How to solve this fredholm equation $u(x)=x+\int_0^2e^{x+t}u(t)dt$?

Question

$u(x)=x+\int_0^2e^{x+t}u(t)dt$

I tried adm and couldnt find a geometric series. ($e^x(1+e^2),e^{2x}(1+e^2)(e^2-1),e^{3x}(1+e^2)(e^2-1)(e^2-1)....etc $ and also tried noise term phenomenon and direct computation method which gave me $u(x)=x+e^x\frac{(e^2+1)}{3/2+e^4/2)}$ and it didnt work.

Answer 1

$$u(x)=x+e^x\int_{0}^{2} e^t u(t) dt~~~(1)$$ Let $\int_{0}^{2} e^t u(t) dt=A$, then $$u(x)=x+Ae^x.$$ For consistency, $$\int_{0}^{2} e^t (t+Ae^t) dt =A$$ $$\int_{0}^{2}te^t dt+A\int_{0}^{2} e^{2t} dt=A$$ $$[te^t-e^t+\frac{A}{2} e^{2t}]_{0}^{2}=A \implies A=\frac{2(1+e^2)}{3-e^4}.$$ So the solution of (1) is $$u(x)=x+\frac{2(1+e^2)}{3-e^4}e^x$$

Answer 2

The equation shows $u(x)=x+ce^{x}$ for some constant $c$. (Pull out $e^{x}$ from the integral to see why this is true). Put this back into the equation to get the value of $c$. You get $c=\frac {2(e^{2}+1)} {3-e^{4}}$ so the solution is $u(x)=x+\frac {2(e^{2}+1)} {3-e^{4}} e^{x}$.