Answer
$C=x^2+xe^{-y}-y$
Work Step by Step
We are given:
$$(1+2xe^y)dx-(e^y+x)dy=0$$
We have that:
$$M(x,y)=1+2xe^y \wedge N(x,y)=e^y+x$$
then
$$M'(x,y)=2xye^y$$
$$N'(x,y)=-1$$
So this equation is not exist.
But we have:
$$\frac{M'_y-N'_y}{M}=\frac{1+2xe^y}{1+2xe^y}=1$$
Integrating factor:
$$I=e^{-\int dy}=e^{-y}$$
Multiplying the equation by $e^{-y}$
$$(e^{-y}+2x)dx-(1+xe^{-y})dy=0$$
There exists a potential function $\phi$ satisfies:
$$\frac{\partial \phi}{\partial x}=M \wedge \frac{\partial \phi}{\partial y}=N$$
$$\frac{\partial \phi}{\partial x}=e^{-y}+2x \wedge \frac{\partial \phi}{\partial y}=-(1+xe^{-y})$$
Integrating $\frac{\partial \phi}{\partial x}=e^{-y}+2x $ we get
$$\phi=x^2+xe^{-y}+f(y)$$
Since $\frac{\partial \phi}{\partial y}=-xe^{-y}-1$
We obtain:
$$f'(y)=-1 \rightarrow f(y)=-yC$$
$C$ is constant of integration
The potential solution is:
$$\phi=x^2+xe^{-y}-y$$
Hence general solution is $C=x^2+xe^{-y}-y$
