University of Florida/Egm4313/s12.teamboss/R3

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Report 3

File:Contents.JPG

Problem R3.1: Using the Method of Undetermined Coefficients and the Modification Rule to find a solution[edit]

Statement[edit]

Given the double root and the excitation , with the initial conditions find the solution .
Plot this solution and the solution to the same problem except with the excitation .

Solution[edit]

The solution is composed of a general solution and a particular solution so that .
Using the given double root, we can find the equation for the characteristic equation which leads to the homogeneous solution:

(1.0)


Homogeneous solution:

(1.1)


(1.2)


First, by using the Modification Rule, we find that the general equation associated with the given double root is:

(1.3)


We need to find the particular solution to the excitation . In analyzing the excitation, it is found that the particular solution looks like this:

(1.4)


Now, we need to find the values for the constants , by taking the first and second derivatives of the particular solution:

(1.4)


(1.5)


(1.6)


Now, plug these into the homogeneous solution (1.2) and simplify:

(1.7)


Comparing the coefficients for and allows us to solve for the unknown coefficients:

(1.8)


(1.9)


(1.10)


(1.11)


Plug into (1.4):

(1.12)


Now that we have the particular solution, we can use the initial conditions to solve for the unknown constants in the general solution, , and then we will be able to solve for the final solution :

(1.13)


Take derivative of (1.13):

(1.14)


Using the given initial conditions and equations (1.13) and (1.14):

(1.15)


(1.16)


Plug these constants back into the solution (1.13) to obtain the final solution:

(1.17)

The solution to the same problem but with excitation is:

(1.18)


This is the plot for (1.17) and (1.18):

R3.1PLOT.jpg

Author[edit]

Solved and Typed By - --Egm4313.s12.team1.wyattling 22:00, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.armanious 22:39, 21 February 2012 (UTC)




Problem R3.2 Perturbation Method for Double Real Root[edit]

Statement[edit]

Perturbation method for double real root:
Developing the 2nd homogeneous solution for the case of double real root as a limiting case of distinct roots (see Sec7 p. 7-5 ). Consider two distinct real roots of the form:



1) Find the homogeneous L2-ODE-CC having the above distinct roots.
2) Show that the following is a homogeneous solution:


The fraction in (3) p.7-5, for small , is a finite difference formula that approximates the derivative



In fact,


3)Find the limit of the homogeneous solution in (3) p.7-5 as [epsilon goes to 0] (think l'Hopital's rule)

4)Take the derivative of with respect to
5)Compare the results in Parts (3) and (4), and relate to the result by variation of parameters.
6)Numerical experiment: Compute (3) p.7-5 using at and with , and compare to get the value obtained from the exact 2nd homogeneous solution.

Solution[edit]

In order to find the corresponding L2-ODE-CC, the characteristic equation corresponding to the given solution must be found:

(2.0)

(2.1)

Therefore the corresponding L2-ODE-CC is:

(2.2)

Note: if ε is equal to zero, the characteristic equation of 2.2 has a double real root at λ.

To show that the following is a homogeneous solution, the first and second derivatives must be taken:

(2.3)

(2.4)

(2.5)

Using these values in 2.2 yields:

(2.6)

Rearranging and simplifying yields:

(2.7)

All of the terms in the above equation cancel to yield:

(2.8)

Using l'Hopital's rule:

(2.9)

Simplifying further:

(2.10)

This ultimately yields:

(2.11)

The following should also be taken into consideration:

(2.12)

Clearly, the results of 2.11 and 2.12 are equivalent. This shows that is an appropriate solution to a homogeneous L2-ODE-CC having one double root.

To test this, test values will be used to solve for the approximate (2.12) and exact (2.13) solutions of the ODE. For this test, and

(2.13)

(2.14)

If the above derivations are true, then the following approximation must also be true:

(2.15)

Plotting both sides of 2.15 as a function of x shows that this is a valid approximation for most values of x.

Figure 3.2-1

Fig3.2.1.png


Author[edit]

Solved and Typed By - Egm4313.s12.team1.armanious 05:51, 21 February 2012 (UTC)
Reviewed By - --Egm4313.s12.team1.durrance 02:53, 22 February 2012 (UTC)




Problem R3.3 Finding Solution of ODE with Polynomial Excitation[edit]

Statement[edit]

Find the complete solution for , with the initial conditions



Plot the solution

Solution[edit]

First we create the characteristic equation in standard form:

(3.0)


Then, by setting it equal to zero, we can find what equals:

(3.1)


(3.2)


Given two, distinct, real roots, the general solution looks like this:

(3.3)

By using the method of undetermined coefficients, the excitation is analyzed to yield a particular solution:
In assessing a polynomial with a second power, the form of the particular solution will look like this:

(3.5)


It's derivative would look like this:

(3.6)


And the second derivative to follow would then become:

(3.7)


Based on the coefficients, the following system of equations exists:

(3.8)


(3.9)


(3.10)


The results of this set of equations make the coefficients of A's:




The resulting particular equation looks like this:

(3.11)


By adding the particular and general solutions, we get the complete solution:

(3.12)


We consider the initial conditions by taking the first derivative of the complete solution:

(3.13)



By plugging in 0 for x, 1 for y, and 0 for y', we can solve for the constants :

(3.14)


(3.15)


Solving the equations proves that :
The resulting complete solution with consideration for initial conditions then becomes:

(3.12)


y plotted looks like this:

R33graph.jpg


Author[edit]

Solved and Typed By -Egm4313.s12.team1.silvestri 15:56, 19 February 2012 (UTC)
Reviewed By -Egm4313.s12.team1.armanious 03:02, 22 February 2012 (UTC)




Problem R3.4 Solving for Particular Solution in Summation Form[edit]

Statement[edit]

From R3.4 in Sec 3 p. 7-11; Use the Basic Rule 1 and the Sum Rule to show that the appropriate particular solution for

(4.1)

is of the form

(4.2)

i.e.,

(4.3)

Basic Rule: Select from the table and determine the coefficients by substituting in

(4.4)

Sum Rule: If is the sum of the terms in the 1st column of table 2.1 then is the sum of the corresponding terms in the 2nd column of this table.

Table 2.1

Table2.1.JPG


Solution[edit]

According to the table the Homogeneous equation has two values of the form . Using the Basic Rule this means that there is a particular solution of the form for each .

(4.5)

So the particular solution for this should be:

(4.6)

Where . Which simplifies to:

(4.7)

For the second :

(4.8)

So the particular solution for this should be:

(4.9)

Where . Which simplifies to:

(4.10)

Using the sum rule:

(4.11)

Now using the Sum Rule which just states if there are two values in any form on the left side of the table, then the particular solution is the sum of the solutions for on the right side of the table.

So the particular solution for where

(4.12)

Author[edit]

Solved and Typed By - User:Egm4313.s12.team1.stewart 22:47, 21 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:57, 22 February 2012 (UTC)




Problem R3.5: Finding Particular Solutions by Combining Linear Coefficient Series Expansions[edit]

Statement[edit]

Given:

(5.0)


(5.1)



Find:
-Coefficients of particular solution using series expansion and matrix back-substitution
-Solution using initial conditions
Plot solution.

Solution[edit]

Using Sec7b-1 p.7-13 Eq. (1):

(5.2)


Which is a combined series expansion for the general coefficient series expansion of Sec7b-1 p.7-12 Eq. (4):

(5.3)


We can find a simultaneous linear system of equations to solve for the coefficients of .

Note: by combining the summations in Eq. (5.2) by setting the upper bounds of the summations to a common value 3, the equation can be simplified to:

(5.4)



Therefore providing a system of equations by equating coefficients from the summations to the :

(5.5)


Eqs. (5.5) can be verified by solving for the coefficients by Eq. (5.2) to prove that the summations were combined correctly:

(5.6)



By summing the terms in (5.6) and grouping like terms, then equating it to the , we find the coefficient equations are the same as Eqs. (5.5):

(5.7)



The linear system of equations in Eqs. (5.5) can be placed in matrix form:

(5.8)



Solving this system of equations by back-substitution yields the values of the coefficients for the :

(5.9)



Substituting these coefficients into Eq. (5.1):

(5.10)



The general solution is the sum of the homogeneous (found in R3.3) and particular solutions:

(5.11)



Considering the initial value condition in the general solution:

(5.12)



Taking the first derivative of Eq. (5.11) to consider the initial value condition :

(5.13)



(5.14)



Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:

(5.15)



This yields our final solution by substituting the coefficients into Eq. (5.11): Solving Eq. (5.12) and Eq. (5.13) yield the coefficients:

(5.16)



A plot of the solution:

Figure 3.5-1


Author[edit]

Solved and Typed By - --Egm4313.s12.team1.durrance 22:56, 21 February 2012 (UTC)--
Reviewed By - Egm4313.s12.team1.silvestri 03:28, 22 February 2012 (UTC)




Problem R3.6: Superimposing Particular Solutions[edit]

Statement[edit]

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

(6.0)

(6.1)

The particular solution had been found in R3.3 p.7-11. Find the particular solution , and then obtain the solution for the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6.

Compare the result with that obtained in R3.5.

Solution[edit]

Beginning with equation (6.1), we find that a particular solution has the form

(6.2)

where . That is,

(6.3)

Differentiating twice:

(6.4)

(6.5)

Substitute (6.3-5) into (6.1) to obtain


(6.6)

Rearranging terms with respect to power:


(6.7)

In matrix form:

(6.8)

Solving for , using MATLAB, yields

(6.9)

which means that

(6.10)

From R3.3:

(6.11)

Summing for the final solution:

(6.12)


(6.13)

For initial condition

(6.14)

Similarly, for initial condition

(6.15)

Equations (6.14) and (6.15) yield the following matrix equation:

(6.16)

Solving (6.16) in MATLAB yields

(6.17)

Therefore the combined solution of is


(6.18)

Which can be simplified to Final Equation

(6.19)

Author[edit]

Solved and Typed By - Egm4313.s12.team1.essenwein 00:40, 18 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.silvestri 04:29, 22 February 2012 (UTC)




Problem R3.7 : Verifying series representation for method of undetermined coefficients[edit]

Statement[edit]

Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities.
These equalities are:

(7.0)


(7.1)



Solution[edit]

This transition occurs through a mid-level variable change. In this particular case, for (7.0), j-2 is represented by k:

(7.3)


We can then represent the variable k with a j. This new summation looks like this:

(7.4)


Expanding both sides, (7.0 and 7.4) yields the same result:

(7.5)



We follow the same process for equality (7.1):
This transition occurs through a mid-level variable change. In this particular case, for (7.1), j-1 is represented by k:

(7.6)


We can then represent the variable k with a j. This new summation looks like this:

(7.7)


Expanding both sides, (7.1 and 7.7)yields the same result:

(7.8)



Author[edit]

Solved and Typed By - Egm4313.s12.team1.silvestri 17:20, 19 February 2012 (UTC)
Reviewed By - --128.227.113.77 17:05, 22 February 2012 (UTC)




Problem R3.8: Finding general solutions to Non-homogeneous Linear ODEs using the Method of Undetermined Coefficients and the Basic Rule[edit]

Statement[edit]

Find a general solution for the following two problems:

Kreyszig 2011 p.84 Problem 5[edit]

Given[edit]

Homogeneous solution:

Solution[edit]

The solution is composed of a general solution and a particular solution so that .
First, we will find the general solution, , by finding the roots of the characteristic equation:

(8.0)


Which means that the characteristic equation has a double root of .
Based on the double root, then, the general solution is:

(8.1)


Next, we need to find the particular solution, and based on analysis of the excitation, , we find that the particular solution is the following:

(8.2)


Now, we need to find the values for the constants , by taking the first and second derivatives of the particular solution (8.2):

(8.2)


(8.3)


(8.4)


Now, plug these into the homogeneous solution and simplify:

(8.5)


(8.6)


Pull out of equation (8.6) and simplify:

(8.7)


(8.8)


Now, solve for the unknown coefficients:

(8.9)


(8.10)


Plug into (8.2):

(8.11)


Now, we have both the particular solution and the general solution, which allows for us to solve for the final solution,

(8.12)

Kreyszig 2011 p.84 Problem 6[edit]

Given[edit]

Homogeneous solution:

Solution[edit]

The solution is composed of a general solution and a particular solution so that .
First, we will find the general solution, , by finding the roots of the characteristic equation, using the quadratic equation:

(8.13)


(8.14)


The roots of the characteristic equation are complex conjugates, meaning that the general solution, , is the following:

(8.15)


Next, we need to find the particular solution, and based on analysis of the excitation, , we find that the particular solution is the following:

(8.16)


Now, we need to find the values for the constants , by taking the first and second derivatives of the particular solution (8.16):

(8.16)


(8.17)


(8.18)


Now, we plug these into the homogeneous solution and simplify but we find that everything cancels out, meaning the unknown constants, are both equal to . This means that the final solution is equal to just the general solution:

(8.19)

Author[edit]

Solved and Typed By ---Egm4313.s12.team1.wyattling 22:10, 20 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.rosenberg 03:07, 22 February 2012 (UTC)




Problem R3.9 Finding general solutions to Non-homogeneous Linear ODEs[edit]

Statement[edit]

K 2011 page 85 problems 13 and 14
Problem 13: Find the complete solution for , with the initial conditions


Problem 14: Find the complete solution for , with the initial conditions


Solution[edit]

Problem 13:
In order to solve the equation we must use the definition . We start with the given equation:

(9.1)


From this we get the homogeneous characteristic equation:

(9.2)


Factoring the characteristic equation gives us:

(9.3)


This give us the roots:

(9.4)


Plugging in the roots gives us the general homogeneous solution:

(9.5)


Now we must solve for the particular solution. So far we know:

(9.6)


Now using the sum rule, , gives us:

(9.7)


Now we must take the first and second derivatives of equation 9.7:

(9.8)


(9.9)


Substituting 9.8 and 9.9 back into our original equation gives us:

(9.10)


Simplifying equation 9.10 give us:

(9.11)


From this we can deduce that:

(9.12)


Thus we get that:

(9.13)


We now have our particular equation:

(9.14)


Thus:

(9.15)


Now we can solve for and using the given initial values.
We are given that at , we can now plug this into our equation to get:

(9.16)


This simplifies to:

(9.17)


Giving us that:

(9.18)


For our second condition, , we must we must take the derivative of our general solution:

(9.19)


At our initial condition gives us:

(9.20)


Thus:

(9.21)


Combining 9.18 and 9.21 gives us that:

(9.22)


Plugging these coefficients in gives us our final solution:

(9.23)





Problem 14:
In order to solve the equation we must use the definition . We start with the given equation:

(9.24)


From this we get the homogeneous characteristic equation:

(9.25)


Factoring the characteristic equation gives us:

(9.26)


This give us the double root:

(9.27)


Plugging in the roots gives us the general homogeneous solution:

(9.28)


Now we must solve for the particular solution. So far we know:

(9.29)


Now we must take the first and second derivatives of equation 9.29:

(9.30)


(9.31)


Substituting 9.30 and 9.31 back into our original equation gives us:

(9.32)


From this we can deduce that:

(9.33)


Thus we get that:

(9.34)


We now have our particular equation:

(9.35)


Thus:

(9.36)


Now we can solve for and using the given initial values.
We are given that at , we can now plug this into our equation to get:

(9.37)


Giving us that:

(9.38)


For our second condition, , we must we must take the derivative of our general solution:

(9.39)


At our initial condition we have:

(9.40)


Thus:

(9.41)


Plugging these coefficients in gives us our final solution:

(9.42)


Author[edit]

Solved and Typed By - Egm4313.s12.team1.rosenberg 03:04, 22 February 2012 (UTC)
Reviewed By - Egm4313.s12.team1.essenwein 02:39, 22 February 2012 (UTC)




Contributing Members[edit]

Team Contribution Table
Problem Number Lecture Assigned To Solved By Typed By Proofread By
3.1 R3.1 in Sec 3 p. 7-5 Wyatt Ling Wyatt Ling Wyatt Ling George Armanious
3.2 R3.2 in Sec 3 p. 7-5 George Armanious George Armanious George Armanious Jesse Durrance
3.3 R3.3 in Sec 3 p. 7-11 Emotion Silvestri Emotion Silvestri Emotion Silvestri George Armanious
3.4 R3.4 in Sec 3 p. 7-11 Chris Stewart Chris Stewart Chris Stewart Emotion Silvestri
3.5 R3.5 in Sec 7 p. 7-17 (Sakai) Jesse Durrance Jesse Durrance Jesse Durrance Emotion Silvestri
3.6 R3.6 in Sec 7 p. 7-17 (Sakai) Eric Essenwein Eric Essenwein Eric Essenwein Emotion Silvestri
3.7 R3.7 in Sec 7 p. 7-17 (Sakai) Emotion Emotion Emotion Wyatt Ling
3.8 R3.8 in Sec 7b p. 7-18 Wyatt Ling Wyatt Ling Wyatt Ling Steven Rosenberg
3.9 R3.9 in Sec 7b p. 7-18 Steven Rosenberg Steven Rosenberg Steven Rosenberg Eric Essenwein