I want to determine the solution of the Fredholm integral equation $$\int_0^1x\xi u(\xi )\, d\xi -u(x)=x , \ 0\leq x\leq 1$$
We have the equation $$u(x)=-x+\int_0^1x\xi u(\xi )\, d\xi \Rightarrow -x+x\int_0^1\xi u(\xi )\, d\xi$$ This is an equation of second order, correct?
But how can we solve that equation? Could you give me a hint?
I tried to differentiate this given equation to get an ivp, but at the end I got the initial equation.
How could I start?
$$u(x)=-x+x\int_0^1\xi u(\xi )\, d\xi$$ $\int_0^1\xi u(\xi )\, d\xi$ is a definite integral with constant bounds. Thus it is a constant, say $C$. It is not any constant insofar $u(x)$ is not any function but is the sought function solution of the integral equation.
$$\int_0^1\xi u(\xi )\, d\xi=C$$ At this step the value of $C$ is still unknown. We will determine it so that the integral equation be satisfied : $$u(x)=-x+x\,C$$ $$u(x)=(C-1)x$$ $$C=\int_0^1\xi u(\xi )\, d\xi=\int_0^1\xi(C-1)\xi\, d\xi=(C-1)\int_0^1\xi^2\, d\xi=\frac{C-1}{2}$$ $$C==\frac{C-1}{2}$$ $$C=-1$$ $$u(x)=-x+x(-1)=-2x$$