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

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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
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...

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