Find $g(x)$ from the following condition: ${g(x)}=(\int_{0}^{1}{e}^{x+t}{g(t)}dt)+x$

Question

Find $g(x)$ from the following condition: $${g(x)}=\left(\int_{0}^{1}{e}^{x+t}{g(t)}dt\right)+x$$ I have tried to solve it by applying Newton-Leibnitz formula and solving the linear differential equation with the help of integrating factor, but I am getting $g(x)$ as $1+x$, it does not satisfy the original equation, Please tell me where I have gone wrong and suggest a better approach to this question.

Answer 1

I suggest another approach. We know: \begin{equation} {g(x)}=\left(\int_{0}^{1}{e}^{t}{g(t)}dt\right){e}^{x}+x =C{e}^{x}+x, \end{equation} where $C=\int_{0}^{1}{e}^{t}{g(t)}dt$. We only need to discover the value of $C$. But: \begin{equation} C=\int_{0}^{1}{e}^{x}{g(x)}dx =\int_{0}^{1}(C{e}^{2x}+xe^x)dx =C\int_{0}^{1}{e}^{2x}dx+\int_{0}^{1}xe^xdx \end{equation} Therefore: \begin{equation} C\left(1-\int_{0}^{1}{e}^{2x}dx\right)=\int_{0}^{1}xe^xdx \end{equation} which can be easily solved.

Answer 2

Write $h(x)=g(x)-x$. Solve the differential equation $$h(x)=e^x\int_0^1e^t(h(t)+t)dt$$ by integrating $\frac{h'(x)}{h(x)}=1.$

the soulotion to the equation is $h(x)=e^{x+c}$ for some constant $c$. To find the constant solve the equation $$e^{x+c}=e^x\int_0^1e^t(e^{t+c}+t)dt$$