University of Florida/Egm4313/s12.team11.R3

Report 3


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

Problem 3.1Edit

Solved by Luca Imponenti

Problem StatementEdit

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 SolutionEdit

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 SolutionEdit

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 SolutionEdit

 ,

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 SolutionEdit

 ,

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 SolutionEdit

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

                     

PlotEdit

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)

 

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

Problem 3.2Edit

Solved by Gonzalo Perez

Problem StatementEdit

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

GivenEdit

Consider two distinct roots of the form:

  and  

(where   is perturbation).

Part 1Edit

GivenEdit

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

SolutionEdit

 

 

 

 


                     (1)


Part 2Edit

GivenEdit

Show that   is a homogeneous solution. (2)

SolutionEdit

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 3Edit

GivenEdit

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

SolutionEdit

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 4Edit

GivenEdit

Take the derivative of   with respect to lambda.

SolutionEdit

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 5Edit

GivenEdit

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

SolutionEdit

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

                            

Part 6Edit

GivenEdit

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.

SolutionEdit

             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 3.3Edit

Solved by Jonathan Sheider

Problem StatementEdit

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

GivenEdit

 

 

SolutionEdit

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:

 

CheckingEdit

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 3.4Edit

Solved by Daniel Suh

Problem StatementEdit

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

SolutionEdit

Finding the Particular Solution with Basic Rule and Sum RuleEdit

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 1Edit

Using the Method of Undetermined Coefficients, we find that
 
 
 


Plugging   into   gives
 


Solving for coefficients
 
 
 


Results in
 
 
 


 

Solving for Particular Solution 2Edit

Using the Method of Undetermined Coefficients, we find that
 
 
 


Plugging   into   gives
 


Solving for coefficients results in
 
 
 
 
 
 


 

Combining Particular SolutionsEdit

Plugging   and   into
  gives


                

Comparing to Summation FormEdit

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)


ResultEdit

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 3.5Edit

Solved by: Andrea Vargas

Problem StatementEdit

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

SolutionEdit

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 solutionEdit

This yields the particular solution:

                                                             

Homogeneous SolutionEdit

 
 
Then, we can find the characteristic equation:
 
 
Then the solution for the homogeneous equation becomes:

                                                                                       

General SolutionEdit

Using the given initial conditions   we find the overall solution:

 
 
 
Using the initial conditions to solve for   and  

 
 
 

The general solution becomes

                                                      

PlotEdit

Below is a plot of the solution:

 

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

Problem 3.6Edit

Solved by Francisco Arrieta

Problem StatementEdit

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.

GivenEdit

First particular solution
 
Initial Conditions
 
 

SolutionEdit

Particular SolutionEdit

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 SolutionEdit

 
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 SolutionEdit

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 3.7Edit


Solved By Kyle Gooding

Problem StatementEdit


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

GivenEdit

 

(2)
 

SolutionsEdit

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 3.8Edit

solved by Luca Imponenti

Kreyszig 2011 pg.84 problem 5Edit

Problem StatementEdit

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

 

Homogeneous SolutionEdit

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 SolutionEdit

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 SolutionEdit

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 6Edit

Problem StatementEdit

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

 

Homogeneous SolutionEdit

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 SolutionEdit

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 SolutionEdit

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

 

 

    

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

Problem 3.9Edit

Solved by Gonzalo Perez

Problem StatementEdit

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

GivenEdit

(K 2011 pg.85 #13)Edit

  (1)

Initial conditions are:

 

SolutionEdit

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:


              


GivenEdit

(K 2011 pg.85 #14)Edit

  (1)

Initial conditions are:

 

SolutionEdit

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)