Solving Systems of Inhomogeneous Linear ODEs

In summary, the conversation is about determining the y_particular solution for a given problem and the confusion between using the method of undetermined coefficients and the method of parameter variation. It is suggested to start over and read a worked example in a book to better understand the concept of solving a system of inhomogeneous linear ODE.
  • #1
terryds
392
13

Homework Statement



2m82rr9.png

Determine the y_particular solution

Homework Equations



2gt92lw.png


The Attempt at a Solution



I've tried this for hours but still don't get the correct value.
This is what I get:
2yngsqv.jpg
The question is the same as the one I found from http://www4.ncsu.edu/eos/users/s/slc/www/COURSES/File2.pdf (Example 6.11.2).
The correct solution is as follows.
24nnrdh.png


I get the coefficient for e^t is 1/8 and 1/16, but it says that the correct one is 1/4 and 0.

I really have no idea where I got wrong, please help me[UPDATE]

It turns out that the method of parameter variation REALLY yields DIFFERENT PARTICULAR SOLUTION from method of undetermined coefficients,

Undetermined Coefficient: http://www4.ncsu.edu/eos/users/s/slc/www/COURSES/File2.pdf (Example 6.11.2)

Variation of Parameters: http://www.wolframalpha.com/input/?i=solve+x'+=+-x+++4y+++e^t,+y'+=+x-y+1So which one is wrong, the undetermined coefficient or variation of parameter for this problem??
Or is it possible that those all are correct answers?

Please help. I'm so confused.
 
Last edited:
Physics news on Phys.org
  • #2
I've not looked at your shown work but am responding to your updated comment directly. Recall that your general solution is the particular solution plus the range of all possible homogeneous solutions. Thus adding any of the homogeneous solutions to the particular solution yields another particular solution. They are both right if the particular plus homogeneous solutions yields the same (correct) set of general solutions.

It is a generalization of the process of integration... to find the indefinite integral you find an anti-derivative (particular) and then add the homogeneous solution to d/dx = 0 namely an arbitrary constant. Note that any anti-derivative may be used as they all differ by a constant. For the more general linear diffeq. all of your particular solutions differ by some solution to the homogeneous eqn.
 
  • #3
terryds said:

Homework Statement



View attachment 212489
Determine the y_particular solution

Homework Equations



View attachment 212490

The Attempt at a Solution



I've tried this for hours but still don't get the correct value.
This is what I get:
View attachment 212491The question is the same as the one I found from http://www4.ncsu.edu/eos/users/s/slc/www/COURSES/File2.pdf (Example 6.11.2).
The correct solution is as follows.
View attachment 212492

I get the coefficient for e^t is 1/8 and 1/16, but it says that the correct one is 1/4 and 0.

I really have no idea where I got wrong, please help me[UPDATE]

It turns out that the method of parameter variation REALLY yields DIFFERENT PARTICULAR SOLUTION from method of undetermined coefficients,

Undetermined Coefficient: http://www4.ncsu.edu/eos/users/s/slc/www/COURSES/File2.pdf (Example 6.11.2)

Variation of Parameters: http://www.wolframalpha.com/input/?i=solve+x'+=+-x+++4y+++e^t,+y'+=+x-y+1So which one is wrong, the undetermined coefficient or variation of parameter for this problem??
Or is it possible that those all are correct answers?

Please help. I'm so confused.

Have you tried substituting your solutions into the DEs, to see if they work? That is always the first step you should do.
 
  • #4
terryds said:
Please help. I'm so confused.

Honestly, I think you should start over. Do yourself a favour and read a worked example in a book, say Boyce & DiPrima, an try to apply these steps to your problem. I think the whole concept of solving a system of inhomogeneous linear ODE is somewhat obscure to you, and that ist the source of your confusion.
 

Related to Solving Systems of Inhomogeneous Linear ODEs

1. What is "Variation of Parameters" in mathematics?

Variation of Parameters is a method used to find a particular solution to a non-homogeneous linear differential equation. It involves finding a solution by varying the parameters of a general solution to the corresponding homogeneous equation.

2. How does "Variation of Parameters" work?

The method of Variation of Parameters works by first finding the general solution to the corresponding homogeneous equation. Then, a particular solution is found by varying the parameters in the general solution according to a specific formula. This particular solution is then added to the general solution to obtain the overall solution to the non-homogeneous equation.

3. When is "Variation of Parameters" used?

"Variation of Parameters" is used when solving non-homogeneous linear differential equations. It is particularly useful when the non-homogeneous term is in the form of a polynomial, exponential, or trigonometric function.

4. What are the advantages of using "Variation of Parameters"?

One advantage of using "Variation of Parameters" is that it is a systematic and straightforward method for finding particular solutions to non-homogeneous differential equations. It also allows for the use of initial or boundary conditions to find a unique solution.

5. Are there any limitations to "Variation of Parameters"?

One limitation of "Variation of Parameters" is that it does not work for all types of non-homogeneous equations. It is also more complicated and time-consuming compared to other methods, such as the method of undetermined coefficients. In some cases, it may be difficult to find the general solution to the corresponding homogeneous equation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
7
Views
552
  • Calculus and Beyond Homework Help
Replies
7
Views
761
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
897
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
987
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Back
Top