Solve $\frac{a^2}{x^2 f'(x)}-\log{{(1-f(x))}}=S^2 +1$

Question

Solve $$ \frac{a^2}{x^2 f'(x)}-\log{{(1-f(x))}}=S^2 +1, $$ for $x\geq 1$, $f(x)<1$ where $a,S\in \mathbb{R}$ and $f'(x)=\frac{d f(x)}{dx}$.

My effort: It is equivalent to $$ {a^2}-(x^2 f'(x))\log{{(1-f(x))}}-(S^2 +1)(x^2 f'(x))=0, $$ or $$ {a^2}- f'(x)x^2(\log{{(1-f(x))}}-S^2 +1)=0, $$

I am not sure what to do with the $\log(1-f(x))$. Any idea?

Answer 1

To make life easier, first let $$\log(1-f)= y \implies f=1-e^{y}\implies f'=e^y \,y'$$ making the equation to be $$\frac{a^2 e^{-y}}{x^2 y'}+y+S^2+1=0$$ which is, as Robert Israel commented, is separable. $$\frac{a^2}{x^2}=-(y+S^2+1)\,e^y\,y'$$ Integrate both sides $$\frac {a^2}x+C=e^y \left(S^2+y\right)$$ If you want to go further to get $y$, let $z=S^2+y$ to make$$e^{S^2}\left(\frac {a^2}x+C\right)=z \,e^z$$ which has a solution $$z=W\left(\frac{e^{S^2} \left(a^2+C x\right)}{x}\right)$$ where appears Lambert function.

Now, go back to $y$ and then to $f$ for an analytical expression.