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, $t=10^{10}$.