1.The solution of the differential equation y(1+logx)dxdy−xlogx=0y(1 + \log x)\frac{dx}{dy} - x\log x = 0y(1+logx)dydx−xlogx=0 isa.xlogx=y+cx\log x = y + cxlogx=y+cb.xlogx=ycx\log x = ycxlogx=ycc.y(1+logx)=cy(1 + \log x) = cy(1+logx)=cd.logx−y=c\log x - y = clogx−y=cLogin to continueOnly logged in users canattempt or see the solution.
