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## Please give suggestions about solving a double integral equation

I have been struggling for a while with a problem I've encountered in my research and I really need help. It is a rather peculiar integral equation
$$f_0(x)=S(x)+\int_{x}^b \frac{\Sigma_s}{2}\exp(-\Sigma_t(x'-x))f_1(x',x'')dx'+\int_{-b}^x \frac{\Sigma_s}{2}\exp(-\Sigma_t(x-x'))f_1(x',x'')dx'$$
in which
$$f_1(x',x'')=\int_{x'}^b \frac{\Sigma_s}{2}\exp(-\Sigma_t(x''-x'))f_0(x'')dx''+\int_{-b}^x' \frac{\Sigma_s}{2}\exp(-\Sigma_t(x'-x''))f_0(x'')dx''$$
and the source $$S(x)=C_0+C_1\cosh(\Sigma_tx)$$
It comes from some new theory concerning Monte carlo simulations for neutron transport.

Since I know nothing about the solution function $$f_0(x)$$, I need an educated guess about its form. If the form is known I can transform into a differential equation and solve for the coefficients. I've tried some stuff but it just wouldn't work.

What may be of help is that the solution to an equation with a single integration which looks like
$$g(x)=C_0+\int_{x'}^b \frac{\Sigma_s}{2}\exp(-\Sigma_t(x'-x))g(x')dx'+\int_{-b}^x \frac{\Sigma_s}{2}\exp(-\Sigma_t(x-x'))g(x')dx'$$
can be written as
$$g(x)=A_0+A_1\cosh(\sqrt{\Sigma_t(\Sigma_t-\Sigma_s)}$$
I've obtained this results by transorming into a differential equation and solving that.