How do I find the interval for solving this differential equation?

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Discussion Overview

The discussion revolves around finding the interval for solving a differential equation, specifically focusing on the implications of the logarithmic term in the equation and the conditions for its validity.

Discussion Character

  • Technical explanation, Debate/contested, Homework-related

Main Points Raised

  • One participant divides through the differential equation and identifies a term involving $\ln{t}$, questioning the interval endpoints.
  • Another participant points out that the restriction $t > 0$ arises from the logarithmic term $\ln(t)$.
  • There is a repeated mention of the term "baloated," which appears to be a misunderstanding or typo regarding the exponential integration step.
  • A participant questions the relevance of introducing the exponential function $e^{p}$ into the discussion, suggesting it may not align with the problem's requirements.
  • There is a reference to the function $Li_2(x)$, but its connection to the main topic is unclear.

Areas of Agreement / Disagreement

Participants express uncertainty about the terminology used and the implications of the logarithmic term, but there is no clear consensus on the interpretation of the interval endpoints or the relevance of certain mathematical steps.

Contextual Notes

The discussion includes unclear terminology and potential misunderstandings regarding the integration steps and their implications for the solution interval.

Who May Find This Useful

Students or individuals studying differential equations, particularly those interested in the conditions for validity of solutions involving logarithmic functions.

karush
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View attachment 8690
doing #1

ok first I divided thru

$$y' + \frac{\ln{t}}{t-3}y=\frac{2t}{t-3}$$

but the $$\exp\int p(t) \, dt$$ step kinda baloated?

ok I see the denominator has $t-3$ so presume 3 is one of the interval ends
but why 0. ?

book answwer is

View attachment 8691
 

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"Baloated"?

The reason for the restriction x> 0 is the "ln(x)" term.
 
Country Boy said:
"Baloated"?

The reason for the restriction x> 0 is the "ln(x)" term.

you probably mean $ln{t}$

yeah how would this really work $e^{p}$
View attachment 8693
 
I am not sure what you are asking. Do you know what "$Li_2(x)$" is?
 
karush said:
yeah how would this really work $e^{p}$
But your problem statement said not to solve the equation. So why are you introducing the [math]e^p[/math] into your solution?

-Dan
 

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