两边对x求导 得:
e^(xy)*(y+xy')+y'(lnx)+y/x=2cos2x
[xe^(xy)+lnx]y'=2cos2x-y/x-ye^(xy)
x[xe^(xy)+lnx]y'=2xcos2x-y-xye^(xy)
dy/dx=y'=[2xcos2x-y-xye^(xy)]/x[xe^(xy)+lnx]