25 & 0\\  25 & 0 & 0\end{bmatrix}

\begin{bmatrix} a_2\\ a_1\\ a_0\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ -2\end{bmatrix}\!$$

solving by back subsitution leads to $$a_2=-\frac{2}{25}, a_1=\frac{8}{125}, a_0=\frac{4}{125}\!$$

so the second particular solution is,
 * $$y_{p2}=-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

General Solution
The general solution is the summation of the homogeneous and particular solutions

$$y=y_h+y_{p1}+y_{p2}\!$$

$$y=c_1e^{5x}+c_2xe^{5x}+\frac{7}{2}x^2e^{5x}-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

$$y=e^{5x}(c_1+c_2x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

Applying the first initial condition $$y(0)=4\!$$

$$y(0)=c_1+\frac{4}{125}=4\!$$

$$c_1=\frac{496}{125}\!$$

Second initial condition $$y'(0)=-5\!$$

$$y'=\frac{d}{dx}y=e^{5x}[5(c_1+c_2x+\frac{7}{2}x^2)+c_2+7x]-\frac{4}{25}x+\frac{8}{125}\!$$

$$y'=e^{5x}[\frac{35}{2}x^2+(5c_2+7)x+5c_1+c_2]-\frac{4}{25}x+\frac{8}{125}\!$$

$$y'(0)=5c_1+c_2+\frac{8}{125}=-5\!$$

$$c_2=-\frac{3113}{125}\!$$

The general solution to the differential equation is therefore

$$y(x)=e^{5x}(\frac{496}{125}-\frac{3113}{125}x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$

Plot
Below is a plot of this solution and the solution to in the example on p.7-3

our solution $$y(x)=e^{5x}(\frac{496}{125}-\frac{3113}{125}x+\frac{7}{2}x^2)-\frac{2}{25}x^2+\frac{8}{125}x+\frac{4}{125}\!$$ (shown in red)

example on p.7-3 $$y(x)=e^{5x}(4-25*x+*\frac{7}{2}x^2)\!$$ (shown in blue)

Egm4313.s12.team11.imponenti 22:31, 20 February 2012 (UTC)


text: File:Plot3 1.jpg

title: File:Plot3 1.jpg