I came across this first order differential equation $$ f'(x) = \left( \frac{1}{x} + \frac{g'(x)}{g(x)} \right) f(x) - c \frac{g'(x)}{g(x)} \textrm{,}$$ where $g(x)$ is this logistic type function $$ g(x) = \frac{h}{1+e^{-(x-a)}} + b $$ and $a$, $b$, $c$ and $h$ are constants.
1st try:
First I tried integrating both sides and got $$ f(x) = \int \frac{f(x)}{x} \mathrm{d}x + \left[ cx - f(x) \right] \ln g(x) + C $$ or $$ \int \frac{f(x)}{x} \mathrm{d}x = \left[ 1 + \ln g(x) \right] f(x) - c x \ln g(x) + C $$ and in both cases the leftover integral term seems to be quite problematic.
2nd try:
After that I remembered that I should probably be using the integrating factor method and looked up that the solution should be $$ f(x) = \frac{1}{\mu(x)} \left( \int \mu(x) \left( - c \frac{g'(x)}{g(x)}\right) \mathrm{d} x + C\right) $$ with $$ \mu(x) = \exp \left[ -\int \left( \frac{g'(x)}{g(x)} + \frac{1}{x} \right) \mathrm{d}x \right] = \frac{1}{x g(x)}$$ so finally $$ f(x) = x g(x) \left( \int \frac{1}{x g(x)} \left( - c \frac{g'(x)}{g(x)}\right) \mathrm{d} x + C \right) = - c x g(x) \left( \int \frac{g'(x)}{g^2(x)} \frac{\mathrm{d} x}{x} + C\right) \textrm{.}$$
The evaluation of the integral $$ \int \frac{g'(x)}{g^2(x)} \frac{\mathrm{d} x}{x} = \int \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{h}{1+e^{-(x-a)}} + b \right) \left( \frac{h}{1+e^{-(x-a)}} + b \right)^{-2} \frac{\mathrm{d} x}{x} $$ $$ = h \int \left( \frac{ e^{-(x-a)}}{(1+e^{-(x-a)})^2} \right) \left( \frac{h}{1+e^{-(x-a)}} + b \right)^{-2} \frac{\mathrm{d} x}{x} $$ seems quite though and I haven't made much progress.
Is there a way to evaluate this integral or is there possibly an easier way to solve the original differential equation?
EDIT: I just realized that $$ g'(x) = [g(x) - b] - [g(x) - b]^2 = g(x) - b - g^2(x) + 2g(x)b - b^2 $$ so that $$ \frac{g'(x)}{g^2(x)} = \frac{1 + 2b}{g(x)} - \frac{b + b^2}{g^2(x)} - 1 $$ and I would only need to know how to evaluate $$ \int \frac{1}{g(x)} \frac{\mathrm{d} x}{x} \quad \textrm{and} \quad \int \frac{1}{g^2(x)} \frac{\mathrm{d} x}{x} \textrm{.} $$ Could there be a closed form solution or maybe just a series representation or something?
Whenever I see $\displaystyle\frac{g'(x)}{g(x)}$, my immediate instinct is to write that as $[\ln g(x)]'$. Here it helps, as follows:
$f'(x) = [\,\displaystyle\frac{1}{x} + (\ln g)'\,]\,f(x) - c\,(\ln g)'$
$f' = \displaystyle\frac{f}{x} + (f-c)\,(\ln g)'$
Now define a new function $u(x) = f(x) - c$. Then,
$u' = \displaystyle\frac{u + c}{x} + (\ln g)'\,u$
$x\,u' = u + c + x\,(\ln g)'\,u = c + \underbrace{[1 + x\,(\ln g)']}_{a(x)}\,u$
$x\,u'(x) - a(x)\,u(x) = c$
The homogeneous equation is:
$\displaystyle\frac{u'(x)}{u(x)} = \displaystyle\frac{a(x)}{x}$
So...
$[\,\ln u(x)\,]' = \displaystyle\frac{a(x)}{x} = \displaystyle\frac{1}{x} + [\,\ln g(x)\,]'$
and
$\ln u = \ln x + \ln g + \ln A$
($A$ is an integration constant)
$u(x) = A\,x\,g(x)$
I'll let you continue from here. You could use the method of variation of parameters to find a particular solution $u_{p}(x)$ of
$x\,u'(x) - a(x)\,u(x) = c$
then use $f(x) = A\,x\,g(x) + u_{p}(x) + c$ as your final general solution.
Edit: I decided to complete the solution so...
Method of variation of parameters. We have a general solution of the homogeneous equation, $u(x) = A\,x\,g(x)$. Now, try a particular solution of the form $u_p(x) = A(x)\,x\,g(x)$. Then,
$x\,u'(x) - a(x)\,u(x) = c$
gives us
$A'(x)\,x^2\,g(x) = c$
so
$A(x) = \displaystyle\int^x \frac{c\,dt}{t^2\,g(t)}$
and
$u_p(x) = A(x)\,x\,g(x) = c\,x\,g(x) \displaystyle\int^x \frac{dt}{t^2\,g(t)}$
and the final general solution is
$f(x) = A\,x\,g(x) + c\,x\,g(x) \displaystyle\int^x \frac{dt}{t^2\,g(t)} + c$
where $A$ is a constant.