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

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In summary, the conversation discusses solving a differential equation by dividing through and using the $\exp\int p(t)\,dt$ step. The restriction x>0 is due to the "ln(x)" term and the book answer refers to a possible mistake in the solution process. The conversation also mentions $Li_2(x)$ and introduces $e^p$ into the solution, which goes against the initial problem statement.
  • #1
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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  • #2
"Baloated"?

The reason for the restriction x> 0 is the "ln(x)" term.
 
  • #3
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
 
  • #4
I am not sure what you are asking. Do you know what "$Li_2(x)$" is?
 
  • #5
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 \(\displaystyle e^p\) into your solution?

-Dan
 

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

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