User:Egm4313.s12.team17/Report 5
Contents
 1 Problem R5.1  Radius of Convergence
 1.1 Part 1 Given
 1.2 Part 1 Problem Statement
 1.3 Part 1 Solution
 1.4 Part 2 Given
 1.5 Part 2 Problem Statement
 1.6 Part 3 Given
 1.7 Part 3 Problem Statement
 1.8 Part 3 Solution
 1.9 Part 4 Given
 1.10 Part 4 Problem Statement
 1.11 Part 4 Solution
 1.12 Part 5 Given
 1.13 Part 5 Problem Statement
 1.14 Part 5 Solution
 1.15 Author & Proofreaders
 2 Problem R5.2  Linear Independent Testing
 3 Problem R5.3  Linear Independent Test using Gramain
 4 Problem R5.4  Particular Solution Verification
 5 Problem R5.5  L2ODECC with Linear Independent Testing
 6 Problem R5.6  Solving for the Unknown Coefficients
 7 Problem R5.7  Projection on a Basis for Vectors
 8 Problem R5.8  Integration
 9 Problem R5.9  Projection on a Basis
 10 Contributing Team Members
Problem R5.1  Radius of Convergence[edit]
Part 1 Given[edit]
Find for the following series

(Eq.1.1.1)
Part 1 Problem Statement[edit]
Find for the following series.
Part 1 Solution[edit]
Using the following formula:

(Eq.1.1.2)
Where for a power series:

(Eq.1.1.3)

(Eq.1.1.4)
Where:

(Eq.1.1.5)

(Eq.1.1.6)
Plugging equations 1.1.5 and 1.1.6 into equation 1.1.2:

(Eq.1.1.7)
Part 2 Given[edit]

(Eq.1.2.1)
Part 2 Problem Statement[edit]

(Eq.1.2.2)

(Eq.1.2.3)
Where:

(Eq.1.2.4)

(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

(Eq.1.2.6)
Part 3 Given[edit]

(Eq.1.3.1)
Part 3 Problem Statement[edit]
Find for the given Taylor series.
Part 3 Solution[edit]
Performing a Maclaurin series expansion:

(Eq.1.3.2)

(Eq.1.3.3)
Where:

(Eq.1.3.4)

(Eq.1.3.5)
Using Eq.1.1.2, plug in Eq.1.3.4 and Eq.1.3.5

(Eq.1.3.6)
Part 4 Given[edit]

(Eq.1.4.1)
Part 4 Problem Statement[edit]
Find for the given Taylor series.
Part 4 Solution[edit]
Performing a Maclaurin series expansion:

(Eq.1.4.2)
Where:

(Eq.1.4.3)

(Eq.1.4.4)
Plugging Eq.1.4.3 and Eq.1.4.4 into Eq.1.1.2:

(Eq.1.4.5)
Part 5 Given[edit]

(Eq.1.5.1)
Part 5 Problem Statement[edit]
Find for the given Taylor series.
Part 5 Solution[edit]
Performing a Taylor series expansion about , results in the following series:

(Eq.1.5.2)
This simplifies to:

(Eq.1.5.3)
Therefore and are then:

(Eq.1.5.4)

(Eq.1.5.5)
Plugging Eq.1.5.4 and Eq.1.5.5 into Eq.1.1.2

(Eq.1.4.6)
Author & Proofreaders[edit]
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[edit]
Part 1 Given[edit]

(Eq.2.1.1)

(Eq.2.1.2)

(Eq.2.1.3)

(Eq.2.1.4)
Part 1 Problem Statement[edit]
Determine whether the following pair of functions are linearly independent using the Wronskian.
Part 1 solution[edit]
The Wronskian of the two functions is written as

(Eq.2.1.5)
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

(Eq.2.1.6)

(Eq.2.1.7)
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

(Eq.2.1.8)

Therefore Eq.2.1.1 and Eq.2.1.2 are Linearly Independent.
Using the second set of functions, Eq.2.1.3 and Eq.2.1.4, the corresponding derivatives are

(Eq.2.1.9)
and

(Eq.2.1.10)
Plugging Eq.2.1.3,Eq.2.1.4,Eq.2.1.9, and Eq.2.1.10 into Eq.2.1.5 gives:

(Eq.2.1.11)
These equations cannot be simplified so

(Eq.2.1.12)

Thus the two functions are linearly independent.
Part 2 Given[edit]
Same as part 1 given
Part 2 Problem Statement[edit]
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[edit]
The gramian is given by

(Eq.2.2.1)
Where

(Eq.2.2.2)
If the Gramian, Eq.2.2.1

(Eq.2.2.3)
Then the two functions are linearly independent
Now to calculate the matrix values:

(Eq.2.2.4)
Which then equals

(Eq.2.2.5)
The other terms are as follows:

(Eq.2.2.6)

(Eq.2.2.7)

(Eq.2.2.8)
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

(Eq.2.2.9)
Eq.2.2.9 then results in:

(Eq.2.2.10)

Thus the two functions are linearly independent.
Using Eq.2.2.2, the following scalar product values are calculated.

(Eq.2.2.11)

(Eq.2.2.12)

(Eq.2.2.13)

(Eq.2.2.14)
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

(Eq.2.2.15)
Which finally becomes

(Eq.2.2.16)

Thus the two functions are linearly independent.
Author & Proofreaders[edit]
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[edit]
Given[edit]

(Eq.3.1)

(Eq.3.2)
Problem Statement[edit]
Verify that and are linearly independent using the Gramian
Solution[edit]
The Grammian is given as:

(Eq.3.3)
According to (3) on p.89

(Eq.3.4)
Calculating the dot products using Eq.3.1 and Eq.3.2:

(Eq.3.5)

(Eq.3.6)

(Eq.3.7)

(Eq.3.8)
Plugging Eq.3.5, Eq.3.6, Eq.3.7, and Eq.3.8 into Eq.3.3 and finding the determinant gives:

(Eq.3.9)

Thus the two functions are linearly independent.
Author & Proofreaders[edit]
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[edit]
Given[edit]

(Eq.4.1)

(Eq.4.2)

(Eq.4.3)
Problem Statement[edit]
Show that Eq. 4.1 is indeed the overall particular solution of the L2OECC, Eq. 4.2
Discuss the choice of for and in
Solution[edit]
The particular solution for Eq. 4.2 is:
Since the excitation is Eq. 4.3, the particular solution is then

(Eq.4.4)
If we expand Eq.4.1

(Eq.4.5)
We have :
Now we can see that the particular solution for a simple excitation satisfies Eq. 4.2:

(Eq.4.6)
Because of linearity, when each particular solution is known, then we see that the overall particular solution is:

(Eq.4.7)

Therefore, Eq.4.1 is indeed the particular solution to Eq.4.2.
Explanation
When the excitation to an equation, is:
We first think of using a particular solution of:
However if we were to take the derivatives of the particular solution and substitute back into the equation, we have:
Notice that has appeared. And it has the same coefficient 'A' as well. When equating common coefficients we see that 'A' must equal 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 . So,the particular solution would be:
In conclusion, if the excitation, includes a sine or cosine, the particular solution must be in the form of:
where 'N' and 'M' are constants.
Author & Proofreaders[edit]
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  L2ODECC with Linear Independent Testing[edit]
Part 1 Given[edit]

(Eq.5.1.1)

(Eq.5.1.2)
Part 1 Problem Statement[edit]
Show that and are linearly independent using the Wronskian and the Gramian (integrate over 1 period).
Part 1 Solution[edit]
Wronskian Method
By Definition

(Eq.5.1.3)
In this problem:

(Eq.5.1.4)

(Eq.5.1.5)
Taking the derivative of Eq.5.1.4 and Eq.5.1.5 respectively:

(Eq.5.1.6)

(Eq.5.1.7)
Plugging equations Eq.5.1.4 through Eq.5.1.7 into Eq.5.1.3

(Eq.5.1.8)

Thus the functions are linearly independent.
Gramian Method
By definition:

(Eq.5.1.9)
Where

(Eq.5.1.10)
In this problem:

(Eq.5.1.11)

(Eq.5.1.12)

(Eq.5.1.13)
Applying Eqs.5.1.11 through Eq.5.1.13 to Eq.5.1.10:

(Eq.5.1.14)

(Eq.5.1.15)

(Eq.5.1.16)

(Eq.5.1.17)
Plugging in values form Eq.5.1.14 to Eq.5.1.17 into Eq.5.1.9:

(Eq.5.1.18)

Thus the functions are linearly independent.
Part 2 Given[edit]
Same as Part 1 Given
Part 2 Problem Statement[edit]
Find 2 equations for the two unknowns M,N, and solve for M,N.
Part 2 Solution[edit]
The ODE to be solved is:

(Eq.5.2.1)
The particular solution will take the following form:

(Eq.5.2.2)
Take the first and second derivatives of Eq.5.2.2:

(Eq.5.2.3)

(Eq.5.2.4)
Plugging Eq.5.2.2 and Eq.5.2.3 into Eq.5.2.1:

(Eq.5.2.5)

(Eq.5.2.6)
By equating like terms, the two equations for solving for M and N are:

(Eq.5.2.7)

(Eq.5.2.8)
Solving the system of equations created by Eq.5.2.7 and Eq.5.2.8:

(Eq.5.2.9)

(Eq.5.2.10)
Part 3 Given[edit]
Initial conditions
Part 3 Problem Statement[edit]
Find the overall solution and plot the solution over 3 periods
Part 3 Solution[edit]
For the overall solution:

(Eq.5.3.1)
To find the homogeneous solution, solve:

(Eq.5.3.2)
The characteristic equation of Eq.5.3.2 is:

(Eq.5.3.3)
To solve the characteristic equation, apply the quadratic formula:

(Eq.5.3.4)

(Eq.3.3.5)

(Eq.5.3.6)

(Eq.5.3.7)
Therefore the homogeneous solution takes the following form:

(Eq.5.3.8)
Taking the first derivative of Eq.5.3.8:

(Eq.5.3.9)
Apply the boundary conditions and to Eq.5.3.8 and Eq.5.3.9:

(Eq.5.3.10)

(Eq.5.3.11)
Solving Eq.5.3.10 and Eq.5.3.11:

(Eq.5.3.12)

(Eq.5.3.13)
The homogeneous solution becomes:

(Eq.5.3.14)
Therefore from Eq.5.3.1 the overall solution is:

(Eq.5.3.15)
Graph
Matlab Code
clear all; %R5.5 graph the complete solution over 3 periods %y(x)=0.045cos7x0.016sin7x+2/7e^(5x)+5/7e^(2x) x = 0:0.2:(3*0.8976);%Domnain over 3 periods y = 0.045*cos(7*x)0.016*sin(7*x)+(2/7)*exp(5*x)+(5/7)*exp(2*x); plot(x,y); %Plot graph %Add labels title('R5.5 Graph'); xlabel('x'); ylabel('y(x)');
Author & Proofreaders[edit]
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[edit]
Given[edit]

(Eq.6.1)

(Eq.6.2)

(Eq.6.3)
Initial conditions
Problem Statement[edit]
Determine the overall solution that corresponds to the initial conditions.
Plot the general solution over 3 periods.
Solution[edit]
Combining the given homogeneous and particular solutions gives us the overall form:

(Eq.6.4)
Applying the initial condition to Eq.6.4 gives us:

(Eq.6.5)
Which is simplified below:

(Eq.6.6)

(Eq.6.7)

(Eq.6.8)
Taking the derivative of Eq.6.4 gives us

(Eq.6.9)
Inserting the initial condition y'(0)=0 into Eq.6.9 we get:

(Eq.6.10)
Which is further simplified to:

(Eq.6.11)
Since :

(Eq.6.12)
We then get the relation:

(Eq.6.13)
Taking the derivative of Eq.6.9 gives us.

(Eq.6.14)
We then insert Eq.6.3, Eq.6.9, and Eq.6.14 into the equation
:

(Eq.6.15)
Which is then simplified:

(Eq.6.16)
Dividing Eq.6.16 by yields:

(Eq.6.17)
We then split Eq.6.17 into a system of equations to determine the value of the coefficients:

(Eq.6.18)

(Eq.6.19)
From Eq.6.18 and Eq.6.19 we can deduce that and .
Inserting the value of M into Eq.6.13 yields:

(Eq.6.20)

(Eq.6.21)
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:

(Eq.6.22)
Making the overall solution:

(Eq.6.23)
Below is a plot of the solution:
Graph
Matlab Code
clear all; %R6 graph the complete solution over 3 periods %y(x)=xe^{2x}((1/2)*sin(3x))+ e^{2x}(1*cos(3x)+(2/3)*sin(3x)) x=0:0.1:6.3;%Domnain over 3 periods from 0 to 6.3 for n=1:1:64; plot(x(n),x(n)*exp(2*x(n))*(1/3)*sin(3*x(n))+exp(2*x(n))*(cos(3*x(n))+(2/3)*sin(3*x(n))),'r:+');hold all end%This loop plots the desired graph %Add labels title('R5.6 Graph'); xlabel('x'); ylabel('y(x)');
Author & Proofreaders[edit]
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[edit]
Given[edit]

(Eq.7.1)

(Eq.7.2)

(Eq.7.3)
Part 1 Problem Statement[edit]
Find using the Gram Matrix
Part 1 Solution[edit]
The Grammian is given as:

(Eq.7.1.1)
According to (3) on p.89

(Eq.7.1.2)
Calculating the dot products:

(Eq.7.1.3)
Plugging Equations 7.1.3 into 7.1.1 gives:

(Eq.7.1.4)
Next we have to get an equation for d, as shown in (5) on pg.810 *notice v is given under heading of problem 5.7

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

(Eq.7.1.6)
From the notes (1) on pg.811

(Eq.7.1.6)
This gives:

(Eq.7.1.7)
So and are:

(Eq.7.1.8)
Part 2 Problem Statement[edit]
(2)Verify the result by using (1)(2) p.7c34 in (2)and rely on the nonzero determinant of the matrix of components of relative to the basis as discussed on p.7c34
Part 2 Solution[edit]
The equations to be entered into the A matrix are as follows:

(Eq.7.2.1)
According to the notes if these two equations are put into a matrix and the determinant is then they are linearly independent. Entering them into a matrix gives:

(Eq.7.2.2)
Solving:

(Eq.7.2.3)
Thus and are linearly independent.
Use the following equation to check the answer from the previous part.

(Eq.7.2.4)

(Eq.7.2.4)

(Eq.7.2.5)
Author & Proofreaders[edit]
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[edit]
Given[edit]

(Eq. 8.1)
With
Integration by Parts

(Eq.8.2)
General Binomial Theorem

(Eq.8.3)
Where
Problem Statement[edit]
Find the indefinite integral below with using the integration by parts and the General Binomial Theorem
Solution[edit]
Using the integration by parts formula, Eq.8.2, we set and so we have the following:
Substituting the above in to Eq. 8.3 we have:

(Eq.8.4)
Now we can set and Eq. 8.4 becomes:

(Eq.8.5)
Where
Can be solved by first using long division on the integrand:
Next integrate term by term
So when we have:

(Eq.8.6)
Now we substitute into Eq. 8.4 which gives us:

(Eq.8.7)
And we have the integral:
Which can also be solved by first using long division on the integrand , so:
Integrating term by term we have that:
So Eq. 8.7 becomes:

(Eq.8.8)
So when :

(Eq.8.9)
Author & Proofreaders[edit]
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[edit]
Part 1 Given[edit]

(Eq.9.1.1)

(Eq.9.1.2)

(Eq.9.1.3)

(Eq.9.1.4)

(Eq.9.1.5)
Part 1 Problem Statement[edit]
Project equation 9.1.2 on the polynomial basis such that it takes the form shown below for and :

(Eq.9.1.6)
Plot equations 9.1.2 and 9.1.6 to show the uniform approximation and convergence.
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[edit]
Let .
Therefore, equation 9.1.5 becomes:

(Eq.9.1.7)
Set up a Gram matrix for the basis functions (eq. 9.1.7):

(Eq.9.1.8)

(Eq.9.1.9)
Let and .

(Eq.9.1.10)

(Eq.9.1.11)
Therefore,

(Eq.9.1.12)
Rhs matrix will take the form shown below:

(Eq.9.1.13)
Therefore,

(Eq.9.1.14)

(Eq.9.1.15)

(Eq.9.1.16)

(Eq.9.1.17)

(Eq.9.1.18)

(Eq.9.1.19)

(Eq.9.1.20)
Using the equation shown below, solve for the d values:

(Eq.9.1.21)

(Eq.9.1.22)
Therefore,

(Eq.9.1.23)

(Eq.9.1.24)
This result will make equation 9.4.6 the following:

(Eq.9.1.25)

(Eq.9.1.26)
Let .
Therefore, equation 9.1.5 becomes:

(Eq.9.1.27)
Set up a Gram matrix for the basis functions (eq. 9.1.27):

(Eq.9.1.28)

(Eq.9.1.29)

(Eq.9.1.30)

(Eq.9.1.31)

(Eq.9.1.32)

(Eq.9.1.33)

(Eq.9.1.34)
Therefore,

(Eq.9.1.35)
Generate a rhs matrix:

(Eq.9.1.36)

(Eq.9.1.37)

(Eq.9.1.38)

(Eq.9.1.39)

(Eq.9.1.40)

(Eq.9.1.41)

(Eq.9.1.42)
Using the equation 9.1.21, to solve for the d values:

(Eq.9.1.43)

(Eq.9.1.44)

(Eq.9.1.45)
This result will make equation 9.4.6 the following:

(Eq.9.1.46)

(Eq.9.1.47)
The plot below shows the projected and actual excitation.
Below is the MATLAB code used to generate the above graph:
x=[.75:0.01:3]; r=log(x+1); r0=0.5711+1x.^0; r1=0.0940+0.5912*x; plot(x,r,x,r0,x,r1) xlabel('xaxis'); ylabel('raxis');
Taylor series expansion for equation 9.1.4 are shown below:

(Eq.9.1.48)

(Eq.9.1.49)
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:
x=[.75:0.01:3]; r=log(x+1); r0=0.5711+1x.^0; r1=0.0940+0.5912*x; t0=x; t1=xx.^2/2; subplot(211) plot(x,r,x,r0,x,t0) xlabel('xaxis'); ylabel('raxis'); title('Approximation when n=0') subplot(212) plot(x,r,x,r1,x,t1) xlabel('xaxis'); ylabel('raxis'); title('Approximation when n=1')
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[edit]
Information from part 1.
Part 2 Problem Statement[edit]
Find a that solves the following equation for :

(Eq.9.2.1)
Then plot the solutions, truncated Taylor series, and numerical solution.
Part 2 Solution[edit]
For .

(Eq.9.2.2)

(Eq.9.2.3)
The homogenous solution must satisfy the following:

(Eq.9.2.4)
Therefore, the characteristic equation is:

(Eq.9.2.5)

(Eq.9.2.6)

(Eq.9.2.7)
Distinct real roots makes the general solution of the homogenous take the following form:

(Eq.9.2.8)
Therefore,

(Eq.9.2.9)
The particular solution must satisfy the following:

(Eq.9.2.10)
Using Method of Undetermined Coefficients, the particular solution will have the form:

(Eq.9.2.11)
Therefore,

(Eq.9.2.12)
Take the first and second derivative of equation 9.2.12:

(Eq.9.2.13)

(Eq.9.2.14)
Substitute equations 9.2.12, 9.2.13, and 9.2.14 into equation 9.2.10:

(Eq.9.2.15)
Therefore,

(Eq.9.2.16)
The particular solution will be:

(Eq.9.2.17)
Superimpose equation 9.2.9 and 9.2.17:

(Eq.9.2.18)
Take the first of equation 9.2.18:

(Eq.9.2.19)
Use equation 9.1.3 to solve for the unknowns:

(Eq.9.2.20)

(Eq.9.2.21)
Therefore,

(Eq.9.2.22)

(Eq.9.2.23)
The final solution for :

(Eq.9.2.24)
For .

(Eq.9.2.25)

(Eq.9.2.26)
The homogenous solution will be the same has equation 9.2.9:

(Eq.9.2.27)
The particular solution must satisfy the following:

(Eq.9.2.28)
Using Method of Undetermined Coefficients, the particular solution will take the form of equation 9.2.11:

(Eq.9.2.29)
Take the first and second derivative of equation 9.2.29:

(Eq.9.2.30)

(Eq.9.2.31)
Substitute equations 9.2.29, 9.2.30, and 9.2.31 into equation 9.2.28:

(Eq.9.2.32)
Solve the coefficients by setting like terms equal to one another.
For :

(Eq.9.2.33)
For :

(Eq.9.2.34)
Therefore,

(Eq.9.2.35)

(Eq.9.2.36)
The particular solution will be:

(Eq.9.2.37)
Superimpose equation 9.2.27 and 9.2.37:

(Eq.9.2.38)
Take the first of equation 9.2.38:

(Eq.9.2.39)
Use equation 9.1.3 to solve for the unknowns:

(Eq.9.2.40)

(Eq.9.2.41)
Therefore,

(Eq.9.2.42)

(Eq.9.2.43)
The final solution for :

(Eq.9.2.44)
For Taylor series expansion for .

(Eq.9.2.45)

(Eq.9.2.46)
The homogenous solution will be the same has equation 9.2.9:

(Eq.9.2.47)
The particular solution must satisfy the following:

(Eq.9.2.48)
Using Method of Undetermined Coefficients, the particular solution will take the form of equation 9.2.11:

(Eq.9.2.49)
Take the first and second derivative of equation 9.2.49:

(Eq.9.2.50)

(Eq.9.2.51)
Substitute equations 9.2.49, 9.2.50, and 9.2.51 into equation 9.2.48:

(Eq.9.2.52)
Solve the coefficients by setting like terms equal to one another.
For :

(Eq.9.2.53)
For :

(Eq.9.2.54)
Therefore,

(Eq.9.2.55)

(Eq.9.2.56)
The particular solution will be:

(Eq.9.2.57)
Superimpose equation 9.2.47 and 9.2.57:

(Eq.9.2.58)
Take the first of equation 9.2.58:

(Eq.9.2.59)
Use equation 9.1.3 to solve for the unknowns:

(Eq.9.2.60)

(Eq.9.2.61)
Therefore,

(Eq.9.2.62)

(Eq.9.2.63)
The final solution for Taylor series expansion with :

(Eq.9.2.64)
For Taylor series expansion for .

(Eq.9.2.65)

(Eq.9.2.66)
The homogenous solution will be the same has equation 9.2.9:

(Eq.9.2.67)
The particular solution must satisfy the following:

(Eq.9.2.68)
Using Method of Undetermined Coefficients, the particular solution will take the form of equation 9.2.11:

(Eq.9.2.69)
Take the first and second derivative of equation 9.2.69:

(Eq.9.2.70)

(Eq.9.2.71)
Substitute equations 9.2.69, 9.2.70, and 9.2.71 into equation 9.2.68:

(Eq.9.2.72)
Solve the coefficients by setting like terms equal to one another.
For :

(Eq.9.2.73)
For :

(Eq.9.2.74)
For :

(Eq.9.2.75)
Therefore,

(Eq.9.2.76)

(Eq.9.2.77)

(Eq.9.2.78)
The particular solution will be:

(Eq.9.2.79)
Superimpose equation 9.2.67 and 9.2.79:

(Eq.9.2.80)
Take the first of equation 9.2.80:

(Eq.9.2.81)
Use equation 9.1.3 to solve for the unknowns:

(Eq.9.2.82)

(Eq.9.2.83)
Therefore,

(Eq.9.2.84)

(Eq.9.2.85)
The final solution for Taylor series expansion with :

(Eq.9.2.86)
Below is the MATLAB code used to create the ODE for equation 9.1.1 with excitation equation 9.1.2:
function pdot = ODE45(t,p) pdot = zeros(2,1); pdot(1) = p(2); pdot(2) = 3*p(2)  2*p(1) + log10(1+t); end
Below is the MATLAB code used to generate the above graph:
[t,p] = ode45('ODE45',[0.75,3],[1 0]); x=[.75:0.01:3]; y0=3.2019*exp(2.*x)+3.0249*exp(x)+0.2856; y1=5.0235*exp(2.*x)+4.1201*exp(x)+0.2956.*x+0.3964; yt0=5.0419*exp(2.*x)+3.7048*exp(x)+0.5.*x+0.75; yt1=7.2827*exp(2.*x)+6.6156*exp(x)0.25.*x.^20.5.*x0.125; subplot(211) plot(t,p(:,1),x,y0,x,yt0) xlabel('xaxis'); ylabel('yaxis'); title('Solution when n=0') subplot(212) plot(t,p(:,1),x,y1,x,yt1) xlabel('xaxis'); ylabel('yaxis'); title('Solution when n=1')
The solutions when :
 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 .
The solutions when :
 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 , since .
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[edit]
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[edit]
Team Member  Contribute  Proofread 

Allan Axelrod  Problem 6  Problem 4, 7, and 8 
Michael Deaver  Problem 9 and Formatting  Problem 2, 5, and 6 
Max Hintz  Problem 2, 3, and 7  
Kelvin Li  Problem 4 and 8  Problem 3 
Thomas Wheeler  Problem 1 and 5  Problem 4 
Chen Ying  Formatting  Problem 1, 8, and 9 