Relationship between two solutions

[Milligram]

Let [itex] t \in \mathbb{R} [/itex] be large.
Let [itex] f [/itex] be a function over [itex] [0,t] [/itex] satisfying [itex] f(0) = 1 [/itex] and [itex] f'(x) = e^{-f(x)} [/itex] for all [itex] x[/itex].
Let [itex] g [/itex] be a function over [itex] [0,t] [/itex] satisfying [itex] g(0) = 1 [/itex] and [itex] g'(x) = (1 - g(x)/t^2)^{t^2} [/itex] for all [itex] x[/itex]. Note that [itex] g'(x) \sim e^{-g(x)}[/itex].

Without solving the two differential equations and finding out [itex] f [/itex] and [itex] g [/itex] (which can be done at least approximately), can the fact that [itex] f(0)=g(0)=1 [/itex] and [itex] g'(x) \sim e^{-g(x)} [/itex] be used to show that [itex] f(x) \sim g(x) [/itex] for all [itex] x[/itex] in [itex] [0,t][/itex] ?

 

[HallsofIvy]

Let [itex] t \in \mathbb{R} [/itex] be large.
Let [itex] f [/itex] be a function over [itex] [0,t] [/itex] satisfying [itex] f(0) = 1 [/itex] and [itex] f'(x) = e^{-f(x)} [/itex] for all [itex] x[/itex].
Let [itex] g [/itex] be a function over [itex] [0,t] [/itex] satisfying [itex] g(0) = 1 [/itex] and [itex] g'(x) = (1 - g(x)/t^2)^{t^2} [/itex] for all [itex] x[/itex].

This makes no sense. If g is a function of x only, its derivative cannot depend upon both x and t. Did you mean [itex]g'(x) = (1 - g(x)/x^2)^{x^2} [/itex]?

Note that [itex] g'(x) \sim e^{-g(x)}[/itex].

Without solving the two differential equations and finding out [itex] f [/itex] and [itex] g [/itex] (which can be done at least approximately), can the fact that [itex] f(0)=g(0)=1 [/itex] and [itex] g'(x) \sim e^{-g(x)} [/itex] be used to show that [itex] f(x) \sim g(x) [/itex] for all [itex] x[/itex] in [itex] [0,t][/itex] ?

 

[Milligram]

HallsofIvy, I meant for [itex] g(x) [/itex] to depend on [itex]x[/itex] and [itex]t[/itex]. You should think of [itex]t[/itex] as being a large constant, say, [itex] t=10^{10} [/itex].