Intermediate engineering analysis/Problem set 2

Problem R2.1Edit


Part 1Edit


Problem StatementEdit


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

SolutionEdit


Characteristic Equation:Edit


 
 

  


Non-Homogeneous L2-ODE-CCEdit


  
Homogeneous Solution:Edit


 
 
Since there is no excitation,
 

  
Substituting the given initial conditions:Edit


 

  

 

  

Solving these two equations for   and   yields:

  
Final SolutionEdit


  

 

Part 2Edit


Problem StatementEdit


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.

SolutionsEdit


 

  


 

  


 

  


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

Report 2, Problem 2Edit

Problem StatementEdit

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

 

with initial conditions   ,and  

Characteristic EquationEdit

 

 

 

Homogeneous SolutionEdit

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

                        

PlotEdit

 

 

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

Problem R2.3Edit

Problem Statement (K 2011 p.59 pb. 3)Edit

Find a general solution. Check your answer by substitution.

GivenEdit

 

SolutionEdit

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

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

GivenEdit

 

SolutionEdit

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 R2.4Edit

K 2011 p.59 pb. 5Edit

Problem StatementEdit

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

GivenEdit

 

SolutionEdit

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. 6Edit

Problem StatementEdit

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

GivenEdit

 

SolutionEdit

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 2.5Edit

Problem StatementEdit

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

ODE SolutionsEdit

Two Real Roots:  
Double Real Roots:  

Reverse EngineeringEdit

 

Case 1:  
 
 

Case 2:  
 
 

Problem SolutionsEdit

R2.5 K2011 p.59 pbs.16Edit

 
 
 


ODE Form:

                                                             

R2.5 K2011 p.59 pbs.17Edit

 
 
 


ODE Form:

                                                             


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

Problem 2.6Edit

Solved by: Andrea Vargas

Problem StatementEdit

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

FigureEdit

 

SolutionEdit

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 SolutionEdit

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 SeriesEdit

Problem StatementEdit

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

SolutionEdit

 
 
                                       


 
 
                                       


 
 
                                       

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


Problem R2.8Edit


Problem StatementEdit


Find a general solution. Check your answer by substitution.

Problem 8Edit


 

Let:
 

Characteristic EquationEdit


 
Using the quadratic equation to find roots we get:
 
 
Therefore:

 
Check By SubstitutionEdit


 
 
 
Substituting   into the original equation, the result is

  

Problem 15Edit


 
Let:  

Characteristic EquationEdit


 
Using the quadratic equation to find roots we get:
 
 
Therefore:

 
Check By SubstitutionEdit


 

 

 

Substituting   into the original equation, the result is

  

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

Report 2, Problem 9Edit

Problem StatementEdit

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 EquationEdit

  with  

 

 

Homogeneous SolutionEdit

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

 

where  

 

Solving for A and BEdit

first initial condition  

 

 

 

second initial condition  

 

 

 

 

 

so the solution to our L2-ODE-CC is

                       

Solution to R2.6Edit

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

 

Characteristic Equation and RootsEdit

 

 

We have a real double root  

Homogeneous SolutionEdit

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

                  

PlotsEdit

Solution to this EquationEdit

 

 

Superimposed GraphEdit

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