The general solution of the differential equation ${e^x}dy + \left( {y{e^x} + 2x} \right)dx = 0$is

A. $x{e^y} + {x^2} = C$

B. $x{e^y} + {y^2} = C$

C. $y{e^x} + {x^2} = c$

D. $y{e^y} + {x^2} = C$

Option C is correct

The given differential equation is:
${e^x}dy + \left( {y{e^x} + 2x} \right)dx = 0$
$\Rightarrow$ ${e^x}\frac{{dy}}{{dx}} + y{e^x} + 2x = 0$

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

This is a linear differential equation of the form

$\frac{{dy}}{{dx}} + Py = Q$, where $P = 1$ and $Q = - 2x{e^{ - x}}$.

Now, I.F $= {e^{\int {Pdx} }} = {e^{\int {dx} }} = {e^x}$

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

$y({\rm{I}}.{\rm{F}}.) = \int {({\rm{Q}} \times {\rm{I}}.{\rm{F}}.)} dx + {\rm{C}}$
$\Rightarrow$ $y{e^x} = \int {\left( { - 2x{e^{ - x}} \cdot {e^x}} \right)} dx + {\rm{C}}$

$\Rightarrow$ $y{e^x} = - \int 2 xdx + {\rm{C}}$
$\Rightarrow$ $y{e^x} = - {x^2} + {\rm{C}}$
$\Rightarrow$ $y{e^x} + {x^2} = {\rm{C}}$
Hence, the correct answer is C.