Answer
y is a solution
Work Step by Step
$y=x\ln x-4x$
Find $y^{\prime} $ (product rule for the first term)
$[x(\displaystyle \ln x)]^{\prime}=x(\ln x)^{\prime}+(x)^{\prime}\ln x=x(\frac{1}{x})+\ln x$
$=1+\ln x$
$(4x)^{\prime}=4$
$y^{\prime}=1+\ln x-4=-3+\ln x$
Substitute $y $ and $y^{\prime}$ and
see if the differential equation is satisfied
$(x+y)-xy^{\prime}=\\=x+x\ln x-4x-x(-3+\ln x)$
$=x+x\ln x-4x+3x-x\ln x\\=0$
It is, so,
y is a solution.
