text:
User:Egm4313.s12.team17/Report 5

=Problem R5.1 - Radius of Convergence=

Part 1 Given
Find $$R_c$$ for the following series

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$$  \displaystyle r(x) = \sum_{k=0}^\infty (k+1)k x^k $$     (Eq.1.1.1)
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Part 1 Problem Statement
Find R_c\, for the following series.

Part 1 Solution
Using the following formula:
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$$  \displaystyle R_{c}=\left[\lim_{k \to \infty}\left|\frac{d_{k+1}}{d_{k}}\right|\right ]^{-1} $$     (Eq.1.1.2) Where for a power series:
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$$  \displaystyle r(x)=\sum_{k=0}^\infty d_k x^k $$     (Eq.1.1.3)
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$$  \displaystyle r(x) = \sum_{k=0}^\infty (k+1)k x^k $$     (Eq.1.1.4) Where:
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$$  \displaystyle d_k=(k+1)k $$     (Eq.1.1.5)
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$$  \displaystyle d_{k+1}=(k+2)(k+1) $$     (Eq.1.1.6) Plugging equations 1.1.5 and 1.1.6 into equation 1.1.2:
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$$  \displaystyle R_{c}=\left[\lim_{k \to \infty}\left|\frac{(k+2)(k+1)}{(k+1)k}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left|\frac{(k+2)}{k}\right|\right ]^{-1}=1 $$     (Eq.1.1.7)
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Part 2 Given

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$$  \displaystyle r(x) = \sum_{k=0}^\infty \frac{(-1)^k}{\gamma^k}x^{2k} $$     (Eq.1.2.1)
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$$  \displaystyle \gamma = constant $$
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Part 2 Problem Statement

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$$  \displaystyle r(x) = \sum_{k=0}^\infty \frac{(-1)^k}{\gamma^k}x^{2k} $$     (Eq.1.2.2)
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$$  \displaystyle \frac{(-1)^k}{\gamma^k}x^{2k}=\frac{(-1)^k}{\gamma^k}(x^{k})^2=\sqrt{\frac{(-1)^k}{\gamma^k}}x^{k} $$     (Eq.1.2.3)
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Where:
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$$  \displaystyle d_k= \sqrt{\frac{(-1)^k}{\gamma^k}} $$     (Eq.1.2.4)
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$$  \displaystyle d_{k+1}=\sqrt{\frac{(-1)^{k+1}}{\gamma^{k+1}}}=\sqrt{\frac{(-1)^{k}(-1)^{1}}{\gamma^{k}\gamma^{1}}} $$     (Eq.1.2.5) Using the same equation in Part 1, plug Eq.1.2.4 and Eq.1.2.5 into Eq.1.1.2
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$$  \displaystyle R_{c}=\left[\lim_{k \to \infty}\left|\frac{\sqrt{\frac{(-1)^{k}(-1)^{1}}{\gamma^{k}\gamma^{1}}}}{\sqrt{\frac{(-1)^k}{\gamma^k}}}\right|\right ]^{-1} = \left[\lim_{k \to \infty}\left|\frac{\sqrt{\frac{(-1)^{1}}{\gamma^{1}}}}{1}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left|\frac{1}{\sqrt{|\gamma|}}\right|\right ]^{-1}=\sqrt{|\gamma|} $$     (Eq.1.2.6)
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Part 3 Given

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$$  \displaystyle sin x $$ (Eq.1.3.1)
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$$  \displaystyle \hat x = 0 $$
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Part 3 Problem Statement
Find R_c\, for the given Taylor series.

Part 3 Solution
Performing a Maclaurin series expansion:
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$$  \displaystyle f(x)=sin(x)=\sum_{k=0}^{\infty }\frac{[(-1)^k]x^{1+2k}}{(1+2k)!} $$     (Eq.1.3.2)
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$$  \displaystyle \frac{[(-1)^k]x^{1+2k}}{(1+2k)!}= \frac{[(-1)^k]x^{1}x^{2k}}{(1+2k)!}= \frac{[(-1)^k]x^{1}(x^{k})^{2}}{(1+2k)!} = \sqrt{\frac{(-1)^kx}{(1+2k)!}}x^{k} $$     (Eq.1.3.3) Where:
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$$  \displaystyle d_{k+1} = \sqrt{\frac{(-1)^{k+1}x}{(1+2(k+1))!}}=\sqrt{\frac{(-1)^{k}(-1)^{1}x}{(1+2(k+1))!}} $$     (Eq.1.3.4)
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$$  \displaystyle d_{k}=\sqrt{\frac{(-1)^kx}{(1+2k)!}} $$     (Eq.1.3.5) Using Eq.1.1.2, plug in Eq.1.3.4 and Eq.1.3.5
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$$  \displaystyle R_{c}=\left[\lim_{k \to \infty}\left|\frac{\sqrt{\frac{(-1)^{k}(-1)^{1}x}{(1+2(k+1))!}}}{\sqrt{\frac{(-1)^kx}{(1+2k)!}}}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left|\frac{\sqrt{\frac{-1}{(2k+3)(2k+2)}}}{1}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left| \sqrt{\frac{1}{(4k^2+8k+3)}}\right|\right ]^{-1}=(\frac{1}{\infty})^{-1}=\infty $$     (Eq.1.3.6)
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Part 4 Given

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$$  \displaystyle \log(1+x) $$     (Eq.1.4.1)
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$$  \displaystyle \hat x = 0 $$
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Part 4 Problem Statement
Find R_c\, for the given Taylor series.

Part 4 Solution
Performing a Maclaurin series expansion:
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$$  \displaystyle f(x)=log(1+x)=\sum_{k=1}^{\infty }\frac{(-1)^{k-1}x^{k}}{k} $$     (Eq.1.4.2) Where:
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$$  \displaystyle d_{k+1} = \frac{(-1)^{k+1-1}}{k+1}=\frac{-1^k}{k+1} $$     (Eq.1.4.3)
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$$  \displaystyle d_{k}=\frac{(-1)^{k-1}}{k}=\frac{(-1)^k(-1)^{-1}}{k}=\frac{-1^{k}}{-k} $$     (Eq.1.4.4) Plugging Eq.1.4.3 and Eq.1.4.4 into Eq.1.1.2:
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$$  \displaystyle R_{c}=\left[\lim_{k \to \infty}\left|\frac{\frac{-1^k}{k+1}}{\frac{-1^{k}}{-k}}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left|\frac{\frac{1}{k+1}}{\frac{-1}{k}}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left|\frac{-k}{k+1}\right|\right ]^{-1}=1 $$     (Eq.1.4.5)
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Part 5 Given

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$$  \displaystyle \log(1+x) $$     (Eq.1.5.1)
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$$  \displaystyle \hat x = 1 $$
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Part 5 Problem Statement
Find R_c\, for the given Taylor series.

Part 5 Solution
Performing a Taylor series expansion about $$\hat x$$, results in the following series:
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$$  \displaystyle f(x)=\log(2)+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^{2}+\frac{1}{24}(x-1)^3-\frac{1}{64}(x-1)^4... $$     (Eq.1.5.2) This simplifies to:
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$$  \displaystyle f(x)=\log(2)+\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2^kk}(x-1)^k $$     (Eq.1.5.3)
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Therefore $$d_k$$ and $$d_(k+1)$$ are then:
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$$  \displaystyle d_{k+1} = \frac{(-1)^{k+1-1}}{2^(k+1)(k+1)}=\frac{-1^k}{2^k*2(k+1)} $$     (Eq.1.5.4)
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$$  \displaystyle d_{k}=\frac{(-1)^{k-1}}{2^k*k}=\frac{(-1)^k(-1)^{-1}}{2^k*k}=\frac{-1^{k}}{-k*2^k} $$     (Eq.1.5.5)
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Plugging Eq.1.5.4 and Eq.1.5.5 into Eq.1.1.2
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$$  \displaystyle R_{c}=\left[\lim_{k \to \infty}\left|\frac{\frac{-1^k}{2^k*2(k+1)}}{\frac{-1^{k}}{-k*2^k}}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left|\frac{\frac{1}{2(k+1)}}{\frac{-1}{k}}\right|\right ]^{-1}=\left[\lim_{k \to \infty}\left|\frac{-k}{2(k+1)}\right|\right ]^{-1}=2 $$     (Eq.1.4.6)
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Author & Proofreaders
Author:Egm4313.s12.team17.wheeler.tw 05:06, 23 March 2012 (UTC)

Proofreader:Egm4313.s12.team17.ying 02:32, 30 March 2012 (UTC)

Editor:Egm4313.s12.team17.ying 02:32, 30 March 2012 (UTC)

=Problem R5.2 - Linear Independent Testing=

Part 1 Given

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$$  \displaystyle f(x) = x^2 $$     (Eq.2.1.1)
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$$  \displaystyle g(x) = x^4 $$     (Eq.2.1.2)
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$$  \displaystyle f(x) = cos x $$ (Eq.2.1.3)
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$$  \displaystyle g(x) = sin 3x $$     (Eq.2.1.4)
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Part 1 Problem Statement
Determine whether the following pair of functions are linearly independent using the Wronskian.

Part 1 solution
The Wronskian of the two functions is written as

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$$  \displaystyle W(f,g) = f(x)g'(x) - g(x)f'(x) $$     (Eq.2.1.5)
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which is the determinant of the functions were they in a 2x2 matrix. If W(f,g) = 0 then the two functions are linearly dependent.

Using the first set of functions, Eq.2.1.1 and Eq.2.1.2, the corresponding derivatives are

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$$  \displaystyle f'(x) = 2x $$     (Eq.2.1.6)
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$$  \displaystyle g'(x) = 4x^3 $$     (Eq.2.1.7)
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and placing Eq.2.1.1, Eq.2.1.2, Eq.2.1.6 and Eq.2.1.7 into Eq.2.1.5 we get

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$$  \displaystyle (x^2)(4x^3) - (x^4)(2x) = 4x^5 - 2x^5 = 2x^5 \ne 0 $$     (Eq.2.1.8)
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Therefore Eq.2.1.1 and Eq.2.1.2 are Linearly Independent.
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Using the second set of functions, Eq.2.1.3 and Eq.2.1.4, the corresponding derivatives are

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$$  \displaystyle f'(x) = -sin(x) $$     (Eq.2.1.9)
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and

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$$  \displaystyle g'(x) = 3cos(3x) $$     (Eq.2.1.10)
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Plugging Eq.2.1.3,Eq.2.1.4,Eq.2.1.9, and Eq.2.1.10 into Eq.2.1.5 gives:

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$$  \displaystyle [cos(x)][cos(3x)] - [sin(3x)][-sin(x)] $$     (Eq.2.1.11)
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These equations cannot be simplified so

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$$  \displaystyle [cos(x)][cos(3x)] - [sin(3x)][-sin(x)] \ne 0 $$     (Eq.2.1.12)
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Thus the two functions are linearly independent.
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Part 2 Given
Same as part 1 given

Part 2 Problem Statement
Verify Eq.2.1.1, Eq.2.1.2, Eq.2.1.3, and Eq.2.1.4 are linearly independent using the Gramian over the interval [a,b] = [-1,1].

Part 2 Solution
The gramian is given by

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$$  \displaystyle \Gamma (f,g) = \begin{bmatrix} \left \langle f,f \right \rangle \left \langle f,g \right \rangle \\ \left \langle g,f \right \rangle \left \langle g,g \right \rangle \end{bmatrix} $$     (Eq.2.2.1)
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Where

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$$  \displaystyle
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\left \langle f,g \right \rangle = \int\limits_{a}^{b} f(x)g(x) dx $$ (Eq.2.2.2)
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If the Gramian, Eq.2.2.1

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$$  \displaystyle
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\Gamma (f,g) \ne 0 $$     (Eq.2.2.3)
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Then the two functions are linearly independent

Now to calculate the matrix values:
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$$  \displaystyle \left \langle f,f \right \rangle = \int\limits_{-1}^{1} x^4 dx $$ (Eq.2.2.4)
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Which then equals
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$$  \displaystyle \frac{1}{5}[(1^5) - (-1^5)] = \frac{2}{5} $$     (Eq.2.2.5)
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The other terms are as follows:

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$$  \displaystyle \left \langle f,g \right \rangle = \int\limits_{-1}^{1} x^6 dx = \frac{2}{7}
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$$     (Eq.2.2.6)
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$$  \displaystyle \left \langle g,f \right \rangle = \int\limits_{-1}^{1} x^6 dx = \frac{2}{7}
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$$     (Eq.2.2.7)
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$$  \displaystyle \left \langle g,g \right \rangle = \int\limits_{-1}^{1} x^8 dx = \frac{2}{9}
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$$     (Eq.2.2.8)
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Entering Eq.2.2.5, Eq.2.2.6, Eq.2.2.7, Eq.2.2.8 into the Gramian Matrix, Eq.2.2.1 results in

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$$  \displaystyle \Gamma (f,g) = \begin{bmatrix} \left (\frac{2}{5} \right) & 	\left (\frac{2}{7} \right) \\ \left (\frac{2}{7} \right) & \left (\frac{2}{9} \right) \end{bmatrix} $$     (Eq.2.2.9)
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Eq.2.2.9 then results in:

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$$  \displaystyle \frac{2}{5}*\frac{2}{9} - \frac{2}{7}*\frac{2}{7} = \frac{4}{45} - \frac{4}{49} \ne 0 $$     (Eq.2.2.10)
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Thus the two functions are linearly independent.
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Using Eq.2.2.2, the following scalar product values are calculated.
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$$  \displaystyle \left \langle f,f \right \rangle = \int\limits_{-1}^{1} cos(x)*cos(x)dx = 1.008 $$     (Eq.2.2.11)
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$$  \displaystyle \left \langle f,g \right \rangle = \int\limits_{-1}^{1} cos(x)*sin(3x) dx = 0
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$$     (Eq.2.2.12)
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$$  \displaystyle \left \langle g,f \right \rangle = \int\limits_{-1}^{1} sin(3x)*cos(x)dx = 0
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$$     (Eq.2.2.13)
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$$  \displaystyle \left \langle g,g \right \rangle = \int\limits_{-1}^{1} sin^2(3x) dx = .9826
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$$     (Eq.2.2.14)
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Entering Eq.2.2.11, Eq.2.2.12, Eq.2.2.13, and Eq.2.2.14 into the Eq.2.2.1 results in
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$$  \displaystyle \Gamma (f,g) = \begin{bmatrix} 1.008 & 0 \\ 0 & .9826 \end{bmatrix} $$     (Eq.2.2.15)
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Which finally becomes

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$$  \displaystyle (1.008)(.9826) - (0)(0) = 0.990 \ne 0 $$     (Eq.2.2.16)
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Thus the two functions are linearly independent.
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Author & Proofreaders
Author:Egm4313.s12.team17.hintz 19:09, 23 March 2012 (UTC)

Proofreader:Egm4313.s12.team17.deaver.md 06:32, 30 March 2012 (UTC)

Editor:Egm4313.s12.team17.ying 03:58, 30 March 2012 (UTC)

=Problem R5.3 - Linear Independent Test using Gramain=

Given

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$$  \displaystyle \mathbf b_1 = 2\mathbf e_1 + 7\mathbf e_2 $$     (Eq.3.1)
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$$  \displaystyle \mathbf b_1 = 1.5\mathbf e_1 + 3\mathbf e_2 $$     (Eq.3.2)
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Problem Statement
Verify that b_1\, and b_2\, are linearly independent using the Gramian

Solution
The Grammian is given as:

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$$  \displaystyle \Gamma (b_1,b_2) = \begin{bmatrix} \left \langle b_1,b_1 \right \rangle \left \langle b_1,b_2 \right \rangle \\ \left \langle b_2,b_1 \right \rangle \left \langle b_2,b_2 \right \rangle \end{bmatrix} $$     (Eq.3.3)
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According to (3) on p.8-9

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$$  \displaystyle \left \langle b_1,b_1 \right \rangle = (b_1 \cdot b_2) $$     (Eq.3.4)
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Calculating the dot products using Eq.3.1 and Eq.3.2:

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$$  \displaystyle \left \langle b_1,b_1 \right \rangle = 4 + 49 = 53 $$     (Eq.3.5)
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$$  \displaystyle \left \langle b_1,b_2 \right \rangle = 3 + 21 = 24 $$     (Eq.3.6)
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$$  \displaystyle \left \langle b_2,b_1 \right \rangle = 3 + 21 = 24 $$     (Eq.3.7)
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$$  \displaystyle \left \langle b_2,b_2 \right \rangle = 2.25 + 9 = 11.25 $$     (Eq.3.8)
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Plugging Eq.3.5, Eq.3.6, Eq.3.7, and Eq.3.8 into Eq.3.3 and finding the determinant gives:

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$$  \displaystyle
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596.25 - 576 = 20.25 = \Gamma \ne 0

$$     (Eq.3.9)
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Thus the two functions are linearly independent.
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Author & Proofreaders
Author:Egm4313.s12.team17.hintz 19:48, 21 March 2012 (UTC)

Proofreader:Egm4313.s12.team17.Li 17:21, 25 March 2012 (UTC)

Editor:Egm4313.s12.team17.ying 04:46, 30 March 2012 (UTC)

=Problem R5.4 - Particular Solution Verification=

Given

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$$  \displaystyle y_{p}(x)= \sum_{i=0}^n y_{p,i}(x) $$     (Eq.4.1)
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$$  \displaystyle y^{\prime\prime}+p(x)y'+q(x)y=r(x) $$     (Eq.4.2)
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$$  \displaystyle r(x)=\sum_{i=0}^n r_{i}(x) $$     (Eq.4.3)
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Problem Statement
Show that Eq. 4.1 is indeed the overall particular solution of the L2-OE-CC, Eq. 4.2

Discuss the choice of y_p(x)\, for and  in y_p(x)\,

Solution
The particular solution for Eq. 4.2 is:
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$$  \displaystyle y_{p}(x)=r(x) $$
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Since the excitation is Eq. 4.3, the particular solution is then
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$$  \displaystyle y_{p}(x)=r(x)=\sum_{i=0}^n r_{i}(x)=r_1(x)+r_2(x) \cdots + r_n(x) $$     (Eq.4.4)
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If we expand Eq.4.1

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$$  \displaystyle y_{p}(x)= \sum_{i=0}^n y_{p,i}(x)=y_{p,1}(x)+y_{p,2}(x)+ \cdots + y_{p,n}(x) $$     (Eq.4.5) We have : :{| style="width:100%" border="0" $$  \displaystyle r_1(x)=y_{p,1} $$
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$$  \displaystyle r_2(x)=y_{p,2} $$ :{| style="width:100%" border="0" $$  \displaystyle \vdots $$ :{| style="width:100%" border="0" $$  \displaystyle r_{n}(x)=y_{p,n} $$ Now we can see that the particular solution $$y_{p,i}(x)$$ for a simple excitation $$r_i(x)$$ satisfies Eq. 4.2: :{| style="width:100%" border="0" $$  \displaystyle y_{p,i}''+p(x)y'_{p,i}+q(x)y_{p,i}=r_i(x) $$     (Eq.4.6)
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Because of linearity, when each particular solution $$ y_{p,i} $$ is known, then we see that the overall particular solution is: :{| style="width:100%" border="0" $$  \displaystyle y_{p}(x)= \sum_{i=0}^n y_{p,i}(x) $$     (Eq.4.7)
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Therefore, Eq.4.1 is indeed the particular solution to Eq.4.2.
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Explanation

When the excitation to an equation, $$y^{\prime\prime}+p(x)y'+q(x)y=r(x)$$ is: :{| style="width:100%" border="0" $$  \displaystyle r(x)=kcoswx $$ We first think of using a particular solution of:
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$$  \displaystyle y_{p}(x)= Asinwx $$ However if we were to take the derivatives of the particular solution and substitute back into the equation, we have:
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$$  \displaystyle -w^2Asin(wx)-wAsin(wx)p(x)+Acos(wx)q(x)= Kcoswx $$ Notice that $$sinwx$$ has appeared. And it has the same coefficient 'A' as well. When equating common coefficients we see that 'A' must equal 0,$$ A=0$$.
 However this can not happen since the cosine will drop out.Therefore, we can see that because of the sine that is shown, our initial guess was wrong.
 In order to counter the mistake, we need to add a sine to $$y_p(x)$$. So,the particular solution would be:
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$$  \displaystyle y_{p}(x)= Acoswx+Bsinwx $$ In conclusion, if the excitation,$$ r(x)$$ includes a sine or cosine, the particular solution must be in the form of:
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$$  \displaystyle y_{p}(x)= Mcoswx+Nsinwx $$ where 'N' and 'M' are constants.
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Author & Proofreaders
Author:Egm4313.s12.team17.Li 21:32, 26 March 2012 (UTC)

Proofreader 1:Egm4313.s12.team17.wheeler.tw 12:15, 28 March 2012 (UTC)

Proofreader 2:Egm4313.s12.team17.axelrod.a 17:23, 29 March 2012 (UTC)

Editor:Egm4313.s12.team17.ying 05:42, 30 March 2012 (UTC)

=Problem R5.5 - L2-ODE-CC with Linear Independent Testing=

Part 1 Given

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$$  \displaystyle cos(7x) $$     (Eq.5.1.1)
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$$  \displaystyle sin(7x) $$     (Eq.5.1.2)
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Part 1 Problem Statement
Show that cos(7x)\, and sin(7x)\, are linearly independent using the Wronskian and the Gramian (integrate over 1 period).

Part 1 Solution
Wronskian Method

By Definition
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$$  \displaystyle W(f,g):= \det \begin{bmatrix} f &g \\ f'&g' \end{bmatrix}=fg'-gf' $$     (Eq.5.1.3) In this problem:
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$$  \displaystyle f(x)=\cos7x $$     (Eq.5.1.4)
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$$  \displaystyle g(x)=\sin7x $$     (Eq.5.1.5) Taking the derivative of Eq.5.1.4 and Eq.5.1.5 respectively:
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$$  \displaystyle f'(x)=-7\sin7x $$     (Eq.5.1.6)
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$$  \displaystyle g'(x)=7\cos7x $$     (Eq.5.1.7) Plugging equations Eq.5.1.4 through Eq.5.1.7 into Eq.5.1.3
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$$  \displaystyle W(f,g)=(\cos7x)(7\cos7x)-(\sin7x)(-7\sin7x)=7cos^27x+7sin^27x \ne 0 $$     (Eq.5.1.8)
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Thus the functions are linearly independent.
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Gramian Method

By definition:
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$$  \displaystyle \Gamma (f,g) := \det \begin{bmatrix}\langle f,f \rangle & \langle f,g \rangle\\ \langle g,f \rangle & \langle g,g \rangle\end{bmatrix} $$     (Eq.5.1.9) Where
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$$  \displaystyle \langle f,g \rangle := \int_a^b f(x)g(x)\,dx $$     (Eq.5.1.10) In this problem:
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$$  \displaystyle f(x)=\cos7x $$     (Eq.5.1.11)
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$$  \displaystyle g(x)=\sin7x $$     (Eq.5.1.12)
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$$  \displaystyle T=\frac{2 \pi}{7}=0.8976 $$     (Eq.5.1.13) Applying Eqs.5.1.11 through Eq.5.1.13 to Eq.5.1.10:
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$$  \displaystyle \langle f,f \rangle := \int_0^{0.8976} f(x)f(x)\,dx = \int_0^{0.8976} \cos7x \cos7x \,dx = \int_0^{0.8976} \cos^27x\,dx = 0.449 =\frac{\pi}{7} $$     (Eq.5.1.14)
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$$  \displaystyle \langle f,g \rangle := \int_0^{0.8976} f(x)g(x)\,dx = \int_0^{0.8976} \cos7x \sin7x \,dx = 0 $$     (Eq.5.1.15)
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$$  \displaystyle \langle g,f \rangle := \int_0^{0.8976} f(x)g(x)\,dx = \int_0^{0.8976} \cos7x \sin7x \,dx = 0 $$     (Eq.5.1.16)
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$$  \displaystyle \langle g,g \rangle := \int_0^{0.8976} g(x)g(x)\,dx = \int_0^{0.8976} \sin7x \sin7x \,dx = \int_0^{0.8976} \sin^27x\,dx = 0.449 = \frac{\pi}{7} $$     (Eq.5.1.17) Plugging in values form Eq.5.1.14 to Eq.5.1.17 into Eq.5.1.9:
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$$  \displaystyle \Gamma (f,g) := \det \begin{bmatrix}\langle f,f \rangle & \langle f,g \rangle\\ \langle g,f \rangle & \langle g,g \rangle\end{bmatrix} = \det \begin{bmatrix} \frac{\pi}{7} &  0 \\  0  &  \frac{\pi}{7} \end{bmatrix} = \frac{\pi^2}{49} \ne 0 $$     (Eq.5.1.18)
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Thus the functions are linearly independent.
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Part 2 Given
Same as Part 1 Given

Part 2 Problem Statement
Find 2 equations for the two unknowns M,N, and solve for M,N.

Part 2 Solution
The ODE to be solved is:
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$$  \displaystyle y''-3y'-10y=3\cos7x $$     (Eq.5.2.1) The particular solution will take the following form:
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$$  \displaystyle y_p(x)=M\cos7x + N\sin7x $$     (Eq.5.2.2) Take the first and second derivatives of Eq.5.2.2:
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$$  \displaystyle y'_p(x)=-7M\sin7x + 7N\cos7x $$     (Eq.5.2.3)
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$$  \displaystyle y''_p(x)=-49M\cos7x - 49N\sin7x $$     (Eq.5.2.4) Plugging Eq.5.2.2 and Eq.5.2.3 into Eq.5.2.1:
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$$  \displaystyle -49M\cos7x - 49N\sin7x-3(-7M\sin7x + 7N\cos7x)-10(M\cos7x + N\sin7x)=3\cos7x $$     (Eq.5.2.5)
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$$  \displaystyle -49M\cos7x-49N\sin7x+21M\sin7x-21N\cos7x-10M\cos7x-10N\sin7x=3\cos7x $$     (Eq.5.2.6) By equating like terms, the two equations for solving for M and N are:
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$$  \displaystyle -49M-21N-10M=3 $$     (Eq.5.2.7)
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$$  \displaystyle -49N+21M-10N=0 $$     (Eq.5.2.8) Solving the system of equations created by Eq.5.2.7 and Eq.5.2.8:
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$$  \displaystyle M=-0.045 $$     (Eq.5.2.9)
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$$  \displaystyle N=-0.016 $$     (Eq.5.2.10)
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Part 3 Given
Initial conditions
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$$  \displaystyle y(0) = 1 $$
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$$  \displaystyle y'(0) = 0 $$
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Part 3 Problem Statement
Find the overall solution $$y(x)$$ and plot the solution over 3 periods

Part 3 Solution
For the overall solution:
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$$  \displaystyle y(x)=y_p(x)+y_h(x) $$     (Eq.5.3.1) To find the homogeneous solution, solve:
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$$  \displaystyle y''-3y'-10y=0 $$     (Eq.5.3.2) The characteristic equation of Eq.5.3.2 is:
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$$  \displaystyle \lambda^2-3\lambda-10=0 $$     (Eq.5.3.3) To solve the characteristic equation, apply the quadratic formula:
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$$  \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (Eq.5.3.4)
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$$  \displaystyle \lambda_{1,2}=\frac{-(-3)\pm\sqrt{(-3)^2-4(-10)}}{2} $$     (Eq.3.3.5)
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$$  \displaystyle \lambda_1=5 $$     (Eq.5.3.6)
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$$  \displaystyle \lambda_2=-2 $$     (Eq.5.3.7) Therefore the homogeneous solution takes the following form:
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$$  \displaystyle y_h(x)=c_1e^{5x}+c_2e^{-2x} $$     (Eq.5.3.8) Taking the first derivative of Eq.5.3.8:
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$$  \displaystyle y'_h(x)=5c_1e^{5x}-2c_2e^{-2x} $$     (Eq.5.3.9) Apply the boundary conditions $$y(0)=1$$ and $$y'(0)=0$$ to Eq.5.3.8 and Eq.5.3.9:
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$$  \displaystyle 1=c_1+c_2 $$     (Eq.5.3.10)
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$$  \displaystyle 0=5c_1-2c_2 $$     (Eq.5.3.11) Solving Eq.5.3.10 and Eq.5.3.11:
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$$  \displaystyle c_1=\frac{2}{7} $$     (Eq.5.3.12)
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$$  \displaystyle c_2=\frac{5}{7} $$     (Eq.5.3.13) The homogeneous solution becomes:
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$$  \displaystyle y_h(x)=\frac{2}{7}e^{5x}-\frac{5}{7}e^{-2x} $$     (Eq.5.3.14) Therefore from Eq.5.3.1 the overall solution is:
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$$  \displaystyle y(x)=y_p(x)+y_h(x)=-0.045\cos7x-0.016\sin7x+\frac{2}{7}e^{5x}+\frac{5}{7}e^{-2x} $$     (Eq.5.3.15) Graph
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Matlab Code

Author & Proofreaders
Author:Egm4313.s12.team17.wheeler.tw 05:10, 23 February 2012 (UTC)

Proofreader:Egm4313.s12.team17.deaver.md 09:25, 27 February 2012 (UTC)

Editor:Egm4313.s12.team17.ying 06:25, 30 February 2012 (UTC)

=Problem R5.6 - Solving for the Unknown Coefficients=

Given

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$$  \displaystyle y_p=xe^{-2x}(Mcos(3x)+Nsin(3x)) $$     (Eq.6.1)
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$$  \displaystyle y_h=e^{-2x}(Acos(3x)+Bsin(3x)) $$     (Eq.6.2)
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$$  \displaystyle y''+4y'+13y=2e^{-2x}cos(3x) $$     (Eq.6.3) Initial conditions
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$$  \displaystyle y(0) = 1 $$
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$$  \displaystyle y'(0) = 0 $$
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Problem Statement
Determine the overall solution $$y(x)$$ that corresponds to the initial conditions.

Plot the general solution over 3 periods.

Solution
Combining the given homogeneous and particular solutions gives us the overall form:
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$$  \displaystyle y(x)=xe^{-2x}(Mcos(3x)+Nsin(3x))+ e^{-2x}(Acos(3x)+Bsin(3x)) $$     (Eq.6.4) Applying the initial condition $$y(0)=1$$ to Eq.6.4 gives us:
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$$  \displaystyle y(0)=1=0e^{-2(0)}(Mcos(3(0))+Nsin(3(0)))+ e^{-2(0)}(Acos(3(0))+Bsin(3(0))) $$     (Eq.6.5) Which is simplified below:
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$$  \displaystyle 1=Acos(0)+Bsin(0) $$     (Eq.6.6)
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$$  \displaystyle 1=Acos(0)+Bsin(0) $$     (Eq.6.7)
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$$  \displaystyle 1=A $$     (Eq.6.8) Taking the derivative of Eq.6.4 gives us
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$$  \displaystyle y'(x)=e^{-2x}(Mcos(3x)+Nsin(3x))-2xe^{-2x}(Mcos(3x)+Nsin(3x))-3xe^{-2x}(Msin(3x)) $$ $$  \displaystyle +3xe^{-2x}(Ncos(3x))-2e^{-2x}(Acos(3x)+Bsin(3x))-3e^{-2x}(Asin(3x))+3e^{-2x}(Bcos(3x)) $$     (Eq.6.9) Inserting the initial condition y'(0)=0 into Eq.6.9 we get:
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$$  \displaystyle y'(0)=0=e^{-2(0)}(Mcos(3(0))+Nsin(3(0)))-2(0)e^{-2(0)}(Mcos(3(0))+Nsin(3(0)))-3(0)e^{-2(0)}(Msin(3(0))) $$ $$  \displaystyle +3(0)e^{-2(0)}(Ncos(3(0)))-2e^{-2(0)}(Acos(3(0))+Bsin(3(0)))-3e^{-2(0)}(Asin(3(0)))+3e^{-2(0)}(Bcos(3(0))) $$     (Eq.6.10) Which is further simplified to:
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$$  \displaystyle 0=(M)-2(A)+3(B) $$     (Eq.6.11) Since $$A=1$$:
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$$  \displaystyle 0=(M)-2(1)+3(B) $$     (Eq.6.12) We then get the relation:
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$$  \displaystyle 2=(M)+3(B) $$     (Eq.6.13) Taking the derivative of Eq.6.9 gives us.
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$$  \displaystyle y''(x)=-2e^{-2x}(Mcos(3x)+Nsin(3x))-e^{-2x}(Msin(3x))+e^{-2x}(Ncos(3x)) $$ $$  \displaystyle -2e^{-2x}(Mcos(3x)+Nsin(3x))+4xe^{-2x}(Mcos(3x)+Nsin(3x))+6xe^{-2x}(Msin(3x))-6xe^{-2x}(Ncos(3x)) $$ $$  \displaystyle -3e^{-2x}(Msin(3x))+6xe^{-2x}(Msin(3x))-9xe^{-2x}(Mcos(3x))+3e^{-2x}(Ncos(3x))-6xe^{-2x}(Ncos(3x))-9xe^{-2x}(Nsin(3x)) $$ $$  \displaystyle +4e^{-2x}(Acos(3x)+Bsin(3x))+6e^{-2x}(Asin(3x))-6e^{-2x}(Bcos(3x)) $$ $$  \displaystyle +6e^{-2x}(Asin(3x))-9e^{-2x}(Acos(3x))-6e^{-2x}(Bcos(3x))-9e^{-2x}(Bsin(3x)) $$      (Eq.6.14)
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We then insert Eq.6.3, Eq.6.9, and Eq.6.14 into the equation $$y''+4y'+13y=2e^{-2x}cos(3x)$$ :
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$$  \displaystyle -4e^{-2x}(Mcos(3x)+Nsin(3x))-4e^{-2x}(Msin(3x))-2e^{-2x}(Ncos(3x)) $$ $$  \displaystyle +4xe^{-2x}(Mcos(3x)+Nsin(3x))+12xe^{-2x}(Msin(3x)) $$ $$  \displaystyle -9xe^{-2x}(Mcos(3x))-6xe^{-2x}(Ncos(3x))-9xe^{-2x}(Nsin(3x)) $$ $$  \displaystyle -2e^{-2x}(Acos(3x)+Bsin(3x))+4e^{-2x}(Acos(3x)+Bsin(3x))+6e^{-2x}(Asin(3x))-6e^{-2x}(Bcos(3x)) $$ $$  \displaystyle +6e^{-2x}(Asin(3x))-9e^{-2x}(Acos(3x))-6e^{-2x}(Bcos(3x))-9e^{-2x}(Bsin(3x)) $$ $$  \displaystyle +4(e^{-2x}(Mcos(3x)+Nsin(3x))-2xe^{-2x}(Mcos(3x)+Nsin(3x))-3xe^{-2x}(Msin(3x)) $$ $$  \displaystyle    +3xe^{-2x}(Ncos(3x))-2e^{-2x}(Acos(3x)+Bsin(3x))-3e^{-2x}(Asin(3x))+3e^{-2x}(Bcos(3x))) $$ $$  \displaystyle +13(xe^{-2x}(Mcos(3x)+Nsin(3x))+ e^{-2x}(Acos(3x)+Bsin(3x)))=2e^{-2x}cos(3x) $$     (Eq.6.15)
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 *

Which is then simplified:
 * }
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$$  \displaystyle -6e^{-2x}(Msin(3x))+6e^{-2x}(Ncos(3x))=2e^{-2x}cos(3x) $$     (Eq.6.16)
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 *

Dividing Eq.6.16 by $$e^{-2x}$$ yields:
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$$  \displaystyle -6(Msin(3x))+6(Ncos(3x))=2cos(3x) $$     (Eq.6.17)
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 *

We then split Eq.6.17 into a system of equations to determine the value of the coefficients:
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$$  \displaystyle -6(Msin(3x))=0sin(3x) $$     (Eq.6.18)
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$$  \displaystyle 6(Ncos(3x))=2cos(3x) $$     (Eq.6.19)
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 * }

From Eq.6.18 and Eq.6.19 we can deduce that $$M=0$$ and $$N=\frac{1}{3}$$. Inserting the value of M into Eq.6.13 yields:
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$$  \displaystyle 2=(0)+3(B) $$     (Eq.6.20)
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$$  \displaystyle \frac{2}{3}=(B) $$     (Eq.6.21)
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Inserting the determined values for A,B,M, and N from Eq.6.8, Eq.6.18, Eq.6.21, and Eq.6.19 we can deduce that the overall solution is:
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$$  \displaystyle y(x)=xe^{-2x}(0cos(3x)+\frac{1}{2}sin(3x))+ e^{-2x}(1cos(3x)+\frac{2}{3}sin(3x)) $$     (Eq.6.22)
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 *

Making the overall solution:
 * }
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$$  \displaystyle y(x)=xe^{-2x}(\frac{1}{3}sin(3x))+ e^{-2x}(1cos(3x)+\frac{2}{3}sin(3x)) $$     (Eq.6.23) Below is a plot of the solution: Graph
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Matlab Code

Author & Proofreaders
Author:Egm4313.s12.team17.axelrod.a 04:46, 30 March 2012 (UTC)

Proofreader:Egm4313.s12.team17.deaver.md 07:05, 30 March 2012 (UTC)

Editor:Egm4313.s12.team17.ying 07:14, 30 March 2012 (UTC)

=Problem R5.7 - Projection on a Basis for Vectors=

Given

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$$  \displaystyle V = 4e_1 + 2e_2 = c_1 b_1 + c_2 b_2 $$     (Eq.7.1)
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$$  \displaystyle b_1 = 2e_1 + 7e_2 $$     (Eq.7.2)
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$$  \displaystyle b_2 = 1.5e_1 + 3e_2 $$     (Eq.7.3)
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Part 1 Problem Statement
Find $$ c_1,c_2 $$ using the Gram Matrix

Part 1 Solution
The Grammian is given as:

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$$  \displaystyle \Gamma (b_1,b_2) = \begin{bmatrix} \left \langle b_1,b_1 \right \rangle \left \langle b_1,b_2 \right \rangle \\ \left \langle b_2,b_1 \right \rangle \left \langle b_2,b_2 \right \rangle \end{bmatrix} $$     (Eq.7.1.1)
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According to (3) on p.8-9

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$$  \displaystyle \left \langle b_1,b_1 \right \rangle = (b_1 \cdot b_2) $$     (Eq.7.1.2)
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Calculating the dot products:

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$$  \displaystyle \left \langle b_1,b_1 \right \rangle = 4 + 49 = 53 $$
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$$  \displaystyle \left \langle b_1,b_2 \right \rangle = 3 + 21 = 24 $$

$$  \displaystyle \left \langle b_2,b_1 \right \rangle = 3 + 21 = 24 $$

$$  \displaystyle \left \langle b_2,b_2 \right \rangle = 2.25 + 9 = 11.25 $$
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(Eq.7.1.3)
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Plugging Equations 7.1.3 into 7.1.1 gives:

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$$  \displaystyle
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\Gamma = \begin{bmatrix} 53 & 24\\ 24 & 11.25\\ \end{bmatrix} $$
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(Eq.7.1.4)
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Next we have to get an equation for d, as shown in (5) on pg.8-10 *notice v is given under heading of problem 5.7

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$$  \displaystyle
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d = \{\left \langle b_1,v \right \rangle, \left \langle b_2,v \right \rangle\}^T

$$
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(Eq.7.1.5)
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Evaluating as equation 7.1.2 shows gives:

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$$  \displaystyle
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d = \{22,12\}^T

$$
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(Eq.7.1.6)
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From the notes (1) on pg.8-11

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$$  \displaystyle
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c = \Gamma ^{-1}d $$
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(Eq.7.1.6)
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This gives:

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$$  \displaystyle
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c = \bigg\{\frac{11.25(22)-24(12)}{20.25},\frac{-24(22)+53(12)}{20.25}\bigg\} $$
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(Eq.7.1.7)
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So $$ c_1 $$ and $$ c_2 $$ are:

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$$  \displaystyle
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c_1 = -2 $$

$$  \displaystyle

c_2 = 5.33 $$
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(Eq.7.1.8)
 * }

Part 2 Problem Statement
(2)Verify the result by using (1)-(2) p.7c-34 in (2)and rely on the non-zero determinant of the matrix of components of $$ b_1, b_2 $$ relative to the basis $$ e_1,e_2$$ as discussed on p.7c-34

Part 2 Solution
The equations to be entered into the A matrix are as follows:

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$$  \displaystyle b_1 = 2 e_1 + 7 e_2 $$
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$$  \displaystyle b_2 = 1.5 e_1 + 3 e_2 $$     (Eq.7.2.1)
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 * }

According to the notes if these two equations are put into a matrix and the determinant is $$ \ne 0 $$ then they are linearly independent. Entering them into a matrix gives:

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$$  \displaystyle A = \begin{bmatrix} 2 & 7 \\ 1.5 & 3\\ \end{bmatrix} $$     (Eq.7.2.2)
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Solving:

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$$  \displaystyle 6 - 10.5 = -4.5 \ne 0 $$     (Eq.7.2.3)
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Thus $$ b_1 $$ and $$ b_2 $$ are linearly independent.

Use the following equation to check the answer from the previous part.

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$$  \displaystyle V=4e_1+2e_2\equiv c_1b_1+c_2b_2 $$     (Eq.7.2.4)
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$$  \displaystyle V\equiv (-2)(2e_1+7e_2)+(5.33)(1.5e_1+3e_2) $$     (Eq.7.2.4)
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$$  \displaystyle V\equiv 4e_1+2e_2 $$     (Eq.7.2.5)
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Author & Proofreaders
Author:Egm4313.s12.team17.hintz 23:46, 24 March 2012 (UTC)

Proofreader:Egm4313.s12.team17.axelrod.a 14:35, 29 March 2012 (UTC)

Editor:Egm4313.s12.team17.ying 07:51, 30 March 2012 (UTC)

=Problem R5.8 - Integration=

Given

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$$  \displaystyle \int x^nlog(1+x)dx $$     (Eq. 8.1) With
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$$  \displaystyle n = 0,1 $$ Integration by Parts
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$$  \displaystyle \int udv= uv- \int vdu $$     (Eq.8.2) General Binomial Theorem
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$$  \displaystyle (x+y)^n=\sum_{k=0}^n \binom {n}{k}x^{n-k}y^k $$     (Eq.8.3) Where
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$$  \displaystyle \binom {n}{k}x^{n-k}y^k=\dfrac {n!}{k!(n-k)!}=\dfrac {n(n-1) \cdots (n-k+1)}{k!} $$
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Problem Statement
Find the indefinite integral below with $$ n=0,1$$ using the integration by parts and the General Binomial Theorem

Solution
Using the integration by parts formula, Eq.8.2, we set $$ u=log(1+x)$$ and $$ dv=x^n$$ so we have the following:
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$$  \displaystyle u=log(1+x) $$
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$$  \displaystyle du = \dfrac{1}{x+1}dx $$
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$$  \displaystyle dv=x^n $$
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$$  \displaystyle v=\dfrac {x^{n+1}}{n+1} $$
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Substituting the above in to Eq. 8.3 we have:
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$$  \displaystyle \int x^nlog(1+x)dx= \dfrac{log(1+x)x^{n+1}}{n+1}-\int \dfrac {x^{n+1}}{(n+1)(1+x)}dx $$     (Eq.8.4) Now we can set $$n=0$$ and Eq. 8.4 becomes:
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$$  \displaystyle \int x^nlog(1+x)dx= \dfrac{log(1+x)x^{1}}{1}-\int \dfrac {x^{1}}{(1)(1+x)}dx $$     (Eq.8.5) Where
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$$  \displaystyle \int \dfrac {x^{1}}{(1)(1+x)}dx $$ Can be solved by first using long division on the integrand:
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$$  \displaystyle \int \dfrac {x^{1}}{(1)(1+x)}dx= \int (1-\dfrac {1}{1+x})dx $$ Next integrate term by term
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$$  \displaystyle \int 1dx-\int \dfrac {1}{1+x}dx = x-log(1+x) $$ So when $$ n=0$$ we have:
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$$  \displaystyle \int x^nlog(1+x)dx= log(1+x)x-x+log(x+1) $$     (Eq.8.6)
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Now we substitute $$n=1$$ into Eq. 8.4 which gives us:
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$$  \displaystyle \int x^nlog(1+x)dx= \dfrac{log(1+x)x^{1+1}}{1+1}-\int \dfrac {x^{1+1}}{(1+1)(1+x)}dx=\dfrac{log(1+x)x^{2}}{2}-\int \dfrac {x^{2}}{(2)(1+x)}dx $$     (Eq.8.7) And we have the integral:
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$$  \displaystyle \int \dfrac {x^{2}}{(2)(1+x)}dx $$ Which can also be solved by first using long division on the integrand $$\dfrac {x^2}{1+x}$$, so:
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$$  \displaystyle \int \dfrac {x^{2}}{(2)(1+x)}dx= \dfrac{1}{2} \int \dfrac {x^2}{1+x}dx= \dfrac{1}{2} \int(x+\dfrac{1}{1+x}-1)dx $$ Integrating term by term we have that:
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$$  \displaystyle \dfrac{1}{2} \int(x+\dfrac{1}{1+x}-1)dx=\dfrac{x^2}{4}-\dfrac{x}{2}+\dfrac{1}{2}log(1+x)=\dfrac{1}{4}((x-2)x+2log(1+x)) $$ So Eq. 8.7 becomes:
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$$  \displaystyle \int x^nlog(1+x)dx=\dfrac{log(1+x)x^{2}}{2}-\int \dfrac {x^{2}}{(2)(1+x)}dx=\dfrac{log(1+x)x^{2}}{2}-(\dfrac{1}{4}((x-2)x+2log(1+x))) $$     (Eq.8.8) So when $$n=1$$:
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$$  \displaystyle \int x^nlog(1+x)dx=\dfrac{\log(1+x)x^{2}}{2}-\dfrac{1}{4}[(x-2)x+2log(1+x)] $$     (Eq.8.9)
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Author & Proofreaders
Author:Egm4313.s12.team17.Li 23:47, 23 March 2012 (UTC)

Proofreader 1:Egm4313.s12.team17.axelrod.a 20:45, 25 March 2012 (UTC)

Proofreader 2:Egm4313.s12.team17.ying 08:22, 30 March 2012 (UTC)

Editor:Egm4313.s12.team17.ying 08:22, 30 March 2012 (UTC)

=Problem R5.9 - Projection on a Basis=

Part 1 Given

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$$  \displaystyle y'' - 3y' + 2y = r(x) $$     (Eq.9.1.1)
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$$  \displaystyle r(x) = log(1+x) $$     (Eq.9.1.2)
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$$  \displaystyle y(-3/4) = 1, y'(-3/4) = 0 $$     (Eq.9.1.3)
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$$  \displaystyle r_n(x) = \log(1+x) = \sum_{n=0}^\infty \frac{x^{n+1}}{n+1}(-1)^{n} $$     (Eq.9.1.4)
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$$  \displaystyle \mathbf{b}=\{b_j(x)=x^j,j=0,1,...,n\} $$     (Eq.9.1.5)
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Part 1 Problem Statement
Project equation 9.1.2 on the polynomial basis such that it takes the form shown below for and n=0,1,2 \,:
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$$  \displaystyle r(x)\approx r_n(x)=\sum_{j=0}^n d_jx^j $$     (Eq.9.1.6) Plot equations 9.1.2 and 9.1.6 to show the uniform approximation and convergence.
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Then, plot equations 9.1.4 and 9.1.6, in order, to compare the pros and cons of the projection on polynomial and Taylor series expansion methods for approximating equation 9.1.2.

Part 1 Solution
Let n=0 \,.

Therefore, equation 9.1.5 becomes:
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$$  \displaystyle \mathbf{b}=\{b_0\}=\{x^0\} $$     (Eq.9.1.7)
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Set up a Gram matrix for the basis functions (eq. 9.1.7):
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$$  \displaystyle \mathbf{\Gamma}(\mathbf{b})=\{\langle b_0,b_0 \rangle\} $$     (Eq.9.1.8)
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$$  \displaystyle \langle b_0,b_0 \rangle = \int\limits_{a}^{b}b_0b_0dx $$     (Eq.9.1.9)
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Let a=-3/4 \, and b=3 \,.
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$$  \displaystyle \langle b_0,b_0 \rangle = \int\limits_{-\frac{3}{4}}^{3}x^0(x^0)dx $$     (Eq.9.1.10)
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$$  \displaystyle \langle b_0,b_0 \rangle = \frac{15}{4} = 3.75 $$     (Eq.9.1.11)
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Therefore,
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$$  \displaystyle \mathbf{\Gamma}=\{3.75\} $$     (Eq.9.1.12)
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Rhs matrix will take the form shown below:
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$$  \displaystyle \mathbf{e}=\{\langle b_0,r(x) \rangle, ...,\langle b_n,r(x) \rangle\}^T $$     (Eq.9.1.13)
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Therefore,
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$$  \displaystyle \mathbf{e}=\{\langle b_0,r(x) \rangle\}^T $$     (Eq.9.1.14)
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$$  \displaystyle \langle b_0,r(x) \rangle = \int\limits_{a}^{b}b_0r(x)dx $$     (Eq.9.1.15)
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$$  \displaystyle \langle b_0,r(x) \rangle = \int\limits_{-\frac{3}{4}}^{3}x^0(log(1+x))dx $$     (Eq.9.1.16)
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$$  \displaystyle \langle b_0,r(x) \rangle = \bigg[(x+1)log(x+1)-x\bigg]_{-\frac{3}{4}}^3 $$     (Eq.9.1.17)
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$$  \displaystyle \langle b_0,r(x) \rangle = \frac{17}{2}log(2)-\frac{15}{4} $$     (Eq.9.1.18)
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$$  \displaystyle \langle b_0,r(x) \rangle = 2.1418 $$     (Eq.9.1.19)
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$$  \displaystyle \mathbf{e}=\{2.1418\}^T $$     (Eq.9.1.20)
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Using the equation shown below, solve for the d values:
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$$  \displaystyle \mathbf{\Gamma}\mathbf{d}=\mathbf{e} $$     (Eq.9.1.21)
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 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{d}=\mathbf{\Gamma}^{-1}\mathbf{e} $$     (Eq.9.1.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \mathbf{d}=\{3.75\}^{-1}\{2.1418\}^T $$     (Eq.9.1.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \mathbf{d}=0.5711 $$     (Eq.9.1.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This result will make equation 9.4.6 the following:
 * {| style="width:100%" border="0"

$$  \displaystyle r_0(x)=\sum^0_{j=0}d_jx_j= d_0x^0 $$     (Eq.9.1.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle r_0(x)=0.5711 $$     (Eq.9.1.26)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Let n=1 \,.

Therefore, equation 9.1.5 becomes:
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{b}=\{b_0,b_1\}=\{x^0,x^1\} $$     (Eq.9.1.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Set up a Gram matrix for the basis functions (eq. 9.1.27):
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{\Gamma}(\mathbf{b})= \begin{bmatrix} \langle b_0,b_0 \rangle & \langle b_0,b_1 \rangle \\ \langle b_1,b_0 \rangle & \langle b_1,b_1 \rangle \\ \end{bmatrix} $$     (Eq.9.1.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_0,b_1 \rangle = \langle b_1,b_0 \rangle = \int\limits_{-\frac{3}{4}}^{3}x^0(x^1)dx $$     (Eq.9.1.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_0,b_1 \rangle = \langle b_1,b_0 \rangle = \frac{1}{2}\bigg[x^2\bigg]^3_{-\frac{3}{4}} $$     (Eq.9.1.30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_0,b_1 \rangle = \langle b_1,b_0 \rangle = \frac{135}{32} = 4.2188 $$     (Eq.9.1.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,b_1 \rangle = \int\limits_{-\frac{3}{4}}^{3}x^1(x^1)dx $$     (Eq.9.1.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,b_1 \rangle = \frac{1}{3}\bigg[x^3\bigg]^3_{-\frac{3}{4}} $$     (Eq.9.1.33)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,b_1 \rangle = \frac{585}{64} = 9.1406 $$     (Eq.9.1.34)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \mathbf{\Gamma}= \begin{bmatrix} 3.7500 & 4.2188 \\ 4.2188 & 9.1406 \\ \end{bmatrix} $$     (Eq.9.1.35)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Generate a rhs matrix:
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{e}=\{\langle b_0,r(x) \rangle\,\langle b_1,r(x) \rangle\}^T $$     (Eq.9.1.36)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,r(x) \rangle = \int\limits_{a}^{b}b_1r(x)dx $$     (Eq.9.1.37)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,r(x) \rangle = \int\limits_{-\frac{3}{4}}^{3}x^1(log(1+x))dx $$     (Eq.9.1.38)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,r(x) \rangle = \bigg[\bigg(\frac{x^2}{2}-\frac{1}{2}\bigg)log(x+1)-\frac{x(x-2)}{4}\bigg]_{-\frac{3}{4}}^3 $$     (Eq.9.1.39)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,r(x) \rangle = \frac{121}{16}log(2)-\frac{15}{64} $$     (Eq.9.1.40)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \langle b_1,r(x) \rangle = 5.0076 $$     (Eq.9.1.41)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \mathbf{e}=\{2.1418,5.0076\}^T $$     (Eq.9.1.42)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the equation 9.1.21, to solve for the d values:
 * {| style="width:100%" border="0"

$$  \displaystyle \mathbf{d}=\frac{1}{det\mathbf{\Gamma}} \begin{bmatrix} 9.1406 & -4.2188 \\ -4.2188 & 3.7500 \\ \end{bmatrix} \begin{bmatrix} 2.1418 \\ 5.0076 \\ \end{bmatrix} $$     (Eq.9.1.43)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \mathbf{d}=\frac{1}{16.479} \begin{bmatrix} 9.1406 & -4.2188 \\ -4.2188 & 3.7500 \\ \end{bmatrix} \begin{bmatrix} 2.1418 \\ 5.0076 \\ \end{bmatrix} $$     (Eq.9.1.44)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \mathbf{d}= \begin{bmatrix} -0.0940 \\ 0.5912 \\ \end{bmatrix} $$     (Eq.9.1.45)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

This result will make equation 9.4.6 the following:
 * {| style="width:100%" border="0"

$$  \displaystyle r_1(x)=\sum^1_{j=0}d_jx_j= d_0x^0 + d_1x^1 $$     (Eq.9.1.46)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle r_1(x)=-0.0940+0.5912x $$     (Eq.9.1.47)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

The plot below shows the projected and actual excitation.

Below is the MATLAB code used to generate the above graph:

Taylor series expansion for equation 9.1.4 are shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle t_0=x $$     (Eq.9.1.48)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle t_1=x-\frac{x^2}{2} $$     (Eq.9.1.49)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The plots below show the comparison of the two methods for approximating equation 9.1.2.

Below is the MATLAB code used to generate the above graph:

The Taylor series expansion is great for approximating a complex function around a specific point. Once, the Taylor series gets farther away from the centered point, it will begin to become inaccurate compared to the true value of the actual function. The Taylor series will require more terms to reduce the error for a larger x range.

Projection on polynomial basis will generate an approximation that evenly distributes the error for a given x range. In order to get a high accurate approximation with this method, you will have to increase the number of terms used which leads to more unknowns. More unknowns means that it will require more time to solve and that will result in higher cost when money is involved.

Part 2 Given
Information from part 1.

Part 2 Problem Statement
Find a y_n(x) \, that solves the following equation for n=0,1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y_n''+ay_n'+by_n= r_n(x) $$     (Eq.9.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Then plot the solutions, truncated Taylor series, and numerical solution.

Part 2 Solution
For n=0 \,.
 * {| style="width:100%" border="0"

$$  \displaystyle y_0''-3y_0'+2y_0= r_0(x) $$     (Eq.9.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_0=y_{0,h}+y_{0,p} $$     (Eq.9.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The homogenous solution must satisfy the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{0,h}''-3y_{0,h}'+2y_{0,h}= 0 $$     (Eq.9.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda^2-3\lambda+2= 0 $$     (Eq.9.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (\lambda-2)(\lambda-1)=0 $$     (Eq.9.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle \lambda_1=2,\lambda_2=1 $$     (Eq.9.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Distinct real roots makes the general solution of the homogenous take the following form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h=d_1e^{\lambda_1 x}+d_2e^{\lambda_2 x} $$ (Eq.9.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{0,h}=d_1e^{2 x}+d_2e^{ x} $$ (Eq.9.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution must satisfy the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{0,p}''-3y_{0,p}'+2y_{0,p}= 0.5711 $$     (Eq.9.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using Method of Undetermined Coefficients, the particular solution will have the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p=\sum^n_{j=0}c_jx^j $$     (Eq.9.2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{0,p}=c_0x^0 $$     (Eq.9.2.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first and second derivative of equation 9.2.12:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{0,p}^\prime=0 $$     (Eq.9.2.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{0,p}^{\prime\prime}=0 $$     (Eq.9.2.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute equations 9.2.12, 9.2.13, and 9.2.14 into equation 9.2.10:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_0= 0.5711 $$     (Eq.9.2.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_0= 0.2856 $$     (Eq.9.2.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution will be:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{0,p}=0.2856 $$     (Eq.9.2.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Superimpose equation 9.2.9 and 9.2.17:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{0}=d_1e^{2 x}+d_2e^{ x}+0.2856 $$     (Eq.9.2.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first of equation 9.2.18:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{0}^\prime=2d_1e^{2 x}+d_2e^{ x} $$ (Eq.9.2.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Use equation 9.1.3 to solve for the unknowns:
 * {| style="width:100%" border="0"

$$  \displaystyle 1=d_1e^{-1.5}+d_2e^{-0.75}+0.2856 $$     (Eq.9.2.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=2d_1e^{-1.5}+d_2e^{-0.75} $$     (Eq.9.2.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_1=-3.2019 $$     (Eq.9.2.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_2=3.0249 $$     (Eq.9.2.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution for n=0 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{0}=-3.2019e^{2 x}+3.0249e^{ x}+0.2856 $$     (Eq.9.2.24)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$  \displaystyle y_1''-3y_1'+2y_1= r_1(x) $$     (Eq.9.2.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_1=y_{1,h}+y_{1,p} $$     (Eq.9.2.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The homogenous solution will be the same has equation 9.2.9:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1,h}=d_1e^{2 x}+d_2e^{ x} $$ (Eq.9.2.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution must satisfy the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1,p}''-3y_{1,p}'+2y_{1,p}= 0.5912x-0.0940 $$     (Eq.9.2.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using Method of Undetermined Coefficients, the particular solution will take the form of equation 9.2.11:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1,p}=c_0x^0+c_1x^1 $$     (Eq.9.2.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first and second derivative of equation 9.2.29:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1,p}^\prime=c_1 $$     (Eq.9.2.30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{1,p}^{\prime\prime}=0 $$     (Eq.9.2.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute equations 9.2.29, 9.2.30, and 9.2.31 into equation 9.2.28:
 * {| style="width:100%" border="0"

$$  \displaystyle -3c_1+2c_0+2c_1x=0.5912x-0.0940 $$     (Eq.9.2.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solve the coefficients by setting like terms equal to one another.

For x^0 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle -3c_1+2c_0=-0.0940 $$     (Eq.9.2.33)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_1=0.5912 $$     (Eq.9.2.34)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_1=0.2956 $$     (Eq.9.2.35)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_0=0.2956 $$     (Eq.9.2.36)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution will be:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1,p}=0.2956x+0.3964 $$     (Eq.9.2.37)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Superimpose equation 9.2.27 and 9.2.37:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1}=d_1e^{2 x}+d_2e^{ x}+0.2956x+0.3964 $$     (Eq.9.2.38)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first of equation 9.2.38:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1}^\prime=2d_1e^{2 x}+d_2e^{ x}+0.2956 $$     (Eq.9.2.39)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Use equation 9.1.3 to solve for the unknowns:
 * {| style="width:100%" border="0"

$$  \displaystyle 1=d_1e^{-1.5}+d_2e^{-0.75}-0.2217+0.3964 $$     (Eq.9.2.40)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=2d_1e^{-1.5}+d_2e^{-0.75}+0.2956 $$     (Eq.9.2.41)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_1=-5.0235 $$     (Eq.9.2.42)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_2=4.1201 $$     (Eq.9.2.43)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution for n=1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{1}=-5.0235e^{2 x}+4.1201e^{ x}+0.2956x+0.3964 $$     (Eq.9.2.44)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For Taylor series expansion for n=0 \,.
 * {| style="width:100%" border="0"

$$  \displaystyle y_{t_0}''-3y_{t_0}'+2y_{t_0}= r_{t_0}(x) $$     (Eq.9.2.45)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{t_0}=y_{{t_0},h}+y_{{t_0},p} $$     (Eq.9.2.46)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The homogenous solution will be the same has equation 9.2.9:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_0},h}=d_1e^{2 x}+d_2e^{ x} $$ (Eq.9.2.47)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution must satisfy the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_0},p}''-3y_{{t_0},p}'+2y_{{t_0},p}=x $$     (Eq.9.2.48)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using Method of Undetermined Coefficients, the particular solution will take the form of equation 9.2.11:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_0},p}=c_0x^0+c_1x^1 $$     (Eq.9.2.49)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first and second derivative of equation 9.2.49:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_0},p}^\prime=c_1 $$     (Eq.9.2.50)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{{t_0},p}^{\prime\prime}=0 $$     (Eq.9.2.51)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute equations 9.2.49, 9.2.50, and 9.2.51 into equation 9.2.48:
 * {| style="width:100%" border="0"

$$  \displaystyle -3c_1+2c_0+2c_1x=x $$     (Eq.9.2.52)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solve the coefficients by setting like terms equal to one another.

For x^0 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle -3c_1+2c_0=0 $$     (Eq.9.2.53)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_1=1 $$     (Eq.9.2.54)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_1=0.5 $$     (Eq.9.2.55)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_0=0.75 $$     (Eq.9.2.56)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution will be:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_0},p}=0.5x+0.75 $$     (Eq.9.2.57)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Superimpose equation 9.2.47 and 9.2.57:
 * {| style="width:100%" border="0"

$$  \displaystyle y_=d_1e^{2 x}+d_2e^{ x}+0.5x+0.75 $$     (Eq.9.2.58)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first of equation 9.2.58:
 * {| style="width:100%" border="0"

$$  \displaystyle y_^\prime=2d_1e^{2 x}+d_2e^{ x}+0.5 $$     (Eq.9.2.59)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Use equation 9.1.3 to solve for the unknowns:
 * {| style="width:100%" border="0"

$$  \displaystyle 1=d_1e^{-1.5}+d_2e^{-0.75}-0.375+0.75 $$     (Eq.9.2.60)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=2d_1e^{-1.5}+d_2e^{-0.75}+0.5 $$     (Eq.9.2.61)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_1=-5.0419 $$     (Eq.9.2.62)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_2=3.7048 $$     (Eq.9.2.63)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution for Taylor series expansion with n=0 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y_=-5.0419e^{2 x}+3.7048e^{ x}+0.5x+0.75 $$     (Eq.9.2.64)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

For Taylor series expansion for n=1 \,.
 * {| style="width:100%" border="0"

$$  \displaystyle y_{t_1}''-3y_{t_1}'+2y_{t_1}= r_{t_1}(x) $$     (Eq.9.2.65)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{t_1}=y_{{t_1},h}+y_{{t_1},p} $$     (Eq.9.2.66)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The homogenous solution will be the same has equation 9.2.9:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_1},h}=d_1e^{2 x}+d_2e^{ x} $$ (Eq.9.2.67)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution must satisfy the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_1},p}''-3y_{{t_1},p}'+2y_{{t_1},p}=x-\frac{x^2}{2} $$     (Eq.9.2.68)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using Method of Undetermined Coefficients, the particular solution will take the form of equation 9.2.11:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_1},p}=c_0x^0+c_1x^1+c_2x^2 $$     (Eq.9.2.69)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first and second derivative of equation 9.2.69:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_1},p}^\prime=c_1+2c_2x $$     (Eq.9.2.70)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{{t_1},p}^{\prime\prime}=2c_2 $$     (Eq.9.2.71)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute equations 9.2.69, 9.2.70, and 9.2.71 into equation 9.2.68:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_2-3c_1-6c_2x+2c_0+2c_1x2c_2x^2=x-\frac{x^2}{2} $$     (Eq.9.2.72)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solve the coefficients by setting like terms equal to one another.

For x^0 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_2-3c_1+2c_0=0 $$     (Eq.9.2.73)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle -6c_2+2c_1=1 $$     (Eq.9.2.74)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^2 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_2=-\frac{1}{2} $$     (Eq.9.2.75)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_2=-0.25 $$     (Eq.9.2.76)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_1=-0.25 $$     (Eq.9.2.77)
 * style="width:95%" |
 * style="width:95%" |
 *

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

$$  \displaystyle c_0=-0.125 $$     (Eq.9.2.78)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The particular solution will be:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{{t_1},p}=-0.25x^2-0.25x-0.125 $$     (Eq.9.2.79)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Superimpose equation 9.2.67 and 9.2.79:
 * {| style="width:100%" border="0"

$$  \displaystyle y_=d_1e^{2 x}+d_2e^{ x}-0.25x^2-0.25x-0.125 $$     (Eq.9.2.80)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first of equation 9.2.80:
 * {| style="width:100%" border="0"

$$  \displaystyle y_^\prime=2d_1e^{2 x}+d_2e^{ x}-0.5x-0.25 $$     (Eq.9.2.81)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Use equation 9.1.3 to solve for the unknowns:
 * {| style="width:100%" border="0"

$$  \displaystyle 1=d_1e^{-1.5}+d_2e^{-0.75}-0.5625+0.1875+0.125 $$     (Eq.9.2.82)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=2d_1e^{-1.5}+d_2e^{-0.75}+0.375-0.25 $$     (Eq.9.2.83)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_1=-7.2827 $$     (Eq.9.2.84)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle d_2=6.6156 $$     (Eq.9.2.85)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution for Taylor series expansion with n=1 \,:
 * {| style="width:100%" border="0"

$$  \displaystyle y_=-7.2827e^{2 x}+6.6156e^{ x}-0.25x^2-0.5x-0.125 $$     (Eq.9.2.86)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Below is the MATLAB code used to create the ODE for equation 9.1.1 with excitation equation 9.1.2:

Below is the MATLAB code used to generate the above graph:

The solutions when n=0 \,:
 * The Taylor series solution is very close to the numerical solution, while the projection solution isn't that accurate.
 * The projection solution seems to start diverging at about x=0.5 \,.

The solutions when n=1 \,:
 * The projection solution is very close to the numerical solution, while the Taylor series solution isn't that accurate.
 * The Taylor series expansion function starts diverging after x=1 \,, since R_c=1 \,.

Conclusion:
 * As the number of terms increase, the Taylor series solution begins to take effects from the radius of divergence.
 * The projection solution becomes more accurate as the number of terms increase, since the solution is using the extra terms to mold into the numerical solution for a given x range.

Author & Proofreaders
Author: Egm4313.s12.team17.deaver.md 05:36, 30 March 2012 (UTC)

Proofreader: Egm4313.s12.team17.ying 09:41, 30 March 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 05:36, 30 March 2012 (UTC)

= Contributing Team Members =

title: User:Egm4313.s12.team17/Report 5


text:
User talk:Eshipley

read week 3 activities and next task Engsihn 3:43 PM, 15 March 2012.

 * Good job! Can you tell me how you uploaded your picture to Wikiversity?

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

title: User talk:Eshipley


text:
MakerBot/How to print STL files with letters or Numbers in it

Print STL Files with letters

 * 1. In ReplicatorG, go to the File drop down menu in the toolbar, select “Open.”
 * 2. In the “Open” window, go to “Desktop” and open the STL file you want to print.
 * 3. The picture of the file should appear on ReplicatorG. To view the model of the object, use the “View”, “Move”, and “Rotate” commands located on the right side of the ReplicatorG window.
 * 4 Go to mirror and Click on the "reflection over the x axis"
 * 5. Click “Generate GCode.”
 * 6. We made changes to the model so, a "Save model?" window will appear. Click yes.

 * 7. In the GCode Generator window, select SF35-cupcake-HBP as a Skeinforge profile. Uploading an STL File onto Replicator G Procedure Step 7.jpg
 * 8. Select “Use raft” check box if you want to create a raft. Uploading an STL File onto Replicator G Procedure Step 8.jpg
 * 9. Click Generate GCode at the bottom of the GCode Generator window. Uploading an STL File onto Replicator G Procedure Step 9.1.jpg
 * 10. Click on the gcode tab. In the ##th line that reads: M04 temp 220 (temperature 220) change the first number (220) to (180).


text:
Illumination/Howard Community College/Spring2012

Please delete everything in italics after filling it out.

''Project documentation focuses on how the team organized itself to tackle the unknown. This is a creative process. The details learned should be captured in tutorials. Project documentation should focus on how the team organized itself to step out into the unknown. Failure is common. The single most effective way to increase engineering respect is to document failures in a creative way that stimulates, inspires and helps others continue the project. Projects are never done. There are always loose ends. There are always decisions that might have been made more clearly. There are always other alternatives that might have been considered. None of this stuff appears in a tutorial. It does appear in the team documentation.''

Problem Statement
What is the final, negotiated problem statement?

Team Members
''Put each team members name here with links to their user pages here. Real names do not have to be used.''
 * Navid Jahanpour
 * Chris Nunez
 * Imane Badra Tate

Summary
Put an overall, short one paragraph summary here.

Poster
''Put a graphic in wikimedia, include the graphic here or link to it here. The graphic should be suitable for creating a traditional project poster.''

Story
''Tell a detailed story of the project. Describe how split up, what the obstacles were, what testing was done, what informal decisions were made, what assumptions were made, what the results were.''

Decision List
List all formal decisions made with links to their documentation such as a decision tree or decision matrix.

Material List

 * 1. List materials used, quantity, size, cost.
 * 2. Describe what needs to be purchased in the future to continue working on this project.

Software List
''You installed different software packages or used already installed software. Describ the programs here.''

Time
''Put one number here. Hours. Total hours that everyone worked on the project. Do not estimate. Do not talk about it. ADD.''

Tutorials
''All projects create new tutorials of technical details future participants are going to want to know. They are going to be separate pages that are linked to here.''

Next Steps
List specific details, advice, describe the current problems that the next team faces associated with the project here.


text:
Engineering Projects/Power projects/Howard Community College/Spring2012/p2550TK

Problem Statement
1. Repairing a ATX power supply, which is already converted to a LAB one

2. Converting more ATX power supplies to LAB ones

3. Converting other electrical boards and power supplies to LAB power supplies

Team Members

 * Keyvan
 * Javiereduturcios

Summary
We've got an ATX power supply which is already converted to a LAB power supply, but it does not work. We are going to figure out what is wrong with it, and repair it. After opening up the power and tracing voltage, we realized that repairing this stuff is so complicated and is beyond of our knowledge and abilities, so we gave it up. We converted two more ATX power supplies to LAB ones. Also, we made two more LAB power supplies with useless electrical boards which was a more challenging experience. We also converted a rectangular power supply to a LAB power supply.

Story
We've got an ATX power supply which is already converted to a LAB power supply, but it does not work. We are going to figure out what is wrong with it, and repair it. We opened it up, plugged it in and measured the output voltage from the points that the wires were connected to the board. we got same results. we checked to connection of the green and a black wire. It was okay. We started tracing input voltage from the power cable to switch and to a part of the power which we guess it was transformer. There was no problem. We figured out the issue is more complicated than we thought before. It needs more knowledge and skills for solving this issue, so we gave it up. We converted two more ATX power supplies to LAB ones. In more detailed, we took power supplies from junk computers. we opened them up and cut of their sockets and extra length of wires. we got one switch, so we used it in one of the power supplies. we connected the green wire and a black wire to the switch and soldered the connections, then we covered the connections and stripped wires by electrical tape. For the other power supply we just shorted the green and a black wires. Then, we connected a 5 ohm power resistor to a red and a black wires in order to discharge capacitors. After that, we connected all remaining same-color wires to each other; blacks, oranges, reds, whites and yellows, and the blue. Next, we drilled 5 holes and attached 5 binding posts to them. Then, we connected black wires to the black binding post; oranges to the next post; reds next to oranges; yellow and white next to reds; and blue to the last binding post. Next, we soldered the connections and covered by electrical tape. Then, we closed the power and labeled the binding posts. Finally, we tested the output voltages by a multimeter. Then, we made a circuit with the power supply, a 1.5 ohm power resistor, and a multimeter that was put on the current feature. We connected the 3.3 volt binding post to the resistor and the resistor to the red lead of the multimeter and its black lead to black binding post. We measured the current. we did same for other binding posts and made a test protocol for the power supply. we got the maximum 10 amps from the power supply.

Also, we converted a rectangular power supply to a LAB one. First, we opened it up and made its wires shorter. Then, We drilled holes for binding posts on its case. We cut off the gray and the purple wires because they were useless. Then, we shorted the green wire to a black wire to make the power run. Then, we gathered all remaining same color wires and connected them to each other.

Decision List

 * 1. Work with a square power supply
 * 2. Work with a rectangle power supply
 * 3. Designing a box for a board to become a power supply.

Material List

 * 1. ATX power supplies (can be taken from junk computers or purchased online)

 * 2. Multimeter

 * 3. Drill

 * 4. Wire stripper

 * 5. Binding posts (quantity depends on how many power supplies are planed to be converted),(each power supply needs at most 5)

 * 6. Drill

 * 7. Screwdrivers( Philips screwdrivers in different sizes)

 * 8. Electrical Tape

 * 9. Soldering Iron and Solder

 * 10. 5 ohms Power resistors( or any combinations of two power resistors which produce the same resistance)

 * 11. Computer Power Cable

 * 12. Switches (optional)

 * 13. 1.5 ohm(or if available 1 ohm) power resistor for measuring the maximum current produce by power supplies

 * 14. Useless electrical boards, which producing DC voltages, for converting to LAB power supplies

 * 15. A small carton or plastic box that fits to the electrical board

 * 16. Aluminum foil

 * 17. wide-clear tape

Software List
Not any specific software

Time
45 hours

Tutorials
Converting an ATX power supply to a LAB one

Another Tutorial for converting ATX power supplies to LAB ones
 * Testing is the power supply works
 * Testing if the power supply does not work
 * Tutorial to drill holes on the power supply shield
 * Tutorial on charging the drill's battery
 * Tutorial on removing the power supply from the computer
 * Tutorial on measuring the voltage with the multimeter
 * Tutorial on how to convert ATX Power Supply to lab power supply
 * Tutorial on how to solder
 * Tutorial on how to put the cables together
 * Tutorial on how to turn the power supply
 * Tutorial on reducing the size of the the purple wire safety
 * Tutorial on reducing the size of the gray wire in a safe way
 * Tutorial on preparing the multimeter for start measuring the voltage

Next Steps
Tips for doing this project:

1. Solder all connections carefully. w/o soldering, wires will be easily disconnected, so the power supply will stop working.

2. Use 2-3 different size of drills from thin to thick. First, make holes with the thin one and then make them bigger by the thicker one. Also, each 2 or 3 seconds, cool down the drill by oil or water. If you do not cool them down, the drills get too hot and start melting, so they lose their sharpness.

3. Use heating gun and shrink tube for covering all stripped wires and connections.

4. Be careful not to touch Capacitors or other interior parts of the power supply's board, otherwise, the possibility of electrical shock threaten you.

5. When using multimeter to determine the current of a circuit, make sure that you change the red lead's jack, otherwise, the multimeter will be damaged.

6. Also, when you measure the current of a circuit, a 1 ohm power resistor would be better. In addition, make sure that resistors when getting hot, their resistance increase, so it will affect your current calculations.

7. When working with electrical boards, make sure that you do not touch any capacitors or wires w/o gloves even when it is unplugged because capacitors can keep some power which can harm you.

8. When working on electrical boards, make sure that you immediately unplug the circuit when a part makes abnormal noise or smoke. Then, look for the possible issue(s).


text:
Engineering Projects/StrandBeest/Howard Community College/spring2012/p2550YFC

Problem Statement
Assemble working StrandBeest and achieve power.

Team Members

 * 9484syoon
 * Brianwcallaghan
 * m'kay m'kay

Summary
The three of us worked on this project during the previous 4 weeks as well. Parts were cut and a design was agreed upon. Still remaining is to settle on a working crank shaft and power the movement with wind or some other means of power.

Story
We began with confidence. Three of us from the previous 4 weeks stayed on the project, so we were all familiar with where we stood. The parts were cut and the overall design was done. In the first 4 weeks, one of us got the numbers right and helped the design and the other made CAD drawings and learned the CNC machine to cut the parts. This go round, without Matt, we split up the work as such. Brian was to build the legs and Soon and Filagot both were to work on the crank shaft. The legs were built over the course of the next 2 weeks. 6 sets of 2. The crank shaft remained a problem. At this time, one member of the team became ill and missed 2 consecutive classes leading up to spring break. When he returned, no progress had been made. This was the last week of the project, so the final result of the 2nd 4 weeks of StrandBeest resulted in the build of 6 leg sets. One team member, who built the legs did experiment with some crank ideas. Here is a video of the experiment which failed. Coat hanger is too flimsy for crank shaft. Last but not least the leg build member made a tutorial on how tho assemble the leg sets. You can find a link below on this page.

Decision List
Decisions were made early on about who would take what role in the process. However, after that communication was poor. As a result, decisions were made independently. Each call in the tutorial was made by Brian Callaghan.

Material List

 * 1) Wooden parts cut on CNC machine
 * 2) Crazy Glue
 * 3) 2 inch cotter pins for backup
 * 4) plactic grommets and pins
 * 5) Drill
 * 6) 1/8 inch drill bit

Software List

 * 1) Picasa 3 was used to make the poster and edit the photos in the tutorial.  Picasa 3 is a free photo editing software that is owned by Google.
 * 2) AutoCAD was used only to open drawings of the parts for reference.

Time
44 hours

Tutorials
The following tutorial shows the step by step assembly of a pair of StrandBeest legs on a single frame that can demonstrate the movement. A later tutorial will show the assembly of these leg sets into a Beest.

 * Leg Assembly Tutorial

Next Steps
The crank shaft is the main obstacle. The rest of the body and legs will have to be built around it, so until we have a design, we are at an impasse. Circles may be the way to go.


text:
User:Gjaksec/ENES100/MyWork

''This is designed for a person working on a project for 4 weeks as part of a 3 credit class. Please copy this form to your user page. Then delete everything in italics when filling out this form.''

Write problem/project Goal
What is your wording of the overall problem/project goal?

Design and build a "Core Control" machine.

My First Task
What are you going to do for the team this first weekend?

Over this first weekend, I am going do research the project and prepare to present the project to the class on Tuesday.

Summary of actual work over first weekend
How is what you did different than what you planned?

Instead of just doing research, I also started the 123D drawing.

Week1 Narrative
Tell a detailed story describing what you did for your team over the weekend.

Over this past weekend, I researched the Core Control system in order to be knowledgeable so I could help present it on Tuesday. I also began drawing the 3D image of the design.

My Second Task
Record what you are planning on doing for the team during the second weekend.

Over this next weekend I plan on finishing the 123D drawing.

Summary of actual work over second weekend
How is what you did different than what you planned?

I was supposed to do the 123D drawing, which I did, but I also created a tutorial for how to draw the Core Control.

Week2 Narrative
Tell a detailed story describing what you did for your team over the weekend.

I finished the drawing, and am going to upload the pictures to wikiversity tomorrow. I also created a tutorial for how to draw the Core Control system in 123D.

My Third task
Record what you are planning on doing for the team during the third weekend.

Continue drafting tutorials.

Summary of actual work over third weekend
How is what you did different than what you planned?

I was going to do a tutorial on how to build the Gatling gun, but then decided to do a 123D tutorial instead.

Week3 Narrative
Tell a detailed story describing what you did for your team over the weekend.

I completed the tutorial on how to draw the Core Control system.

My Fourth task
''Record what you are planning on doing for the team during the final weekend of project work. ''

Continue making tutorials

Summary of actual work over fourth weekend
How is what you did different than what you planned?

I was not planning on working on the team page, but I did add onto the tutorial section.

Week4 Narrative
Tell a detailed story describing what you did for your team over the weekend.

Over the weekend, I completed a tutorial on the basics of 123D. Also I added some pictures to the Tutorial for the Core Control.

Complete Team Page
Fill out the Team Form (should have already copied the form, created the team page, linked to it and started filling it out).

title: User:Gjaksec/ENES100/MyWork


text: User:Mwnuk/PhotoDump

title: User:Mwnuk/PhotoDump


text:
Engineering Projects/Illuminate Clothing/Howard Community College/Spring2012

Project documentation focuses on how the team organized itself to tackle the unknown. This is a creative process. The details learned should be captured in tutorials. Project documentation should focus on how the team organized itself to step out into the unknown. Failure is common. The single most effective way to increase engineering respect is to document failures in a creative way that stimulates, inspires and helps others continue the project. Projects are never done. There are always loose ends. There are always decisions that might have been made more clearly. There are always other alternatives that might have been considered. None of this stuff appears in a tutorial. It does appear in the team documentation. Contents [hide]

1 Electronic Sections Expected 1.1 Problem Statement 1.2 Team Members 1.3 Summary 1.4 Poster 1.5 Story 1.6 Decision List 1.7 Material List 1.8 Software List 1.9 Time 1.10 Tutorials 1.11 Next Steps

Electronic Sections Expected[edit] Problem Statement[edit]

What is the final, negotiated problem statement? Team Members[edit]

Put each team members name here with links to their user pages here. Real names do not have to be used. Summary[edit]

Put an overall, short one paragraph summary here. Poster[edit]

Put a graphic in wikimedia, include the graphic here or link to it here. The graphic should be suitable for creating a traditional project poster. Story[edit]

Tell a detailed story of the project. Describe how split up, what the obstacles were, what testing was done, what informal decisions were made, what assumptions were made, what the results were. Decision List[edit]

List all formal decisions made with links to their documentation such as a decision tree or decision matrix. Material List[edit]

1. List materials used, quantity, size, cost. 2. Describe what needs to be purchased in the future to continue working on this project.

Software List[edit]

You installed different software packages or used already installed software. Describ the programs here. Time[edit]

Put one number here. Hours. Total hours that everyone worked on the project. Do not estimate. Do not talk about it. ADD. Tutorials[edit]

All projects create new tutorials of technical details future participants are going to want to know. They are going to be separate pages that are linked to here. Next Steps[edit]

List specific details, advice, describe the current problems that the next team faces associated with the project here.