I want to solve this differential equation : $$ {(1-k) \over f(x)}+ {(2k-1) \over x} = {1 \over f(x)}{df(x) \over dx} $$
$f(x)$is ordinary function of $x$ $\in$ $(0,\infty)$. The range of $f(x)$ is $(0,\infty)$.
k is a constant in $(0,1)$.
I want to solve it towards $f(x)$. I am looking for what method is needed for solving this and the individual steps.
If you multiply through by $f$ and rearrange the terms, you get the equation $$f'(x) - \frac{(2k-1)}{x}f(x) = 1-k$$ which can be solved using an integrating factor.
See http://mathworld.wolfram.com/IntegratingFactor.html for more details about integrating factors and how to use them.