text:
Physics (A-level)/Capacitors and exponential decay

Capacitance
A capacitor is a passive electrical component capable of storing electrical charge. This produces electrical potential energy. A capacitor consists of two conducting plates separated by an insulator. Capacitance is defined as the charge stored per unit voltage, deriving the equation: C=Q/V Where C is the capacitance measured in Faradays(F), Q is the charge measured in Amps(A) and V is the voltage of potential difference measured in volts(V). 1F is very large value. More common magnitudes are: picofarads(pF), nanofarads(nF) and microfarads(μF). Micheal Faraday

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User:MrFidaAliEngr/My Friends

I trust in the friendship based on humanitarian principles in order to inculcate cooperation and coordination for humanitarian and environmental benefits. I owe a lot to my friends who have always been supporting me in many ways since my birth. Some of them have been saying good prayers to Allah Almighty for me. May Allah bless them all!

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

title: User:MrFidaAliEngr/My Friends


text:
User talk:Mdmc

2/10/2012  Michael McGill

My purpose is to understand the evolution of the thought process as reguards the necessity of speech, and the content of, ramifications of, sound increments/emotional composition of these verbal units. The anthropological examination of the oldest human skull(s) have determined that speech capabilities were available. I would conjecture before, our modern break down of composite structure, firstly an object would demand expression, a possibility for another to share the experience, meaning exchanged? What was the original silence? What was so out of the ordinary: safety, emotional attachment, personal demarcation, unattributable consequences of ancient times?

title: User talk:Mdmc


text:
User:Egm4313.s12.team17/Report 3

=Problem R3.1 - Method of Undetermined Coefficients=

Given

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$$  \displaystyle y'' - 10y' + 25y = r(x) $$     (Eq.1.1)
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$$  \displaystyle r(x) = 7e^{5x} - 2x^2 $$     (Eq.1.2)
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$$  \displaystyle y(0) = 4, y'(0) = -5 $$     (Eq.1.3)
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Problem Statement
Find the solution for Eq. 1.1 using Eq. 1.2 (excitation) and Eq. 1.3 (initial conditions).

Solution
Determine the characteristics equation for Eq. 1.1:
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$$  \displaystyle \lambda^2-10\lambda+25=0 $$     (Eq.1.4)
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Take Eq. 1.4 and solve for the roots:
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$$  \displaystyle (\lambda-5)(\lambda-5)=0 $$     (Eq.1.5)
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$$  \displaystyle \lambda=5 $$     (Eq.1.6)
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Therefore, the homogenous solution of a double real solution takes the form:
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$$  \displaystyle y_h(x)=(c_1+c_2x)e^{\lambda x} $$ (Eq.1.7)
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Substitute Eq. 1.6 into Eq. 1.7 will results in:
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$$  \displaystyle y_h(x) = (c_1 + c_2x)e^{5x} $$     (Eq.1.8)
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Set the excitation as the following:
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$$  \displaystyle r(x)=r_1(x)+r_2(x) $$     (Eq.1.9)
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Solve for the particular solution by using the Sum rule for the Method of Undetermined Coefficients:
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$$  \displaystyle y_p(x) = y_{p1}(x) - y_{p2}(x) $$     (Eq.1.10)
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The first particular solution will take the form:
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$$  \displaystyle y_{p1}(x)=Ce^{\gamma x} $$ (Eq.1.11)
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The \gamma \, term in Eq. 1.11 will have to equal 5 because it has to have the same coefficient as r_1(x) \,.

Since the first particular solution is a solution to the homogenous ODE, then it has to be multiply by x^2 \, (Modification Rule):
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$$  \displaystyle y_{p1}(x)=Cx^2e^{5 x} $$ (Eq.1.12)
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The second particular solution will take the form:
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$$  \displaystyle y_{p2}(x)=\sum_{j=0}^n K_j x^j $$     (Eq.1.13)
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Set n = 2 \,, since the second excitation is x^2 \,:
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$$  \displaystyle y_{p2}(x)=\sum_{j=0}^2 K_j x^j $$     (Eq.1.14)
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$$  \displaystyle y_{p2}(x)= K_2x^2+K_1x+K_0 $$     (Eq.1.15)
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Therefore, Eq. 1.10 becomes:
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$$  \displaystyle y_{p} = Cx^2e^{5x} - (K_2x^2 + K_1x + K_0) $$     (Eq.1.16)
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Taking the first and second derivatives of the particular solution (Eq. 1.16) yields:
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$$  \displaystyle y_{p}' = C(5x^2e^{5x} + 2xe^{5x}) - (2K_2x + K_1) $$     (Eq.1.17)
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$$  \displaystyle y_{p}'' = C(25x^2e^{5x}+20xe^{5x}+2e^{5x}) - 2K_2 $$     (Eq.1.18)
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Then plug in Eq. 1.16, 1.17, and 1.18 into :
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$$  \displaystyle C(25x^2e^{5x}+20xe^{5x}+2e^{5x})-2K_2-C(20xe^{5x}+50x^2e^{5x})+20K_2x+10K_1+25Cx^2e^{5x}-25(K_2x^2+K_1x+K_0)=7e^{5x}-2x^2 $$     (Eq.1.19)
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After simplifying Eq. 1.19:
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$$  \displaystyle 2Ce^{5x}-2K_2+20K_2x+10K_1-25(K_2x^2+K_1x+K_0)=7e^{5x}-2x^2 $$     (Eq.1.20)
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Using coefficient matching, the following constants can be solved for:
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$$  \displaystyle 2Ce^{5x}=7e^{5x} $$     (Eq.1.21)
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$$  \displaystyle -25K_2x^2=-2x^2 $$     (Eq.1.22)
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$$  \displaystyle -25K_1+20K_2x=0 $$     (Eq.1.23)
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$$  \displaystyle 10K_1-2K_2-25K_0=0 $$     (Eq.1.24)
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Solving for the unknown variables yield values of:
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$$  \displaystyle C=7/2 $$     (Eq.1.25)
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$$  \displaystyle K_2=2/25=0.08 $$     (Eq.1.26)
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$$  \displaystyle K_1=8/125=0.064 $$     (Eq.1.27)
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$$  \displaystyle K_0=12/625=0.0192 $$     (Eq.1.28)
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Plug these values back into the particular solution (Eq. 1.16):
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$$  \displaystyle y_{p}(x) = 3.5x^2e^{5x} - (0.08x^2 + 0.064x + 0.0192) $$     (Eq.1.29)
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The final solution general solution is the following:
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$$  \displaystyle y(x)=y_h(x)+y_p(x) $$     (Eq.1.30)
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Substitute Eq. 1.8 and 1.30 into Eq. 1.31:
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$$  \displaystyle y(x)=c_1e^{5x} + c_2xe^{5x}+ 3.5x^2e^{5x} - (0.08x^2 + 0.064x + 0.0192) $$     (Eq.1.31)
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Use the initial conditions (Eq. 1.3) to solve for c_1 \, and c_2 \,:
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$$  \displaystyle 4=c_1+0.0192 $$     (Eq.1.32)
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$$  \displaystyle c_1=3.98 $$     (Eq.1.33)
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$$  \displaystyle -5=5c_1+c_2-0.064 $$     (Eq.1.34)
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$$  \displaystyle c_2=24.84 $$     (Eq.1.35)
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Therefore, the final answer is:
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$$  \displaystyle y(x) = 3.98e^{5x} - 24.84xe^{5x} + 3.5x^2e^{5x} - (0.08x^2 + 0.064x + 0.0192) $$     (Eq.1.36)
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Can use back substitution to check answer.

Author & Proofreaders
Author:Egm4313.s12.team17.ying 03:14, 18 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.deaver.md 04:04, 22 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 04:04, 22 February 2012 (UTC)

=Problem R3.2 -Perturbation Method for Double Real Root=

Part 1 Given

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$$  \displaystyle \lambda_1=\lambda, \lambda_2=\lambda+\epsilon $$     (Eq.2.1.1)
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Part 1 Problem Statement
Find the homogeneous L2-ODE-CC having the given distinct roots(Eq. 2.1.).

Part 1 Solution
Let :
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$$  \displaystyle \lambda_1=x $$     (Eq.2.1.2)
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$$  \displaystyle \lambda_2=x+\epsilon $$     (Eq.2.1.3)
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Rearrange Eq. 2.1.2 and 2.1.3, so that there is a zero on the right hand side:
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$$  \displaystyle \lambda_1-x=0 $$     (Eq.2.1.4)
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$$  \displaystyle \lambda_2-(x+\epsilon)=0 $$     (Eq.2.1.5)
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Now, multiply Eq. 2.1.4 and 2.1.5:
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$$  \displaystyle (\lambda_1-x)(\lambda_2-(x+\epsilon))=0 $$     (Eq.2.1.6)
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Set :
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$$  \displaystyle (\gamma-x)(\gamma-(x+\epsilon))=0 $$     (Eq.2.1.7)
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Expand Eq. 2.1.7:
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$$  \displaystyle \gamma^2-\gamma(2x+\epsilon)+x(x+\epsilon)=0 $$     (Eq.2.1.8)
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Now, set and :
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$$  \displaystyle y^{\prime\prime}-y^\prime(2\lambda+\epsilon)+y\lambda(\lambda+\epsilon)=0 $$     (Eq.2.1.9)
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Part 2 Given
Part 1 givens.

Part 2 Problem Statement
Show that the following is a homogeneous solution:
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$$  \displaystyle y_h=\frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon} $$     (Eq.2.2.1)
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Part 2 Solution
Take the first and second derivative of Eq. 2.2.1:
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$$  \displaystyle y_h^\prime=\frac{(\lambda+\epsilon)e^{(\lambda+\epsilon)x}-\lambda e^{\lambda x}}{\epsilon} $$     (Eq.2.2.2)
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$$  \displaystyle y_h^{\prime\prime}=\frac{(\lambda+\epsilon)^2e^{(\lambda+\epsilon)x}-\lambda^2e^{\lambda x}}{\epsilon} $$     (Eq.2.2.3)
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Substitute Eq. 2.2.1, 2.2.2, and 2.2.3 into Eq. 2.1.9:
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$$  \displaystyle \frac{(\lambda+\epsilon)^2e^{(\lambda+\epsilon)x}-\lambda^2e^{\lambda x}}{\epsilon}-((\frac{(\lambda+\epsilon)e^{(\lambda+\epsilon)x}-\lambda e^{\lambda x}}{\epsilon})(2\lambda+\epsilon))+(\frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon})\lambda(\lambda+\epsilon)=0 $$     (Eq.2.2.4)
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$$  \displaystyle [\lambda^2+2\lambda \epsilon +\epsilon^2 -(2\lambda^2+3\lambda\epsilon +\epsilon^2)+\lambda^2+\lambda\epsilon]e^{(\lambda+\epsilon)x}+e^{\lambda x}[-\lambda^2 +\lambda\epsilon -\lambda^2-\lambda\epsilon+2\lambda^2]=0 $$     (Eq.2.2.5)
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Therefore, the Eq. 2.2.1 is a homogeneous solution for Eq. 2.1.9.
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Part 3 Given
Part 1 and 2.

Part 3 Problem Statement
Find the limit of Eq. 2.2.1 as.

Part 3 Solution
The limit is shown below:
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$$  \displaystyle \lim_{\epsilon \rightarrow 0} \frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon} $$     (Eq.2.3.1)
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L'Hospital's Rule is the following:
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$$  \displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a} \frac{f^\prime(x)}{g^\prime(x)} $$     (Eq.2.3.2)
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Therefore,
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$$  \displaystyle \lim_{\epsilon \rightarrow 0} \frac{e^{(\lambda+\epsilon)x}-e^{\lambda x}}{\epsilon}=\lim_{\epsilon \rightarrow 0} \frac{xe^{(\lambda+\epsilon)x}-0}{1} $$     (Eq.2.3.3)
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Now, taking the limit of Eq. 2.3.3:
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$$  \displaystyle \lim_{\epsilon \rightarrow 0} \frac{xe^{(\lambda+\epsilon)x}-0}{1}=xe^{\lambda x} $$ (Eq.2.3.4)
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Part 4 Given
Part 1, 2, and 3 information.

Part 4 Problem Statement
Take the derivative of with respect to lambda.

Part 4 Solution

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$$  \displaystyle \frac{d}{d\lambda}(e^{\lambda x})=xe^{\lambda x} $$ (Eq.2.4.1)
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Part 5 Given
Part 1, 2, 3, and 4 information.

Part 5 Problem Statement
Compare Eq. 2.4.1 and 2.3.4. Also, relate to the result by variation of parameters.

Part 5 Solution

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$$  \displaystyle \frac{d}{d\lambda}(e^{\lambda x})= \lim_{\epsilon \rightarrow 0} \frac{xe^{(\lambda+\epsilon)x}-0}{1} = xe^{\lambda x} $$ (Eq.2.5.1)
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Part 6 Given
Part 1, 2, 3, and 5 information.

Part 6 Problem Statement
Compute Eq. 2.2.1 using 5 for lambda and 0.001 for the perturbation, while comparing it to Eq. 2.4.1.

Part 6 Solution
Eq. 2.2.1 becomes:
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$$  \displaystyle y_h=\frac{e^{5.001x}-e^{5x}}{0.001} $$     (Eq.2.6.1)
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$$  \displaystyle y_h=1000(e^{5.001x}-e^{5x}) $$     (Eq.2.6.2)
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Eq. 2.4.1 becomes:
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$$  \displaystyle \frac{d}{d\lambda}(e^{\lambda x})=xe^{5x} $$     (Eq.2.6.3)
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In order for the two equation in part 6 to not be approximately equal, the value of x has to be extremely large. The following graph shows the error between them for different values of x.

Author & Proofreaders
Author: Egm4313.s12.team17.deaver.md 14:36, 22 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.ying 15:20, 22 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 14:36, 22 February 2012 (UTC)

=Problem R3.3 - Method of Undetermined Coefficients=

Given

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$$  \displaystyle y^{\prime\prime}-3y^\prime+2y=4x^2 $$     (Eq.3.1)
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$$  \displaystyle y(0)=1,y^\prime(0)=0 $$     (Eq.3.2)
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Problem Statement
Find the solution for Eq. 3.1 and plot it.

Solution
First find the solution to the homogeneous case:
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$$  \displaystyle y^{\prime\prime}-3y^\prime+2y=0 $$     (Eq.3.3) Use the following to generate the characteristic equation:
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$$  \displaystyle y^{\prime\prime}=\lambda^2 $$
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$$  \displaystyle y^{\prime}=\lambda $$

$$  \displaystyle y=1 $$     (Eq.3.4)
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Substitute Eq. 3.4 into Eq. 3.3:
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$$  \displaystyle \lambda^2-3\lambda +2=0 $$     (Eq.3.5)
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To solve the characteristic equation, apply the quadratic formula:
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$$  \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (Eq.3.6)
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$$  \displaystyle \lambda=\frac{-(-3)\pm\sqrt{(-1)^2-4(1)(2)}}{2(1)} $$     (Eq.3.7)
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$$  \displaystyle \lambda=\frac{3\pm\sqrt{1}}{2} $$     (Eq.3.8)
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$$  \displaystyle \lambda_{1}=\frac{3+ \sqrt{1}}{2}=2 $$     (Eq.3.9)
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$$  \displaystyle \lambda_{2}=\frac{3-\sqrt{1}}{2}=1 $$     (Eq.3.10)
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Since both roots are real, the general solution will take the form:
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$$  \displaystyle y = C_1 e^{\lambda_1 x}+C_2 e^{\lambda_2 x} $$ (Eq.3.11)
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Then, the general solution to the homogeneous case will be the following:
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$$  \displaystyle y_h = C_1 e^{2 x}+C_2 e^{ x} $$ (Eq.3.12)
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Now find the particular solution for:
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$$  \displaystyle y^{\prime\prime}-3y^\prime+2y=4x^2 $$     (Eq.3.13)
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The particular solution will take the following form:
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$$  \displaystyle y_{p}(x)=\sum_{j=0}^{n}c_j x^j $$     (Eq.3.14)
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Applying equation 3.14 to the equation 3.13, the particular solution will take the following form:
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$$  \displaystyle y_{p}(x)= c_2 x^2+ c_1 x+ c_0 $$     (Eq.3.15)
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Take the first and second derivative of equation 3.15:
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$$  \displaystyle y_{p}^\prime(x)=2 c_2 x+ c_1 $$     (Eq.3.16)
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$$  \displaystyle y_{p}^{\prime\prime}(x)=2 c_2 $$     (Eq.3.17)
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 *

 * }

Substitute equations 3.15, 3.16, and 3.17 into equation 3.13:
 * {| style="width:100%" border="0"

$$  \displaystyle 2 c_2 -3(2 c_2 x+ c_1 )+2(c_2 x^2+ c_1 x+ c_0 )=4x^2 $$     (Eq.3.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using equation 3.18 to relate variables, results in the following equations:
 * {| style="width:100%" border="0"

$$  \displaystyle 2 c_2 x^2=4x^2 $$     (Eq.3.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -6 c_2 x+2 c_1 x=0 $$     (Eq.3.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2 c_2 -3 c_1 +2 c_0 =0 $$     (Eq.3.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving the system of equations consisting of 3.19, 3.20, and 3.21, the following values are obtained:
 * {| style="width:100%" border="0"

$$  \displaystyle c_2 = 2, $$ $$ \displaystyle c_1 = 6, $$ $$ \displaystyle c_0 = 7 $$     (Eq.3.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p}(x)=2x^2+6x+7 $$     (Eq.3.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The complete solution comprises of the homogeneous solution plus the particular solution:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=y_p(x)+y_h(x) $$     (Eq.3.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y(x)=2x^2+6x+7+C_1 e^{2x} + C_2 e^{x} $$     (Eq.3.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the initial conditions (Eq. 3.2) to solve for the other two unknown variables:
 * {| style="width:100%" border="0"

$$  \displaystyle y(0)=2(0)^2+6(0)+7+C_1 e^{2(0)} + C_2 e^0=1 $$     (Eq.3.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 7+C_1+C_2=1 $$     (Eq.3.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle C_1+C_2=-6 $$     (Eq.3.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the derivative of equation 3.25:
 * {| style="width:100%" border="0"

$$  \displaystyle y^{\prime} (x)=4x+6+2C_1 e^{2x} + C_2 e^{x} $$     (Eq.3.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using the initial conditions (Eq. 3.2) to solve for the other two unknown variables:
 * {| style="width:100%" border="0"

$$  \displaystyle y'(0)=4(0)+6+2C_1e^{2(0)}+C_2e^{0}=0 $$     (Eq.3.30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6+2C_1 + C_2 =0 $$     (Eq.3.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2C_1 + C_2 =-6 $$     (Eq.3.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving equations 3.28 and 3.32:
 * {| style="width:100%" border="0"

$$  \displaystyle C_1=0, $$ $$  \displaystyle C_2=-6 $$     (Eq.3.33)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y(x)=2x^2+6x+7- 6 e^{x} $$     (Eq.3.34)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Check
To check the solution take the first and second derivative of equation 3.34:
 * {| style="width:100%" border="0"

$$  \displaystyle y'(x)=4x+6-6 e^{x} $$     (Eq.3.35)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''(x)=4-6 e^{x} $$     (Eq.3.36)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Plug equations 3.34, 3.35, and 3.36 into 3.1:
 * {| style="width:100%" border="0"

$$  \displaystyle 4-6 e^{x}-3(4x+6-6 e^{x})+2(2x^2+6x+7- 6 e^{x})=4x^2 $$     (Eq.3.37)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 4x^2=4x^2 $$     (Eq.3.38)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Author & Proofreaders
Author:Egm4313.s12.team17.wheeler.tw 07:14, 15 February 2012 (UTC)

Proofreader:Egm4313.s12.team17.axelrod.a 12:13, 17 February 2012 (UTC)

Editor:Egm4313.s12.team17.deaver.md 04:43, 22 February 2012 (UTC)

=Problem R3.4 - Method of Undetermined Coefficients=

Given

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

$$  \displaystyle y''-3y'+2y=4x^2-6x^5 $$     (Eq.4.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Problem Statement
Use the basic and sum rule from the Method of Undetermined Coefficients to show that the appropriate particular solution for Eq. 4.1 is of the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}(x)= \sum_{j=0}^n c_{j}x^j $$     (Eq.4.2) with n = 5 \,.
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solution
In Eq. 4.1 there are two excitation:
 * {| style="width:100%" border="0"

$$  \displaystyle r_{1}(x)=4x^2 $$     (Eq.4.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And

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

$$  \displaystyle r_{2}(x)=-6x^5 $$     (Eq.4.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The general form for Eq. 4.3 and 4.4 are Eq. 4.2. Where n \, is equal to the highest order of x\,.

Therefore we have the summation for the particular solutions to be:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p1}(x)= \sum_{j=0}^2 d_j x^j $$     (Eq.4.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p2}(x)= \sum_{j=0}^5 k_j x^j $$     (Eq.4.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final particular solution is:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p}(x)= y_{p1}(x)+y_{p2}(x) $$     (Eq.4.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_p=\sum_{j=0}^2 d_j x^j+\sum_{j=0}^5 k_j x^j $$     (Eq.4.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_p=d_0+d_1x+d_2x^2+k_0+k_1x+k_2x^2+k_3x^3+k_4x^4+k_5x^5 $$     (Eq.4.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Note that the unknown variables are real numbers; therefore:
 * {| style="width:100%" border="0"

$$  \displaystyle c_0=d_0+k_0 $$     (Eq.4.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_1=d_1+k_1 $$     (Eq.4.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_2=d_2+k_2 $$     (Eq.4.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_3=k_3 $$     (Eq.4.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_4=k_4 $$     (Eq.4.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_5=k_5 $$     (Eq.4.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_p=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 $$     (Eq.4.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using summation to simplify Eq. 4.16 yields the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p=\sum_{j=0}^5 c_{j}x^j $$     (Eq.4.17)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

which is equivalent to Eq. 4.2 with n = 5 \,. Also, when solving for the coefficients both excitation values are used.

Author & Proofreaders
Author: Egm4313.s12.team17.Li Kelvin Li 10:11, 21 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.wheeler.tw 14:45, 21 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 15:53, 22 February 2012 (UTC)

= Problem R3.5 - Nonhomogeneous ODE=

Part 1 Given

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

$$  \displaystyle y''-3y'+2y=4x^2-6x^5 $$     (Eq.5.1.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Part 1 Problem Statement
Obtain equations for the coefficients of.

Part 1 Solution
The order of the given equation is defined as the highest power of x\, in Eq. 5.1.1.

Therefore, the order of Eq. 5.1.1 is 5 and the coefficients of the solution may be found using the below sum rule:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=\sum_{j=0}^s c_j x^j $$     (Eq.5.1.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where s\, in Eq. 5.1.2 is equal to the order of the Eq. 5.1.1 which was defined above to be 5:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=\sum_{j=0}^5 c_j x^j $$     (Eq.5.1.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq. 5.1.3 may be substituted into Eq. 5.1.1 for y\, and below are the corresponding substitutions for y'\, and y''\,:
 * {| style="width:100%" border="0"

$$  \displaystyle y'(x)=\sum_{j=0}^5 j c_j x^{(j-1)}=\sum_{j=1}^5 j c_j x^{(j-1)} $$     (Eq.5.1.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''(x)=\sum_{j=0}^5 j(j-1) c_j x^{(j-2)}=\sum_{j=2}^5 j(j-1) c_j x^{(j-2)} $$     (Eq.5.1.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting Eq. 5.1.3, Eq. 5.1.4, and Eq. 5.1.5 into Eq. 5.1.1 yields the below equation:
 * {| style="width:100%" border="0"

$$  \displaystyle ( \sum_{j=2}^5 j(j-1) c_j x^{(j-2)} ) -3( \sum_{j=1}^5 j c_j x^{(j-1)} )+2( \sum_{j=0}^5 c_j x^j )=4x^2-6x^5 $$     (Eq.5.1.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

To obtain equation for the coefficients of x\,. The j\, values of Eq. 5.1.6 are adjusted for each summation such that the exponent of x will be equal to 1:
 * {| style="width:100%" border="0"

$$  \displaystyle ( \sum_{j=3}^3 j(j-1) c_j x^{(j-2)} ) -3( \sum_{j=2}^2 j c_j x^{(j-1)} )+2( \sum_{j=1}^1 c_j x^j ) $$     (Eq.5.1.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which is further simplified below:
 * {| style="width:100%" border="0"

$$  \displaystyle (3(3-1) c_3 x^{(3-2)} ) -3( 2 c_2 x^{(2-1)} )+2(c_1 x^1 ) $$     (Eq.5.1.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 3(2c_3x^1)-3(2c_2x^1)+2c_1x^1 $$     (Eq.5.1.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6c_3x^1-6c_2x^1+2c_1x^1 $$     (Eq.5.1.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2(3c_3-3c_2+c_1)x^1 $$     (Eq.5.1.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And therefore the coefficients of x^1\, are:
 * {| style="width:100%" border="0"

$$  \displaystyle 2(3c_3-3c_2+c_1)=0 $$     (Eq.5.1.12)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

To obtain equation for the coefficients of x^2\,. The j\, values of Eq. 5.1.6 are adjusted for each summation such that the exponent of x will be equal to 2:
 * {| style="width:100%" border="0"

$$  \displaystyle ( \sum_{j=4}^4 j(j-1) c_j x^{(j-2)} ) -3( \sum_{j=3}^3 j c_j x^{(j-1)} )+2( \sum_{j=2}^2 c_j x^j ) $$     (Eq.5.1.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which is further simplified below:
 * {| style="width:100%" border="0"

$$  \displaystyle ( 4(4-1) c_4 x^{(4-2)} ) -3( 3 c_3 x^{(3-1)} )+2(c_2 x^2 ) $$     (Eq.5.1.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle ( 4)3 c_4 x^{(4-2)} -3( 3 c_3 x^{(2)} )+2c_2 x^2 $$     (Eq.5.1.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 12c_4 x^{(2)} -9 c_3 x^{(2)} +2c_2 x^2 $$     (Eq.5.1.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (12c_4 -9 c_3 +2c_2 )x^2 $$     (Eq.5.1.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And therefore the coefficients of x^2\, are:
 * {| style="width:100%" border="0"

$$  \displaystyle 12c_4 -9 c_3 +2c_2=4 $$     (Eq.5.1.18)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

To obtain equation for the coefficients of x^3\,. The j\, values of Eq. 5.1.6 are adjusted for each summation such that the exponent of x will be equal to 3:
 * {| style="width:100%" border="0"

$$  \displaystyle ( \sum_{j=5}^5 j(j-1) c_j x^{(j-2)} ) -3( \sum_{j=4}^4 j c_j x^{(j-1)} )+2( \sum_{j=3}^3 c_j x^j ) $$     (Eq.5.1.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which is further simplified below:
 * {| style="width:100%" border="0"

$$  \displaystyle ( 5(5-1) c_5 x^{(5-2)} ) -3(  4 c_4 x^{(4-1)} )+2( c_3 x^3 ) $$     (Eq.5.1.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle ( 5(4) c_5 x^{(3)} ) -12 c_4 x^{(3)} +2 c_3 x^3 $$     (Eq.5.1.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 20 c_5 x^{(3)} -12 c_4 x^{(3)} +2 c_3 x^3 $$     (Eq.5.1.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2(10 c_5 -6 c_4 +c_3) x^3 $$     (Eq.5.1.23)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And therefore the coefficients of x^3\, are:
 * {| style="width:100%" border="0"

$$  \displaystyle 2(10 c_5 -6 c_4 +c_3)=0 $$     (Eq.5.1.24)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

To obtain equation for the coefficients of x^5\,. The j\, values of Eq. 5.1.6 are adjusted for each summation such that the exponent of x will be equal to 5:
 * {| style="width:100%" border="0"

$$  \displaystyle ( \sum_{j=7}^5 j(j-1) c_j x^{(j-2)} ) -3( \sum_{j=6}^5 j c_j x^{(j-1)} )+2( \sum_{j=5}^5 c_j x^j ) $$     (Eq.5.1.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since the bounds of the summations in Eq. 5.1.25 are exceeded in two cases those are simplified to zero as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle 2( \sum_{j=5}^5 c_j x^j ) $$     (Eq.5.1.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which is further simplified below:
 * {| style="width:100%" border="0"

$$  \displaystyle 2( c_5 x^5 ) $$     (Eq.5.1.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2c_5 x^5 $$     (Eq.5.1.28)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

And therefore the coefficients of x^5\, are:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_5=-6 $$     (Eq.5.1.29)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Part 2 Given
The solutions from Part 1 and part 1 given.

Part 2 Problem Statement
Verify the coefficients found in part 1 using long hand expansion.

Part 2 Solution
The order of the given equation is defined as the highest power of x\, in Eq. 5.1.1.

Therefore, the order of Eq. 5.1.1 is 5 and the coefficients of the solution may be found using the below sum rule:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=\sum_{j=0}^s c_j x^j $$     (Eq.5.2.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Where s\, in Eq. 5.2.1 is equal to the order of the Eq. 5.1.1 which was defined above to be 5:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=\sum_{j=0}^5 c_j x^j $$     (Eq.5.2.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq. 5.2.2 may be substituted into Eq. 5.1.1 for y\, and below are the corresponding substitutions for y'\, and y''\,:
 * {| style="width:100%" border="0"

$$  \displaystyle y'(x)=\sum_{j=0}^5 j c_j x^{(j-1)}=\sum_{j=1}^5 j c_j x^{(j-1)} $$     (Eq.5.2.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y''(x)=\sum_{j=0}^5 j(j-1) c_j x^{(j-2)}=\sum_{j=2}^5 j(j-1) c_j x^{(j-2)} $$     (Eq.5.2.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substituting Eq.5.2.2, Eq.5.2.3, and Eq.5.2.4 into Eq.5.1 yields the below equation:
 * {| style="width:100%" border="0"

$$  \displaystyle ( \sum_{j=2}^5 j(j-1) c_j x^{(j-2)} ) -3( \sum_{j=1}^5 j c_j x^{(j-1)} )+2( \sum_{j=0}^5 c_j x^j )=4x^2-6x^5 $$     (Eq.5.2.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

As only the terms in each summation of the same order may be added, the summations in Eq. 5.2.5 will be expanded and then added to each other:
 * {| style="width:100%" border="0"

$$  \displaystyle ( 2(2-1)c_2 x^{(2-2)} + 3(3-1)c_3 x^{(3-2)} + 4(4-1)c_4 x^{(4-2)} + 5(5-1)c_5 x^{(5-2)} ) -3( 1c_1x^{(1-1)}+2c_2x^{(2-1)}+3c_3x^{(3-1)}+4c_4x^{(4-1)}+5c_5x^{(5-1)} ) $$ $$  \displaystyle +2( c_0x^0+c_1x^1+c_2x^2+c_3x^3+c_4x^4+c_5x^5 )=4x^2-6x^5 $$     (Eq.5.2.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Which is further simplified below:
 * {| style="width:100%" border="0"

$$  \displaystyle ( 2(1)c_2 x^{(0)} + 3(2)c_3 x^{(1)} + 4(3)c_4 x^{(2)} + 5(4)c_5 x^{(3)} ) -3( 1c_1x^{0}+2c_2x^{1}+3c_3x^{2}+4c_4x^{3}+5c_5x^{4} )+2( c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5 ) $$ $$  \displaystyle =4x^2-6x^5 $$     (Eq.5.2.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle ( 2c_2 + 6c_3 x + 12c_4 x^{(2)} + 20c_5 x^{(3)} ) -( 1c_1+2c_2x+3c_3x^{2}+4c_4x^{3}+5c_5x^{4} )+(2c_0+2c_1x+2c_2x^2+2c_3x^3+2c_4x^4+2c_5x^5 )=4x^2-6x^5 $$     (Eq.5.2.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle ( 2c_2 + 6c_3 x + 12c_4 x^{(2)} + 20c_5 x^{(3)} ) -( 3c_1+6c_2x+9c_3x^{2}+12c_4x^{3}+15c_5x^{4} )+( 2c_0+2c_1x+2c_2x^2+2c_3x^3+2c_4x^4+2c_5x^5 )=4x^2-6x^5 $$     (Eq.5.2.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle (2c_2-3c_1+2c_0)+(6c_3-4c_2+2c_1)x+(12c_4-9c_3+2c_2)x^2+(20c_5-12c_4+2c_3)x^3+(-15c_5+2c_4)x^4+2c_5x^5=(0)+(0)x $$ $$  \displaystyle +(4)x^2+(0)x^3+(0)x^4+(-6)x^5 $$     (Eq.5.2.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Eq. 5.2.10 may now be separated into 6 different equations corresponding to the exponent of x:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_2-3c_1+2c_0=0 $$     (Eq.5.2.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6c_3-6c_2+2c_1=0 $$     (Eq.5.2.12)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$  \displaystyle 12c_4-9c_3+2c_2=4 $$     (Eq.5.2.13)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$  \displaystyle 20c_5-12c_4+2c_3=0 $$     (Eq.5.2.14)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

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

$$  \displaystyle -15c_5+2c_4=0 $$     (Eq.5.2.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2c_5=-6 $$     (Eq.5.2.16)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Eq. 5.2.12, Eq.5.2.13, Eq.5.2.14, and Eq.5.2.16 are equivalent to Eq.5.1.12, Eq.5.1.18, Eq.5.1.24, and Eq.5.1.29 respectively.

Part 3 Given
The solutions from the preceding Parts of Problem 5 and part 1 given.

Part 3 Problem Statement
Put the system of equations obtained in Part 2 Eq. 5.2.11, Eq. 5.2.12, Eq. 5.2.13, Eq. 5.2.14, Eq. 5.2.15, and Eq. 5.2.16 into matrix form for the following coefficients $$c_0$$, $$c_1$$ , $$c_2$$ , $$c_3$$ , $$c_4$$ , and $$ c_5 $$.

Part 3 Solution
For x^0\,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_2-3c_1+2c_0=0 $$     (Eq.5.3.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6c_3-6c_2+2c_1=0 $$     (Eq.5.3.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 12c_4-9c_3+2c_2=4 $$     (Eq.5.3.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^3\,:
 * {| style="width:100%" border="0"

$$  \displaystyle 20c_5-12c_4+2c_3=0 $$     (Eq.5.3.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^4\,:
 * {| style="width:100%" border="0"

$$  \displaystyle -15c_5+2c_4=0 $$     (Eq.5.3.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^5\,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_5=-6 $$     (Eq.5.3.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The exponent of x\, will correspond to the rows of the matrix while the columns of the matrix will correspond to the subscripts of the above coefficients As such, the matrix form of the system of equations is shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0\\ 0 & 0 & 0 & 2 & -12 & 20\\ 0 & 0 & 0 & 0 & 2 & -15\\ 0 & 0 & 0 & 0 & 0 & 2 \end{bmatrix} * \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 4 \\ 0 \\ 0 \\ -6 \end{bmatrix}
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |

$$     (Eq.5.3.7)
 *

 * }

Part 4 Given
The solutions from the preceding Parts of Problem 5 and part 1 given.

Part 4 Problem Statement
Use Back Substitution to solve for the following coefficients $$c_0$$, $$c_1$$ , $$c_2$$ , $$c_3$$ , $$c_4$$ , and $$ c_5 $$ using the algorithm dictated by http://www.mathwords.com/b/back_substitution.htm on Eq. 5.3.7

Part 4 Solution

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

$$  \displaystyle \begin{bmatrix} 2 & -3 & 2 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0\\ 0 & 0 & 0 & 2 & -12 & 20\\ 0 & 0 & 0 & 0 & 2 & -15\\ 0 & 0 & 0 & 0 & 0 & 2 \end{bmatrix} * \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 4 \\ 0 \\ 0 \\ -6 \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq.5.4.1)
 *

 * }

From the website, http://www.mathwords.com/b/back_substitution.htm, we may simplify Eq. 5.4.1 into a single augmented matrix:
 * {| style="width:100%" border="0"

$$  \displaystyle \left[ \begin{array}{cccccc|c} 2 & -3 & 2 & 0 & 0 & 0 & 0\\ 0 & 2 & -6 & 6 & 0 & 0 & 0\\ 0 & 0 & 2 & -9 & 12 & 0 & 4\\ 0 & 0 & 0 & 2 & -12 & 20 & 0\\ 0 & 0 & 0 & 0 & 2 & -15 & 0\\ 0 & 0 & 0 & 0 & 0 & 2 & -6 \end{array} \right ]
 * style="width:95%" |
 * style="width:95%" |

$$     (Eq.5.4.2)
 *

 * }

Which can then be separated back into the system of equations derived in Part 3.

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

$$  \displaystyle 2c_2-3c_1+2c_0=0 $$     (Eq.5.4.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6c_3-6c_2+2c_1=0 $$     (Eq.5.4.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 12c_4-9c_3+2c_2=4 $$     (Eq.5.4.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^3\,:
 * {| style="width:100%" border="0"

$$  \displaystyle 20c_5-12c_4+2c_3=0 $$     (Eq.5.4.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^4\,:
 * {| style="width:100%" border="0"

$$  \displaystyle -15c_5+2c_4=0 $$     (Eq.5.4.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^5\,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_5=-6 $$     (Eq.5.4.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solving for $$ c_5 $$ in Eq. 5.4.8 will allow us to substitute and solve for $$ c_4 $$ in Eq. 5.4.7 and so on as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle c_5=-3 $$     (Eq.5.4.9)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now $$ c_5 $$ can be substituted into Eq. 5.4.7 to solve for $$ c_4 $$ as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle -15(-3)+2c_4=0 $$     (Eq.5.4.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 45+2c_4=0 $$     (Eq.5.4.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2c_4=-45 $$     (Eq.5.4.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_4=-22.5 $$     (Eq.5.4.13)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now $$ c_4 $$ and $$ c_5 $$ can be substituted into Eq. 5.4.6 to solve for $$ c_3 $$ as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle 20(-3)-12(-22.5)+2c_3=0 $$     (Eq.5.4.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -60+270+2c_3=0 $$     (Eq.5.4.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 210+2c_3=0 $$     (Eq.5.4.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2c_3=-210 $$     (Eq.5.4.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_3=-105 $$     (Eq.5.4.18)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now $$ c_3 $$ and $$ c_4 $$ can be substituted into Eq. 5.4.5 to solve for $$ c_2 $$ as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle 12(-22.5)-9(-105)+2c_2=4 $$     (Eq.5.4.19)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -270+945+2c_2=4 $$     (Eq.5.4.20)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 675+2c_2=4 $$     (Eq.5.4.21)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2c_2=-671 $$     (Eq.5.4.22)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_2=-335.5 $$     (Eq.5.4.23)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now $$ c_2 $$ and $$ c_3 $$ can be substituted into Eq. 5.4.4 to solve for $$ c_1 $$ as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle 6(-105)-6(-335.5)+2c_1=0 $$     (Eq.5.4.24)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -630+2013+2c_1=0 $$     (Eq.5.4.25)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 1383+2c_1=0 $$     (Eq.5.4.26)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2c_1=-1383 $$     (Eq.5.4.27)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_1=-691.5 $$     (Eq.5.4.28)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Now $$ c_1 $$ and $$ c_2 $$ can be substituted into Eq. 5.4.3 to solve for $$ c_0 $$ as shown below:
 * {| style="width:100%" border="0"

$$  \displaystyle 2(-335.5)-3(-691.5)+2c_0=0 $$     (Eq.5.4.29)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle -671+2074.5+2c_0=0 $$     (Eq.5.4.30)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 1403.5+2c_0=0 $$     (Eq.5.4.31)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 2c_0=-1403.5 $$     (Eq.5.4.32)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle c_0=-701.75 $$     (Eq.5.4.33)
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 *

 * }

And now we have obtained solutions for all of the constants: $$ c_0 $$, $$ c_1 $$ , $$ c_2 $$ , $$ c_3 $$ , $$ c_4 $$ , and $$ c_5 $$.

Part 5 Given
The solutions from the preceding parts of Problem 5 and part 1 given.

Part 5 Problem Statement
Find the solution y(x)\, using the initial conditions y(0)=1\, and and then graph y(x)\,.

Part 5 Solution
The particular solution is the following:
 * {| style="width:100%" border="0"

$$  \displaystyle y_p(x)=-701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5 $$     (Eq.5.5.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

The final solution is in the form:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=y_h(x)+y_p(x) $$     (Eq.5.5.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Substitute Eq. 3.12 and 5.5.1 into Eq. 5.5.2:
 * {| style="width:100%" border="0"

$$  \displaystyle y(x)=-701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5+C_1e^{2x}+C_2e^x $$     (Eq.5.5.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Take the first derivative of Eq. 5.5.3:
 * {| style="width:100%" border="0"

$$  \displaystyle y^\prime(x)=-691.5-671x-315x^2-90x^3-15x^4+2C_1e^{2x}+C_2e^{x} $$     (Eq.5.5.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Solve for the unknown constants that are remaining by using the initial condition:
 * {| style="width:100%" border="0"

$$  \displaystyle 1=-701.75+C_1+C_2 $$     (Eq.5.5.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 0=-691.5+2C_1+C_2 $$     (Eq.5.5.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle C_1=-11.25 $$     (Eq.5.5.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle C_2=714 $$     (Eq.5.5.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y(x)=-701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5-11.25e^{2x}+714e^{x} $$     (Eq.5.5.9)
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 02:55, 19 February 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.hintz 07:50, 20 February 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.ying 08:12, 20 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 08:22, 22 February 2012 (UTC)

= Problem R3.6 - Method of Undetermined Coefficients=

Given
We are given the following equations:
 * {| style="width:100%" border="0"

$$  \displaystyle y''-3y'+2y=4x^{2}-6x^{5} $$     (Eq.6.1)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y(0)=1,y^\prime(0)=0 $$     (Eq.6.2)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y^{\prime\prime}_{p1}-3y^{\prime}_{p1}+2y_{p1}=r_1(x):=4x^2 $$     (Eq.6.3)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y^{\prime\prime}_{p2}-3y^{\prime}_{p2}+2y_{p2}=r_2(x):=-6x^5 $$     (Eq.6.4)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Problem Statement
Solve for the second particular solution, then use the solution from Problem 3 to solve for the final solution.

Solution
Determine the characteristic equation for Eq. 6.1:
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda^{2}-3\lambda+2=0 $$     (Eq.6.5)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Obtain the two roots (by means of factoring):
 * {| style="width:100%" border="0"

$$  \displaystyle \lambda_{1}=2, \lambda_{2}=1 $$     (Eq.6.6)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Since, the roots are real distinct roots which the homogeneous solution will have the general form:
 * {| style="width:100%" border="0"

$$  \displaystyle y_h(x)=C_1e^{\lambda_1 x}+C_2e^{\lambda_2 x} $$ (Eq.6.7)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_h(x)=C_1e^{2 x}+C_2e^{ x} $$ (Eq.6.8)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Generate the second particular solution:
 * {| style="width:100%" border="0"

$$  \displaystyle y_{p2}=\sum_{j=0}^5 c_{j}x^{j} $$     (Eq.6.9)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p2}'=\sum_{j=1}^5 c_{j}(j)x^{j-1} $$     (Eq.6.10)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle y_{p2}''=\sum_{j=2}^5 c_{j}(j)(j-1)x^{j-2} $$     (Eq.6.11)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Plug Eq. 6.9. 6.10, 6.11 into Eq. 6.4:
 * {| style="width:100%" border="0"

$$  \displaystyle \sum_{j=2}^5 c_{j}(j)(j-1)x^{j-2}-3\sum_{j=1}^5 c_{j}(j)x^{j-1}+2\sum_{j=0}^5 c_{j}x^{j} = -6x^{5} $$     (Eq.6.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Expand Eq. 6.12:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_2+6c_3x+12c_4x^2+20c_5x^3-3(c_1+2c_2x+3c_3x^2+4c_4x^3+5c_5x^4)+2(c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5) = -6x^{5} $$     (Eq.6.12)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Using coefficient matching, the following constants can be solved.

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

$$  \displaystyle 2c_2-3c_1+2c_0=0 $$     (Eq.6.13)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 6c_3-6c_2+2c_1=0 $$     (Eq.6.14)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

$$  \displaystyle 12c_4-9c_3+2c_2=0 $$     (Eq.6.15)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^3\,:
 * {| style="width:100%" border="0"

$$  \displaystyle 20c_5-12c_4+2c_3=0 $$     (Eq.6.16)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^4\,:
 * {| style="width:100%" border="0"

$$  \displaystyle -15c_5+2c_4=0 $$     (Eq.6.17)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

For x^5\,:
 * {| style="width:100%" border="0"

$$  \displaystyle 2c_5=-6 $$     (Eq.6.18)
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Therefore, the constant values are the following:
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$$  \displaystyle c_5=-3 $$     (Eq.6.19)
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$$  \displaystyle c_4=-22.5 $$     (Eq.6.20)
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$$  \displaystyle c_3=-105 $$     (Eq.6.21)
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$$  \displaystyle c_2=-337.5 $$     (Eq.6.22)
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$$  \displaystyle c_1=-697.5 $$     (Eq.6.23)
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$$  \displaystyle c_0=-708.75 $$     (Eq.6.24)
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The second particular solution is the following:
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$$  \displaystyle y_{p2}(x)=-708.75-697.5x-337.5x^2-105x^3-22.5x^4-3x^5 $$     (Eq.6.25)
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The final form of the solution is:
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$$  \displaystyle y(x)=y_{p1}(x)+y_{p2}(x)+y_{h}(x) $$     (Eq.6.26)
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Substitute Eq. 6.25, 6.8, and 3.23 into Eq. 6.26:
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$$  \displaystyle y(x)=2x^2+6x+7-708.75-697.5x-337.5x^2-105x^3-22.5x^4-3x^5+C_1e^{\lambda_1 x}+C_2e^{\lambda_2 x} $$ (Eq.6.27)
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$$  \displaystyle y(x)=-701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5+C_1e^{2x}+C_2e^{x} $$     (Eq.6.27)
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Take the first derivative of Eq. 6.27:
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$$  \displaystyle y^\prime(x)=-691.5-671x-315x^2-90x^3-15x^4+2C_1e^{2x}+C_2e^{x} $$     (Eq.6.28)
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Solve for the unknown constants that are remaining by using the initial condition (Eq. 6.2):
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$$  \displaystyle 1=-701.75+C_1+C_2 $$     (Eq.6.29)
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$$  \displaystyle 0=-691.5+2C_1+C_2 $$     (Eq.6.30)
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$$  \displaystyle C_1=-11.25 $$     (Eq.6.31)
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$$  \displaystyle C_2=714 $$     (Eq.6.32)
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Therefore, the final solution is:
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$$  \displaystyle y(x)=-701.75-691.5x-335.5x^2-105x^3-22.5x^4-3x^5-11.25e^{2x}+714e^{x} $$     (Eq.6.33)
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This is the same result from problem 5.

Author & Proofreaders
Author: Egm4313.s12.team17.hintz 04:19, 20 February 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.axelrod.a 08:10, 21 February 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.Li 07:15, 21 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 09:45, 22 February 2012 (UTC)

=Problem R3.7 - Summation Index Changes=

Part 1 Given

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$$  \displaystyle \sum_{j=2}^5 c_jj(j-1)x^{j-1}=\sum_{j=0}^3 c_{j+2}(j+2)(j+1)x^j $$     (Eq.7.1.1)
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Part 1 Problem Statement
Expand the series on both sides of Eq. 7.1.1 to verify these equalities.

Part 1 Solution
First, we expand the left side of Eq. 7.1.1 yielding:
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$$  \displaystyle \sum_{j=2}^5 c_jj(j-1)x^{j-2}=c_2(2)(2-1)x^0+c_3(3)(3-1)x^1+c_4(4)(4-1)x^2+c_5(5)(5-1)x^3 $$     (Eq.7.1.2)
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After simplifying we have:
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$$  \displaystyle \sum_{j=2}^5 c_jj(j-1)x^{j-2}=2c_2+6c_3x^1+12c_4x^2+20c_5x^3 $$     (Eq.7.1.3)
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Now for the right side of Eq. 7.1.1, we make the substitution $$k=j-2$$ to the left side of Eq. 7.1.1 to obtain:
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$$  \displaystyle \sum_{k=0}^3 c_{k+2}(k+2)(k+1)x^k $$     (Eq.7.1.4)
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If we rename the dummy variable k to j, $$ k \rightarrow j$$ we obtain the right side of Eq. 7.1.1:
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$$  \displaystyle \sum_{j=0}^3 c_{j+2}(j+2)(j+1)x^j $$     (Eq.7.1.5)
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Expanding right side of Eq. 7.1.1:
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$$  \displaystyle \sum_{j=0}^3 c_{j+2}(j+2)(j+1)x^j=c_{0+2}(0+2)(0+1)x^0+c_{1+2}(1+2)(1+1)x^1+c_{2+2}(2+2)(2+1)x^2+c_{3+2}(3+2)(3+1)x^5 $$     (Eq.7.1.5)
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Simplifying Eq. 7.1.5:
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$$  \displaystyle \sum_{j=0}^3 c_{j+2}(j+2)(j+1)x^j=2c_2+6c_3x^1+12c_4x^2+20c_5x^5 $$     (Eq.7.1.6)
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We see that Eq. 7.1.3 and Eq. 7.1.6 are the same, therefore, Eq. 7.1.1 has been verified and is correct.
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Part 2 Given

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$$  \displaystyle \sum_{j=1}^5 c_jjx^{j-1}=\sum_{j=0}^4 c_{j+1}(j+1)x^j $$     (Eq.7.2.1)
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Part 2 Problem Statement
Expand the series on both sides of Eq. 7.2.1 to verify these equalities.

Part 2 Solution
First we expand the left side of Eq. 7.2.1 to obtain the following:
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$$  \displaystyle \sum_{j=1}^5 c_jjx^{j-1}=c_1(1)x^{1-1}+c_2(2)x^{2-1}+c_3(3)x^{3-1}+c_4(4)x^{4-1}+c_5(5)x^{5-1} $$     (Eq.7.2.2)
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Simplifying, we have:
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$$  \displaystyle \sum_{j=1}^5 c_jjx^{j-1}=1c_1+2c_2x^{1}+3c_3x^{2}+4c_4x^{3}+5c_5x^{4} $$     (Eq.7.2.3)
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Now for the right side of Eq. 7.2.1 we make the substitution $$ k=j-1$$ to the let side of Eq. 7.2.1 to obtain:
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$$  \displaystyle \sum_{k=0}^4 c_{k+1}(k+1)x^{k} $$    (Eq.7.2.4)
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Renaming the dummy variable k to j, $$ k \rightarrow j$$ we obtain the right side of Eq. 7.2.1:
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$$  \displaystyle \sum_{j=0}^4 c_{j+1}(j+1)x^{j} $$     (Eq.7.2.4) Expanding right side of Eq. B:
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$$  \displaystyle \sum_{j=0}^4 c_{j+1}(j+1)x^{j}=c_{0+1}(0+1)x^0+c_{1+1}(1+1)x^1+c_{2+1}(2+1)x^2+c_{3+1}(3+1)x^3+c_{4+1}(4+1)x^4 $$     (Eq.7.2.5)
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Simplifying, we have:
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$$  \displaystyle \sum_{j=0}^4 c_{j+1}(j+1)x^{j}=1c_1+2c_2x^{1}+3c_3x^{2}+4c_4x^{3}+5c_5x^{4} $$     (Eq.7.2.6)
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We see that Eq. 7.2.3 and Eq. 7.2.6 are the same, therefore, Eq. 7.2.1 has been verified and is correct.
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Author & Proofreaders
Author: Egm4313.s12.team17.Li Kelvin Li 06:24, 20 February 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.wheeler.tw 02:32, 21 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 10:43, 22 February 2012 (UTC)

=Problem R3.8 - Method of Undetermined Coefficients=

Part Given

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$$  \displaystyle y^{\prime\prime}+4y^\prime+4y=e^{-x}cosx $$     (Eq.8.1.1)
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Part 1 Problem Statement
Find a (real) general solution. State which rule you are using.

Part 1 Solution
First find the homogeneous solution. Use the following to generate the characteristic equation:
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$$  \displaystyle y^{\prime\prime}=\lambda^2 $$, $$  \displaystyle y^{\prime}=\lambda $$, $$  \displaystyle y=1 $$     (Eq.8.1.2)
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Substitute Eq. 8.1.2 into Eq. 8.1.1, the characteristic equation is:
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$$  \displaystyle \lambda^2+4 \lambda +4=0 $$     (Eq.8.1.3) Factoring equation 8.1.3:
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$$  \displaystyle (\lambda + 2)^2 =0 $$     (Eq.8.1.4) The roots for $$ \lambda $$ are:
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$$  \displaystyle \lambda_1 = -2 $$     (Eq.8.1.5)
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$$  \displaystyle \lambda_2 = -2 $$     (Eq.8.1.6) Since, the solution is a double root, the homogeneous solution will take the following form:
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$$  \displaystyle y_h = e^{-2x}(C_1+C_2x) $$     (Eq.8.1.7) Now, using the basic solution to solve for the particular case:
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$$  \displaystyle y_p = e^{-x}(c_0cosx + c_1sinx) $$     (Eq.8.1.8) Take the first and second derivative of equation 8.1.8:
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$$  \displaystyle y_p' = -e^{-x}(c_0cosx+c_1sinx)+ e^{x}(-c_0sinx+c_1cosx) $$     (Eq.8.1.9)
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$$  \displaystyle y_p''= e^{-x}(c_0cosx+c_1sinx)- e^{-x}(-c_0sinx+c_1cosx) - e^{-x}(-c_0sinx+c_1cosx)+e^{-x}(-c_0cosx-c_1sinx) = -2e^{-x}(-c_0sinx + c_1cosx) $$     (Eq.8.1.10) Substitute equations 8.1.8, 8.1.9, 8.1.10 into equation 8.1.1:
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$$  \displaystyle -2e^{-x}(-c_0sinx + c_1cosx)+4(-e^{-x}(c_0cosx+c_1sinx)+ e^{-x}(-c_0sinx+c_1cosx))+4(e^{-x}(c_0cosx + c_1sinx))=e^{- x}cosx $$     (Eq.8.1.11) Solving equation 8.1.11 for $$c_0$$ and $$c_1$$:
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$$  \displaystyle c_0=0 $$     (Eq.8.1.12)
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$$  \displaystyle c_1=0.5 $$     (Eq.8.1.13) Substituting equations 8.1.12 and 8.1.13 into equation 8.1.8, the particular solution becomes:
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$$  \displaystyle y_p=0.5e^{-x}sinx $$     (Eq.8.1.14) Since the general solution is:
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$$  \displaystyle y=y_h+y_p $$     (Eq.8.1.15) Substitute equations 8.1.7 and 8.1.14 into equation 8.1.15:
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$$  \displaystyle y(x)=e^{-2x}(C_1+C_2x) + 0.5e^{-x}sinx $$     (Eq.8.1.16)
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Part 2 Problem Statement
Find a (real) general solution. State which rule you are using.
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$$  \displaystyle y''+y'+(\pi^2 + \frac{1}{4})y = e ^{-\frac{x}{2}}sin\pi x $$ (Eq.8.2.1)
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Part 2 Solution
First find the homogeneous solution. Use the following to generate the characteristic equation:
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$$  \displaystyle y^{\prime\prime}=\lambda^2 $$, $$  \displaystyle y^{\prime}=\lambda $$, $$  \displaystyle y=1 $$     (Eq.8.2.2) Substitute Eq. 8.1.2 into Eq. 8.1.1, the characteristic equation is:
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$$  \displaystyle \lambda^2+ \lambda +(\pi^2 + \frac{1}{4})=0 $$     (Eq.8.2.3) Use the quadratic formula to factor equation 8.2.3:
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$$  \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$     (Eq.8.2.4)
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$$  \displaystyle \lambda=\frac{-(1)\pm\sqrt{(1)^2-4(1)(\pi^2 + \frac{1}{4})}}{2(1)} $$     (Eq.8.2.5)
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$$  \displaystyle \lambda=\frac{-(1)\pm\sqrt{-4 \pi^2}}{2} $$     (Eq.8.2.6)
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$$  \displaystyle \lambda=\frac{-1 \pm 2 \pi i }{2} $$     (Eq.8.2.7)
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$$  \displaystyle \lambda_1= - \frac{1}{2} + \pi i $$ (Eq.8.2.8)
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$$  \displaystyle \lambda_2= - \frac{1}{2} - \pi i $$ (Eq.8.2.9) The general, homogeneous solution will take the following form:
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$$  \displaystyle y_h = e^{- \frac{x}{2}}(C_1cos \pi x+C_2sin \pi x) $$ (Eq.8.2.10) Based on the Basic Rule, the particular solution will take the following form:
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$$  \displaystyle y_p=e^{- \frac{x}{2}}(c_0cos \pi x+c_1sin \pi x) $$ (Eq.8.2.11) Since the equation 8.2.11 is a solution of the homogeneous solution, use the Modification rule to find the particular solution. Multiply equation 8.2.11 by $$x$$:
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$$  \displaystyle y_p=e^{- \frac{x}{2}}(c_0xcos \pi x+c_1xsin \pi x) $$ (Eq.8.2.12) Take the first and second derivative of equation 8.2.12:
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$$  \displaystyle y_p'=-\frac{1}{2}e^{- \frac{x}{2}}(c_0xcos \pi x+c_1xsin \pi x) + e^{- \frac{x}{2}}(c_0cos \pi x- \pi c_0 x sin \pi x +c_1 sin \pi x + \pi c_1 x cos \pi x) $$ (Eq.8.2.13)
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$$  \displaystyle y_p''=\frac{1}{4}e^{- \frac{x}{2}}(c_0xcos \pi x+c_1xsin \pi x) - \frac{1}{2} e^{- \frac{x}{2}}(c_0cos \pi x- \pi c_0 x sin \pi x +c_1 sin \pi x + \pi c_1 x cos \pi x) -\frac{1}{2} e^{-\frac{x}{2}}(c_0 cos \pi x - \pi c_0 x sin \pi x + c_1 sin \pi x + \pi c_1 x cos \pi x) $$ $$ +e^{-\frac{x}{2}}(-c_0 \pi sin \pi x - \pi c_0 sin \pi x - \pi^2 c_0 x cos \pi x + c_1 \pi cos \pi x + \pi c_1 cos \pi x - \pi^2 c_1 x sin \pi x) $$ (Eq.8.2.14) Substituting equations 8.2.12, 8.2.13, and 8.2.14 into 8.2.1:
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$$  \displaystyle \frac{1}{4}e^{- \frac{x}{2}}(c_0xcos \pi x+c_1xsin \pi x) - \frac{1}{2} e^{- \frac{x}{2}}(c_0cos \pi x- \pi c_0 x sin \pi x +c_1 sin \pi x + \pi c_1 x cos \pi x) -\frac{1}{2} e^{-\frac{x}{2}}(c_0 cos \pi x - \pi c_0 x sin \pi x + c_1 sin \pi x + \pi c_1 x cos \pi x) + $$ $$ e^{-\frac{x}{2}}(-c_0 \pi sin \pi x - \pi c_0 sin \pi x - \pi^2 c_0 x cos \pi x + c_1 \pi cos \pi x + \pi c_1 cos \pi x - \pi^2 c_1 x sin \pi x)+ $$ $$ -\frac{1}{2}e^{- \frac{x}{2}}(c_0xcos \pi x+c_1xsin \pi x) + e^{- \frac{x}{2}}(c_0cos \pi x- \pi c_0 x sin \pi x +c_1 sin \pi x + \pi c_1 x cos \pi x) $$ $$ (\pi^2 + \frac{1}{4})e^{- \frac{x}{2}}(c_0xcos \pi x+c_1xsin \pi x)= e ^{-\frac{x}{2}}sin\pi x $$ (Eq.8.2.15) Solving equation 8.2.15 for $$c_0$$ and $$c_1$$
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$$  \displaystyle c_0= - \frac{1}{2 \pi} $$     (Eq.8.2.16)
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$$  \displaystyle c_1=0 $$     (Eq.8.2.17) Substituting equation 8.2.16 and 8.2.17 into equation 8.2.12:
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$$  \displaystyle y_p=-\frac{1}{2 \pi}e^{- \frac{x}{2}}xcos \pi x $$ (Eq.8.2.18) Since the general solution is:
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$$  \displaystyle y=y_h+y_p $$     (Eq.8.2.19) Substitute equations 8.1.7 and 8.1.14 into equation 8.1.15:
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$$  \displaystyle y(x)=e^{- \frac{x}{2}}(C_1cos \pi x+C_2sin \pi x)-\frac{1}{2 \pi}e^{- \frac{x}{2}}xcos \pi x $$ (Eq.8.2.20)
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Author & Proofreaders
Author: Egm4313.s12.team17.wheeler.tw 03:13, 20 February 2012 (UTC) Proofreader: Egm4313.s12.team17.hintz 04:13, 20 February 2012 (UTC) Editor: Egm4313.s12.team17.deaver.md 11:45, 22 February 2012 (UTC)

=Problem R3.9 - Method of Undetermined Coefficients=

Part 1 Given

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$$  \displaystyle 8y'' - 6y' + y = 6coshx $$
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$$  \displaystyle y(0) = 0.2,y'(0) = 0.05 $$     (
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Part 1 Problem Statement
Find the solution to the given ODE and state which rule you are using.

Part 1 Solution
With the given equation above, the general solution of the homogeneous ODE can be solved for:
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$$  \displaystyle 8\lambda^2 - 6\lambda + 1 = (4\lambda - 1)(2\lambda - 1) = 0 $$     (Eq.9.1.1)
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$$  \displaystyle \lambda = \frac{1}{4},\frac{1}{2} $$     (Eq.9.1.2) Which then gives the general solution:
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$$  \displaystyle y_h = c_1e^{x/4} + c_2e^{x/2} $$     (Eq.9.1.3) The excitation $$6coshx$$ can be represented as:
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$$  \displaystyle r(x) = coshx = 6\frac{e^x + e^{-x}}{2} = r_1(x) + r_2(x) $$     (Eq.9.1.4) Then looking at Table 2.1 in the textbook, it gives the necessary terms for the particular solution using the sum rule and basic rule:
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$$  \displaystyle r_1(x) = 3e^{x} \rightarrow y_{p1}(x) = C_1e^{x} $$     (Eq.9.1.5)
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$$  \displaystyle r_2(x) = 3e^{-x} \rightarrow y_{p2}(x) = C_2e^{-x} $$     (Eq.9.1.6) Therefore, the particular solution is:
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$$  \displaystyle y_p(x) = y_{p1} + y_{p2} = C_1e^{x} + C_2e^{-x} $$     (Eq.9.1.7) The first and second derivatives of the particular solution is then:
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$$  \displaystyle y_p' = C_1e^{x} - C_2e^{-x} $$     (Eq.9.1.8)
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$$  \displaystyle y_p'' = C_1e^{x} + C_2e^{-x} $$     (Eq.9.1.9) Then substituting these equations into the particular solution equation $$y''_p + ay'_p + by_p = r(x)$$ where a and b are $$-6/8$$ and $$1/8$$ respectively, gives the equation:
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$$  \displaystyle C_1e^{x} + C_2e^{-x} - \frac{6}{8}(C_1e^{x} - C_2e^{-x}) + \frac{1}{8}(C_1e^{x} + C_2e^{-x}) = \frac{1}{8}(3e^x + 3e^{-x}) $$     (Eq.9.1.10) Combining terms together and multiplying the entire equation by 8, $$C_1$$ and $$C_2$$ are
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$$  \displaystyle 3C_1e^{x} + 15C_2e^{-x} = 3e^x + 3e^{-x} $$     (Eq.9.1.11)
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$$  \displaystyle 3C_1 = 3 $$     (Eq.9.1.12)
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$$  \displaystyle C_1 = 1 $$     (Eq.9.1.13)
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$$  \displaystyle 15C_2= 3 $$     (Eq.9.1.14)
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$$  \displaystyle C_2= \frac{1}{5} $$     (Eq.9.1.15) Therefore the particular solution is:
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$$  \displaystyle y_p'' = e^{x} + \frac{1}{5}e^{-x} $$     (Eq.9.1.16)
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The final solution is the sum of the particular and the homogeneous solution:
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$$  \displaystyle y(x) = c_1e^{x/4} + c_2e^{x/2} + e^{x} + \frac{1}{5}e^{-x} $$     (Eq.9.1.17)
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Now using the initial conditions given, $$c_1$$ and $$c_2$$ can be solved for:
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$$  \displaystyle 0.2 = y(0) = c_1 + c_2 + 1 + \frac{1}{5} $$     (Eq.9.1.18)
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$$  \displaystyle 0.05 = y'(0) = \frac{1}{4}c_1 + \frac{1}{2}c_2 + 1 - \frac{1}{5} $$     (Eq.9.1.19)
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$$  \displaystyle -1 = c_1 + c_2 $$ (Eq.9.1.20)
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$$  \displaystyle -\frac{3}{4} = \frac{1}{4}c_1 + \frac{1}{2}c_2 $$     (Eq.9.1.21)
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Solving for $$c_1$$ in Eq.9.1.20 and substituting it into Eq.9.1.21 and solving for $$c_2$$ gives
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$$  \displaystyle c_2 = -2 $$     (Eq.9.1.22) Back substituting it into Eq.9.1.20 to solve for $$c_1$$ gives:
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$$  \displaystyle c_1 = 1 $$     (Eq.9.1.23)
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Finally, the solution to the homogeneous equation is then:
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$$  \displaystyle y(x) = e^{x/4} - 2e^{x/2} + e^{x} + \frac{1}{5}e^{-x} $$     (Eq.9.1.24)
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Part 2 given

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$$  \displaystyle y'' + 4y' + 4y = e^{-2x}\sin2x $$
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$$  \displaystyle y(0) = 1,y'(0) = -1.5 $$
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Part 2 Problem Statement
Find the solution to the given ODE and state which rule you are using.

Part 2 Solution
With the given equation above, the general solution of the homogeneous ODE can be solved for:
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$$  \displaystyle \lambda^2 + 4\lambda + 4 = (\lambda + 2)(\lambda + 2) = 0 $$     (Eq.9.2.1)
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$$  \displaystyle \lambda = -2,-2 $$     (Eq.9.2.2) The solutions is a double root, therefore the homogeneous solution is then:
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$$  \displaystyle y_h = c_1e^{-2x} + c_2xe^{-2x} $$     (Eq.9.2.3)
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The excitation of this problem is:
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$$  \displaystyle r(x) = e^{-2x}sin2x $$     (Eq.9.2.4) Looking at Table 2.1 in the textbook, it gives the necessary terms for the particular solution using the basic rule:
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$$  \displaystyle r(x) = e^{-2x}sin2x \rightarrow y_{p}(x) = e^{-2x}\Big(Kcos(2x) + Msin(2x)\Big) $$     (Eq.9.2.5) Thus the particular solution is:
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$$  \displaystyle y_p(x) = e^{-2x}\Big(Kcos(2x) + Msin(2x)\Big) $$     (Eq.9.2.6) Taking the first and second derivatives of the particular solution is
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$$  \displaystyle y_p' = -2e^{-2x}\Big((K+M)sin(2x) + (K-M)cos(2x)\Big) $$     (Eq.9.2.7)
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$$  \displaystyle y_p'' = e^{-2x}\Big(8Ksin(2x) - 8Mcos(2x)\Big) $$     (Eq.9.2.8) Then substituting these equations into the particular solution equation $$y''_p + ay'_p + by_p = r(x)$$
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$$  \displaystyle e^{-2x}\Big(8Ksin(2x) - 8Mcos(2x)\Big) + 4\Big(-2e^{-2x}[(K+M)sin(2x) + (K-M)cos(2x)]\Big) + 4\Big(e^{-2x}(Kcos(2x) + Msin(2x))\Big) = e^{-2x}sin2x $$     (Eq.9.2.9)
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Combining similar terms together, Eq. 9.2.9 then becomes
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$$  \displaystyle e^{-2x}\Big((8K - 8K - 8M +4M)sin(2x) + (-8M-8K+8M+4K)cos(2x)\Big) = e^{-2x}sin2x $$     (Eq.9.2.10) Now matching terms on both sides of the equation the following equations are set up:
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Matching Sine terms:
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$$  \displaystyle 8K - 8K - 8M +4M = - 4M = 1 $$     (Eq.9.2.11) Matching Cosine terms:
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$$  \displaystyle -8M-8K+8M+4K = - 8K = 0 $$     (Eq.9.2.12) Solving for constants:
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$$  \displaystyle K=0 $$     (Eq.9.2.13)
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$$  \displaystyle M=-\frac{1}{4} $$     (Eq.9.2.14) Therefore, the particular solution is:
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$$  \displaystyle y_p(x) = -\frac{1}{4}e^{-2x}\sin2x $$     (Eq.9.2.15) The final solution can then be written as:
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$$  \displaystyle y(x) = y_h(x) + y_p(x) = c_1e^{-2x} + c_2xe^{-2x} -\frac{1}{4}e^{-2x}\sin2x $$     (Eq.9.2.16) Using the initial conditions given above, $$y(0) = 1,$$ $$y'(0) = -1.5$$, a system of equations can be set up to solve for the constants $$c_1$$ and $$c_2$$ :
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$$  \displaystyle y(0) = c_1 + 0 = 1 $$     (Eq.9.2.17)
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$$  \displaystyle y'(0) = -2c_1 + c_2 - \frac{1}{2} = -1.5 $$     (Eq.9.2.18)
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$$  \displaystyle c_1 = 1 $$     (Eq.9.2.19)
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$$  \displaystyle c_2 = 1 $$     (Eq.9.2.20)
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The final solution is therefore
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$$  \displaystyle y(x) = e^{-2x} + xe^{-2x} -\frac{1}{4}e^{-2x}\sin2x $$     (Eq.9.2.21)
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Author & Proofreaders
Author:Egm4313.s12.team17.ying 03:14, 19 February 2012 (UTC)

Proofreader: Egm4313.s12.team17.Li 07:00, 20 February 2012 (UTC)

Editor: Egm4313.s12.team17.deaver.md 16:15, 22 February 2012 (UTC)

=Problem R3.10 - General Formulation for Polynomial Excitation=

Given
Given the equation:
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$$  \displaystyle y''+ay'+by= \sum_{j=0}^{n}k_jx^j $$
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The equation for the coefficients of $$x^0$$ :
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$$  \displaystyle 2c_2+ac_1+bc_0=k_0 $$
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Problem Statement
Obtain equations for $$x^1$$, $$x^2$$ , $$x^{n-2}$$ , $$x^{n-1}$$ , and $$x^{n-1}$$ and insert those equations as well as the given equation for $$x^0$$ into a tri-diagonal matrix.

Solution
The order of the given equation is defined as the highest power of $$ x $$ in Eq. 10.1 below:
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$$  \displaystyle y''+ay'+by=\sum_{j=0}^{n}k_jx^j $$     (Eq.10.1) Therefore, the order of Eq. 10.1 is $$n$$ and the coefficients of the solution may be found using the below sum rule:
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$$  \displaystyle y(x)=\sum_{j=0}^s c_j x^j $$     (Eq.10.2) Where $$ s $$ in Eq. 5.1.2 is equal to the order of the Eq. 10.1 which was defined above to be $$n$$ :
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$$  \displaystyle y(x)=\sum_{j=0}^n c_j x^j $$     (Eq.10.3) Eq. 10.3 may be substituted into Eq. 10.1 for $$ y $$ and below are the corresponding substitutions for $$ y' $$ and $$ y'' $$ :
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$$  \displaystyle y'(x)=\sum_{j=0}^n j c_j x^{(j-1)}=\sum_{j=1}^n j c_j x^{(j-1)} $$     (Eq.10.4)
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$$  \displaystyle y''(x)=\sum_{j=0}^n j(j-1) c_j x^{(j-2)}=\sum_{j=2}^n j(j-1) c_j x^{(j-2)} $$     (Eq.10.5) Substituting Eq. 10.3, Eq. 10.4, and Eq. 10.5 into Eq. 10.1 yields the below equation:
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$$  \displaystyle ( \sum_{j=2}^n j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=1}^n j c_j x^{(j-1)} )+b( \sum_{j=0}^n c_j x^j )=\sum_{j=0}^n k_jx^j $$     (Eq.10.6) To obtain equation for the coefficients of $$x^1$$. The $$ j $$ values of Eq.10.6 are adjusted for each summation such that the exponent of x will be equal to 1:
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$$  \displaystyle ( \sum_{j=3}^3 j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=2}^2 j c_j x^{(j-1)} )+b( \sum_{j=1}^1 c_j x^j )=\sum_{j=1}^1k_jx^j $$     (Eq.10.7) Which is further simplified below:
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$$  \displaystyle (3(3-1) c_3 x^{(3-2)} ) +a( 2 c_2 x^{(2-1)} )+b(c_1 x^1 )=k_1x^1 $$     (Eq.10.8)
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$$  \displaystyle (3)2c_3x^1+a(2c_2x^1)+bc_1x^1=k_1x $$     (Eq.10.9)
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$$  \displaystyle 6c_3x^1+2ac_2x^1+bc_1x^1=k_1x $$     (Eq.10.10)
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$$  \displaystyle 2(3c_3+ac_2+bc_1)x^1=k_1x^1 $$     (Eq.10.11) And therefore the coefficients of $$ x^1 $$ are:
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$$  \displaystyle 2(3c_3+ac_2+bc_1)=k_1 $$     (Eq.10.12) To obtain equation for the coefficients of $$x^2$$. The $$ j $$ values of Eq. 10.6 are adjusted for each summation such that the exponent of x will be equal to 2:
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$$  \displaystyle ( \sum_{j=4}^4 j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=3}^3 j c_j x^{(j-1)} )+b( \sum_{j=2}^2 c_j x^j )=\sum_{j=2}^2 k_2x^2 $$     (Eq.10.13) Which is further simplified below:
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$$  \displaystyle ( 4(4-1) c_4 x^{(4-2)} ) +a( 3 c_3 x^{(3-1)} )+b(c_2 x^2 )=k_2x^2 $$     (Eq.10.14)
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$$  \displaystyle ( (4)3 c_4 x^{(4-2)} ) +a( 3 c_3 x^{(2)} )+bc_2 x^2=k_2x^2 $$     (Eq.10.15)
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$$  \displaystyle 12c_4 x^{(2)} +3a c_3 x^{(2)} +bc_2 x^2 =k_2x^2 $$     (Eq.10.16)
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$$  \displaystyle (12c_4 +3a c_3 +bc_2 )x^2 =k_2x^2 $$     (Eq.10.17) And therefore the coefficients of $$ x^2 $$ are:
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$$  \displaystyle 12c_4 +3a c_3 +bc_2=k_2 $$     (Eq.10.18) To obtain equation for the coefficients of $$x^3$$. The $$ j $$ values of Eq. 10.6 are adjusted for each summation such that the exponent of x will be equal to $$n-2$$ :
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$$  \displaystyle ( \sum_{j=n}^{n} j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=n-1}^{n-1} j c_j x^{(j-1)} )+b( \sum_{j=n-2}^{n-2} c_j x^j )=\sum_{j=n-2}^{n-2} k_{n-2}x^{n-2} $$     (Eq.10.19) Which is further simplified below:
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$$  \displaystyle ( n(n-1) c_{n} x^{(n-2)} ) +a( (n-1) c_{n-1} x^{(n-1-1)} )+b( c_{n-2} x^{n-2} )= k_{n-2}x^{n-2} $$     (Eq.10.20)
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$$  \displaystyle ( n(n-1) c_{n} x^{(n-2)} ) +a( (n-1) c_{n-1} x^{(n-2)} )+b( c_{n-2} x^{n-2} )= k_{n-2}x^{n-2} $$     (Eq.10.21)
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$$  \displaystyle ((n-1)( n c_{n} x^{(n-2)} +(a)c_{n-1} )+bc_{n-2} )x^{n-2}= k_{n-2}x^{n-2} $$     (Eq.10.22) And therefore the coefficients of $$ x^3 $$ are:
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$$  \displaystyle ((n-1)( n c_{n} x^{(n-2)} +a*c_{n-1} )+bc_{n-2} )= k_{n-2} $$     (Eq.10.23) To obtain equation for the coefficients of $$x^{n-1}$$. The $$ j $$ values of Eq. 10.6 are adjusted for each summation such that the exponent of x will be equal to $$n-1$$ :
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$$  \displaystyle ( \sum_{j=n+1}^{n} j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=n}^{n} j c_j x^{(j-1)} )+b( \sum_{j=n-1}^{n-1} c_j x^j )=\sum_{j=n-1}^{n-1} k_j x^j $$     (Eq.10.24) Since, the bounds of a summation in Eq. 10.24 is exceeded in one case that summation is simplified to zero as shown below:
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$$  \displaystyle a( n c_n x^{(n-1)} )+b(c_{n-1} x^{n-1} )=k_{n-1} x^{n-1} $$     (Eq.10.25) Which is further simplified below:
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$$  \displaystyle (an c_n +bc_{n-1}) x^{n-1} =k_{n-1} x^{n-1} $$     (Eq.10.26) And therefore the coefficients of $$ x^{n-1} $$ are:
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$$  \displaystyle (an c_n +bc_{n-1}) =k_{n-1} $$     (Eq.10.27)
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To obtain equation for the coefficients of $$x^n$$. The $$ j $$ values of Eq. 10.6 are adjusted for each summation such that the exponent of x will be equal to $$n$$ :
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$$  \displaystyle ( \sum_{j=n+2}^{n} j(j-1) c_j x^{(j-2)} ) +a( \sum_{j=n+1}^{n} j c_j x^{(j-1)} )+b( \sum_{j=n}^{n} c_j x^j )=\sum_{j=n}^{n} k_j x^j $$     (Eq.10.28) Since, the bounds of the summations in Eq. 10.28 are exceeded in two cases those summation are simplified to zero as shown below:
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$$  \displaystyle b(c_n x^n) =k_n x^n $$     (Eq.10.29) Which is further simplified below:
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$$  \displaystyle (bc_n) x^n =k_n x^n $$     (Eq.10.29) And therefore the coefficients of $$ x^n $$ are:
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$$  \displaystyle bc_n=k_n $$     (Eq.10.30) The following equations Eq. 10.12, Eq. 10.18, Eq. 10.23, Eq. 10.27, and Eq. 10.30 as well as the given equation for the coefficients of $$x^0$$ can have the left hand side of the equations inserted into a tri-diagonal matrix as shown below:
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$$  \displaystyle \begin{bmatrix} b & a & 2 & 0 & 0 & 0\\ 0 & 2b & 2a & 6 & 0 & 0\\ 0 & 0 & b & 3a & 12 & 0\\ 0 & 0 & 0 & b & (n-1)a & n^2-n\\ 0 & 0 & 0 & 0 & b & a*n\\ 0 & 0 & 0 & 0 & 0 & b \end{bmatrix} $$     (Eq.10.31)
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Author & Proofreaders
Author: Egm4313.s12.team17.axelrod.a 20:41, 15 February 2012 (UTC)

Proofreader 1: Egm4313.s12.team17.hintz 21:30, 20 February 2012 (UTC)

Proofreader 2: Egm4313.s12.team17.deaver.md 15:36, 22 February 2012 (UTC))

Editor: Egm4313.s12.team17.deaver.md 15:36, 22 February 2012 (UTC)

= Contributing Team Members =

title: User:Egm4313.s12.team17/Report 3


text: User:Vanished user adhmfdfmykrdyr/ACTA

title: User:Vanished user adhmfdfmykrdyr/ACTA


title: File:Butterfly.HW.1.A.Idea.20120211.pdf


title: File:CORDIC.Background.4.A.20120213.pdf


text:
TAO/Concept

''The content of this page is currently being developed. Please got to Talk:TAO/Concept to contribute. --Simon Lüke 11:59, 16 February 2012 (UTC)''

This page is meant to give you an overview on the TAO project, especially on its relation to the part taking place in Wikiversity. In the TAO project as whole there are lots of activities going on that are not oder not very comprehensively represented by Wikiversity pages, but there could be - for example - a researcher that wants to document experiences and results of such activities in Wikiversity.

Graphic
At current stage this graphic shall be the main resource and reference to discuss the concept. It actually contains too much information to be easily comprehensible, probably not even by a second or third glance. The idea is to develop an overall reference to explain things and to use it by cutting out single aspects in regard of a specific purpose.

To Edit: https://docs.google.com/drawings/d/1B2EwgHpT9Cah-ZrZFptFqlv6fRzyKUyGoe48eeHXKew/edit
 * (Goolge Docs Drawing, which contains the respectively latest version.)

Formats:
 * (This files (in Wikiversity) will be updated from time to time from above Google Doc)

Click to get full resolution of PNG.

Wording
Definition of terms for better understanding:
 * TAO/Wikiversity for the part of the TAO project which is directly related to Wikiversity.
 * More to come after discussion and agreement on the content of the graphic...
 * More to come after discussion and agreement on the content of the graphic...

title: TAO/Concept


text: File:TAO-Wikiversity-Explanation and Wording-A cheats heet.svg