I have the integral equation $$f(x)=\int_0^x (x-t)f(t)dt$$ and want to turn it into an boundary value problem, so I differentiate both sides with respect to $x$ and I get $$f'(x)=\int_0^x f(t)dt$$ Differentiate again and I get $$f''(x)=f(x)$$
However I don't know how to obtain the boundary values for $$f''(x)=f(x)$$, there must be two boundary values of the form $$f(a)+f'(a)=A$$ and $$f(b)+f'(b)=B$$, can you explain how to get them?
The equation $f(x)=\int_0^x (x-t)f(t)dt$ has the unique solution $f=0:$
From the equation we get $f(0)=0$ and from $f'(x)=\int_0^x f(t)dt$ we derive $f'(0)=0.$
$f''(x)=f(x)$ gives
$$f(x)=c_1e^x+c_2e^{-x}.$$
Since $f(0)=0=f'(0)$ we get $c_1=c_2=0.$