Can we solve $u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt$ with using noise terms phenomenon?

Question

question is can $u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt$ be solved via noise terms phenomenon?

I made

$u_0(x)=-1+x+\frac{x^2}2+2e^x $

$u_{k+1}(x)=-\int_0^xu_k(t)dt$, $k\geq0$

$u_1(x)=-\int_0^xu_0(t)dt=-\int_0^x(-1+x+\frac{x^2}2+2e^x)dt=x-\frac{x^2}2-\frac{x^3}6-2(e^x-1)$

so I canceled $\pm \frac {x^2} {2}$ and $\pm2e^x$ from $u_0(x)$ and found the solution as $u(x)=-1+x$ But it doesnt justify the integral equation. Does this mean it cant be solved with noise terms p.? Is this a proper way to check?

Answer 1

$u(x)=-1+x+\frac{x^2}{2}+2e^x-\int_0^xu(t)dt \Rightarrow \frac{du}{dx}=2x+2e^x+1-u$

$\frac{du}{dx}=2x+2e^x+1-u \Rightarrow e^x\frac{du}{dx}+e^xu=2e^xx+2e^{2x}+e^x$

$e^x\frac{du}{dx}+e^xu=2e^xx+2e^{2x}+e^x \Rightarrow e^xdu+e^xudx=d(e^xu)=(2e^xx+2e^{2x}+e^x)dx$

$d(e^xu)=(2e^xx+2e^{2x}+e^x)dx \Rightarrow e^xu+c=\int(2e^xx+2e{^2x}+e^x)dx$

$e^xu+c=\int(2e^xx+2e^{2x}+e^x)dx=e^x+e^{2x}+2(xe^x-e^x)$

$e^xu+c=e^x+e^{2x}+2(xe^x-e^x) \Rightarrow u=2x-1+e^x-\frac{c}{e^x}$

Answer 2

Define: $$U(x)=\int_0^xu(t)dt$$ Then you have: $$U(x)+U'(x)=-1+x+\frac{x^2}2+2e^x$$ Which is just a first order ODE