Integrating Factor Method and Absolute Value Bars

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Discussion Overview

The discussion revolves around the application of the integrating factor method to solve a differential equation, specifically addressing the treatment of absolute value bars in the integrating factor derived from the equation.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant presents a differential equation and derives the integrating factor as \(\mu(y) = |y|\), questioning the implications of the absolute value in the context of solving the equation.
  • Another participant raises a question about the logarithm of a negative number, \(\ln(-1)\), highlighting a potential concern regarding the use of absolute values.
  • A different participant asserts that the logarithm is undefined, reiterating the importance of absolute values in the integrating factor.
  • One participant suggests that it is acceptable to use either \(y\) or \(-y\) when multiplying the entire equation, indicating that this choice does not affect the final result.
  • A later reply indicates understanding of the previous points raised in the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of using absolute values in the integrating factor, indicating that multiple competing views remain without a consensus on the best approach.

Contextual Notes

The discussion does not resolve the implications of dropping the absolute value bars, nor does it clarify the conditions under which this might be valid.

Who May Find This Useful

Readers interested in differential equations, integrating factors, and the implications of absolute values in mathematical solutions may find this discussion relevant.

manenbu
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So there's this equation:
[tex]x^2 y^2 dx + (x^3y-1)dy[/tex]
It has to be solved with the integrating factor method, so I get this:
[tex]\mu(y) = e^{\int \frac{dy}{y}} = e^{\ln{|y|}} = |y|[/tex]

My question is, what do I do with the absolute value bars?
If I just drop them and multiply the entire equation with y, then I can solve the equation and get:
[tex]2x^3 y^3 - 3 y^2 = C[/tex]
Which is the correct answer.
But I'm not sure that dropping it will always be correct, so what should be done here?
 
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what is [itex]\ln(-1)[/itex]?
 
It's undefined, and I know that.
This is the reason you put the bars in the first place, but my question was about the integrating factor itself, should it be y or |y|.
 
Use either y or -y. Since you are multiplying the entire equation by that, it doesn't affect the result.
 
ok I understand!
 

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