Let us consider the following Cauchy problem on $\mathbb{R}_+$: $$ g'(t)=-g^2(t)+g^3(t)+g^4(t)h(g(t)),\quad g(0)=\alpha $$ where $\alpha$ is real and positive and $h$ is analytic in a neighbourhood of $0$. I would like to prove, for $\alpha$ small enough, the existence of a bounded function $y$ such that the unique solution of this Cauchy problem writes: $$ \frac{\alpha}{g(t)}=1+\alpha t-\alpha\log(1+\alpha t)-\alpha y(t),\quad y(0)=0. $$ In the simple case $h\equiv 0$, I can prove the existence of a bounded $y$ by posing $u=\alpha/g$ and plugging the ansatz into the ODE for $u$: $$ u'=\alpha\big(1-\frac{\alpha}u\big). $$ Then I integrate this last equation, get an implicit equation on $u$ and plug the ansatz into it. I can conclude by some standard arguments.

In the general case, I cannot use the same strategy and thus would like to prove boundedness of $y$ directly from its ODE. Already in the case $h\equiv 0$, I do not know how to proceed. Here is the ODE satisfied by $y$ in this simple case: $$ y'(t)=\frac{\alpha^2}{1+\alpha t}\frac{\log(1+\alpha t)+y(t)}{1+\alpha t-\alpha\log(1+\alpha t)-\alpha y(t)}. $$ Does anyone know how to prove boundedness of the solution of this last ODE with initial condition $y(0)=0$?