I have the following differential equation $$ \frac{d}{dx}\left(\mu e^{cx}f(x)\right) = -\mu\left(\frac{a xe^{-cx}}{a x+x-1}\right) $$ that I am trying to integrate to find $f(x)$ with the boundary constraint that $f(1) =1$. The integrating factor $\mu$ is given by $$ \mu = e^{\frac{a c}{1+a}x}\left((a+1)x-1\right)^{\frac{1+a+ac}{(1+a)^2}} $$ If we integrate this with a lower limit of $x$ then $$ \mu e^{ct}f(t)\big|_x = -\int_x \mu\left(\frac{a te^{-ct}}{a t+t-1}\right)dt $$ However, what would be a sensible upper limit so that I can find $f(x) = ?$. The integrating factor has a zero at $x=1/(a+1)$. Would it make sense to integrate up to this boundary?
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I am a bit confused. Why don't you just integrate between $1$ and $x$ so you get to use your initial condition $f(1)=1$? You will get $$\mu(x) e^{ct}f(t)\big|_1^x = -\int_1^x \mu(t)\left(\frac{a te^{-ct}}{a t+t-1}\right)dt$$ which gives $$\mu(x) e^{cx}f(x)-\mu(1) e^{c}1 = -\int_1^x \mu(t)\left(\frac{a te^{-ct}}{a t+t-1}\right)dt$$ and so you get $$f(x) =\frac{\mu(1) e^{c}}{\mu(x) e^{cx}} -\frac{1}{\mu(x) e^{cx}}\int_1^x \mu(t)\left(\frac{a te^{-ct}}{a t+t-1}\right)dt,$$which holds for all $x \ne 1/(a+1)$. I doubt the integral can be solved, since it looks a lot like a Gamma function.
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$\begingroup$ Many thanks for your help, I was wondering if you could take a quick look at the following question, which attempts to solve a similar integral to this? $\endgroup$ – AngusTheMan Aug 21 '18 at 13:41
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$\begingroup$ math.stackexchange.com/questions/2889856/… $\endgroup$ – AngusTheMan Aug 21 '18 at 13:42
