If I have an ODE $Lf(x)=g(x)$, then I can write the solution as $f(x)=\int G(x,y)g(y)dy$ where $G(x,y)$ is the Green's function of $L$ (it is the kernel of the inverse of $L$). Here $g(x)$ is given and $f(x)$ is to be found.
I also read about the Fredholm equation, which is of the same form, $f(x)=\int K(x,y)g(y)dy$, but where now $f$ is given and $g$ is to be found.
My question is about the relation between Green's functions and Fredholm equations. I have read that using the Green's function we are "reducing" the ODE to a Fredholm equation. I don't understand this statement. They look more like complementary problems, with different origins and different applications.
I would appreciate a broad overview of this. Which problem is more fundamental? What is going on here?
One thing I can say is that you can indeed turn one problem into the other.
Say you have the equation $$f''(x)+A(x)f'(x)+B(x)f(x)=g(x)$$ with given initial value conditions $f(a)$ and $f'(a)$.
Define $u(x)=f''(x)$, which leads to $f'(x)=f'(a)+\int_a^x u(t)dt$ and $$f(x)=f(a)+(x-a)f'(a)+\int_a^x\int_a^{x'}u(t)dtdx'=f(a)+(x-a)f'(a)+\int_a^x(x-t)u(t)dt.$$
You can write the original equation as $$u(x)=g(x)+h(x)+\int_a^xK(x,t)u(t)dt,$$ where $h(x)=A(x)f'(a)+B(x)(f(a)+(x-a)f'(a))$ and $K(x,t)=A(x)+B(x)(x-t)$. This is now an integral equation for the unkown function $u(x)$.