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

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SUMMARY

The discussion focuses on finding the interval for solving the differential equation \(y' + \frac{\ln{t}}{t-3}y=\frac{2t}{t-3}\). Participants identify that the denominator \(t-3\) indicates that \(t\) cannot equal 3, while the term \(\ln(t)\) necessitates that \(t > 0\). The conclusion is that the valid interval for \(t\) is \( (0, 3) \) due to these restrictions.

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karush
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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

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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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