Solving Systems of DEs with Undetermined Coefficients

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SUMMARY

The discussion focuses on solving a system of differential equations (DEs) using undetermined coefficients, specifically involving the equation a + b = Ab + (1,0)T, where 'a' is defined as (1,1)T and A is the coefficient matrix [[2, -1], [3, -2]]. The participant attempts to isolate vector 'b' by rearranging the equation but encounters difficulties due to the non-invertibility of the matrix (A - I). The conclusion drawn is that (A - I) acts as a projection onto the line x = y, indicating that the equation (A - I)b = (0,1) has no solution.

PREREQUISITES
  • Understanding of systems of differential equations (DEs)
  • Familiarity with matrix operations, including addition and multiplication
  • Knowledge of undetermined coefficients method in DEs
  • Basic concepts of linear transformations and projections
NEXT STEPS
  • Study the method of undetermined coefficients in greater detail
  • Learn about matrix projections and their geometric interpretations
  • Explore the implications of non-invertible matrices in linear algebra
  • Investigate alternative methods for solving systems of DEs, such as variation of parameters
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations and linear algebra, as well as anyone seeking to deepen their understanding of matrix theory and its applications in solving DEs.

Melawrghk
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Homework Statement


So I'm doing a systems of DEs question with undetermined coefficients. After comparing coefficients, I have this:
a+b=Ab+(1,0)T

where a&b are vectors (2D) and A is the coefficient matrix:
[2 -1]
[3 -2]

I know that 'a' is (1,1)T. I need to find b.

I tried rearranging it, but I don't know if that's right.
(1,1)-(1,0)=Ab-b
(0,1)=(A-1)b

So I found A-I:
[1 -1]
[3 -3]

I was going to invert it and multiply by (0,1), but it's not invertible. What do I do?
 
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Hi Melawrghk! :wink:
Melawrghk said:
… I tried rearranging it, but I don't know if that's right.
(1,1)-(1,0)=Ab-b
(0,1)=(A-1)b

Yes, that's fine! :smile:
I tried rearranging it, but I don't know if that's right.
(1,1)-(1,0)=Ab-b
(0,1)=(A-1)b
Give up! :biggrin:

A - I is a projection onto the line x = y (because it sends (p,q) to (p-q,p-q)), so you can't have (A - I)b = (0,1).
 

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