text:
User:Egm6322.s12.TEAM1.HW/HW3

Problem 1: Verification of Terms
 Solved without assistance

Given
after removing the coefficients of $$ \rho^n \ $$ we get

Solution
After removing the terms in Eq.1.2 fFrom Eq1.1, the remaining initial terms form the following equation: Combining and moving terms to the left hand side of the equation and setting equal to zero, one obtains: Simplify to obtain: Combine like terms to obtain: From Lecture 41, Eqs(1) and (2) the following is known: Substituting 1.7, 1.8 and 1.9 into 1.6, one obtains: Combining like terms and subtracting results in: $$ \therefore \ $$ Recurrence Relation 2 is satisfied using the initial terms

Problem 2: Generate Legendre Polynommials using Recurrence Relation2
 Solved without assistance

Solution
Given F10 class lecture 36, page2, equation 7 as shown here: starting with $$ P_3 \ $$ and using the integer portion of $$ \frac{n}{2} \ $$, thus when $$ n=1 \ $$ , $$ i=0 \ $$ and when $$ n= 2,3 \ $$, $$ i=1 \ $$. Thus we have the following: Repeating above steps for $$ n= 4,5,6 \ $$ we get the following equations and solutions:

Problem 3: Evaluation
 Solved without assistance

Given
and

Find
Evaluate $$ h_0 \ $$ by direct integration, then use Eq 3.1 to deduce Eq 3.2 by induction.

Solution:
First, integrate directly the first term: $$\begin{align} & {{h}_{0}}=\left\langle {{P}_{0}},{{P}_{0}} \right\rangle =\int_{x=-1}^{x=1}{{{P}_{0}}(x)}{{P}_{0}}(x)dx \\ & =\int_{x=-1}^{x=1}{1}*dx=2 \end{align}$$ Then, continue deriving: $$\begin{align} & {{h}_{n}}=\frac{2n-1}{2n+1}{{h}_{n-1}}=\frac{2(n-1)+1}{2(n-1)+3}{{h}_{n-1}} \\ & =\frac{2(n-1)+1}{2(n-1)+3}\times \frac{2(n-2)+1}{2(n-2)+3}{{h}_{n-2}}=\frac{2(n-2)+1}{2(n-1)+3}{{h}_{n-2}} \\ & =\frac{2(n-3)+1}{2(n-1)+3}{{h}_{n-3}} \\ & =\frac{2(n-3)+1}{2(n-1)+3}\cdots \frac{2*(1)+1}{2*(1)+3}\times \frac{2*(0)+1}{2*(0)+3}{{h}_{0}} \\ & =\frac{2*(0)+1}{2(n-1)+3}{{h}_{0}} \\ & =\frac{2}{2n+1} \end{align}$$

Problem 4: Weight Calculation
 Solved without assistance

Find
Show that $$ w_2=1 \ $$ by carrying out the calculation as shown in Eq4.1

Solution:
According to the following equation: $$\displaystyle {{w}_{j}}=\frac{-2}{(n+1){{{{P}'}}_{n}}({{x}_{j}}){{P}_{n+1}}({{x}_{j}})}$$ $$ \displaystyle n=2$$, and Legendre polynomials are: $$\displaystyle {{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1)$$ $$\displaystyle {{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x)$$ $$\displaystyle {{P}_{2}}^{\prime }(x)=3x$$ $${{x}_{1,2}}=\pm \frac{1}{\sqrt{3}}$$ So,

$$\begin{align} & {{w}_{2}}=\frac{-2}{(2+1){{{{P}'}}_{2}}({{x}_{2}}){{P}_{3}}({{x}_{2}})} \\ & =\frac{-2}{(2+1)*3*\frac{1}{\sqrt{3}}*\frac{1}{2}(5{{\left( \frac{1}{\sqrt{3}} \right)}^{3}}-3\frac{1}{\sqrt{3}})} \\ & =\frac{-2}{(2+1)*3*\frac{1}{\sqrt{3}}*\frac{1}{2}*(-\frac{4\sqrt{3}}{9})} \\ & =1 \end{align}$$

Problem 5: Verify table for Gauss Legendre Quadrature
 Solved without assistance

Find: Verify Table
(1) Verify values in table against corresponding numerical values from NIST Digital Library of Math Functions (2) Discuss the relative merits of using the exact, but irrational vaules versus using the numerical values. (3) Extend above table to include $$ n=6 \ $$

Solution:
(1).Verify the values for the roots of the Legendre polynomials:
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{align} P_0 &= 1\\ P_1 &= x \\ P_2 &= \frac{1}{2}(3x^2-1) \\ P_3 &= \frac{1}{2}(5x^3-3x) \\ P_4 &= \frac{1}{8}(35x^4-30x^2+3) \\ P_5 &= \frac{1}{8}(63x^5-70x^3+15x) \\ P_6 &= \frac{1}{16}(231x^6-315x^4+105x^2-5) \end{align} $$
 *
 * }
 * }

To verify the values of the roots in the table just plug them into Legendre polynomials and see whether they are equal to 0.
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{align} P_1 &= x =0\\ P_2 &= \frac{1}{2}(3(\pm 1/\sqrt{3})^2-1)=0 \\ P_3 &= \frac{1}{2}(5(\pm \sqrt{3/5})^3-3(\pm \sqrt{3/5}))=0 \\ P_4 &= \dots =0 \\ P_5 &= \dots =0 \end{align} $$ So, the roots are verified.
 *
 * }
 * }

Verify the weights in the table using:
 * {| style="width:100%" border="0" align="left"

\displaystyle w_j=\frac{-2}{(n+1)P_n'(x_j)P_{n+1}(x_j)} $$ n=1:
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_1=\frac{-2}{(1+1)P_1'(0)P_2(0)}=\frac{-2}{2\cdot 1\cdot (-\frac{1}{2})}=2 $$
 *
 * }
 * }

n=2:
 * {| style="width:100%" border="0" align="left"

\displaystyle w_1=\frac{-2}{(2+1)P_2'(-1/\sqrt{3})P_3(-1/\sqrt{3})}=\frac{-2}{3\cdot (-3/\sqrt{3})(2/3\cdot \sqrt{3})}=1 $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_2=\frac{-2}{3\cdot (3/\sqrt{3})\cdot (-2/3\cdot \sqrt{3})}=1 $$
 *
 * }
 * }

n=3:
 * {| style="width:100%" border="0" align="left"

\displaystyle w1=\frac{-2}{(3+1)P_3'(-\sqrt{3/5})P_4(-\sqrt{3/5})}=\frac{5}{9} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_2=\frac{-2}{(3+1)\cdot P_3'(0)\cdot P_4(0)}=\frac{8}{9} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_3=\frac{5}{9} $$
 *
 * }
 * }

n=4:
 * {| style="width:100%" border="0" align="left"

\displaystyle w_1=\frac{-2}{(4+1)\cdot P_4'(x_1)P_5(x_1)}=\frac{18-\sqrt{30}}{36} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_2=\frac{18+\sqrt{30}}{36} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_3=\frac{18+\sqrt{30}}{36} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_4=\frac{18-\sqrt{30}}{36} $$
 *
 * }
 * }

n=5:
 * {| style="width:100%" border="0" align="left"

\displaystyle w_3=\frac{128}{225} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_1=w_5=\frac{322-13\sqrt{70}}{900} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle w_2=w_4=\frac{322+13\sqrt{70}}{900} $$
 *
 * }
 * }

(2)Verify these exact values in the table against the corresponding numerical values in NIST: There's only a table for 5-point Gauss-Legendre formula:
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{matrix} \pm x_k & w_k \\ 0.00000 \ 00000 \ 00000 & 0.56888\ 88888\ 88889 \\ 0.53846\ 93101\ 05683 & 0.47862\ 86704\ 99366 \\ 0.90617\ 98459\ 38664 & 0.23692\ 68850\ 56189 \end{matrix} $$
 *

 * }
 * }

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

\displaystyle \begin{matrix} x_3=0 \\ w_3=\frac{128}{225}=0.5689 \\ x_2,x_4=\pm \frac{1}{3} \sqrt{5-2 \sqrt{10/7}}=\pm 0.5385\\ w_{2,4}=\frac{322+13\sqrt{70}}{900}=0.4786\\ x_1,x_5=\pm \frac{1}{3}\sqrt{5+2\sqrt{10/7}}=\pm 0.9062\\ w_1,5=\frac{322-13\sqrt{70}}{900}=0.2369 \end{matrix}

$$ So, using numerical values will be more convenient when the expression of the function is comlicated. But exact, irrational values will be more precise.
 *

 * }
 * }

(3)Extend the table to 6.
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{matrix} Number\ of\ points,n & Points,x_i & Weights,n_i\\ 6 & \pm 0.9325 & 0.1715 \\ & \pm 0.6612 & 0.3607 \\ & \pm 0.2386 & 0.4679 \end{matrix}

$$
 *

 * }
 * }

Problem 6: Pose Question to be solved
 Solved without assistance

Given
from NIST Digital Library of Math Functions, the weights are given by the formula: with

Find
Show that can be obtained from equations 6.1 and 6.2

Problem 7: Compute using GL Quadrature
 Solved without assistance

Given
$$\displaystyle A_n = \frac{2n+1}{2} \int_{-1}^{1} f(\mu)P_n(\mu)d\mu $$||
 (7.1) $$\displaystyle f(\mu) = T_0 \sinh(1-\mu^2) $$||
 (7.2)

Find
Compute the first 3 non-zero coefficients using Gauss-Legendre quadrature to within 5% error.

Solution
To know non zero coefficient, We have to define $$\displaystyle A_{2n} = A_{2n+1} = 0?$$ $$\displaystyle <f, P_n> = \int_{-1}^1 f(\mu) P_n(\mu) \, {\rm d}\mu = \int_{-1}^0 f(\mu) P_n(\mu) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_n(\mu) \, {\rm d}\mu $$ If n is odd, $$\displaystyle P_n$$ and $$\displaystyle f(\mu)$$ is $$\displaystyle even$$. Thus, when $$\displaystyle n$$ is odd, we have that $$\displaystyle n = 2k + 1$$ $$\displaystyle <f, P_{2k+1}> = \int_{-1}^0 f(-\mu) (-P_{2k+1}(-\mu)) \, {\rm d}\mu + \int_{0}^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ Let's define $$\displaystyle -\mu = \xi$$ $$\displaystyle <f, P_{2k+1}> = \int_1^0 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu $$ $$\displaystyle <f, P_{2k+1}> = - \int_0^1 f(\xi) P_{2k+1}(\xi) \, {\rm d}\xi + \int_0^1 f(\mu) P_{2k+1}(\mu) \, {\rm d}\mu = 0 $$ Thus $$\displaystyle A_{2k+1} = 0 $$ This shows that when n  = odd number, $$\displaystyle A_n = 0$$ $$\displaystyle \begin{align} &P_0\left(\mu\right)=1 ,\\ &P_2\left(\mu\right)=\frac{1}{2}\left(3\mu^2-1\right) ,\\ &P_4\left(\mu\right)=\frac{1}{8}\left(35\mu^4-30\mu^2+3\right) ,\\ &P_6\left(\mu\right)=\frac{1}{16}\left(231\mu^6-315\mu^4+105\mu^2-5\right) .\\ \end{align} $$

Using the above values and plugging into eq 7.1,
 $$\displaystyle A_n = \frac {2(n) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_n(\mu)d\mu$$ $$\displaystyle A_0 = \frac {2(0) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_0(\mu)d\mu = \frac {1}{2} * 1.492T_0 = 0.746T_0$$ $$\displaystyle A_2 = \frac {2(2) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_2(\mu)d\mu = -0.7997T_0$$ $$\displaystyle A_4 = \frac {2(4) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_4(\mu)d\mu = 0.066713T_0$$

Using Matlab, I compute the coefficient.

Problem 8: Verification
 Solved without assistance

Find
Verify equation 8.1 and show that for $$ i=1,...,n \ $$ and thus

Part1
From the wikipedia, $$\ell_j(x_i) = \prod_{m=0,\, m\neq j}^{k} \frac{x_i-x_m}{x_j-x_m} $$ From above equation, when $$m = j$$, the product skips And if $$i = j$$ the above equation's all terms are $$\frac{x_j-x_m}{x_j-x_m} = 1$$ except where $$x_j = x_m$$ if $$i \neq j$$ then since $$m \neq j$$ doesn't preclude it, one term in the product will be for $$m=i$$, $$\frac{x_i-x_i}{x_j-x_i} = 0$$ Thus, $$\ell_j(x_i) = \delta_{ji} = \begin{cases} 1, & \text{if } j=i  \\ 0, & \text{if } j \ne i \end{cases} $$ Where $$\displaystyle \delta_{ji}$$ is that The value is 1 if they are equal $$ \delta_{1 \, 1} = 1 $$

And 0 otherwise. $$\delta_{1 \, 2} = 0. $$

Part2
$$\displaystyle p^h(x) = \sum_{j=1}^n p(x_j)\ell_j(x) $$ $$\displaystyle d(x)=p(x) - p^h(x) = p(x) - \sum_{j=1}^n p(x_j)\ell_j(x) $$
 $$\displaystyle d(x_i)=p(x_i) - p^h(x_i) = p(x_i) - \sum_{j=1}^n p(x_j)\ell_j(x_i) $$
 $$\displaystyle d(x_i)=p(x_i) - \sum_{j=1}^n p(x_j)\ell_j(x_i) = p(x_i)-[p(x_1)\ell_1(x_i) + p(x_2)\ell_2(x_i)+ p(x_3)\ell_3(x_i)+ .......+ p(x_n)\ell_n(x_i)] $$
 When j = i, $$\displaystyle \ell_j(x_i) = \delta_{ji} = 1$$ $$\displaystyle p(x_i)\ell_i(x_i) = p(x_i)$$ The other terms are always zero. Hence,
 $$\displaystyle d(x_i) = p(x_i) - p(x_i) = 0$$

=Contributing Members & Referenced Lecture=

=Table of Contributions=

title: User:Egm6322.s12.TEAM1.HW/HW3


text:
User:Egm6322.s12.TEAM1.REPORT/REPORT 4

Problem 1: Pose Question to be solved
 Solved without assistance

Problem 2: Pose Question to be solved
 Solved without assistance

Problem 3: Pose Question to be solved
 Solved without assistance

Problem 4: Pose Question to be solved
 Solved without assistance

Problem 5: Pose Question to be solved
 Solved without assistance

Problem 6: Pose Question to be solved
 Solved without assistance

Problem 7: Pose Question to be solved
 Solved without assistance

Solution: ????????????????????
=Contributing Members & Referenced Lecture=

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

title: User:Egm6322.s12.TEAM1.REPORT/REPORT 4


text:
User:Egm6322.s12.TEAM1.HW/HW5

Problem 1: Computing non-zero coefficients in the Legendre Series
 Solved without assistance

Given: Expectation of 1% accuracy
Reference: Problem 7, Report 10 $$\displaystyle f(\mu) = T_0 \sinh(1-\mu^2) $$ $$\displaystyle A_n = \frac {2(n) + 1}{2}\int_{-1}^{1}T_0sinh(1-\mu^2)P_n(\mu)d\mu$$

Find
A) Compute the first non-zero coefficient in the Legendre Series with increasing number of integration points until convergence to within 1% accuracy is achieved. B) Provide the expression for the exact solution if possible (use WA) C) Plot the integration error versus the number of integration points.

part1
According to the homework 10.7, the coefficient $$\displaystyle A_n$$is non-zero term when n = even number. Thus A0, A2, A4, A6.... will be non-zero term. The first non-zero term coefficient is $$\displaystyle A_0 = 0.7459T_0$$ $$I_n(f) = \sum_{j=1}^n w_jf(x_j)\approx A_j = \frac{2j+1}{2} \int_{-1}^1 P_j(x)T_0 \sinh{(1 - x^2)}dx $$ When j =0, the above equation is following that $$\displaystyle A_0 = 0.7459T_0 = I_n(f) = \sum_{j=1}^n w_jf(x_j)$$ Where $$\displaystyle x_j$$ is the roots of polynomials When integration point is 1, $$\displaystyle A_0 = 1/2*\sum_{j=1}^1 w_1f(x_1)=1/2*(1.175*2)=1.175T_0$$ When integration point is 2, $$\displaystyle A_0 = 1/2*(w_1*f(x_1)+w_2*f(x_2))=0.7172T_0$$ When integration point is 3, $$\displaystyle A_0 = 1/2*(w_1*f(x_1)+w_2*f(x_2)+w_3*f(x_3))=0.7505T_0$$ When integration point is 4, $$\displaystyle A_0 = 1/2*(w_1*f(x_1)+w_2*f(x_2)+w_3*f(x_3)+w_4*f(x_4))=0.7459T_0$$ These results show that n=4, the accuracy is almost 0%. Actually the accuracy condition was satisfied by n = 3. Now see the lecture note 45-7. Equation in lecture note 45-7 is that $$E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{f^{2n}(\eta)}{(2n)!}, \eta \in [-1,1]$$ $$I(f) = I_n(f) + E_n(f)$$ $$I(f) - I_n(f) = E_n(f)$$ When n goes to oo, E_n(f) will be zero. Assume the exact A_0 is 0.7459.For the accuracy, 0.007459 will be allowed in $$\displaystyle I(f)-I_n(f)$$ $$\displaystyle E_n(f)= 0.007459$$ using the matlab, the results showed that Thus the answer is n = 3.

part2
By using Wlofram Alpha, exact expression should be

The exact solution is 0.74601369.....(this value was estimated by n=5)

part3
The above table is from wikipedia. $$E_n(f) = \frac{2^{2n+1}(n!)^4}{(2n+1)[(2n)!]^2} \frac{f^{2n}(\eta)}{(2n)!}, \eta \in [-1,1]$$ When $$\displaystyle \eta $$ is 0.9, 9, -0.9,

Problem 2: Prove an Alternative formula for the weights of Gauss-Legendre Quadrature
 Solved without assistance

Find
Prove that Eq2.3 below is valid

Solution:
Recall (2) p.44b-8: $$\displaystyle (1-{{x}^{2}}){{P}_{n}}^{\prime }(x)=-nx{{P}_{n}}(x)+n{{P}_{n-1}}(x)$$ And, RR2 is shown below: $$\displaystyle (n+1){{P}_{n+1}}-(2n+1)x{{P}_{n}}+n{{P}_{n-1}}=0$$ ,which can be transformed like: $$\displaystyle -nx{{P}_{n}}+n{{P}_{n-1}}=(n+1)x{{P}_{n}}-(n+1){{P}_{n+1}}$$ From the two equations above, we can obtain the following equation: $$(n+1)x{{P}_{n}}-(n+1){{P}_{n+1}}=(1-x_{j}^{2}){{{{P}'}}_{n}}({{x}_{j}})$$ When x is selected to be $${{x}_{j}}$$ (the roots of $${{P}_{n}}(x)$$), $$\displaystyle (n+1){{x}_{j}}{{P}_{n}}({{x}_{j}})-(n+1){{P}_{n+1}}({{x}_{j}})=(1-x_{j}^{2}){{{{P}'}}_{n}}({{x}_{j}})$$ where, $${{P}_{n}}({{x}_{j}})=0$$. So, we can get: $$\displaystyle -(n+1){{P}_{n+1}}({{x}_{j}})=(1-x_{j}^{2}){{{{P}'}}_{n}}({{x}_{j}})$$

Problem 3: Plotting a sequence versus an irrational number to observe convergence
 Solved without assistance

Given
A sequence: the irrational number: $$ n \ $$ a constant $$ \sqrt{2} \ $$ and a limit:

Find
Plot the sequence versus the irrational number: $$ n \ $$ with the constant $$ \sqrt{2} \ $$ to observe the convergence in Eq3.2.

Solution
I ran the iterations through an excel spreadsheet using the Smyrna Algorithm found on class notes 48-7 I ran the algorithm for 30 iterations for my own curiosity. I found that after 21 iterations, the irrational number n converged completely ( to 16 decimal places) with the square root of 2. A plot of the convergence was created from the data in the excel spreadsheet and is shown below along with the raw data form the spreadsheet.

Problem 4: Write a matlab program to compute the convergents for $$\sqrt 2$$
 Solved without assistance

Given:

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

\displaystyle \sqrt 2=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{\ddots}}}}} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle C_0:=1 $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle C_1:=1+\frac{1}{2}=\frac{3}{2} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle C_2:=1+\frac{1}{2+\frac{1}{2}}=1+\frac{2}{5}=\frac{7}{5} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle C_3:=1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=1+\frac{1}{2+\frac{2}{5}}=1+\frac{5}{12}=\frac{17}{12} $$
 *
 * }
 * }

Find:
Write a matlab program to compute the convergents $$C_i$$for i=0,1,...,10,and plot these convergents together with the line $$\sqrt 2$$to visualize the convergence
 * {| style="width:100%" border="0" align="left"

\displaystyle \lim_{i \to \infty}C_i=\sqrt 2 $$
 *
 * }
 * }

Solution:
clear

n=10;

f=sqrt(2); f=sym(f); f=char(f); fplot(f,[-2 2],'r','o'); legend('sqrt(2)'); hold on

disp(['C' char(48) '=']) disp(1)

f=1; f=sym(f); f=char(f); fplot(f,[-2 2]);

for x=1:n; s=0; for i=1:x; s=1/(2+s); end s=1+s; s=vpa(s); disp(['C' char(sym(x)) '=']) disp(s) f=s; f=sym(f); f=char(f); fplot(f,[-2 2]); end axis([-2 2 1.39 1.51])

The results are as follows:

From the results above, we can see that with the increasing of fraction terms, the value will be convergent to the actual value: $$\sqrt 2=1.4142135623730950488016887242097$$.

Problem 5: Verify the Last Term for the nth Approximant of Given Equation
 Solved without assistance

Find
A) Verify the last term for the nth approximant as shown in Eq5.1 B) Use Eq5.1 to find the Legendre Polynomials $$ \left \{ P_1,...,P_6 \right \} \ $$ and compare the results to those in R10.2 p44-6

Solution:
A).Verify the last term for the nth approximant as shown in Eq5.1. When n=2, the second approximant is: $$\displaystyle \frac{-\frac{2(n-1)-1}{2(n-1)}}{\frac{4(n-1)-1}{2(n-1)}x}=\frac{-\frac{2(2-1)-1}{2(2-1)}}{\frac{4(2-1)-1}{2(2-1)}x}=\frac{-\frac{1}{2}}{\frac{3}{2}x}$$ When n=3, the second approximant is: $$\displaystyle \frac{-\frac{2(n-1)-1}{2(n-1)}}{\frac{4(n-1)-1}{2(n-1)}x}=\frac{-\frac{2(3-1)-1}{2(3-1)}}{\frac{4(3-1)-1}{2(3-1)}x}=\frac{-\frac{3}{4}}{\frac{7}{4}x}$$ So, the n-th term is verified by the these beginning terms. B). According to continued fraction,the Legendre polynomial P_n(x) is the denominator of the n-th approximant to the continue fraction. So, the first one is: $$\displaystyle {{P}_{1}}(x)=x$$ Secondly: $$\displaystyle \frac{x+\frac{-1/2}{3x/2}}=\frac{{{a}_{1}}\centerdot 3x/2}{3{{x}^{2}}/2+-1/2}$$ So, $$\displaystyle {{P}_{2}}(x)=3{{x}^{2}}/2+-1/2$$ Thirdly: $$\displaystyle \frac{x+\frac{-1/2}{3x/2+\frac{-3/4}{7x/4}}}=\frac{x+\frac{-7x/8}{21{{x}^{2}}/8-3/4}}=\frac{{{a}_{1}}\left( 21{{x}^{2}}/8-3/4 \right)}{21{{x}^{3}}/8-3x/4-7x/8}=\frac{{{a}_{1}}\left( 21{{x}^{2}}/8-3/4 \right)}{21{{x}^{3}}/8-13x/8}$$ So, $$\displaystyle {{P}_{3}}(x)=21{{x}^{3}}/8-13x/8$$ Fourthly: $$\displaystyle \frac{x+\frac{-1/2}{3x/2+\frac{-3/4}{7x/4+\frac{-5/6}{11x/6}}}}=\frac{{{a}_{1}}(231{{x}^{3}}/48-63x/24)}{(231{{x}^{4}}-203{{x}^{2}}+20)/48}$$ So, $$\displaystyle {{P}_{4}}(x)=(231{{x}^{4}}-203{{x}^{2}}+20)/48$$ Fifthly: $$\displaystyle \frac{x+\frac{-1/2}{3x/2+\frac{-3/4}{7x/4+\frac{-5/6}{11x/6+\frac{-7/8}{15x/8}}}}}=\frac{{{a}_{1}}(1155{{x}^{4}}-924{{x}^{2}}+84)/128}{\left( 1155{{x}^{5}}-1309{{x}^{3}}+282x \right)/128}$$

So, $$\displaystyle {{P}_{5}}(x)=\left( 1155{{x}^{5}}-1309{{x}^{3}}+282x \right)/128$$

Sixthly: $$\displaystyle \frac{x+\frac{-1/2}{3x/2+\frac{-3/4}{7x/4+\frac{-5/6}{11x/6+\frac{-7/8}{15x/8+\frac{-9/10}{19x/10}}}}}}=\frac{{{a}_{1}}\left( 4389{{x}^{5}}-4620{{x}^{3}}+924x \right)/256}{\left( 4389{{x}^{6}}-6083{{x}^{4}}+2046{{x}^{2}}-96 \right)/256}$$

So, $$\displaystyle {{P}_{6}}(x)=\left( 4389{{x}^{6}}-6083{{x}^{4}}+2046{{x}^{2}}-96 \right)/256$$ In total, polynomial obtained by continued fraction is shown below. $$\displaystyle {{P}_{1}}(x)=x$$ $$\displaystyle {{P}_{2}}(x)=3{{x}^{2}}/2+-1/2$$ $$\displaystyle {{P}_{3}}(x)=21{{x}^{3}}/8-13x/8$$ $$\displaystyle {{P}_{4}}(x)=(231{{x}^{4}}-203{{x}^{2}}+20)/48$$ $$\displaystyle {{P}_{5}}(x)=\left( 1155{{x}^{5}}-1309{{x}^{3}}+282x \right)/128$$ $$\displaystyle {{P}_{6}}(x)=\left( 4389{{x}^{6}}-6083{{x}^{4}}+2046{{x}^{2}}-96 \right)/256$$ The Legendre polynomial obtained by RR2: $$ \displaystyle \begin{align} P_1 &= x \\ P_2 &= \frac{1}{2}(3x^2-1) \\ P_3 &= \frac{1}{2}(5x^3-3x) \\ P_4 &= \frac{1}{8}(35x^4-30x^2+3) \\ P_5 &= \frac{1}{8}(63x^5-70x^3+15x) \\ P_6 &= \frac{1}{16}(231x^6-315x^4+105x^2-5) \end{align} $$ Compared with P_n(x) obtained in R10.2 P44b-6, the factor of P_n(x) generated by continued fraction is larger and increased faster. However, the orthogonality of these polynomials won't be changed by the difference of factor. For example,

$$\displaystyle \left\langle {{P}_{2}},{{P}_{3}} \right\rangle =\int\limits_{-1}^{1}{\left( 3{{x}^{2}}/2-1/2 \right)\left( 21{{x}^{3}}/8-13x/8 \right)dx}=0$$

=Contributing Members & Referenced Lecture=

=Table of Contributions=

title: User:Egm6322.s12.TEAM1.HW/HW5


text:
User:Egm6322.s12.TEAM1.REPORT/REPORT 6

Problem 1: Pose Question to be solved
 Solved without assistance

Problem 2: Pose Question to be solved
 Solved without assistance

Problem 3: Pose Question to be solved
 Solved without assistance

Problem 4: Pose Question to be solved
 Solved without assistance

Problem 5: Pose Question to be solved
 Solved without assistance

Problem 6: Pose Question to be solved
 Solved without assistance

Problem 7: Pose Question to be solved
 Solved without assistance

Solution: ????????????????????
=Contributing Members & Referenced Lecture=

title: User:Egm6322.s12.TEAM1.REPORT/REPORT 6


text:
User:Egm6322.s12.TEAM1.HW/HW7

Problem 1: Find a coefficient of the 3-term reccurence relation
 Solved without assistance

Find
Find an expression for $$ B_n \ $$ similar to expressions given in Eq1.1 and Eq1.2.

Solution
Knowing the three term recurrence relation is of the form: Also given that the following holds: Allowing $$ n=0 \ $$ and plugging the values of Eq1.4 thru Eq1.6 into Eq1.3, one gets the following: Plugging in for $$ p_1(x) \ $$ and with $$ A_n = \frac{leading\ coef\!ficient\ of\ p_{n+1}}{leading\ coef\!ficient\ of\ p_n } = 1 \ $$ Moving the values to opposite sides of the equivalency and combining like terms results in the following: or more generalized:

Problem 2: Proof of the magnitude of a polynomial
 Solved without assistance

Find
Prove Eq2.1 can be obtained using Eq 2.2.

Solution:
From equation (2.2) we can obtain: $$\begin{align} & {{C}_{n}}=\frac\frac \\ & {{h}_{n}}={{h}_{n-1}}{{C}_{n}}\frac \\ & {{C}_{n-1}}=\frac\frac \\ & {{h}_{n-1}}={{h}_{n-2}}{{C}_{n-1}}\frac \\ \end{align}$$ Then, we continue our derivation:

$$\begin{align} & {{h}_{n}}={{h}_{n-1}}{{C}_{n}}\frac \\ & ={{h}_{n-2}}{{C}_{n-1}}\frac{{C}_{n}}\frac \\ & ={{h}_{n-2}}{{C}_{n-1}}\frac{{C}_{n}}\frac \\ & ={{h}_{n-2}}{{C}_{n-1}}{{C}_{n}}\frac \\ & ={{h}_{n-3}}{{C}_{n-2}}{{C}_{n-1}}{{C}_{n}}\frac \\ & ={{h}_{0}}{{C}_{1}}...{{C}_{n-2}}{{C}_{n-1}}{{C}_{n}}\frac \end{align}$$ So, eqation(2.1) has been obtained.

Problem 3: Re-framing the the 2-term recurrence relation into the 3-term recurrence relation
 Solved without assistance

Find
Put the recurrence relation RR2, Eq3.1, for the Legendre Polynomials in the framework of the recurrence realtion in Eq3.2; i.e. identify the coefficients $$ A_n,B_n,C_n $$

Solution:
According to RR2 equation (3.1): $$\displaystyle(n+1){{P}_{n+1}}-(2n+1)x{{P}_{n}}+n{{P}_{n-1}}=0$$ we have: $$\displaystyle{{P}_{n+1}}=\frac{(2n+1)}{(n+1)}x{{P}_{n}}-\frac{n}{(n+1)}{{P}_{n-1}}$$ Compared with the equation (3.2): $$\begin{align} p_{n+1}(x)=(A_nx+B_n)p_n(x)+C_np_{n-1}(x) \end{align}$$ we can obtain the corresponding coefficients:

$$\displaystyle\begin{align} & {{A}_{n}}=\frac{(2n+1)}{(n+1)} \\ & {{B}_{n}}=0 \\ & {{C}_{n}}=\frac{-n}{(n+1)} \end{align}$$

Problem 4: Obtain an expression for recurrence relation relative ot the magnitude
 Solved without assistance

Find
Reconcile the reccurence relation, Eq4.2, with the recurrence relation in Eq4.1 and obtain the expression for $$ h_n \ $$ in Eq4.3 usng Eq4.2.

Solution
$$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1} $$ $$\displaystyle h_n=h_0\frac{A_0}{A_n}C_1 C_2 \cdots C_n $$ From the legendre polynomials,  $$\displaystyle (n+1)p_{n+1}(x) - (2n+1)xp_n(x)-np_{n-1}(x)$$ $$\displaystyle{{P}_{n+1}}=\frac{(2n+1)}{(n+1)}x{{P}_{n}}+\frac{n}{(n+1)}{{P}_{n-1}}$$ According to the above relationships, we can get $$\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)+C_np_{n-1}(x) $$ Hence, we can define as below:  $$\displaystyle\begin{align} & {{A}_{n}}=\frac{(2n+1)}{(n+1)} \\ & {{B}_{n}}=0 \\ & {{C}_{n}}=\frac{n}{(n+1)} \end{align}$$

Let's look at an example for the relationship between eq(4.1) and eq(4.2). $$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1}=h_0\frac{A_0}{A_n}C_1 C_2 \cdots C_n $$ $$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1}=h_0\frac{(n+1)}{(2n+1)}C_1 C_2 \cdots C_n $$ When n = 1,  $$\displaystyle h_1=\frac{1}{3}h_{0}=h_0\frac{(2)}{(3)}\frac{1}{2}=h_0\frac{(1)}{(3)} $$ When n = 2, $$\displaystyle h_2=\frac{3}{5}h_{1}=\frac{3}{5}\frac{1}{3}h_{0}=h_0\frac{(3)}{(5)}C_1C_2=h_0\frac{(3)}{(5)}\frac{(1)}{(2)}\frac{(2)}{(3)}=\frac{1}{5}h_0 $$ From the above results, hn can be written as below: $$\displaystyle h_n=\frac{2n-1}{2n+1}h_{n-1}$$ And from eq 4.2  $$\displaystyle h_{n-1}=h_0\frac{A_0}{A_{n-1}}C_1 C_2 \cdots C_n=h_0\frac{n}{(2n-1)}\frac{1}{n}$$ Then plugging the above equation into eq (4.1) Finally we can get  $$\displaystyle h_n=\frac{2n-1}{2n+1}h_0\frac{n}{(2n+1)}\frac{1}{n}=\frac{2n-1}{2n+1}h_0\frac{1}{(2n-1)}=\frac{1}{2n+1}h_0$$ By using inner product of the 0th order legendre polynomials, we can get the value of h0. $$\displaystyle h_0 = \int_{-1}^{1} P_0(x)^2dx = \int_{-1}^{1}1dx = 2$$ Plugging the above value, we can express as below: $$\displaystyle h_n=\frac{1}{2n+1}h_0=\frac{2}{2n+1}$$

Problem 5: Derive the Christoffel-Darboux identity for Legendre polynomials
 Solved without assistance

Given:
The 3-term recurrence relation for any sequence of orthogonal polynomials;
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$ for n=0,1,2,..., and $$\,\,p_{-1}(x)=0\,\,$$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle A_n\,,B_n\,,C_n \in \mathcal R\,,with\, A_{n-1}A_nC_n > 0 $$ If the highest coefficient of $$\,\,p_n(x)\,\,$$ is kn, then
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle A_n=\frac{k_{n+1}}{k_n} $$
 *
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$
 *
 * }
 * }

Find:

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

\displaystyle \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k}=\frac{p_{n+1}(x)p_n(y)-p_n(x)p_{n+1}(y)}{A_nh_n(x-y)} $$
 *
 * }
 * }

Solution:
From the three term recurrence relation
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(x)=(A_nx+B_n)p_n(x)-C_np_{n-1}(x) $$ multiply both sides by pn(y):
 *

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

\displaystyle p_{n+1}(x)p_n(y)=(A_nx+B_n)p_n(x)p_n(y)-C_np_n(y)p_{n-1}(x) $$ $$ And for the symmetry between the factors
 *
$$\displaystyle (Equation\;5.1 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(y)p_n(x)=(A_ny+B_n)p_n(x)p_n(y)-C_np_n(x)p_{n-1}(y) $$ $$ Substract equation 5.2 from equation 5.1:
 *
$$\displaystyle (Equation\;5.2 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)=A_n(x-y)p_n(x)p_n(y)-C_n[p_n(y)p_{n-1}(x)-p_n(x)p_{n-1}(y)] $$ For
 *

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

\displaystyle C_n=\frac{A_n}{A_{n-1}}\frac{h_n}{h_{n-1}} $$ Then
 *

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

\displaystyle \begin{align} & \frac{p_{n+1}(x)p_n(y)-p_{n+1}(y)p_n(x)}{A_nh_n(x-y)} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_n(x)p_{n-1}(y)-p_n(y)p_{n-1}(x)}{A_{n-1}(x-y)h_{n-1}} \\ =& \frac{p_n(x)p_n(y)}{h_n}+\frac{p_{n-1}(x)p_{n-1}(y)}{h_{n-1}}+\frac{p_{n-1}(x)p_{n-2}(y)-p_{n-1}(y)p_{n-2}(x)}{A_{n-2}(x-y)h_{n-2}} \\ & \vdots \\ =& \sum_{k=0}^n \frac{p_k(x)p_k(y)}{h_k} \end{align} $$
 *

 * }
 * }

=Contributing Members & Referenced Lecture=

=Table of Contributions=

title: User:Egm6322.s12.TEAM1.HW/HW7


text: File:SpringDashpot.jpg

title: File:SpringDashpot.jpg


text: File:Massinternal.jpg

title: File:Massinternal.jpg


text:
User:Peterrawsthorne/ALD/Ch01

=Chapter 01: Introduction to Agile Learner Design=

Reference

 * 1) Critical technology blog post: Agile Instructional Design
 * 2) The learning generalist blog post: The Agile e-learning design manual
 * 3) The learning generalist blog post: Agile Re-explained

title: User:Peterrawsthorne/ALD/Ch01


text:
User:Egm4313.sp12.team5.dikova

testing

title: User:Egm4313.sp12.team5.dikova


text:
User:Peterrawsthorne/ALD/Ch05

=Chapter 05: Envisioning and Personal Curriculum Mapping=

Reference

 * 1) Critical technology blog: Agile Instructional Design: ENVISION
 * 2) Critical technology blog: Personal Curriculum Mapping

title: User:Peterrawsthorne/ALD/Ch05


text:
User:Egm4313.s12.team19.Horwitz.R2 Code test page

A Code test page 2
 * Bulleted list item

R2.8
Contributed by Vladimir Horwitz.

Problem Statement:
Answer questions 8 and 15 from Kreyszing.

Solution:
Prob. 15)

 * {| style = "width: 73%"

! $$ \ y'' + .54 y' + (.0729 + \pi) y = 0 $$ !! (Eq. 1)
 * $$ \ \lambda^2 + .54 \lambda + .0729 + \pi = 0 $$
 * (Eq. 2)
 * The characteristic equation has these roots.
 * $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 * (Eq. 3, a and b)
 * }
 * $$ \lambda = -.27 + \sqrt {\pi} i, \bar \lambda = -.27 - \sqrt {\pi} i $$
 * (Eq. 3, a and b)
 * }

Due to the fact that this roots are complex, the following equations from Kreyszig 2011, page 57 case 3, can be applied.

 * {| class = wikitable style="width: 73%; text-align:center"

! $$ \ y = e^{-.27} * (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) $$ (Eq. 4)
 * }

 * {| style = "width: 73%"

! And following this, we can say !!
 * $$ y' = -.27 e^{-.27} \ (A \cos (\sqrt \pi x) + B \sin (\sqrt \pi x)) \ + \ e^{-.27} \ (-A \sqrt \pi \sin (\sqrt \pi x) + B \sqrt \pi \cos (\sqrt \pi x)) $$
 *  (Eq. 5)
 * and
 * $$ y'' = .0729 e^{-.27} \ (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) \ + \ e^{-.27} \ (- \pi A \cos \sqrt \pi x \ - \pi B \sin \sqrt \pi x) + \downarrow $$
 * $$ \uparrow -.27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) \ - \ .27 e^{-.27} \ (-A \sqrt \pi \sin \sqrt \pi x \ + B \sqrt \pi \cos \sqrt \pi x) $$
 * (Eq. 6)
 * Then, after major calculator work, we find that
 * $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * (Eq. 7)
 * }
 * Then, after major calculator work, we find that
 * $$ 0 = (0) * e^{-.27} * (A \cos \sqrt \pi x \ + B \sin \sqrt \pi x) $$
 * (Eq. 7)
 * }
 * (Eq. 7)
 * }


text:
User:Peterrawsthorne/ALD/Ch02

=Chapter 02: Examples of Agile Learner Design=

Reference

 * 1) Critical technology blog: Implementation of AID

title: User:Peterrawsthorne/ALD/Ch02


text:
User:Egm4313.s12.team17/Report 2

= Problem 1 - Non-Homogeneous L2-ODE-CC =

Given
Roots of the characteristic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1 = -2, \lambda_2 = +5 $$     (Eq.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle y(0) = 1, y'(0) = 0 $$     (Eq.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Part 1 Problem Statement
Find the standard form and solution in terms of the initial conditions with a general excitation r(x)  for the non-homogeneous L2-ODE-CC.

Part 1 Solution
Given the roots, one can find the characteristic equation with the given general form:
 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda - \lambda_1)(\lambda - \lambda_2) = 0 $$     (Eq.1.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Substituting the corresponding values of and  into Eq. 1.1.1:

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

$$  \displaystyle (\lambda - (-2))(\lambda - 5) = \lambda^2 - 3\lambda - 10 = 0 $$     (Eq.1.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From Eq. 1.1.2, the non-homogeneous L2-ODE-CC is:
 * {| style="width:100%" border="0"

$$  \displaystyle y'' - 3y' - 10y = r(x) $$     (Eq.1.1.3)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *
 * }

To solve for the constants c_1 \, and c_2 \,, a homogeneous solution will be generated and substituted into the overall solution:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x) = y_h(x) + y_p(x) $$     (Eq.1.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle y_h(x) = c_1e^{-2x} + c_2e^{5x} $$     (Eq.1.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Substituting the Eq. 1.1.5 into Eq. 1.1.4:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x) = c_1e^{-2x} + c_2e^{5x} + y_p(x) $$     (Eq.1.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Satisfying the initial conditions:
 * {| style="width:100%" border="0"

$$  \displaystyle y(0) = 1 = c_1 + c_2 + y_p(0) $$     (Eq.1.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle y'(0) = 0 = -2c_1 + 5c_2 + y_p'(0) $$     (Eq.1.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

Multiplying Eq. 1.1.7 by 2, adding 1.1.7 and 1.1.8, and then solve for c_2 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle c_2 = \frac{2}{7} - \frac{1}{7}[2y_p(0) + y_p'(0)] $$     (Eq.1.1.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Multiplying Eq. 1.1.7 by 5, subtracting 1.1.8 and 1.1.7, and then solve for c_1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle c_1 = \frac{5}{7} - \frac{1}{7}[5y_p(0) - y_p'(0)] $$     (Eq.1.1.10)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Because the excitation r(x)  is equal to 0, c_1 \, and c_2 \, can be solved for:
 * {| style="width:100%" border="0"

$$  \displaystyle c_1 = \frac{5}{7} $$     (Eq.1.1.11)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle c_2 = \frac{2}{7} $$     (Eq.1.1.12)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

The final equation is then given by:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x) = \frac{5}{7}e^{-2x} + \frac{2}{7}e^{5x} $$     (Eq.1.1.13)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *
 * }

With the corresponding plot of Eq. 1.1.13:

Part 2 Problem Statement
Admit the two values from p.3-7 (3a) as the two roots of the corresponding characteristic equation and generate three non-standard (and non-homogeneous) L2-ODE-CC.

Part 2 Solution
Three non-standard and non-homogeneous ( r(x) \ne 0 ) L2-ODE-CC:
 * {| style="width:100%" border="0"

$$  \displaystyle 4(\lambda - (-2))(\lambda - 5) = 4\lambda^2 -28\lambda - 40 $$     (Eq.1.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$  \displaystyle 5(\lambda - (-2))(\lambda - 5) = 5\lambda^2 -35\lambda - 50 $$     (Eq.1.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6(\lambda - (-2))(\lambda - 5) = 6\lambda^2 -42\lambda - 60 $$     (Eq.1.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.ying 17:11, 6 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.wheeler.tw 18:00, 6 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 04:55, 8 February 2012 (UTC)

= Problem 2 - Solution for L2-ODE-CC =

Given

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

$$  \displaystyle y'' - 10y' + 25y = r(x) $$     (Eq.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y(0) = 1, y'(0) = 0 $$     (Eq.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

No Excitation:
 * {| style="width:100%" border="0"

$$  \displaystyle r(x) = 0 $$     (Eq.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Problem Statement
For the given L2-ODE-CC (Eq. 2.1), find and plot the solution.

Solution
From section 5 of the lecture notes, pg.5.6 given the homogeneous solution as:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h(x) = c_1e^{5x} + c_2xe^{5x} $$     (Eq.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the solution is:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x) = c_1e^{5x} + c_2xe^{5x} + y_p(x) $$     (Eq.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Satisfying the initial conditions:
 * {| style="width:100%" border="0"

$$  \displaystyle y(0) = 1 = c_1 + y_p(0) $$     (Eq.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y'(0) = 0 = 5c_1 + \cancel{c_2*5xe^{5x}} + c_2 + y_p'(0) $$     (Eq.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Multiplying Eq. 2.6 by 5, subtracting 2.7 from 2.6, and then solve for c_2 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle c_2 = 5 + 5 y_p(0) - y_p'(0) $$     (Eq.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solve for c_1 \, from Eq. 2.6:
 * {| style="width:100%" border="0"

$$  \displaystyle c_1 = 1 - y_p(0) $$     (Eq.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Because the excitation r(x)  is equal to 0, c_1 \, and c_2 \, can be solved for:
 * {| style="width:100%" border="0"

$$  \displaystyle c_1 = 1 $$     (Eq.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_2 = 5 $$     (Eq.2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final equation is then:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x) = e^{5x} + 5xe^{5x} $$     (Eq.2.12)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

The plot of the solution is as follows:

Author & Proofreaders
Author: Egm4313.s12.team17.ying 00:19, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.hintz 04:25, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 05:19, 8 February 2012 (UTC)

= Problem 3 - Homogeneous L-ODE-CC = For problem reference.

Part 1 Given

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

$$  \displaystyle y''+6y'+8.96y=0 $$     (Eq.3.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Problem Statement
Find a general solution for Eq. 3.1.1 and confirm the general solution using substitution.

Part 1 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 3.1.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)

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

$$  \displaystyle \lambda^2+6\lambda+8.96\lambda=0 $$     (Eq.3.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since Eq. 3.1.2 is of the form, we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.3.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When we apply the coefficients of Eq. 3.1.2 (where a=1, b=6 , and c=8.96 ) to Eq. 3.1.3 we get:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-6\pm \sqrt{6^2-4(1)(8.96)}}{2(1)} $$     (Eq.3.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which can be simplified to yield the root(s) of the characteristic equation as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-6\pm \sqrt{36-35.84}}{2} $$     (Eq.3.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-6\pm \sqrt{0.16}}{2} $$     (Eq.3.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-6\pm 0.4}{2} $$     (Eq.3.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=-3\pm 0.2 $$     (Eq.3.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

At this point we can see that there are two roots to the characteristic equation due to the presence of the \pm sign. The two roots will be called \lambda_1 and \lambda_2 and are shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1=-3+0.2 $$     (Eq.3.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_2=-3-0.2 $$     (Eq.3.1.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The simplified roots are shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1=-2.8 $$     (Eq.3.1.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_2=-3.2 $$     (Eq.3.1.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq. 3.1.11 and Eq. 3.1.12 indicate that this equation has two distinct, real roots and as such, the general solution will be in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$     (Eq.3.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where c_1 and c_2 are arbitrary constants and \lambda_1 and \lambda_2 are defined in Eq. 3.1.11 and Eq. 3.1.12, respectively.

The general solution is shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle y=c_1e^{-2.8x}+c_2e^{-3.2x} $$     (Eq.3.1.14)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we're going to check through substitution that the general solution we've obtained in Eq. 3.1.14 is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle y=c_1e^{-2.8x}+c_2e^{-3.2x} $$     (Eq.3.1.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y'=-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x} $$     (Eq.3.1.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''=7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x} $$     (Eq.3.1.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If substituting Eq. 3.1.14, Eq. 3.1.15, and Eq. 3.1.16 into Eq. 3.1.1 and does not violate Eq. 3.1.1, then we can accept Eq. 3.1.14 to be the general solution.
 * {| style="width:100%" border="0"

$$  \displaystyle 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x} + 6(-2.8c_1e^{-2.8x}-3.2c_2e^{-3.2x})+8.96(c_1e^{-2.8x}+c_2e^{-3.2x})=0 $$     (Eq.3.1.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We will simplify this equation until we can conclusively show whether or not the general solution is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle 7.84c_1e^{-2.8x}+10.24c_2e^{-3.2x} - 16.8c_1e^{-2.8x}-19.2c_2e^{-3.2x}+8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x}=0 $$     (Eq.3.1.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -8.96c_1e^{-2.8x}-8.96c_2e^{-3.2x} +8.96c_1e^{-2.8x}+8.96c_2e^{-3.2x}=0 $$     (Eq.3.1.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=0 $$     (Eq.3.1.20)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we have proven that Eq. 3.1.14 is indeed the general solution to Eq. 3.1.1.

Part 2 Given

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

$$  \displaystyle y''+4y'+( \pi^2 + 4 )y=0 $$     (Eq.3.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2 Problem Statement
Find a general solution for Eq. 3.2.1 and confirm the general solution using substitution.

Part 2 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 3.2.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+4\lambda+(\pi^2+4)\lambda=0 $$     (Eq.3.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since Eq. 3.2.2 is of the form we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.3.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When we apply the coefficients of Eq. 3.2.2 (where a=1, b=4 , and c=\pi^2+4 ) to Eq. 3.2.3 we get:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-4\pm \sqrt{4^2-4(1)(\pi^2+4)}}{2(1)} $$     (Eq.3.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which can be simplified to yield the root(s) of the characteristic equation as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-4\pm \sqrt{16-16-4\pi^2}}{2} $$     (Eq.3.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-4\pm \sqrt{-4\pi^2}}{2} $$     (Eq.3.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-4\pm 2 \pi i}{2} $$     (Eq.3.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=-2\pm \pi i $$ (Eq.3.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

At this point we can see that there are two roots to the characteristic equation. The two roots will be called \lambda_1 and \lambda_2 and are shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1=-2+ \pi i $$ (Eq.3.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_2=-2- \pi i $$ (Eq.3.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq. 3.2.9 and Eq. 3.2.10 indicate that this equation has two complex conjugate roots and as such, the general solution will be in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(c_1\cos wx+c_2\sin wx) $$     (Eq.3.2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where c_1 and c_2 are arbitrary constants and a is the real component of the complex conjugate and w is the imaginary component of the complex conjugate roots are defined in Eq. 3.2.9 and Eq. 3.2.10, respectively.

The general solution is shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-2x}(c_1\cos \pi x+c_2\sin \pi x) $$ (Eq.3.2.12)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we're going to check through substitution that the general solution we've obtained in Eq. 3.2.12 is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-2x}(c_1\cos \pi x+c_2\sin \pi x) $$ (Eq.3.2.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y'=-2e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)+e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x) $$ (Eq.3.2.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''=4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)-2e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x)-2e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x) $$ $$  \displaystyle +e^{-2x}(- \pi^2 c_1\cos \pi x- \pi^2 c_2\sin \pi x) $$ (Eq.3.2.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq. 3.2.14 can be further simplified as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle y''=4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)-4e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x)+e^{-2x}(- \pi^2 c_1\cos \pi x- \pi^2 c_2\sin \pi x) $$ (Eq.3.2.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If substituting Eq. 3.2.12, Eq.3.2.13, and Eq. 3.2.15 into Eq. 3.2.1 and does not violate Eq. 3.2.1, then we can accept Eq. 3.1.12 to be the general solution.
 * {| style="width:100%" border="0"

$$  \displaystyle 4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)-4e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x)+e^{-2x}(- \pi^2 c_1\cos \pi x- \pi^2 c_2\sin \pi x)+4(-2e^{-2x}(c_1\cos \pi x+c_2\sin \pi x) $$ $$  \displaystyle    +e^{-2x}(- \pi c_1\sin \pi x+ \pi c_2\cos \pi x))+( \pi^2 + 4 )(e^{-2x}(c_1\cos \pi x+c_2\sin \pi x))=0 $$     (Eq.3.1.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We will simplify this equation until we can conclusively show whether or not the general solution is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle (-4e^{-2x}+4e^{-2x})(c_1\cos \pi x+c_2\sin \pi x)=0 $$     (Eq.3.2.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=0 $$     (Eq.3.2.18)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we have proven that Eq. 3.2.12 is indeed the general solution to Eq. 3.2.1.

Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 15:12, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.Li 16:05, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 06:21, 8 February 2012 (UTC)

= Problem 4 - Homogeneous L-ODE-CC = For problem reference.

Part 1 Given

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

$$  \displaystyle y''+2 \pi y'+\pi^2 y=0 $$     (Eq.4.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Problem Statement
Find a general solution for Eq. 4.1.1 and confirm the general solution using substitution.

Part 1 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 4.1.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+2 \pi\lambda+\pi^2\lambda=0 $$     (Eq.4.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since Eq. 4.1.2 is of the form, we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.4.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When we apply the coefficients of Eq. 4.1.2 (where a=1, b=2\pi , and c=\pi^2 ) to Eq. 4.1.3 we get:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-(2 \pi )\pm \sqrt{(2\pi)^2-4(1)(\pi^2)}}{2(1)} $$     (Eq.4.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which can be simplified to yield the root(s) of the characteristic equation as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-(2 \pi )\pm \sqrt{4\pi^2-4\pi^2}}{2} $$     (Eq.4.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-2 \pi\pm \sqrt{0}}{2} $$     (Eq.4.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=-\pi $$     (Eq.4.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From Eq. 4.1.7 we can observe that there is a real double root to the characteristic equation. As such the general solution will be in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{\lambda x} $$ (Eq.4.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where c_1 and c_2 are arbitrary constants and \lambda is defined in Eq. 4.1.7, above.

The general solution is shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we're going to check through substitution that the general solution we've obtained in Eq. 4.1.9 is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y'=c_2e^{-\pi x}-\pi(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''=-c_2\pi e^{-\pi x}-c_2\pi e^{-\pi x}+\pi^2(c_1+c_2 x)e^{-\pi x} $$ (Eq.4.1.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If substituting Eq. 4.1.9, Eq. 4.1.10, and Eq. 4.1.11 into Eq. 4.1.1 and does not violate Eq. 4.1.1, then we can accept Eq. 4.1.9 to be the general solution.
 * {| style="width:100%" border="0"

$$  \displaystyle -c_2\pi e^{-\pi x}-c_2\pi e^{-\pi x}+\pi^2(c_1+c_2 x)e^{-\pi x}+2 \pi (c_2e^{-\pi x}-\pi(c_1+c_2 x)e^{-\pi x})+\pi^2 ((c_1+c_2 x)e^{-\pi x})=0 $$     (Eq.4.1.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We will simplify this equation until we can conclusively show whether or not the general solution is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle -2c_2\pi e^{-\pi x}+\pi^2(c_1+c_2 x)e^{-\pi x}+2 \pi c_2e^{-\pi x}-2 \pi^2(c_1+c_2 x)e^{-\pi x}+\pi^2 ((c_1+c_2 x)e^{-\pi x})=0 $$     (Eq.4.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \pi^2(c_1+c_2 x)e^{-\pi x}-\pi^2(c_1+c_2 x)e^{-\pi x}=0 $$     (Eq.4.1.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=0 $$     (Eq.4.1.15)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we have proven that Eq. 4.1.9 is indeed the general solution to Eq. 4.1.1.

Part 2 Given

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

$$  \displaystyle 10y''-32y'+25.6y=0 $$     (Eq.4.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2 Problem Statement
Find a general solution for Eq. 4.2.1 and confirm the general solution using substitution.

Part 2 Solution
The first step to deriving the general solution would be to substitute the following into Eq. 4.2.1:

( \lambda^2 for, \lambda for y^{\prime} , and 1 for y \,, which yields the following characteristic equation)

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

$$  \displaystyle 10\lambda^2-32\lambda+25.6\lambda=0 $$     (Eq.4.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since Eq. 4.2.2 is of the form, we can make use of the quadratic equation to solve for the root(s) of the characteristic equation.

Shown below is the general form of the quadratic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-b\pm \sqrt{b^2-4ac}}{2a} $$     (Eq.4.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

When we apply the coefficients of Eq. 4.2.2 (where a=10, b=-32 , and c=25.6 ) to Eq. 4.2.3 we get:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{32\pm \sqrt{(-32)^2-4(10)(25.6)}}{2(10)} $$     (Eq.4.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which can be simplified to yield the root(s) of the characteristic equation as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{32\pm \sqrt{1024-1024}}{20} $$     (Eq.4.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{32\pm \sqrt{0}}{20} $$     (Eq.4.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=1.6 $$     (Eq.4.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From Eq. 4.2.7 we can observe that there is a real double root to the characteristic equation. As such the general solution will be in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{\lambda x} $$ (Eq.4.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where c_1 and c_2 are arbitrary constants and \lambda is defined in Eq. 4.2.7, above.

The general solution is shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we're going to check through substitution that the general solution we've obtained in Eq. 4.2.9 is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle y=(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y'=c_2e^{1.6 x}+1.6(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''=1.6c_2 e^{1.6 x}+1.6 c_2 e^{1.6 x}+1.6^2(c_1+c_2 x)e^{1.6 x} $$ (Eq.4.2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

If substituting Eq. 4.2.9, Eq. 4.2.10, and Eq. 4.2.11 into Eq. 4.2.1 and does not violate Eq. 4.2.1, then we can accept Eq. 4.2.9 to be the general solution.
 * {| style="width:100%" border="0"

$$  \displaystyle 10(1.6c_2 e^{1.6 x}+1.6 c_2 e^{1.6 x}+1.6^2(c_1+c_2 x)e^{1.6 x})-32(c_2e^{1.6 x}+1.6(c_1+c_2 x)e^{1.6 x})+25.6((c_1+c_2 x)e^{1.6 x})=0 $$     (Eq.4.2.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We will simplify this equation until we can conclusively show whether or not the general solution is valid:
 * {| style="width:100%" border="0"

$$  \displaystyle 16c_2 e^{1.6 x}+16 c_2 e^{1.6 x}+25.6(c_1+c_2 x)e^{1.6 x}-32c_2e^{1.6 x}-51.2(c_1+c_2 x)e^{1.6 x}+25.6(c_1+c_2 x)e^{1.6 x}=0 $$     (Eq.4.2.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 16c_2 e^{1.6 x}-16 c_2 e^{1.6 x}+25.6(c_1+c_2 x)e^{1.6 x}-25.6(c_1+c_2 x)e^{1.6 x}=0 $$     (Eq.4.2.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=0 $$     (Eq.4.2.15)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now we have proven that Eq. 4.2.9 is indeed the general solution to Eq. 4.2.1.

Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 15:26, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.hintz 17:31, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 08:09, 8 February 2012 (UTC)

= Problem 5 - Reverse Engineering = For problem reference.

Part 1 Given

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

$$  \displaystyle e^{2.6x}, e^{-4.3x} $$     (Eq.5.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Problem Statement
Find an ODE for the given basis when.

Part 1 Solution
The roots of Eq. 5.1.1 are:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_1 = 2.6 $$     (Eq.5.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_2 = -4.3 $$     (Eq.5.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To obtain the formula of the factored quadratic, we set the roots (Eq. 5.1.2 and 5.1.3) equal to zero:
 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda_1 - 2.6) = 0 $$     (Eq.5.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (\lambda_2 + 4.3) = 0 $$     (Eq.5.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To obtain the original function, we multiply Eq. 5.1.4 and 5.1.5 together to yield:
 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda_1 - 2.6)(\lambda_2 + 4.3) = 0 $$     (Eq.5.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda^2 + (4.3\lambda) -(2.6\lambda) - 11.18 =0 $$     (Eq.5.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Simplifying Eq. 5.1.7:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2 + 1.7\lambda - 11.18=0 $$     (Eq.5.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Converting the characteristic equation to an ODE:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime} + 1.7y^{\prime} - 11.18y=0 $$     (Eq.5.1.9)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part 2 Given

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

$$  \displaystyle e^{-\sqrt{5}x}, x e^{-\sqrt{5}x} $$     (Eq.5.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2 Problem Statement
Find an ODE for the given basis when.

Part 2 Solution
The root of Eq. 5.2.1 is:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda = -\sqrt{5} $$     (Eq.5.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To obtain the formula of the factored quadratic, we set Eq. 5.2.2 equal to zero:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda + \sqrt{5}= 0 $$     (Eq.5.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To obtain the original function, we square Eq. 5.2.3 to yield the answer:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2 +2\sqrt{5}\lambda + 5 =0 $$     (Eq.5.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Converting the characteristic equation to an ODE:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime} +2\sqrt{5}y^\prime + 5y =0 $$     (Eq.5.2.5)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.hintz 10:00, 3 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.Li 19:32, 3 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 08:31, 8 February 2012 (UTC)

= Problem 6 - Spring-Dashpot-Mass System in Series =

Given

 * k \, is the spring constant
 * c \, is the damping constant
 * y(t) is the displacement of the mass

The equation for this system is:
 * {| style="width:100%" border="0"

$$  \displaystyle my_k^{\prime\prime}+\frac{mk}{c}y_k^\prime+ky_k=f(t) $$     (Eq.6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Parameters to this equation can be found by the characteristic equation for a double root if given \lambda.

Problem Statement
Find the values for the coefficients m, k, and c. Using \lambda=-3.

Solution
When \lambda=-3, then the characteristic equation becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle (\lambda + 3)^2=\lambda^2+6\lambda+9=0 $$     (Eq.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Comparing Eq 6.1 and Eq 6.2 we can set the coefficients of each equation equal to each other:
 * {| style="width:100%" border="0"

$$  \displaystyle m = 1 $$     (Eq.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{mk}{c} = 6 $$     (Eq.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle k = 9 $$     (Eq.6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

We can see that coefficients m and k have already been solved.

Using Eq 6.4, c can be solved as well:
 * {| style="width:100%" border="0"

$$  \displaystyle c =\frac{mk}{6} $$     (Eq.6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c = \frac{(1)(9)}{6} $$     (Eq.6.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c = \frac{3}{2} $$     (Eq.6.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle m=1 $$     (Eq.6.9)
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$  \displaystyle k=9 $$     (Eq.6.10)
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$  \displaystyle c=\frac{3}{2} $$     (Eq.6.11)
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 * style="width:7%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now, the equation for this system is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_k^{\prime\prime}+6y_k^\prime+9y_k=f(t) $$     (Eq.6.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.Li Kelvin Li 06:12, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.axelrod.a 15:57, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 10:56, 8 February 2012 (UTC)

= Problem 7 - MacLaurin Series =

Given
By definition a Taylor series about x=a  is defined as:
 * {| style="width:100%" border="0" align="left"

$$  \displaystyle f(t)= \sum_{n=0}^{\infty }\frac{f^{(n)}(a)}{n!}(t-a)^n $$       (Eq.7.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where f ^{(n)} represents the order of the derivative.

For a MacLaurin Series a=0. The formula for a MacLaurin Series becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)= \sum_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}(t)^n $$     (Eq.7.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for e^t.

Part 1 Solution

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

$$  \displaystyle f(t)= e^t $$     (Eq.7.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle f^{(0)}(0)=e^{(0)}=1 $$     (Eq.7.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=1 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(1)}(0)=\frac{d}{dt}e^t=e^0=1 $$     (Eq.7.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=2 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(2)}(0)=\frac{d^2}{dt^2}e^t=e^0=1 $$     (Eq.7.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=3 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(3)}(0)=\frac{d^3}{dt^3}e^t=e^0=1 $$     (Eq.7.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=4 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(4)}(0)=\frac{d^4}{dt^4}e^t=e^0=1 $$     (Eq.7.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=5 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(5)}(0)=\frac{d^5}{dt^5}e^t=e^0=1 $$     (Eq.7.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Apply Eq. 7.1.2 - 7.1.7 to Eq. 7.2:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=\frac{1}{0!}t^0+\frac{1}{1!}t^1+\frac{1}{2!}t^2+\frac{1}{3!}t^3+\frac{1}{4!}t^4+\frac{1}{5!}t^5 $$     (Eq.7.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the Maclaurin Series expansion for f(t)=e^t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120} .... $$ (Eq.7.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Determined from the pattern of Eq. 7.1.8, the summation for a Taylor Series approximation of f(t)=e^t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=e^t=\sum_{n=0}^{\infty }\frac{t^n}{n!} $$     (Eq.7.1.10)
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part 2 Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for \cos t.

Part 2 Solution

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

$$  \displaystyle f(t)= cos(t) $$     (Eq.7.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle f^{(0)}(0)=cos(0)=1 $$     (Eq.7.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=1 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(1)}(0)=\frac{d}{dt}cos(0)=-sin(0)=0 $$     (Eq.7.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=2 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(2)}(0)=\frac{d^2}{dt^2}cos(0)=-cos(0)=-1 $$     (Eq.7.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=3 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(3)}(0)=\frac{d^3}{dt^3}cos(0)=sin(0)=0 $$     (Eq.7.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=4 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(4)}(0)=\frac{d^4}{dt^4}sin(0)=cos(0)=1 $$     (Eq.7.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=5 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(5)}(0)=\frac{d^5}{dt^5}cos(0)=-sin(0)=0 $$     (Eq.7.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Apply Eq. 7.2.2 - 7.2.7 to Eq. 7.2:
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=\frac{1}{0!}t^0+\frac{0}{1!}t^1-\frac{1}{2!}t^2+\frac{0}{3!}t^3+\frac{1}{4!}t^4+\frac{0}{5!}t^5 $$     (Eq.7.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the Maclaurin Series expansion for f(t)=\cos t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}-\frac{t^6}{720}+\frac{t^8}{40320} .... $$ (Eq.7.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Determined from the pattern of Eq. 7.2.8, the summation for a Taylor Series approximation of f(t)=\cos t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=cos(t)=\sum_{n=0}^{\infty }\frac{[(-1)^n]t^{2n}}{(2n)!} $$     (Eq.7.2.10)
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part 3 Problem Statement
Develop the MacLaurin series (Taylor series at t=0) for \sin t.

Part 3 Solution

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

$$  \displaystyle f(t)= sin(t) $$     (Eq.7.3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle f^{(0)}(0)=sin(0)=0 $$     (Eq.7.3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=1 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(1)}(0)=\frac{d}{dt}sin(0)=cos(0)=1 $$     (Eq.7.3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=2 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(2)}(0)=\frac{d^2}{dt^2}sin(0)=-sin(0)=0 $$     (Eq.7.3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=3 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(3)}(0)=\frac{d^3}{dt^3}sin(0)=-cos(0)=-1 $$     (Eq.7.3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=4 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(4)}(0)=\frac{d^4}{dt^4}sin(0)=sin(0)=0 $$     (Eq.7.3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For n=5 :
 * {| style="width:100%" border="0"

$$  \displaystyle f^{(5)}(0)=\frac{d^5}{dt^5}sin(0)=cos(0)=1 $$     (Eq.7.3.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Apply Eq. 7.3.2 - 7.3.7 to Eq. 7.2:
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=\frac{0}{0!}x^0+\frac{1}{1!}x^1+\frac{0}{2!}x^2-\frac{1}{3!}x^3+\frac{0}{4!}x^4+\frac{1}{5!}x^5 $$     (Eq.7.3.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the Maclaurin Series expansion for f(t)=\sin t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(x)=sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040} .... $$     (Eq.7.3.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Determined from the pattern of Eq. 7.3.8, the summation for a Taylor Series approximation of f(t)=\sin t is:
 * {| style="width:100%" border="0"

$$  \displaystyle f(t)=sin(t)=\sum_{n=0}^{\infty }\frac{[(-1)^n]t^{1+2n}}{(1+2n)!} $$     (Eq.7.3.10)
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 * style="width:25%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.wheeler.tw 19:22, 7 February 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.Li 21:12, 7 February 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.hintz 22:31, 7 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 09:21, 8 February 2012 (UTC)

= Problem 8 - Homogeneous L-ODE-CC = For problem reference.

Part 1 Given

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

$$  \displaystyle y^{\prime\prime}+y^\prime+3.25y=0 $$     (Eq.8.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Problem Statement
Find an ODE for the given basis when.

Part 1 Solution
Use the following to generate the characteristic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}=\lambda^2 $$, $$  \displaystyle y^{\prime}=\lambda $$, $$  \displaystyle y=1 $$     (Eq.8.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute Eq. 8.1.2 into Eq. 8.1.1
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+\lambda +3.25=0 $$     (Eq.8.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To solve the characteristic equation, apply the quadratic formula:
 * {| style="width:100%" border="0"

$$  \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (Eq.8.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-(1)\pm\sqrt{(1)^2-4(1)(3.25)}}{2(1)} $$     (Eq.8.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-1\pm\sqrt{-12}}{2} $$     (Eq.8.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-1\pm i \sqrt{12}}{2} $$     (Eq.8.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_{1}=\frac{-1+ i \sqrt{12}}{2} $$     (Eq.8.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_{2}=\frac{-1- i \sqrt{12}}{2} $$     (Eq.8.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since, the roots are in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=a\pm\omega i $$ (Eq.8.1.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the general solution will take the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(C_1\cos\omega x+C_2\sin\omega x) $$ (Eq.8.1.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then, the general solution will be the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y= e^{-\frac{1}{2}x}(C_1\cos\sqrt{12}x+C_2\sin\sqrt{12}x) $$     (Eq.8.1.12)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

In order to check the general solution, take the first and second derivative of Eq. 8.1.12 with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime= -\frac{1}{2}e^{-\frac{1}{2} x }(C_{1}\cos\sqrt{12}x+C_{2}\sin\sqrt{12}x)+e^{-\frac{1}{2}x}(-\sqrt{12}C_{1}\sin\sqrt{12}x+\sqrt{12}C_{2}\cos\sqrt{12}x) $$     (Eq.8.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y^{\prime\prime}=\frac{1}{4}e^{-\frac{1}{2} x }(C_{1}\cos\sqrt{12}x+C_{2}\sin\sqrt{12}x)-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{12}C_{1}\sin\sqrt{12}x+\sqrt{12}C_{2}\cos\sqrt{12}x) $$ $$ -\frac{1}{2}e^{-\frac{1}{2} x }(-\sqrt{12}C_{1}\sin\sqrt{12}x+\sqrt{12}C_{2}\cos\sqrt{12}x)+e^{-\frac{1}{2} x }(-12C_{1}\cos\sqrt{12}x-12C_{2}\sin\sqrt{12}x) $$     (Eq.8.1.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute Eq. 8.1.12, 8.1.13, and 8.1.14 into Eq. 8.1.1.

The final result of that substitution is:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$     (Eq.8.1.15)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part 2 Given

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

$$  \displaystyle y^{\prime\prime}+0.54y^\prime+(0.0729+\pi)y=0 $$     (Eq.8.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2 Problem Statement
Find an ODE for the given basis when.

Part 2 Solution
Use the following to generate the characteristic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}=\lambda^2 $$, $$  \displaystyle y^{\prime}=\lambda $$, $$  \displaystyle y=1 $$     (Eq.8.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute Eq. 8.2.2 into Eq. 8.2.1
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+0.54\lambda +(0.0729+\pi)=0 $$     (Eq.8.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To solve the characteristic equation, apply the quadratic formula:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-(0.54)\pm\sqrt{(0.54)^2-4(1)(0.0729+\pi)}}{2(1)} $$     (Eq.8.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-0.54\pm\sqrt{-\pi}}{2} $$     (Eq.8.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-0.54\pm i\sqrt{\pi}}{2} $$     (Eq.8.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_{1}=-0.27+ i \sqrt{\pi} $$     (Eq.8.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_{2}=-0.27- i \sqrt{\pi} $$     (Eq.8.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since, the roots are in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=a\pm\omega i $$ (Eq.8.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the general solution will take the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(C_1\cos\omega x+C_2\sin\omega x) $$ (Eq.8.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then, the general solution will be the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y= e^{-0.27x}(C_{1}\cos\sqrt{\pi}x+C_{2}\sin\sqrt{\pi}x) $$     (Eq.8.2.11)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

In order to check the general solution, take the first and second derivative of Eq. 8.2.11 with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime= -0.27e^{-0.27x}(C_{1}\cos\sqrt{\pi}x+C_{2}\sin\sqrt{\pi}x)+e^{-0.27x}(-\sqrt{\pi}C_{1}\sin\pi x+\sqrt{\pi}C_{2}\cos \pi x) $$ (Eq.8.2.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y^{\prime\prime}=0.0729e^{-0.27x}(C_{1}\cos\sqrt{\pi}x+C_{2}\sin\sqrt{\pi}x)-0.27e^{-0.27x} (-\sqrt{\pi}C_{1}\sin\sqrt{\pi}x+\sqrt{\pi}C_{2}\cos\sqrt{\pi}x) $$
 * style="width:95%" |
 * style="width:95%" |

$$ -0.27e^{-0.27x}(-\sqrt{\pi}C_{1}\sin\pi x+\sqrt{\pi}C_{2}\cos\pi x)+e^{-0.27x}(-\pi C_{1}\cos\sqrt{\pi}x-\pi C_{2}\sin\sqrt{\pi}x) $$     (Eq.8.2.13)
 *

 * }

Substitute Eq. 8.2.11, 8.2.12, and 8.2.13 into Eq. 8.2.1.

The final result of that substitution is:
 * {| style="width:100%" border="0"

$$  \displaystyle 0=0 $$     (Eq.8.2.14)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.wheeler.tw 19:40, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.deaver.md 10:23, 8 February 2012 (UTC) Editor: Egm4313.s12.team17.deaver.md 10:23, 8 February 2012 (UTC)

= Problem 9 - Three Cases of Damping=

Part 1 Given
Initial Conditions:
 * {| style="width:100%" border="0"

$$  \displaystyle y(0)=1,y^\prime(0)=0 $$     (Eq.9.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

No excitation:
 * {| style="width:100%" border="0"

$$  \displaystyle r(x)=0 $$     (Eq.9.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda^2+4\lambda+13=0 $$     (Eq.9.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Problem Statement
Find the solution for the L2-ODE-CC that correspond to Eq. 9.1.3 and plot the solution.

Part 1 Solution
Solve for the roots in Eq. 9.1.3 by using the quadratic formula:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-4\pm \sqrt{4^2-4(1)(13)}}{2(1)} $$     (Eq.9.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-4\pm \sqrt{-36}}{2} $$     (Eq.9.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=-2\pm3i $$     (Eq.9.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since, the roots are in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=a\pm\omega i $$ (Eq.9.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the general solution will take the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(A\cos\omega x+B\sin\omega x) $$ (Eq.9.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the values from Eq. 9.1.6, substitute them into Eq. 9.1.8:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-2x}(A\cos3 x+B\sin3 x) $$ (Eq.9.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the derivative of Eq. 9.1.9 with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime=e^{-2x}(-2A\cos3 x - 3A\sin3 x - 2B\sin3 x + 3B\cos3 x) $$ (Eq.9.1.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solve for A \, and B \, by using the initial conditions (Eq. 9.1.1):
 * {| style="width:100%" border="0"

$$  \displaystyle 1=e^{-2(0)}(A\cos3(0)+B\sin3(0)) $$     (Eq.9.1.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle A=1 $$     (Eq.9.1.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=e^{-2(0)}(-2A\cos3 (0) - 3A\sin3 (0) - 2B\sin3 (0) + 3B\cos3 (0)) $$     (Eq.9.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=-2A + 3B $$     (Eq.9.1.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle B=\frac{2}{3} $$     (Eq.9.1.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the solution for this L2-ODE-CC that uses the initial condition shown above is the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-2x}(\cos3 x+\frac{2}{3}\sin3 x) $$ (Eq.9.1.16)
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 *

 * }

In order to check if this solution is correct, we can substitute it in the L2-ODE-CC standard form.

The standard form of the L2-ODE-CC:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}+4y^\prime+13y=r(x) $$     (Eq.9.1.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Use Eq. 9.1.2 to simplify Eq. 9.1.17:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime\prime}+4y^\prime+13y=0 $$     (Eq.9.1.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the derivative of Eq. 9.1.16 twice with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime}=-\frac{13}{3}e^{-2x}\sin3x $$     (Eq.9.1.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y^{\prime\prime}=\frac{26}{3}e^{-2x}\sin3x-13e^{-2x}\cos3x $$     (Eq.9.1.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Next, take Eq. 9.1.16,9.1.19, and 9.1.20 and plug it into Eq. 9.1.18:
 * {| style="width:100%" border="0"

$$  \displaystyle [\frac{26}{3}e^{-2x}\sin3x - 13e^{-2x}\cos3x] + 4[-\frac{13}{3}e^{-2x}\sin3x] + 13[e^{-2x}(\cos3 x+\frac{2}{3}\sin3 x)] = 0 $$     (Eq.9.1.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \frac{26}{3}e^{-2x}\sin3x - 13e^{-2x}\cos3x - \frac{52}{3}e^{-2x}\sin3x + 13e^{-2x}\cos3 x + \frac{26}{3}e^{-2x}\sin3 x = 0 $$     (Eq.9.1.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

From Eq. 9.1.22, we can see that everything negates one another; therefore, Eq. 9.1.16 is a correct.

The plot of Eq. 9.1.16 is shown below:

Part 2 Given

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

$$  \displaystyle y = \frac{5}{7}e^{-2x} + \frac{2}{7}e^{5x} $$     (Eq.1.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_k^{\prime\prime}+6y_k^\prime+9y_k=f(t) $$     (Eq.6.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y=e^{-2x}(\cos3 x+\frac{2}{3}\sin3 x) $$ (Eq.9.1.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 2 Problem Statement
Take the equations that are shown in the given and superpose them.

Part 2 Solution
Need to solve for the solution for Eq. 6.12 that meet the initial condition from part 1 (Eq. 9.1.1).

Therefore, convert Eq. 6.12 into the characteristic equation:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^2+6\lambda+9=0 $$     (Eq.9.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then, use the quadratic formula to get the roots:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=\frac{-6\pm \sqrt{6^2-4(1)(9)}}{2(1)} $$     (Eq.9.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=\frac{-6\pm \sqrt{0}}{2} $$     (Eq.9.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda=-3 $$     (Eq.9.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since, the roots are in the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda=a $$     (Eq.9.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the general solution will take the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{ax}(A+B x) $$ (Eq.9.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the values from Eq. 9.2.2, substitute them into Eq. 9.2.4:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-3x}(A+B x) $$ (Eq.9.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the derivative of Eq. 9.1.9 with respect to x \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime=-3e^{-3x}(A+B x)+Be^{-3x} $$     (Eq.9.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solve for A \, and B \, by using the initial conditions (Eq. 9.1.1):
 * {| style="width:100%" border="0"

$$  \displaystyle 1=e^{-3(0)}(A+B (0)) $$     (Eq.9.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle A=1 $$     (Eq.9.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=-3e^{-3(0)}(A+B (0))+Be^{-3(0)} $$     (Eq.9.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle B=3 $$     (Eq.9.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the solution for this L2-ODE-CC that uses the initial condition shown above is the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y=e^{-3x}(1+3 x) $$ (Eq.9.2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Now, the superpose plot shown below:

Overdamping is the distinct real roots ODE. (Eq.1.1.13)

Critical damping is real double root ODE. (Eq.9.2.11)

Underdamping is the complex conjugate roots ODE. (Eq.9.1.16)

Author & Proofreaders
Author: Egm4313.s12.team17.deaver.md 11:57, 7 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.ying 15:23, 8 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 11:57, 8 February 2012 (UTC)

= References =

= Contributing Team Members =

title: User:Egm4313.s12.team17/Report 2


text:
User:Vladimir.Code test page 2

This is the format of a normal equation.
 * {| style = "width: 73%"

!  $$ \Sigma F = m a = m {d^2 y \over d t^2} $$ !!  (Eq. 1)
 * }

This is the format of the final solution.
 * {| class = wikitable style="width: 73%; text-align:center"

!  $$ \Sigma F = m a = m {d^2 y \over d t^2} $$  (Eq. 1)
 * }

This is the format for the inserted picture. Change the size, width and file type as needed.

This is Grant's user wiki page:

Egm4313.s12.team19.Uppercue

title: User:Vladimir.Code test page 2