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Simplified question*:

Given $f(t)$ that satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$ we denote $F=\exp\left(f\left(t\right)\right)$. Let $H(t)$ be a solution of $$ \dot{H}=F $$ Can we approximate H by F?

Specifically, I want to show that $$ \lim_{s\to\infty}\frac{H^{-1}(s)}{F^{-1}(s+C)}=1 $$ for some constant C. $F^{-1}$ and $H^{-1}$ are the inverse functions of $F$ and $H$, respectively.

If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

* thanks to Willie Wong helpful suggestion.

The original question (to make to motivation more clear):

I have the following differential equation: $$\dot{g}(t)=\exp\left(-f\left(g\left(t\right)\right)\right)$$ and I know that $f(t)$ satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.

I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.

The motivation behind this solution is $$ \exp\left(f\left(g(t)\right)\right)\approx\exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g\left(t\right)\right)\right]\right) $$ since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve $$ \exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g(t)\right)\right]\right)\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g\left(t\right)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$ for some constatnt C.

Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

Simplified question*:

Given $f(t)$ that satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$ we denote $F=\exp\left(f\left(t\right)\right)$. Let $H(t)$ be a solution of $$ \dot{H}=F $$ Can we approximate H by F?

Specifically, I want to show that $$ \lim_{s\to\infty}\frac{H^{-1}(s)}{F^{-1}(s+C)}=1 $$ for some constant C. $F^{-1}$ and $H^{-1}$ are the inverse functions of $F$ and $H$, respectively.

If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

* thanks to Willie Wong helpful suggestion.

The original question (to make the motivation more clear):

I have the following differential equation: $$\dot{g}(t)=\exp\left(-f\left(g\left(t\right)\right)\right)$$ and I know that $f(t)$ satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.

I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.

The motivation behind this solution is $$ \exp\left(f\left(g(t)\right)\right)\approx\exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g\left(t\right)\right)\right]\right) $$ since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve $$ \exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g(t)\right)\right]\right)\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g\left(t\right)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$ for some constatnt C.

Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

I have the following differential equation: $$\dot{g}(t)=\exp\left(-f\left(g\left(t\right)\right)\right)$$ and I know that $f(t)$ satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.

I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.

The motivation behind this solution is $$ \exp\left(f\left(g(t)\right)\right)\approx\exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g\left(t\right)\right)\right]\right) $$ since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve $$ \exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g(t)\right)\right]\right)\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g\left(t\right)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$ for some constatnt C.

Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

Simplified question*:

Given $f(t)$ that satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$ we denote $F=\exp\left(f\left(t\right)\right)$. Let $H(t)$ be a solution of $$ \dot{H}=F $$ Can we approximate H by F?

Specifically, I want to show that $$ \lim_{s\to\infty}\frac{H^{-1}(s)}{F^{-1}(s+C)}=1 $$ for some constant C. $F^{-1}$ and $H^{-1}$ are the inverse functions of $F$ and $H$, respectively.

If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

* thanks to Willie Wong helpful suggestion.

The original question (to make to motivation more clear):

I have the following differential equation: $$\dot{g}(t)=\exp\left(-f\left(g\left(t\right)\right)\right)$$ and I know that $f(t)$ satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.

I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.

The motivation behind this solution is $$ \exp\left(f\left(g(t)\right)\right)\approx\exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g\left(t\right)\right)\right]\right) $$ since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve $$ \exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g(t)\right)\right]\right)\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g\left(t\right)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$ for some constatnt C.

Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

I have the following differential equation: $$\dot{g}(t)=\exp(-f(g(t))$$ and I know that $f(t)$ satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.

I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.

The motivation behind this solution is $$ \exp\left(f(g(t)\right)\approx\exp\left(f(g(t)+\log\left( f'\left(g(t)\right)\right)\right) $$ since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve $$ \exp(f(g(t)+\log\left( f'\left(g(t)\right)\right))\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g(t)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$ for some constatnt C.

Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.

I have the following differential equation: $$\dot{g}(t)=\exp\left(-f\left(g\left(t\right)\right)\right)$$ and I know that $f(t)$ satisfies $f'(t)>0$, $f'(t)=\omega\left(t^{-1}\right)$, $\log\left(f'(t)\right)=o\left(f(t)\right)$.

I would like to show that the solution to this equation is $$g(t)=f^{-1}(\log(t+C))+h(t)$$ where $h(t)=o\left( f^{-1}(\log(t)) \right)$, $f^{-1}(t)$ is the inverse function of $f(t)$ and C is some constant.

The motivation behind this solution is $$ \exp\left(f\left(g(t)\right)\right)\approx\exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g\left(t\right)\right)\right]\right) $$ since $\log\left(f'(t)\right)=o\left(f(t)\right)$, and therefore we can solve $$ \exp\left(f\left(g\left(t\right)\right)+\log\left[ f'\left(g(t)\right)\right]\right)\dot{g}(t)=1 \Rightarrow \frac{d}{dt}\exp\left(f(g\left(t\right)\right)=1 \Rightarrow f(g(t))=\log(t+C)$$ for some constatnt C.

Can I rigorously show this result? If necessary I can add additional assumptions on $f(t)$, though I would like to keep it as general as possible.