I need to find $x$ such that the following equation works, $x(s)-\int_o^1 2stx(t)dt=sin(\pi s)$ ,where $ s\in [0,1], y\in C[0,1], k\in C([0,1]^2)$. I know that it can be resolved using operator theory, but i wanted to find the solution with simpler means. Can somebody give me a hint ?
Thanks.
$$x(s)-\int_0^1 2st\,x(t)\,dt=x(s)-2s\int_0^1 t\,x(t)\,dt=\sin(\pi s)$$ means $$x(s)=\sin(\pi s)+2s\int_0^1 t\,x(t)\,dt=\sin(\pi s)+2As,\tag1$$ where $$A=\int_0^1 t\,x(t)\,dt\tag2.$$ If we substitute (1) into (2), we get the equation $$A=\int_0^1 t\,\sin(\pi t)\,dt+2A\int_0^1 t^2\,dt.$$ Can you finish from here?