University of Florida/Egm4313/s12.team11.R2

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Problem R2.1[edit | edit source]


Part 1[edit | edit source]


Problem Statement[edit | edit source]


Given the two roots and the initial conditions:




Find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation .
Consider no excitation:

Plot the solution

Solution[edit | edit source]


Characteristic Equation:[edit | edit source]




 


Non-Homogeneous L2-ODE-CC[edit | edit source]


 
Homogeneous Solution:[edit | edit source]




Since there is no excitation,

 
Substituting the given initial conditions:[edit | edit source]



 


 

Solving these two equations for and yields:

 
Final Solution[edit | edit source]


 

2.1fig1.jpg

Part 2[edit | edit source]


Problem Statement[edit | edit source]


Generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the 2 values in (3a) p.3-7 as the 2 roots of the corresponding characteristic equation.

Solutions[edit | edit source]



 



 



 


--Egm4313.s12.team11.gooding 02:01, 7 February 2012 (UTC)

Report 2, Problem 2[edit | edit source]

Problem Statement[edit | edit source]

Find and plot the solution for the homogeneous L2-ODE-CC

with initial conditions ,and

Characteristic Equation[edit | edit source]

Homogeneous Solution[edit | edit source]

The solution to a L2-ODE-CC with real double root is given by

First initial condition

Second initial condition

The solution to our L2-ODE-CC is

                       

Plot[edit | edit source]

PlotR2 2.jpg

Egm4313.s12.team11.imponenti 00:30, 8 February 2012 (UTC)

Problem R2.3[edit | edit source]

Problem Statement (K 2011 p.59 pb. 3)[edit | edit source]

Find a general solution. Check your answer by substitution.

Given[edit | edit source]

Solution[edit | edit source]

We can write the above differential equation in the following form:

Let

The characteristic equation of the given DE is

Now, in order to solve for , we can use the quadratic formula:

Therefore, we have:

and

Thus, we have found that the general solution of the DE is actually:

Check:

To check if is indeed the solution of the given DE, we can differentiate the

what we found to be the general solution.

Substituting the values of and in the given equation, we get:

and thus:

Therefore, the solution of the given DE is in fact:



Problem Statement (K 2011 p.59 pb. 4)[edit | edit source]

Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given[edit | edit source]

Solution[edit | edit source]

The characteristic equation of this ODE is therefore:

Evaluating the discriminant:




Therefore the equation has two complex conjugate roots and a general homogenous solution of the form:





Where:

And finally we find the general homogenous solution:


                                                  


Check:

We found that:




Differentiating to obtain and respectively:


                                   



                 


Substituting these equations into the original ODE yields:











Therefore, the solution is correct.

--Gonzalo Perez

Problem R2.4[edit | edit source]

K 2011 p.59 pb. 5[edit | edit source]

Problem Statement[edit | edit source]

Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given[edit | edit source]



Solution[edit | edit source]

The characteristic equation of this ODE is therefore:



Evaluating the discriminant:



Therefore the equation has a real double root and a general homogenous solution of the form:

[1]

And finally we find the general homogenous solution:

                                                       


Checking:

We found that:



Differentiating to obtain and respectively:

                                                  


And,


                                                 


Substituting these equations into the original ODE yields:









Therefore this solution is correct.

  1. Kreyszig 2011, p.54-57.

K 2011 p.59 pb. 6[edit | edit source]

Problem Statement[edit | edit source]

Find a general solution to the given ODE. Check your answer by substituting into the original equation.

Given[edit | edit source]



Solution[edit | edit source]

Initially we modify the original ODE to put it in the form of a second-order homogenous linear ODE with constant coefficients:



Dividing both sides by 10:



The characteristic equation of this ODE is now therefore:



Evaluating the discriminant:



Therefore the equation has a real double root and a general homogenous solution of the form:

[1]

And finally we find the general homogenous solution:

                                                       


Checking:

We found that:



Differentiating to obtain and respectively:

                                                  


And,


                                                 


Substituting these equations into the original ODE yields:









Therefore this solution is correct.

  1. Kreyszig 2011, p.54-57.

Report 2, Problem 2.5[edit | edit source]

Problem Statement[edit | edit source]

Problem 2.5 Find an Ordinary Differential Equation. Use for the given basis.

ODE Solutions[edit | edit source]

Two Real Roots:
Double Real Roots:

Reverse Engineering[edit | edit source]



Case 1:



Case 2:

Problem Solutions[edit | edit source]

R2.5 K2011 p.59 pbs.16[edit | edit source]





ODE Form:

                                                            

R2.5 K2011 p.59 pbs.17[edit | edit source]





ODE Form:

                                                            


Created by [Daniel Suh] 20:57, 7 February 2012 (UTC)

Problem 2.6[edit | edit source]

Solved by: Andrea Vargas

Problem Statement[edit | edit source]

For the following spring-dashpot-mass system (in series) find the values for the parameters knowing that the system has the double real root

Figure[edit | edit source]

R2.6.jpg

Solution[edit | edit source]

Previously, we have derived the following equation for such a system:
(From Sec 1 (d), (3) p.1-5)

We can write this equation in standard form by diving through by :

Here, we can take the coefficients of and as and :



Next,considering the double real root:

We can find the characteristic equation to be:

Which is in the form:


Then, we know that and :

Setting and from the first equation equal to these, we obtain:

                                                             



Clearly, there is an infinite amount of solutions to this problem because we have 2 equations but 3 unknowns. This can be solved by fixing one of the values and finding the other two.

Example of Solution[edit | edit source]

An example of fixing one of the constants to find the other two is provided here. By solving the simple equations above, we can illustrate how to find . We had:

If we fix the mass to . We find:



Then,



Finally, we obtain:

                                                             


--Andrea Vargas 21:44, 7 February 2012 (UTC)


Problem 2.7: McLaurin Series[edit | edit source]

Problem Statement[edit | edit source]

Develop the McLaurin Series (Taylor Series at t=0) for

Solution[edit | edit source]

                                      


                                      


                                      

--Egm4313.s12.team11.arrieta 17:07, 6 February 2012 (UTC)


Problem R2.8[edit | edit source]


Problem Statement[edit | edit source]


Find a general solution. Check your answer by substitution.

Problem 8[edit | edit source]



Let:

Characteristic Equation[edit | edit source]



Using the quadratic equation to find roots we get:


Therefore:


Check By Substitution[edit | edit source]





Substituting into the original equation, the result is

 

Problem 15[edit | edit source]



Let:

Characteristic Equation[edit | edit source]



Using the quadratic equation to find roots we get:


Therefore:


Check By Substitution[edit | edit source]





Substituting into the original equation, the result is

 

Egm4313.s12.team11.gooding 03:41, 7 February 2012 (UTC)

Report 2, Problem 9[edit | edit source]

Problem Statement[edit | edit source]

Find and plot the solution for the L2-ODE-CC corresponding to

with

and initial conditions ,

In another figure, superimpose 3 figs.:(a)this fig. (b) the fig. in R2.6 p.5-6, and (c) the fig. in R2.1 p.3-7

Quadratic Equation[edit | edit source]

with

Homogeneous Solution[edit | edit source]

The solution to a L2-ODE-CC with two complex roots is given by

where

Solving for A and B[edit | edit source]

first initial condition

second initial condition

so the solution to our L2-ODE-CC is

                      

Solution to R2.6[edit | edit source]

After solving for the constants and we have the following homogeneous equation

Characteristic Equation and Roots[edit | edit source]

We have a real double root

Homogeneous Solution[edit | edit source]

We know the homogeneous solution to a L2-ODE-CC with a double real root to be

Assuming object starts from rest

,

Plugging in and applying our first initial condition

Taking the derivative and applying our second condition

Giving us the final solution

                 

Plots[edit | edit source]

Solution to this Equation[edit | edit source]

Plotr2 9.jpg

Superimposed Graph[edit | edit source]

Our solution: shown in blue

Equation for fig. in R2.1 p.3-7: shown in red

Equation for fig. in R2.6 p.5-6: shown in green

R2superposed.jpg

Egm4313.s12.team11.imponenti 03:38, 8 February 2012 (UTC)