Type of the differential equation is what you have to identify. Here if you divide the given equation by x lnx, it becomes dydx+1xlnx×y=2x which when compared to standard form of linear differential equation matches it. Here P(x) = 1xlnx
Q(x)=2x
Solution will be y X I.F = ∫ Q(I.F.)dx + C
I.F. = e∫P(x)dx
So I.F = lnx.
So general solution will be
y×lnx=∫2xlnxdx+C
ylnx=(lnx)2+C
So n = 2.