Hi i need an help to solve this problem Let $h,g$ an entire functions such that $$(e^{h(z)}+1)(g(z)+g'(z))=1, g(0)=1,g(1)=2-e^{-1}$$ I need $g(2)$
I have noticed that $ (e^zg(z))'=\frac{e^z}{1+e^{h(z)}}$ then $$ g(z)=e^{-z}\int_{0}^z\frac{e^{w}}{1+e^{h(w)}}dw+e^{-z}g(0)$$ $$ g(2)=e^{-2}\int_{0}^2\frac{e^{w}}{1+e^{h(w)}}dw+e^{-2}g(0)$$ $$ g(2)=e^{-2}\int_{0}^2\frac{e^{w}}{1+e^{h(w)}}dw+e^{-2}g(0)$$ $$ g(1)=e^{-1}\int_{0}^1\frac{e^{w}}{1+e^{h(w)}}dw+e^{-1}g(0)$$ $$ g(2)=e^{-2}\left[ \int_{0}^1\frac{e^{w}}{1+e^{h(w)}}dw+ \int_{1}^2\frac{e^{w}}{1+e^{h(w)}}dw\right]+e^{-2}g(0) $$ $$ g(2)=e^{-2}\left[ e^{1}g(1)-g(0)+ \int_{1}^2\frac{e^{w}}{1+e^{h(w)}}dw\right]+e^{-2}g(0) $$ $$ g(2)=e^{-1}g(1)+ e^{-2}\int_{1}^2\frac{e^{w}}{1+e^{h(w)}}dw$$
I don't know how to continue ?
Note that $e^h+1$ avoids one so unless $h$ constant it means that $e^h+1$ must take zero which is impossible by the hypothesis, hence $h$ is constant and so $g+g'=c \ne 0$ or $g'+g''=0$ or $(g'e^x)'=0$ or $g'=ke^{-x}$ or $g=-ke^{-x}+k_1, k_1 \ne 0$ and putting the initial values in we get $k_1-k=1, k_1-k/e=2-1/e$ so $k=1, k_1=2$ and $g(2)=2-1/e^2$