How can I solve this problem?, $$\frac{y^2}{2} + 2 y e^x + (y + e^x) y'(x) = 0$$
I tried to convert to a exact differential equation by multiply it by the integration factor $e^{\int \frac{M_{y}-N_{x}}{N} dx} = e^{\int x dx} = e^{x}$ where $M= \frac{y^2}{2} + 2 y e^x$ and $N= y + e^x$
I got that one families of solutions is $$e^{x}y + e^{2x}+c$$ But I think I did it wrong.
hint
Assuming that the equation you want to solve is $$y^2+4ye^x+2(y+e^x)y'=0$$
put $$y=ze^x$$
it becomes
$$z^2+4z+2(z+1)(z'+z)=0$$ or
$$\frac{2(z+1)}{z^2+2z}z'=-3$$ and $$(\frac 1z+\frac{1}{z+2})z'=-3$$
thus $$z(z+2)=\lambda.e^{-3x}=(z+1)^2-1$$