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
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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