#### Answer

$y=2xe^{x}-e^{x}+ce^{-x}$

#### Work Step by Step

Here, we have $y'+y=4(xe^x)$
The integrating factor is: $I=e^{\int (1) dx}=e^{x}$
Now, $(e^{x})[y'+y]=4(xe^x)(e^{x})$
This implies that
$\int [e^{x}y]' =\int 4xe^{2x} dx $
$\implies e^{x}y=(2x)(e^{2x})-e^{x}+c$
Hence, $y=(2x)(e^{x})-e^{x}+ce^{-x}$