I am trying to solve the integral equation $$ f(x)=\int_0^x (x-y)f(y)dy, \ 0\leq x\leq 1 $$ in the space $C([0,1])$ of continuous functions on $[0,1]$. My reasoning is this: Since $u(y)=(x-y)f(y)$ is continuous, the integral $\int_0^x (x-y)f(y)dy$ is a differentiable function, which means that the solution $f$ will also be differentiable.
Differentiating gives $f'(x)=\int_0^xf(y)dy$. By a similar argument as before $f'$ is differentiable as well, therefore $f''=f$. This equation has the general solution $$ f(x)=a\exp(x)+b\exp(-x) $$ for some constants a,b. Now I can use $f(0)=\int_0^0\ldots=0$ to obtain $a=-b$.
My question is whether these are correct and if they are, and how to proceed in determining the constants...
What you wrote is correct. From $ f'(x)=\int_0^xf(y)dy$ you get $f'(0)=0.$ This gives $a=b=0.$