Relationship between two solutions

[Milligram]

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

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

 

[HallsofIvy]

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

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

Note that g'(x) \sim e^{-g(x)}.

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

 

[Milligram]

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