There is a first-order ODE
$$\frac{dy}{dt}=\frac{a(\ln\frac{1-c}{1-y})^3}{\frac{b-y}{1-y}+\ln\frac{1-c}{1-y}},$$ which is subjected to the initial condition $y(t=0)=y_0$ with $a,b,c$ are all constants. I do know it is hardly to get an explicit expression for $y(t)$. Even so, it should be solvable with the method of separation of variables for an implicit result for $y(t)$, which might include the logarithmic integral function
$$li(x)=\int_0^x\frac{dt}{\ln t}.$$ Thank you for any hint.
I don't know whether this will help, I find the $li$ integral here: $$\frac{dt}{dy}=\frac{1}{a}\frac{1}{\ln^2 \frac{1-c}{1-y}}+\frac{1}{a}\frac{\frac{b-y}{1-y}}{\ln \frac{1-c}{1-y}}$$ let $s=\frac{1-c}{1-y}$, it is $$\frac{dt}{dy}=\frac{dt}{ds}\frac{ds}{dy}=\frac{dt}{ds}\frac{1-c}{(1-y)^2}=\frac{1}{a}\ln ^{-2}s + \frac{1}{a}\frac{b-y}{(1-y)\ln ^3 s}$$ let $r=\frac{1}{s},1-y=(1-c)r$ so $$\frac{dt}{dr}=-\frac{1-c}{a \ln ^2 r}+\frac{1-c}{a \ln ^3 r}+\frac{b+1}{a r \ln ^3 r}$$ the last terms is like $\frac{d \ln ^{-2} y}{dy}$, and the 2nd term can use integral by part to turn to the first one, and use integral by part also change the first one to the the $li$ integral, so let $$T=t-\frac{(1-c)r}{2a\ln r}+\frac{(1-c)r}{2a\ln ^2 r}+\frac{b+1}{2a \ln^2 r}$$ it is $$\frac{dT}{dr}=-\frac{1-c}{2a}\frac{1}{\ln r}$$ and the $li$ function can be found here. Thanks to @lxy to find my mistake :)