University of Florida/Egm4313/s12.team11.R3

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Report 3


Intermediate Engineering Analysis
Section 7566
Team 11
Due date: February 22, 2012.

Contents

Problem 3.1[edit]

Solved by Luca Imponenti

Problem Statement[edit]

Find the solution to the following L2-ODE-CC:

With the following excitation:

And the following initial conditions:

Plot this solution and the solution in the example on p.7-3

Homogeneous Solution[edit]

To find the homogeneous solution we need to find the roots of our equation

We know the homogeneous solution for the case of a real double root with to be

Particular Solution[edit]

For the given excitation we must use the Sum Rule to the particular solution as follows

where and are the solutions to and , respectively

First Particular Solution[edit]

,

from table 2.1, K 2011, pg. 82 we have

but this corresponds to one of our homogeneous solutions so we must use the modification rule to get

Plugging this into the original L2-ODE-CC then substituting;

so and the first particular solution is,

Second Particular Solution[edit]

,

from table 2.1, K 2011, pg. 82 we have

Plugging this into the original L2-ODE-CC then substituting;

grouping like terms we get three equations to solve for the three unknowns, these are written in matrix form



solving by back subsitution leads to

so the second particular solution is,

General Solution[edit]

The general solution is the summation of the homogeneous and particular solutions

Applying the first initial condition

Second initial condition

The general solution to the differential equation is therefore

                    

Plot[edit]

Below is a plot of this solution and the solution to in the example on p.7-3

our solution (shown in red)

example on p.7-3 (shown in blue)

Plot3 1.jpg

Egm4313.s12.team11.imponenti 05:55, 20 February 2012 (UTC)

Problem 3.2[edit]

Solved by Gonzalo Perez

Problem Statement[edit]

Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.

Given[edit]

Consider two distinct roots of the form:

and

(where is perturbation).

Part 1[edit]

Given[edit]

Find the homogeneous L2-ODE-CC having the above distinct roots.

Solution[edit]


                    (1)


Part 2[edit]

Given[edit]

Show that is a homogeneous solution. (2)

Solution[edit]

Let's find the corresponding derivatives:

If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:

      Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.

Part 3[edit]

Given[edit]

Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).

Solution[edit]

Using l'Hopital's Rule,

(this is an indeterminate form).

L'Hopital's Rule states that we can divide this function into two functions, and , and then find their derivatives and attempt to find the limit of . If a limit exists for this, then a limit exists for our original function.

                                            

Part 4[edit]

Given[edit]

Take the derivative of with respect to lambda.

Solution[edit]

Taking the derivative with respect to lambda, we find that:

                                     .

It is important to remember that we must hold as a constant when finding this derivative.

Part 5[edit]

Given[edit]

Compare the results in parts (3) and (4), and relate to the result by using variation of parameters

Solution[edit]

Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:

                           

Part 6[edit]

Given[edit]

Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001, and compare to the value obtained from the exact second homogeneous solution.

Solution[edit]

             After performing these calculations, from (2) we get 148.478.
             And from the exact second homogeneous solution, we get 200.05.

Egm4313.s12.team11.perez.gp 20:32, 22 February 2012 (UTC)

Problem 3.3[edit]

Solved by Jonathan Sheider

Problem Statement[edit]

Find the complete solution to the ODE with the given initial conditions. Plot the solution y(x).

Given[edit]





Solution[edit]

First let us analyze the homogenous solution to the given ODE:



The characteristic equation for this ODE is therefore:



Solving for :





and

Therefore the equation has two real roots and a homogenous solution of the following form (from Kreyszig 2011, p.54-57):



And finally we find the general homogenous solution:

                                                       


Next let us evaluate the particular solution to the given ODE:



The function on the right hand side of this equation implies that the particular solution has the following form based on Table 2.1 (from Kreyszig 2011, p.82):



Therefore for this ODE we have:



Differentiating to obtain and respectively:





Substituting these equations into the original ODE yields:





And we know that the coefficients of the variables on each side of the equation must be equal:



Therefore we find:





Now, solving for and :







And also:





Finally we arrive at the particular solution:

                                                        



By superposition, we can find the complete general solution:



                                                   



Using the given initial conditions, we can solve for and , first by using the initial conditions for :



Differentiating the complete general solution and using the given initial condition for :





Solving this system of equations, by solving the first equation for :



Plugging this into the second equation yields:





And therefore:





Finally we have the complete general solution that evaluates the given initial conditions:

                                                  



A plot of from to is shown using MatLAB:

R3.3.jpg

Checking[edit]

Differentiating the solution to find and respectively:







Substituting into the original ODE yields:











Therefore this solution is correct.

--Egm4313.s12.team11.sheider 21:30, 21 February 2012 (UTC)

Problem 3.4[edit]

Solved by Daniel Suh

Problem Statement[edit]

Use Basic Rule (1) and Sum Rule (3) to show that the appropriate particular solution for
is of the form , with n = 5.

Solution[edit]

Finding the Particular Solution with Basic Rule and Sum Rule[edit]

We know that in standard form, for a particular solution,



Using Basic Rule (1) and Sum Rule (3), we know that


Additionally, we choose


Solving for Particular Solution 1[edit]

Using the Method of Undetermined Coefficients, we find that




Plugging into gives


Solving for coefficients




Results in





Solving for Particular Solution 2[edit]

Using the Method of Undetermined Coefficients, we find that




Plugging into gives


Solving for coefficients results in








Combining Particular Solutions[edit]

Plugging and into
gives


               

Comparing to Summation Form[edit]

The particular solution in the summation form is

if n=5, then



Plugging the particular solution into , the final solution gives

         

(check Problem 3.5 for explanation on how to get this answer)


Result[edit]

As you can see, the two particular solutions of are equal. Thus, using the Basic Rule (1) and Sum rule (3) does give you the correct particular solution for
, in the form of

Created by [Daniel Suh] 20:57, 21 February 2012 (UTC)

Problem 3.5[edit]

Solved by: Andrea Vargas

Problem Statement[edit]

Given

1. Obtain the coefficients of
2.Verify all equations by long-hand expansion. Use the series before adjusting the indexes.

3. Put the system of equations in an upper triangular matrix.

4. Solve for by using back substitution.

5. Using the initial conditions find and plot it

Solution[edit]

1. To obtain the equations for the coefficients we use the following equation from (1) p7-11:


Finding the coefficients of where :


Finding the coefficients of where :


Finding the coefficients of where :


Finding the coefficients of where :



2. To verify all the above equations by long hand expansion we use the following equation from (4) p7-12:

Finding the coefficients when :


Finding the coefficients when :


Finding the coefficients when :


Finding the coefficients when :



Finding the coefficients when :



By collecting these terms we can compare them to the equations of part 1.

Coefficients of :


Coefficients of :


Coefficients of :


Coefficients of :



We can see that we obtain the same system of equations to solve for the coefficients with both methods.

3.Constructing the coefficients matrix:

Then, the system becomes:



4. Solving for the coefficients:







Particular solution[edit]

This yields the particular solution:

                                                            

Homogeneous Solution[edit]



Then, we can find the characteristic equation:


Then the solution for the homogeneous equation becomes:

                                                                                      

General Solution[edit]

Using the given initial conditions we find the overall solution:




Using the initial conditions to solve for and





The general solution becomes

                                                     

Plot[edit]

Below is a plot of the solution:

R3 5.jpg

--Egm4313.s12.team11.vargas.aa 05:56, 21 February 2012 (UTC)

Problem 3.6[edit]

Solved by Francisco Arrieta

Problem 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-ODE-CC (see p. 7-2b)
The particular solution to 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.

Given[edit]

First particular solution
Initial Conditions

Solution[edit]

Particular Solution[edit]

Since the specific excitation , using table 2.1 from K 2011 p.82, the choice for the particular solution is
Then


Plugging these equations back in the original



Using coefficient matching of left hand side to the right hand side of the equation, the following equations are obtained


Which is equivalent to the following matrix


Using back substitution method to solve for every coefficient, starting with


Plugging these values back into gives


                                        


Since these equations are L2-ODE-CC, the superposition principle applies and and the general particular solution becomes


                                         


Homogeneous Solution[edit]

Due to the linearity of this homogeneous equation, a linear combination of two linear independent solutions is also a solution


With solutions


In order to determine the value of , the characteristic equation must be determine from the homogeneous equation


Then the solutions for each distinct linearly independent homogeneous equation becomes


Because of linearity, the linear combination of the previous two equations times two constants that satisfy 2 initial conditions, is also a solution
                                                    


General Solution[edit]

The overall solution for the L2-ODE-CC


Using the initial conditions to solve for and


The general solution becomes
                 

--Egm4313.s12.team11.arrieta 20:26, 19 February 2012 (UTC)


Problem 3.7[edit]


Solved By Kyle Gooding

Problem Statement[edit]


Expand the series on both sides of (1),(2) pg. 7-12b to verify these equalities.
(1)

Given[edit]


(2)

Solutions[edit]

Expanding both sides of (1) results in:



Simplifying:


 The two sums are equal.

Expanding both sides of (2) results in:


Simplifying:


 The two sums are equal.

Egm4313.s12.team11.gooding 23:49, 19 February 2012 (UTC)

Problem 3.8[edit]

solved by Luca Imponenti

Kreyszig 2011 pg.84 problem 5[edit]

Problem Statement[edit]

Find a (real) general solution. State which rule you are using. Show each step of your work.

Homogeneous Solution[edit]

To find the homogeneous solution, , we must find the roots of the equation

We know the homogeneous solution for the case of a double root to be

Particular Solution[edit]

We have the following excitation

From table 2.1, K 2011, pg. 82, we have

Since this does not correspond to our homogeneous solution we can use the Basic Rule (a), K 2011, pg. 81 to solve for the particular solution

where

and

Plugging these equations back into the differential equation

from the above equation it is obvious that and

therefore the particular solution to the differential equation is

General Solution[edit]

The general solution will be the summation of the homogeneous and particular solutions

   

The coefficients and can be readily solved for given either initial conditions or boundary value conditions.

Kreyszig 2011 pg.84 problem 6[edit]

Problem Statement[edit]

Find a (real) general solution. State which rule you are using. Show each step of your work.

Homogeneous Solution[edit]

To find the homogeneous solution, , we must find the roots of the equation

with

We know the homogeneous solution for the case of a double root to be

Particular Solution[edit]

We have the following excitation

From table 2.1, K 2011, pg. 82, we have

Since this corresponds to our homogeneous solution we must use the Modification Rule (b), K 2011, pg. 81 to solve for the particular solution

so

differentiating

and

grouping cosine and sine terms we get

and

next we substitute the above equations into the ODE

after cancelling terms; we can equate cosine and sine coefficients to get two equations

so and

and the particular solution to the ODE is

General Solution[edit]

The general solution will be the summation of the homogeneous and particular solutions

   

Egm4313.s12.team11.imponenti 04:28, 21 February 2012 (UTC)

Problem 3.9[edit]

Solved by Gonzalo Perez

Problem Statement[edit]

Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.

Given[edit]

(K 2011 pg.85 #13)[edit]

(1)

Initial conditions are:

Solution[edit]

The general solution of the homogeneous ordinary differential equation is

We can use this information to determine the characteristic equation:

And proceeding to find the roots,

Thus, .

Solving for the roots, we find that

where the general solution is

.

The solution of of the non-homogeneous ordinary differential equation is

.

Using the Sum rule as described in Section 2.7, the above function translates into the following:

, where Table 2.1 tells us that:

and .

Therefore, .

Now, we can substitute the values () into (1) to get:

Now that we have this equation, we can equate coefficients to find that:

and thus,

We find that the general solution is in fact:

whereas the general solution of the given ordinary differential equation is actually:

Solving for the initial conditions given and first plugging in , we get that:

. (2)

And now we can determine the first order ODE :

The second initial condition that was given to us, can now be plugged in:

(3)

Once we solve (2) and (3), we can get the values:

.

And once we substitute these values, we get the following solution for this IVP:


             


Given[edit]

(K 2011 pg.85 #14)[edit]

(1)

Initial conditions are:

Solution[edit]

The general solution of the homogeneous ordinary differential equation is

We can use this information to determine the characteristic equation:

And proceeding to find the roots,

Solving for the roots, we find that

where the general solution is:

, or:

Now, according to the Modification Rule and Table 2.1 in Section 2.7, we know that we have to multiply by x to get:

, since the solution of is a double root of the characteristic equation.

We can then derive to get :

Deriving once again to solve for , we get the following:

Now, we can substitute the values () into (1) to get:

Now that we have this equation, we can equate coefficients to find that:

and

and finally discover that:

and .

Plugging in these values in , we find that:

And finally, we arrive at the general solution of the given ordinary differential equation:

Solving for the initial conditions given and first plugging in , we get that:

The second initial condition that was given to us, can now be plugged in:

And once we substitute these values, we get the following solution for this IVP:


                


Egm4313.s12.team11.perez.gp 20:34, 22 February 2012 (UTC)