1. The problem statement, all variables and given/known data f '(x)-cf(x)+g(x)=0 2. Relevant equations f(X)=1 f(x)=e-c(X-x) + ∫xX e-c(r-x) * g(r)dr X is a constant. 3. The attempt at a solution Ok, so the part of this problem that is confusing me is the integral within the function, f(x). It is an exponential function times an unknown function, g, which also is in the differential equation so my first thought was to use integration by parts. Before I get to that though, let me put down how far I've gotten into solving this quandary. Plugging f(x) into the DE I've got: f '(x)= ce-c(X-x) + e-c(X-x)g(X)-g(x) Therefore: f '(x)-cf(x)+g(x)=0 ==> ce-c(X-x) + e-c(X-x)g(X)-g(x)-c(e-c(X-x) + ∫xX e-c(r-x) * g(r)dr)+g(x)=0 The +-g(x) and +-ce-c(X-x) cancel to leave us with: e-c(X-x)g(X) - c∫xX e-c(r-x) * g(r)dr=0 Or equivalently; e-c(X-x)g(X) = c∫xX e-c(r-x) * g(r)dr Now back to my idea of integrating by parts. Again, it's just an unknown function and an exponential so it seems easy but I'm getting lost. I set: u=g(r) v'=e-c(r-x) such that u=g(r) v=-(e-c(r-x))/c u'=g'(r) v'=e-c(r-x) Then my uv-∫u'v would mean e-c(X-x)g(X) = c([-g(r)(e-c(r-x))/c]|xX + ∫xX g'(r)(e-c(r-x))/c dr ______________________________________________________________ Getting here at least means that I have g'(r) in my integral instead of g(r) but, from this point I don't know what to do. I'm not even sure my math is right up to this point. I feel like if I try to integrate by parts again then I will just get caught in an infinite loop. Was this even the correct method in the first place? Any help or guidance would be greatly appreciated.