Answer
$y=2xe^{x}-e^{x}+ce^{-x}$
Work Step by Step
The standard form of the given equation is:
$y'+y=4xe^x$ ....(1)
The integrating factor is: $e^{\int (1) dx}=e^x$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [e^xy]' =\int 4xe^{2x} dx$
or, $ye^x=2xe^{2x}-e^{2x}+c$...(2)
Thus, equation (2) becomes:
$y=2xe^{x}-e^{x}+ce^{-x}$
