Solve the following integral equations: $$ \int_0^xu(y)\, dy=\frac{1}{3}xu(x) \tag 1 \label 1 $$ and $$ \int_0^xe^{-x}u(y)\, dy=e^{-x}+x-1. \tag 2 \label 2 $$
Concerning $\eqref 1$, I read that it can be solved by differentiation. Differentiation on both sides gives $$ u(x)=\frac{1}{3}(u(x)+xu'(x))\Leftrightarrow u'(x)=\frac{2}{x}u(x) $$ and then by separation of this ODE the solution is $u(x)=Cx^2$.
How can I solve $\eqref 2$?
Hint:
$$\int_0^xe^{-x}u(y)\, dy=e^{-x}+x-1$$ So $$ e^{-x}\int_0^xu(y)\, dy=e^{-x}+x-1$$ and therefore, $$ \ \int_0^xu(y)\, dy=1+\frac{x-1}{e^{-x}} $$