Positive solution of non-linear ODE with logarithm I have problem with following ODE:
$$ \dot{x} = f(x) -\alpha \ln(\frac{\gamma x}{1 - \lambda x}) - \alpha \beta e ^ {-(\beta + \lambda)t}$$
where every parameter is greater than $0$. And $f$ is bounded, positive, increasing , Lipschitz function. I am only inerested when it has non negative solution but I am not interested if it exists. Do you know any useful theorem or general strategy for this kind of equations ?
 A: The logarithmic term only exists for $0<x<\frac1λ$. Approaching these boundaries from the inside of the interval, the logarithmic term grows to infinity pointing into the interval, that is, for $x\to0^+$ $-\ln x$ grows to $+\infty$ and for $x\to\frac1λ-0$ the contribution $\ln(1-λx)=-\ln\frac1{1-λx}$ falls to $-\infty$.  In consequence, the lines $x=0$ and $x=\frac1λ$ are always repelling, independent of the other terms on the right side, as these are bounded.

In summary, the domain of the right side is $(t,x)\in\Bbb R\times(0,\frac1λ)$ and any solution to an IVP with IC $(t_0,x_0)$ contains $[t_0,\infty)$ in its maximal domain.

A: Assume $\alpha,\beta,\lambda,\gamma\neq0$ for the key case:
Hint:
$\dfrac{dx}{dt}=f(x)-\alpha\ln\dfrac{\gamma x}{1-\lambda x}-\alpha\beta e^{-(\beta+\lambda)t}$
$\left(f(x)-\alpha\ln\dfrac{\gamma x}{1-\lambda x}-\alpha\beta e^{-(\beta+\lambda)t}\right)\dfrac{dt}{dx}=1$
Let $u=e^{-(\beta+\lambda)t}$ ,
Then $t=-\dfrac{\ln u}{\beta+\lambda}$
$\dfrac{dt}{dx}=-\dfrac{1}{(\beta+\lambda)u}\dfrac{du}{dx}$
$\therefore\left(f(x)-\alpha\ln\dfrac{\gamma x}{1-\lambda x}-\alpha\beta u\right)\left(-\dfrac{1}{(\beta+\lambda)u}\dfrac{du}{dx}\right)=1$
$\left(u+\dfrac{1}{\beta}\ln\dfrac{\gamma x}{1-\lambda x}-\dfrac{f(x)}{\alpha\beta}\right)\dfrac{du}{dx}=\dfrac{(\beta+\lambda)u}{\alpha\beta}$
This belongs to an Abel equation of the second kind.
