The general solution of the differential equation

The given differential equation is:

$e^{x} d y+\left(y e^{x}+2 x\right) d x=0$

$\Rightarrow e^{x} \frac{d y}{d x}+y e^{x}+2 x=0$

$\Rightarrow \frac{d y}{d x}+y=-2 x e^{-x}$

This is a linear differential equation of the form

$\frac{d y}{d x}+P y=Q$, where $P=1$ and $Q=-2 x e^{-x} .$

Now, I.F $=e^{\int P d x}=e^{\int d x}=e^{x}$

The general solution of the given differential equation is given by,

$y($ I.F. $)=\int($ Q $\times$ I.F. $) d x+\mathrm{C}$

$\Rightarrow y e^{x}=\int\left(-2 x e^{-x} \cdot e^{x}\right) d x+\mathrm{C}$

$\Rightarrow y e^{x}=-\int 2 x d x+\mathrm{C}$

$\Rightarrow y e^{x}=-x^{2}+\mathrm{C}$

$\Rightarrow y e^{x}+x^{2}=\mathrm{C}$

Hence, the correct answer is C.