Solving Systems of Inhomogeneous Linear ODEs

terryds
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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:
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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.
 
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.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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