JEE Main 2008Annual, Paper 1MathematicsEasyDifferential Equations163× since 2002The solution of the differential equation dydx=x+yx{{dy} \over {dx}} = {{x + y} \over x}dxdy=xx+y satisfying the condition y(1)=1y(1)=1y(1)=1 is :AOption A. y=lnx+xy = \ln x + xy=lnx+xBOption B. y=xlnx+x2y = x\ln x + {x^2}y=xlnx+x2COption C. y=xe(x−1) y = x{e^{\left( {x - 1} \right)}}\,y=xe(x−1)DOption D. y=x lnx+xy = x\,\ln x + xy=xlnx+xCheckSkip · show solution
