How many equations are needed to solve this system?

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Discussion Overview

The discussion revolves around a system of linear differential equations that participants are attempting to solve using the method of undetermined coefficients. The focus is on the number of equations relative to the unknowns and the process of finding solutions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a system of equations and expresses confusion about solving it, noting that there are two equations and three unknowns.
  • Another participant identifies a potential contradiction in the proposed solution for one of the variables, suggesting a mistake in either the system or the solution.
  • A correction is made regarding the formulation of one of the equations, indicating a mistake in the original post.
  • One participant suggests a substitution method to simplify the system, proposing to express some variables in terms of others to reduce the number of equations.
  • Another participant asserts that there are four equations and four unknowns, implying that a unique solution should be obtainable if the determinant of the coefficient matrix is non-zero.
  • A later reply confirms that the initial confusion has been resolved, indicating that the system can indeed be solved.
  • Some participants reiterate the observation that there are four equations in total, which appears to contradict the initial claim of having two equations and three unknowns.

Areas of Agreement / Disagreement

Participants express differing views on the number of equations and unknowns, with some asserting there are four equations while others initially believed there were only two equations with three unknowns. The discussion reflects uncertainty and correction of earlier claims without reaching a consensus on the initial confusion.

Contextual Notes

There are unresolved aspects regarding the interpretation of the equations and the conditions under which a unique solution can be obtained, particularly concerning the determinant of the coefficient matrix.

enceladus_
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This system originates from systems of linear Diff Eqs. I am required to use the method of undetermined coefficients.

-a2 + 5b2 - a1 = 0
-a2 + b2 -b1 -2 = 0
-a1 + 5b1 +a2 + 1 = 0
-a1 + b1 + b2 = 0

I can't figure out how I am able to solve this. In each case, I have two equations and three unknowns. I know the system is correct.

The answers are:

a2 = -3, b2 = -2/3, a1 = -1/3, b1 = 1/3.

Thanks in advance.
 
Last edited:
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You can solve your third equation directly:

-a1 + 5b1 +a1 + 1 = 0 → 5b1 + 1 = 0 → b1 = -1/5

This contradicts the answer for b1 you stated. There is either a mistake in the system or in the solution.

Junaid Mansuri
 
Last edited:
It should be:

-a1 +5b1 + a2 + 1 = 0

My mistake.
 
An easy approach to solve this system is through substitution. Solve for a1 in equation 4 (in terms of the b's). Solve for a2 in equation 2 (in terms of the b's). Substitute the expressions for a1 and a2 into equations 1 and 3 and you will then have a system of two equations with only b1 and b2 in them which is easy to solve. Then you can use the known values of b1 and b2 to directly find a1 and a2 from the expressions you had before. I think this method would be easier than elimination.
 
Last edited:
enceladus_ said:
This system originates from systems of linear Diff Eqs. I am required to use the method of undetermined coefficients.

-a2 + 5b2 - a1 = 0
-a2 + b2 -b1 -2 = 0
-a1 + 5b1 +a2 + 1 = 0
-a1 + b1 + b2 = 0

I can't figure out how I am able to solve this. In each case, I have two equations and three unknowns. I know the system is correct.

The answers are:

a2 = -3, b2 = -2/3, a1 = -1/3, b1 = 1/3.

Thanks in advance.

It looks like you have four equations in four unknowns (a1, a2, b1, and b2). Some of the variables just have 0 coefficients in each of the equations. As long as the determinant of the matrix of coefficients for the system is non-zero, you can obtain a unique solution for the system.
 
Many thanks, I was able to solve it!
 
enceladus_ said:
This system originates from systems of linear Diff Eqs. I am required to use the method of undetermined coefficients.

-a2 + 5b2 - a1 = 0
-a2 + b2 -b1 -2 = 0
-a1 + 5b1 +a2 + 1 = 0
-a1 + b1 + b2 = 0

I can't figure out how I am able to solve this. In each case, I have two equations and three unknowns. I know the system is correct.

The answers are:

a2 = -3, b2 = -2/3, a1 = -1/3, b1 = 1/3.

Thanks in advance.

I see four equations in four unknowns?
 
mathman said:
I see four equations in four unknowns?

You're late to the party. See post #5 above.
 

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